Does battery voltage actually get lower when connected to a load, or does it just appear to do so?
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Sorry if Im wording this question strangely. I am using a 3.7V battery and my microcontroller monitors the voltage and goes to sleep if my battery voltage is too low. The issue is that it reads a lower voltage than the battery shows if I disconnect it and check it with my multimeter. For example, my microcontroller would read 3.65V when my multimeter would read my disconnected battery at 3.8V. Is my microcontroller reading the voltage incorrectly or should I treat the with-load voltage that my microcontroller is readings as the actual voltage?
voltage batteries
asked Dec 13 '18 at 19:17
Tapatio SombreroTapatio Sombrero
15217
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add a comment |
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Sorry if Im wording this question strangely. I am using a 3.7V battery and my microcontroller monitors the voltage and goes to sleep if my battery voltage is too low. The issue is that it reads a lower voltage than the battery shows if I disconnect it and check it with my multimeter. For example, my microcontroller would read 3.65V when my multimeter would read my disconnected battery at 3.8V. Is my microcontroller reading the voltage incorrectly or should I treat the with-load voltage that my microcontroller is readings as the actual voltage?
voltage batteries
asked Dec 13 '18 at 19:17
Tapatio SombreroTapatio Sombrero
15217
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1
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Are you measuring the battery (connected vs disconnected) with the multimeter both times, or are you comparing something internal from the microcontroller? Use the multimeter to make the measurement while the controller is connected if you can.
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– mbrig
Dec 13 '18 at 20:06
7
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A car battery has over 13V when not connected, yet drops to 10.5V while starting the engine. Which voltage is correct? Both.
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– Mast
Dec 14 '18 at 4:19
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Just going to add a note. Some batteries, such as lithium ion, are pretty well modeled by the series resistance concept. There is a true cell voltage which is inaccessible, and it is equal to Vout + Iout * ESR. When you remove the load, the voltage recovers quickly. But with lead acid or alkaline batteries, it may take a lot longer to recover to the final open-circuit voltage after removing the load. In other words, it is more complicated than a voltage source in series with a resistor. There is some kind of long time-constant voltage recovery that occurs.
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– mkeith
Dec 16 '18 at 10:05
add a comment |
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Sorry if Im wording this question strangely. I am using a 3.7V battery and my microcontroller monitors the voltage and goes to sleep if my battery voltage is too low. The issue is that it reads a lower voltage than the battery shows if I disconnect it and check it with my multimeter. For example, my microcontroller would read 3.65V when my multimeter would read my disconnected battery at 3.8V. Is my microcontroller reading the voltage incorrectly or should I treat the with-load voltage that my microcontroller is readings as the actual voltage?
voltage batteries
asked Dec 13 '18 at 19:17
Tapatio SombreroTapatio Sombrero
15217
$endgroup$
Sorry if Im wording this question strangely. I am using a 3.7V battery and my microcontroller monitors the voltage and goes to sleep if my battery voltage is too low. The issue is that it reads a lower voltage than the battery shows if I disconnect it and check it with my multimeter. For example, my microcontroller would read 3.65V when my multimeter would read my disconnected battery at 3.8V. Is my microcontroller reading the voltage incorrectly or should I treat the with-load voltage that my microcontroller is readings as the actual voltage?
voltage batteries
voltage batteries
asked Dec 13 '18 at 19:17
Tapatio SombreroTapatio Sombrero
15217
asked Dec 13 '18 at 19:17
Tapatio SombreroTapatio Sombrero
15217
asked Dec 13 '18 at 19:17
Tapatio SombreroTapatio Sombrero
15217
asked Dec 13 '18 at 19:17
Tapatio SombreroTapatio Sombrero
15217
asked Dec 13 '18 at 19:17
Tapatio SombreroTapatio Sombrero
15217
15217
1
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Are you measuring the battery (connected vs disconnected) with the multimeter both times, or are you comparing something internal from the microcontroller? Use the multimeter to make the measurement while the controller is connected if you can.
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– mbrig
Dec 13 '18 at 20:06
7
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A car battery has over 13V when not connected, yet drops to 10.5V while starting the engine. Which voltage is correct? Both.
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– Mast
Dec 14 '18 at 4:19
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Just going to add a note. Some batteries, such as lithium ion, are pretty well modeled by the series resistance concept. There is a true cell voltage which is inaccessible, and it is equal to Vout + Iout * ESR. When you remove the load, the voltage recovers quickly. But with lead acid or alkaline batteries, it may take a lot longer to recover to the final open-circuit voltage after removing the load. In other words, it is more complicated than a voltage source in series with a resistor. There is some kind of long time-constant voltage recovery that occurs.
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– mkeith
Dec 16 '18 at 10:05
add a comment |
1
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Are you measuring the battery (connected vs disconnected) with the multimeter both times, or are you comparing something internal from the microcontroller? Use the multimeter to make the measurement while the controller is connected if you can.
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– mbrig
Dec 13 '18 at 20:06
7
$begingroup$
A car battery has over 13V when not connected, yet drops to 10.5V while starting the engine. Which voltage is correct? Both.
$endgroup$
– Mast
Dec 14 '18 at 4:19
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Just going to add a note. Some batteries, such as lithium ion, are pretty well modeled by the series resistance concept. There is a true cell voltage which is inaccessible, and it is equal to Vout + Iout * ESR. When you remove the load, the voltage recovers quickly. But with lead acid or alkaline batteries, it may take a lot longer to recover to the final open-circuit voltage after removing the load. In other words, it is more complicated than a voltage source in series with a resistor. There is some kind of long time-constant voltage recovery that occurs.
$endgroup$
– mkeith
Dec 16 '18 at 10:05
1
1
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Are you measuring the battery (connected vs disconnected) with the multimeter both times, or are you comparing something internal from the microcontroller? Use the multimeter to make the measurement while the controller is connected if you can.
$endgroup$
– mbrig
Dec 13 '18 at 20:06
$begingroup$
Are you measuring the battery (connected vs disconnected) with the multimeter both times, or are you comparing something internal from the microcontroller? Use the multimeter to make the measurement while the controller is connected if you can.
$endgroup$
– mbrig
Dec 13 '18 at 20:06
7
7
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A car battery has over 13V when not connected, yet drops to 10.5V while starting the engine. Which voltage is correct? Both.
$endgroup$
– Mast
Dec 14 '18 at 4:19
$begingroup$
A car battery has over 13V when not connected, yet drops to 10.5V while starting the engine. Which voltage is correct? Both.
$endgroup$
– Mast
Dec 14 '18 at 4:19
$begingroup$
Just going to add a note. Some batteries, such as lithium ion, are pretty well modeled by the series resistance concept. There is a true cell voltage which is inaccessible, and it is equal to Vout + Iout * ESR. When you remove the load, the voltage recovers quickly. But with lead acid or alkaline batteries, it may take a lot longer to recover to the final open-circuit voltage after removing the load. In other words, it is more complicated than a voltage source in series with a resistor. There is some kind of long time-constant voltage recovery that occurs.
$endgroup$
– mkeith
Dec 16 '18 at 10:05
$begingroup$
Just going to add a note. Some batteries, such as lithium ion, are pretty well modeled by the series resistance concept. There is a true cell voltage which is inaccessible, and it is equal to Vout + Iout * ESR. When you remove the load, the voltage recovers quickly. But with lead acid or alkaline batteries, it may take a lot longer to recover to the final open-circuit voltage after removing the load. In other words, it is more complicated than a voltage source in series with a resistor. There is some kind of long time-constant voltage recovery that occurs.
$endgroup$
– mkeith
Dec 16 '18 at 10:05
add a comment |
9 Answers
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Yes, it does get lower.
The effect you see is called internal resistance:
A practical electrical power source which is a linear electric circuit may <...> be represented as an ideal voltage source in series with an impedance. This impedance is termed the internal resistance of the source.
Put simply, a battery is not an ideal voltage source. A typical battery (i. e. non-ideal voltage source) will look like this:
What you are measuring is voltage between terminals A and B. According to Ohm's Law:
$$ U_{AB} = E * frac{R}{R+r} $$
When there is no circuit, you can imagine your voltmeter's internal series resistance $ R_{volt} $ taking the role of $ R $. However, $ R_{volt} $ is usually so large (tens or hundreds of megaohms) compared to $ r $ (usually fractions of an ohm) that $ frac{R_{volt}}{R_{volt}+r} $ tends to 1, hence measured open-circuit voltage tends to the battery's internal (true) voltage $ E $.
When there is a closed circuit with equivalent series resistance of $ R $, you will be able to see that the measured voltage $ U_{AB} $ drops proportionally to $ R $, in accordance with the above formula.
So, the voltage drop is real — the measured voltage is what your load gets. The more current it draws from the battery, the lower is voltage it gets.
answered Dec 14 '18 at 6:22
intelfxintelfx
41114
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When the battery is open you are measuring an open cell voltage. When the battery is in the system it's closed cell voltage under load. You are dropping some voltage across the internal impedance of the battery because your system is drawing current when the measurement is being made (so at the terminals the voltage is indeed lower). So both measurements MCU and multimeter are correct, the difference is that the multimeter is >1Mohm load while the MCU is much lower (since probably drawing at least mAs of power).
There may be another effect at play. Batteries do exhibit a recovery phenomenon where if left open cell with no load some of the voltage will recover after a time interval.
answered Dec 13 '18 at 19:23
Gonzik007Gonzik007
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So I want my microcontroller to go to sleep when my battery is at 3.6V. The connected closed cell votlage under load seems to be 0.2V less than the open cell voltage. Should I sleep when my microcontroller reads 3.4V? Or just sleep when it reads 3.6V even though it would still show 3.8V open cell?
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– Tapatio Sombrero
Dec 13 '18 at 20:56
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That is one option. You are correlating your open and closed cell voltages and assuming linear mapping. However, keep in mind that various cells may have different internal impedance. If you are mass producing a product corner/data studies must be done to figure out what the mapping should be. Another option is to place the system in the lowest power state possible (turn off all peripherals and functions expect MCU at some low power state). If you can then measure voltage under a very light load your MCU and multi-meter readings should be closer and have less error.
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– Gonzik007
Dec 13 '18 at 22:40
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Also make sure to check your battery chemistry. Make sure that there is no hard requirement for safe shutdown. You want to make sure that your measurement error can never place the cell into some unsafe state. With the voltages you list, and assuming lithium battery you should be ok, but just thought it worth mentioning.
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– Gonzik007
Dec 13 '18 at 22:41
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@fishinear - "The OP gets notifications of all comments" That's not how it works (assuming you mean the question's "OP"). You commented on an answer, not on the question, yet your comment was implicitly directed to the question's OP - Tapatio. See this Meta.SE post where the accepted answer says "On a question, you will receive notifications automatically for comments on the question only, not the answers" which confirms my experience. Without using "@", the answer writer Gonzik007 was notified of your comment, not the question's OP. HTH
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– SamGibson
Dec 16 '18 at 13:40
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@fishinear - "pretty much all CPUs I know of have brown-out detectors" Yes, but those reset the MCU, they don't cause it to enter sleep mode. Different things. That's why your comment that "Normally you'd want to put the CPU to sleep just before it will stop functioning because of too low voltage [...] most micro controllers do that themselves already, so no need for you to do it" was confusing, as it seemed to suggest that MCUs put themselves to sleep in a low-voltage condition. Now you've clarified that you meant that they reset due to BOR - thanks for that.
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– SamGibson
Dec 16 '18 at 13:41
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Every battery has a certain amount of output resistance. What happens if current flows through a resistor? Yes, a voltage drop! So the more current you draw from the battery, the lower the output voltage is.
answered Dec 13 '18 at 19:23
Stefan WyssStefan Wyss
2,0891313
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This is true of all power supplies
Indeed, batteries sag their voltage on being loaded. So does everything else.
The main culprit is Ohm's Law, E=IR, where voltage drop across any conductor is proportional to its amperage drawn.
Part of a battery's sag is chemical, but part is simply the Ohm's Law resistance of its internal components.
Let's suppose you have a mad gaming rig with 4 paralleled video cards, the combo pulls 1000 watts when gaming. But it's just sitting at the Windows Homescreen and only pulling 100 watts. The power cables are carrrying 20A@5V, and dropping 0.01 volts, so the cards get 4.99 volts. (The wires are 2000 Siemens == 1/2000 Ohms.)
At this light load, the AC power supply is inefficient and poor power factor, so it is drawing 240VA or 2 amps off the 120V mains. The branch circuit wiring back to the panel is dropping 0.4 volts. Conductance is 5 Siemens == 1/5 ohm.
Now you fire up your most demanding game. Pulling 200A at 5V, the resistive losses alone inside your PC's wiring jumps to 0.1 volts. So the cards get 4.90 volts. That's a drop.
Meanwhile, the power supply draws 10A (1200VA) from the AC mains. Wiring voltage drop predictably increases to 2.0 volts, so voltage at the power suppy is 118V. Most likely a switching power supply pulls a skitch more current to compensate, otherwise its output voltage would sag also.
No current is being drawn on safety ground, so it's not dropping. Measured from ground, neutral is 1 volt and hot is 119 volts. And we can use this to affirm correct wiring. It's like the pointer bar on a torque wrench, it doesn't bend.
Of course, similar drops are happening all the way back to the power plant. There, increased load (in amps) sags the voltage because of the internal resistance of the generator, but also due to turbine horsepower. VA=W. If A increases beyond spec, V must decrease in proportion so W can remain within the turbine's ability. Having the turbine bog and slow down is not an option, because it's AC power and must stay in sync.
answered Dec 15 '18 at 2:12
HarperHarper
6,299826
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All batteries have a memory effect when unloaded such that they return slowly to near the previous voltage after a short burst load. There is also momentary quick drop in voltage due to a load of ESR*I = Δ V.
So both measurements must be taken at the same time to check calibration for errors and consider the amount of hysteresis thresholds required to prevent oscillation of sleep, wake-up cycles.
The memory effect time constant can be several to many minutes depending on the "no-load" leakage current after a load.
Because of these combined effects which might be computed for a given cell (ΔV= ESR * V/Rload + t/ESR * C2) the cut-off voltage is often lowered to capture the charge stored in memory capacitance C2 as long you know it returns to the safe Vmin threshold. Battery rapid aging occurs for the amount of time below its Vmin threshold.
Review the battery datasheet for details.
answered Dec 13 '18 at 19:27
Sunnyskyguy EE75Sunnyskyguy EE75
67.4k22397
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There is drop in voltage due internal resistance of the battery coming into play so you will see the voltage drop by a value of i * r ( where i is the current flowing and r is the internal resistance of the battery)
answered Dec 16 '18 at 5:12
Vinu PhilipVinu Philip
111
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A new battery will have much less loaded voltage drop than you have.
An old, worn out, or damaged Lithium battery has a much higher internal resistance than a new battery. It is damaged if it has been fully charged for longer than a few months, if it has been discharged too low or if it has had too many charge-discharge cycles.
answered Dec 14 '18 at 0:28
AudioguruAudioguru
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While all the other answers are great and tell what I would tell you too (the battery voltage actually get's lower when there is a load), I'd like to add something:
The reason which is presented that the voltage drops is the "internal resistance". I want to mention that the model of an internal resistance is JUST A MODEL that works very well in modelling the properties of a voltage source, while at the same time being simple and easy to calculate.
In reality, it's more complicated. The resistance of the internal components of the battery that current has to pass through (I intentionally don't call those "internal resistance", because this is a term from the above mentioned model) plays a role, but it's not the only role.
In most batteries there is a chemical reaction going on that separates charges at some border layer. This chemical reaction follows the laws of statistical physics. It comes to halt when the (chemical equilibrium is reached. The separation of charges generates the voltage that you can measure, and this voltage is a factor in the chemical equilibrium (the higher the voltage is, the less separation occurs to create a new pair of separated charges).
When you attach a load now, you take away charges at constant intervals (because there is an electrical current). If the system reaches the equilibrium situation now, the amount of separated charges and voltage will be less (because more charges have to be created).
answered Dec 16 '18 at 9:52
QuantumwhispQuantumwhisp
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Battery voltage doesn't usually drop just because there's a load connected. But the measured voltage does tend to drop
Here's what you've got to know about voltage measurements
A voltmeter uses a resistor with a very high resistance. Ideally, it's infinite. The voltmeter measures the voltage across this resistor.
So when you connect a battery to your voltmeter, the internal resistance of the battery is insignificant compared to the voltmeter's resistance. So most of the voltage drop occurs across the voltmeter's resistance and not the battery's internal resistor. Hence, you measure the right voltage.
However, your microcontroller may have a resistance that's not too high.
If the battery had an internal resistance of say, 1 milli ohm, and the voltmeter was using a 24000 ohm resistor, this error is expected.
answered Dec 14 '18 at 8:09
Pradyoth ShandilyaPradyoth Shandilya
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There's no need to make things more confusing by talking about multimeter design. An ideal (infinite-resistance) voltmeter would experience the same effect. The issue is not comparing voltmeter resistance to the battery's internal resistance, but comparing the battery's internal resistance to the load resistance.
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– Sneftel
Dec 14 '18 at 8:37
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The multimeter's design explains why there's a difference in the readings. That's something anyone dealing with basic electronics must know
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– Pradyoth Shandilya
Dec 14 '18 at 8:41
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No, it doesn't explain that. There's a difference in the readings because the voltage is different. Even if there were no multimeter being used, the actual voltage across the battery is actually different when a load is connected.
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– Sneftel
Dec 14 '18 at 10:21
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9 Answers
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$begingroup$
Yes, it does get lower.
The effect you see is called internal resistance:
A practical electrical power source which is a linear electric circuit may <...> be represented as an ideal voltage source in series with an impedance. This impedance is termed the internal resistance of the source.
Put simply, a battery is not an ideal voltage source. A typical battery (i. e. non-ideal voltage source) will look like this:
What you are measuring is voltage between terminals A and B. According to Ohm's Law:
$$ U_{AB} = E * frac{R}{R+r} $$
When there is no circuit, you can imagine your voltmeter's internal series resistance $ R_{volt} $ taking the role of $ R $. However, $ R_{volt} $ is usually so large (tens or hundreds of megaohms) compared to $ r $ (usually fractions of an ohm) that $ frac{R_{volt}}{R_{volt}+r} $ tends to 1, hence measured open-circuit voltage tends to the battery's internal (true) voltage $ E $.
When there is a closed circuit with equivalent series resistance of $ R $, you will be able to see that the measured voltage $ U_{AB} $ drops proportionally to $ R $, in accordance with the above formula.
So, the voltage drop is real — the measured voltage is what your load gets. The more current it draws from the battery, the lower is voltage it gets.
answered Dec 14 '18 at 6:22
intelfxintelfx
41114
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add a comment |
$begingroup$
Yes, it does get lower.
The effect you see is called internal resistance:
A practical electrical power source which is a linear electric circuit may <...> be represented as an ideal voltage source in series with an impedance. This impedance is termed the internal resistance of the source.
Put simply, a battery is not an ideal voltage source. A typical battery (i. e. non-ideal voltage source) will look like this:
What you are measuring is voltage between terminals A and B. According to Ohm's Law:
$$ U_{AB} = E * frac{R}{R+r} $$
When there is no circuit, you can imagine your voltmeter's internal series resistance $ R_{volt} $ taking the role of $ R $. However, $ R_{volt} $ is usually so large (tens or hundreds of megaohms) compared to $ r $ (usually fractions of an ohm) that $ frac{R_{volt}}{R_{volt}+r} $ tends to 1, hence measured open-circuit voltage tends to the battery's internal (true) voltage $ E $.
When there is a closed circuit with equivalent series resistance of $ R $, you will be able to see that the measured voltage $ U_{AB} $ drops proportionally to $ R $, in accordance with the above formula.
So, the voltage drop is real — the measured voltage is what your load gets. The more current it draws from the battery, the lower is voltage it gets.
answered Dec 14 '18 at 6:22
intelfxintelfx
41114
$endgroup$
add a comment |
$begingroup$
Yes, it does get lower.
The effect you see is called internal resistance:
A practical electrical power source which is a linear electric circuit may <...> be represented as an ideal voltage source in series with an impedance. This impedance is termed the internal resistance of the source.
Put simply, a battery is not an ideal voltage source. A typical battery (i. e. non-ideal voltage source) will look like this:
What you are measuring is voltage between terminals A and B. According to Ohm's Law:
$$ U_{AB} = E * frac{R}{R+r} $$
When there is no circuit, you can imagine your voltmeter's internal series resistance $ R_{volt} $ taking the role of $ R $. However, $ R_{volt} $ is usually so large (tens or hundreds of megaohms) compared to $ r $ (usually fractions of an ohm) that $ frac{R_{volt}}{R_{volt}+r} $ tends to 1, hence measured open-circuit voltage tends to the battery's internal (true) voltage $ E $.
When there is a closed circuit with equivalent series resistance of $ R $, you will be able to see that the measured voltage $ U_{AB} $ drops proportionally to $ R $, in accordance with the above formula.
So, the voltage drop is real — the measured voltage is what your load gets. The more current it draws from the battery, the lower is voltage it gets.
answered Dec 14 '18 at 6:22
intelfxintelfx
41114
$endgroup$
Yes, it does get lower.
The effect you see is called internal resistance:
A practical electrical power source which is a linear electric circuit may <...> be represented as an ideal voltage source in series with an impedance. This impedance is termed the internal resistance of the source.
Put simply, a battery is not an ideal voltage source. A typical battery (i. e. non-ideal voltage source) will look like this:
What you are measuring is voltage between terminals A and B. According to Ohm's Law:
$$ U_{AB} = E * frac{R}{R+r} $$
When there is no circuit, you can imagine your voltmeter's internal series resistance $ R_{volt} $ taking the role of $ R $. However, $ R_{volt} $ is usually so large (tens or hundreds of megaohms) compared to $ r $ (usually fractions of an ohm) that $ frac{R_{volt}}{R_{volt}+r} $ tends to 1, hence measured open-circuit voltage tends to the battery's internal (true) voltage $ E $.
When there is a closed circuit with equivalent series resistance of $ R $, you will be able to see that the measured voltage $ U_{AB} $ drops proportionally to $ R $, in accordance with the above formula.
So, the voltage drop is real — the measured voltage is what your load gets. The more current it draws from the battery, the lower is voltage it gets.
answered Dec 14 '18 at 6:22
intelfxintelfx
41114
edited Dec 14 '18 at 11:30
answered Dec 14 '18 at 6:22
intelfxintelfx
41114
answered Dec 14 '18 at 6:22
intelfxintelfx
41114
answered Dec 14 '18 at 6:22
intelfxintelfx
41114
41114
add a comment |
add a comment |
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When the battery is open you are measuring an open cell voltage. When the battery is in the system it's closed cell voltage under load. You are dropping some voltage across the internal impedance of the battery because your system is drawing current when the measurement is being made (so at the terminals the voltage is indeed lower). So both measurements MCU and multimeter are correct, the difference is that the multimeter is >1Mohm load while the MCU is much lower (since probably drawing at least mAs of power).
There may be another effect at play. Batteries do exhibit a recovery phenomenon where if left open cell with no load some of the voltage will recover after a time interval.
answered Dec 13 '18 at 19:23
Gonzik007Gonzik007
2,8161024
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1
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So I want my microcontroller to go to sleep when my battery is at 3.6V. The connected closed cell votlage under load seems to be 0.2V less than the open cell voltage. Should I sleep when my microcontroller reads 3.4V? Or just sleep when it reads 3.6V even though it would still show 3.8V open cell?
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– Tapatio Sombrero
Dec 13 '18 at 20:56
$begingroup$
That is one option. You are correlating your open and closed cell voltages and assuming linear mapping. However, keep in mind that various cells may have different internal impedance. If you are mass producing a product corner/data studies must be done to figure out what the mapping should be. Another option is to place the system in the lowest power state possible (turn off all peripherals and functions expect MCU at some low power state). If you can then measure voltage under a very light load your MCU and multi-meter readings should be closer and have less error.
$endgroup$
– Gonzik007
Dec 13 '18 at 22:40
$begingroup$
Also make sure to check your battery chemistry. Make sure that there is no hard requirement for safe shutdown. You want to make sure that your measurement error can never place the cell into some unsafe state. With the voltages you list, and assuming lithium battery you should be ok, but just thought it worth mentioning.
$endgroup$
– Gonzik007
Dec 13 '18 at 22:41
1
$begingroup$
@fishinear - "The OP gets notifications of all comments" That's not how it works (assuming you mean the question's "OP"). You commented on an answer, not on the question, yet your comment was implicitly directed to the question's OP - Tapatio. See this Meta.SE post where the accepted answer says "On a question, you will receive notifications automatically for comments on the question only, not the answers" which confirms my experience. Without using "@", the answer writer Gonzik007 was notified of your comment, not the question's OP. HTH
$endgroup$
– SamGibson
Dec 16 '18 at 13:40
1
$begingroup$
@fishinear - "pretty much all CPUs I know of have brown-out detectors" Yes, but those reset the MCU, they don't cause it to enter sleep mode. Different things. That's why your comment that "Normally you'd want to put the CPU to sleep just before it will stop functioning because of too low voltage [...] most micro controllers do that themselves already, so no need for you to do it" was confusing, as it seemed to suggest that MCUs put themselves to sleep in a low-voltage condition. Now you've clarified that you meant that they reset due to BOR - thanks for that.
$endgroup$
– SamGibson
Dec 16 '18 at 13:41
|
show 4 more comments
$begingroup$
When the battery is open you are measuring an open cell voltage. When the battery is in the system it's closed cell voltage under load. You are dropping some voltage across the internal impedance of the battery because your system is drawing current when the measurement is being made (so at the terminals the voltage is indeed lower). So both measurements MCU and multimeter are correct, the difference is that the multimeter is >1Mohm load while the MCU is much lower (since probably drawing at least mAs of power).
There may be another effect at play. Batteries do exhibit a recovery phenomenon where if left open cell with no load some of the voltage will recover after a time interval.
answered Dec 13 '18 at 19:23
Gonzik007Gonzik007
2,8161024
$endgroup$
1
$begingroup$
So I want my microcontroller to go to sleep when my battery is at 3.6V. The connected closed cell votlage under load seems to be 0.2V less than the open cell voltage. Should I sleep when my microcontroller reads 3.4V? Or just sleep when it reads 3.6V even though it would still show 3.8V open cell?
$endgroup$
– Tapatio Sombrero
Dec 13 '18 at 20:56
$begingroup$
That is one option. You are correlating your open and closed cell voltages and assuming linear mapping. However, keep in mind that various cells may have different internal impedance. If you are mass producing a product corner/data studies must be done to figure out what the mapping should be. Another option is to place the system in the lowest power state possible (turn off all peripherals and functions expect MCU at some low power state). If you can then measure voltage under a very light load your MCU and multi-meter readings should be closer and have less error.
$endgroup$
– Gonzik007
Dec 13 '18 at 22:40
$begingroup$
Also make sure to check your battery chemistry. Make sure that there is no hard requirement for safe shutdown. You want to make sure that your measurement error can never place the cell into some unsafe state. With the voltages you list, and assuming lithium battery you should be ok, but just thought it worth mentioning.
$endgroup$
– Gonzik007
Dec 13 '18 at 22:41
1
$begingroup$
@fishinear - "The OP gets notifications of all comments" That's not how it works (assuming you mean the question's "OP"). You commented on an answer, not on the question, yet your comment was implicitly directed to the question's OP - Tapatio. See this Meta.SE post where the accepted answer says "On a question, you will receive notifications automatically for comments on the question only, not the answers" which confirms my experience. Without using "@", the answer writer Gonzik007 was notified of your comment, not the question's OP. HTH
$endgroup$
– SamGibson
Dec 16 '18 at 13:40
1
$begingroup$
@fishinear - "pretty much all CPUs I know of have brown-out detectors" Yes, but those reset the MCU, they don't cause it to enter sleep mode. Different things. That's why your comment that "Normally you'd want to put the CPU to sleep just before it will stop functioning because of too low voltage [...] most micro controllers do that themselves already, so no need for you to do it" was confusing, as it seemed to suggest that MCUs put themselves to sleep in a low-voltage condition. Now you've clarified that you meant that they reset due to BOR - thanks for that.
$endgroup$
– SamGibson
Dec 16 '18 at 13:41
|
show 4 more comments
$begingroup$
When the battery is open you are measuring an open cell voltage. When the battery is in the system it's closed cell voltage under load. You are dropping some voltage across the internal impedance of the battery because your system is drawing current when the measurement is being made (so at the terminals the voltage is indeed lower). So both measurements MCU and multimeter are correct, the difference is that the multimeter is >1Mohm load while the MCU is much lower (since probably drawing at least mAs of power).
There may be another effect at play. Batteries do exhibit a recovery phenomenon where if left open cell with no load some of the voltage will recover after a time interval.
answered Dec 13 '18 at 19:23
Gonzik007Gonzik007
2,8161024
$endgroup$
When the battery is open you are measuring an open cell voltage. When the battery is in the system it's closed cell voltage under load. You are dropping some voltage across the internal impedance of the battery because your system is drawing current when the measurement is being made (so at the terminals the voltage is indeed lower). So both measurements MCU and multimeter are correct, the difference is that the multimeter is >1Mohm load while the MCU is much lower (since probably drawing at least mAs of power).
There may be another effect at play. Batteries do exhibit a recovery phenomenon where if left open cell with no load some of the voltage will recover after a time interval.
answered Dec 13 '18 at 19:23
Gonzik007Gonzik007
2,8161024
answered Dec 13 '18 at 19:23
Gonzik007Gonzik007
2,8161024
answered Dec 13 '18 at 19:23
Gonzik007Gonzik007
2,8161024
answered Dec 13 '18 at 19:23
Gonzik007Gonzik007
2,8161024
2,8161024
1
$begingroup$
So I want my microcontroller to go to sleep when my battery is at 3.6V. The connected closed cell votlage under load seems to be 0.2V less than the open cell voltage. Should I sleep when my microcontroller reads 3.4V? Or just sleep when it reads 3.6V even though it would still show 3.8V open cell?
$endgroup$
– Tapatio Sombrero
Dec 13 '18 at 20:56
$begingroup$
That is one option. You are correlating your open and closed cell voltages and assuming linear mapping. However, keep in mind that various cells may have different internal impedance. If you are mass producing a product corner/data studies must be done to figure out what the mapping should be. Another option is to place the system in the lowest power state possible (turn off all peripherals and functions expect MCU at some low power state). If you can then measure voltage under a very light load your MCU and multi-meter readings should be closer and have less error.
$endgroup$
– Gonzik007
Dec 13 '18 at 22:40
$begingroup$
Also make sure to check your battery chemistry. Make sure that there is no hard requirement for safe shutdown. You want to make sure that your measurement error can never place the cell into some unsafe state. With the voltages you list, and assuming lithium battery you should be ok, but just thought it worth mentioning.
$endgroup$
– Gonzik007
Dec 13 '18 at 22:41
1
$begingroup$
@fishinear - "The OP gets notifications of all comments" That's not how it works (assuming you mean the question's "OP"). You commented on an answer, not on the question, yet your comment was implicitly directed to the question's OP - Tapatio. See this Meta.SE post where the accepted answer says "On a question, you will receive notifications automatically for comments on the question only, not the answers" which confirms my experience. Without using "@", the answer writer Gonzik007 was notified of your comment, not the question's OP. HTH
$endgroup$
– SamGibson
Dec 16 '18 at 13:40
1
$begingroup$
@fishinear - "pretty much all CPUs I know of have brown-out detectors" Yes, but those reset the MCU, they don't cause it to enter sleep mode. Different things. That's why your comment that "Normally you'd want to put the CPU to sleep just before it will stop functioning because of too low voltage [...] most micro controllers do that themselves already, so no need for you to do it" was confusing, as it seemed to suggest that MCUs put themselves to sleep in a low-voltage condition. Now you've clarified that you meant that they reset due to BOR - thanks for that.
$endgroup$
– SamGibson
Dec 16 '18 at 13:41
|
show 4 more comments
1
$begingroup$
So I want my microcontroller to go to sleep when my battery is at 3.6V. The connected closed cell votlage under load seems to be 0.2V less than the open cell voltage. Should I sleep when my microcontroller reads 3.4V? Or just sleep when it reads 3.6V even though it would still show 3.8V open cell?
$endgroup$
– Tapatio Sombrero
Dec 13 '18 at 20:56
$begingroup$
That is one option. You are correlating your open and closed cell voltages and assuming linear mapping. However, keep in mind that various cells may have different internal impedance. If you are mass producing a product corner/data studies must be done to figure out what the mapping should be. Another option is to place the system in the lowest power state possible (turn off all peripherals and functions expect MCU at some low power state). If you can then measure voltage under a very light load your MCU and multi-meter readings should be closer and have less error.
$endgroup$
– Gonzik007
Dec 13 '18 at 22:40
$begingroup$
Also make sure to check your battery chemistry. Make sure that there is no hard requirement for safe shutdown. You want to make sure that your measurement error can never place the cell into some unsafe state. With the voltages you list, and assuming lithium battery you should be ok, but just thought it worth mentioning.
$endgroup$
– Gonzik007
Dec 13 '18 at 22:41
1
$begingroup$
@fishinear - "The OP gets notifications of all comments" That's not how it works (assuming you mean the question's "OP"). You commented on an answer, not on the question, yet your comment was implicitly directed to the question's OP - Tapatio. See this Meta.SE post where the accepted answer says "On a question, you will receive notifications automatically for comments on the question only, not the answers" which confirms my experience. Without using "@", the answer writer Gonzik007 was notified of your comment, not the question's OP. HTH
$endgroup$
– SamGibson
Dec 16 '18 at 13:40
1
$begingroup$
@fishinear - "pretty much all CPUs I know of have brown-out detectors" Yes, but those reset the MCU, they don't cause it to enter sleep mode. Different things. That's why your comment that "Normally you'd want to put the CPU to sleep just before it will stop functioning because of too low voltage [...] most micro controllers do that themselves already, so no need for you to do it" was confusing, as it seemed to suggest that MCUs put themselves to sleep in a low-voltage condition. Now you've clarified that you meant that they reset due to BOR - thanks for that.
$endgroup$
– SamGibson
Dec 16 '18 at 13:41
1
1
$begingroup$
So I want my microcontroller to go to sleep when my battery is at 3.6V. The connected closed cell votlage under load seems to be 0.2V less than the open cell voltage. Should I sleep when my microcontroller reads 3.4V? Or just sleep when it reads 3.6V even though it would still show 3.8V open cell?
$endgroup$
– Tapatio Sombrero
Dec 13 '18 at 20:56
$begingroup$
So I want my microcontroller to go to sleep when my battery is at 3.6V. The connected closed cell votlage under load seems to be 0.2V less than the open cell voltage. Should I sleep when my microcontroller reads 3.4V? Or just sleep when it reads 3.6V even though it would still show 3.8V open cell?
$endgroup$
– Tapatio Sombrero
Dec 13 '18 at 20:56
$begingroup$
That is one option. You are correlating your open and closed cell voltages and assuming linear mapping. However, keep in mind that various cells may have different internal impedance. If you are mass producing a product corner/data studies must be done to figure out what the mapping should be. Another option is to place the system in the lowest power state possible (turn off all peripherals and functions expect MCU at some low power state). If you can then measure voltage under a very light load your MCU and multi-meter readings should be closer and have less error.
$endgroup$
– Gonzik007
Dec 13 '18 at 22:40
$begingroup$
That is one option. You are correlating your open and closed cell voltages and assuming linear mapping. However, keep in mind that various cells may have different internal impedance. If you are mass producing a product corner/data studies must be done to figure out what the mapping should be. Another option is to place the system in the lowest power state possible (turn off all peripherals and functions expect MCU at some low power state). If you can then measure voltage under a very light load your MCU and multi-meter readings should be closer and have less error.
$endgroup$
– Gonzik007
Dec 13 '18 at 22:40
$begingroup$
Also make sure to check your battery chemistry. Make sure that there is no hard requirement for safe shutdown. You want to make sure that your measurement error can never place the cell into some unsafe state. With the voltages you list, and assuming lithium battery you should be ok, but just thought it worth mentioning.
$endgroup$
– Gonzik007
Dec 13 '18 at 22:41
$begingroup$
Also make sure to check your battery chemistry. Make sure that there is no hard requirement for safe shutdown. You want to make sure that your measurement error can never place the cell into some unsafe state. With the voltages you list, and assuming lithium battery you should be ok, but just thought it worth mentioning.
$endgroup$
– Gonzik007
Dec 13 '18 at 22:41
1
1
$begingroup$
@fishinear - "The OP gets notifications of all comments" That's not how it works (assuming you mean the question's "OP"). You commented on an answer, not on the question, yet your comment was implicitly directed to the question's OP - Tapatio. See this Meta.SE post where the accepted answer says "On a question, you will receive notifications automatically for comments on the question only, not the answers" which confirms my experience. Without using "@", the answer writer Gonzik007 was notified of your comment, not the question's OP. HTH
$endgroup$
– SamGibson
Dec 16 '18 at 13:40
$begingroup$
@fishinear - "The OP gets notifications of all comments" That's not how it works (assuming you mean the question's "OP"). You commented on an answer, not on the question, yet your comment was implicitly directed to the question's OP - Tapatio. See this Meta.SE post where the accepted answer says "On a question, you will receive notifications automatically for comments on the question only, not the answers" which confirms my experience. Without using "@", the answer writer Gonzik007 was notified of your comment, not the question's OP. HTH
$endgroup$
– SamGibson
Dec 16 '18 at 13:40
1
1
$begingroup$
@fishinear - "pretty much all CPUs I know of have brown-out detectors" Yes, but those reset the MCU, they don't cause it to enter sleep mode. Different things. That's why your comment that "Normally you'd want to put the CPU to sleep just before it will stop functioning because of too low voltage [...] most micro controllers do that themselves already, so no need for you to do it" was confusing, as it seemed to suggest that MCUs put themselves to sleep in a low-voltage condition. Now you've clarified that you meant that they reset due to BOR - thanks for that.
$endgroup$
– SamGibson
Dec 16 '18 at 13:41
$begingroup$
@fishinear - "pretty much all CPUs I know of have brown-out detectors" Yes, but those reset the MCU, they don't cause it to enter sleep mode. Different things. That's why your comment that "Normally you'd want to put the CPU to sleep just before it will stop functioning because of too low voltage [...] most micro controllers do that themselves already, so no need for you to do it" was confusing, as it seemed to suggest that MCUs put themselves to sleep in a low-voltage condition. Now you've clarified that you meant that they reset due to BOR - thanks for that.
$endgroup$
– SamGibson
Dec 16 '18 at 13:41
|
show 4 more comments
$begingroup$
Every battery has a certain amount of output resistance. What happens if current flows through a resistor? Yes, a voltage drop! So the more current you draw from the battery, the lower the output voltage is.
answered Dec 13 '18 at 19:23
Stefan WyssStefan Wyss
2,0891313
$endgroup$
add a comment |
$begingroup$
Every battery has a certain amount of output resistance. What happens if current flows through a resistor? Yes, a voltage drop! So the more current you draw from the battery, the lower the output voltage is.
answered Dec 13 '18 at 19:23
Stefan WyssStefan Wyss
2,0891313
$endgroup$
add a comment |
$begingroup$
Every battery has a certain amount of output resistance. What happens if current flows through a resistor? Yes, a voltage drop! So the more current you draw from the battery, the lower the output voltage is.
answered Dec 13 '18 at 19:23
Stefan WyssStefan Wyss
2,0891313
$endgroup$
Every battery has a certain amount of output resistance. What happens if current flows through a resistor? Yes, a voltage drop! So the more current you draw from the battery, the lower the output voltage is.
answered Dec 13 '18 at 19:23
Stefan WyssStefan Wyss
2,0891313
answered Dec 13 '18 at 19:23
Stefan WyssStefan Wyss
2,0891313
answered Dec 13 '18 at 19:23
Stefan WyssStefan Wyss
2,0891313
answered Dec 13 '18 at 19:23
Stefan WyssStefan Wyss
2,0891313
2,0891313
add a comment |
add a comment |
$begingroup$
This is true of all power supplies
Indeed, batteries sag their voltage on being loaded. So does everything else.
The main culprit is Ohm's Law, E=IR, where voltage drop across any conductor is proportional to its amperage drawn.
Part of a battery's sag is chemical, but part is simply the Ohm's Law resistance of its internal components.
Let's suppose you have a mad gaming rig with 4 paralleled video cards, the combo pulls 1000 watts when gaming. But it's just sitting at the Windows Homescreen and only pulling 100 watts. The power cables are carrrying 20A@5V, and dropping 0.01 volts, so the cards get 4.99 volts. (The wires are 2000 Siemens == 1/2000 Ohms.)
At this light load, the AC power supply is inefficient and poor power factor, so it is drawing 240VA or 2 amps off the 120V mains. The branch circuit wiring back to the panel is dropping 0.4 volts. Conductance is 5 Siemens == 1/5 ohm.
Now you fire up your most demanding game. Pulling 200A at 5V, the resistive losses alone inside your PC's wiring jumps to 0.1 volts. So the cards get 4.90 volts. That's a drop.
Meanwhile, the power supply draws 10A (1200VA) from the AC mains. Wiring voltage drop predictably increases to 2.0 volts, so voltage at the power suppy is 118V. Most likely a switching power supply pulls a skitch more current to compensate, otherwise its output voltage would sag also.
No current is being drawn on safety ground, so it's not dropping. Measured from ground, neutral is 1 volt and hot is 119 volts. And we can use this to affirm correct wiring. It's like the pointer bar on a torque wrench, it doesn't bend.
Of course, similar drops are happening all the way back to the power plant. There, increased load (in amps) sags the voltage because of the internal resistance of the generator, but also due to turbine horsepower. VA=W. If A increases beyond spec, V must decrease in proportion so W can remain within the turbine's ability. Having the turbine bog and slow down is not an option, because it's AC power and must stay in sync.
answered Dec 15 '18 at 2:12
HarperHarper
6,299826
$endgroup$
add a comment |
$begingroup$
This is true of all power supplies
Indeed, batteries sag their voltage on being loaded. So does everything else.
The main culprit is Ohm's Law, E=IR, where voltage drop across any conductor is proportional to its amperage drawn.
Part of a battery's sag is chemical, but part is simply the Ohm's Law resistance of its internal components.
Let's suppose you have a mad gaming rig with 4 paralleled video cards, the combo pulls 1000 watts when gaming. But it's just sitting at the Windows Homescreen and only pulling 100 watts. The power cables are carrrying 20A@5V, and dropping 0.01 volts, so the cards get 4.99 volts. (The wires are 2000 Siemens == 1/2000 Ohms.)
At this light load, the AC power supply is inefficient and poor power factor, so it is drawing 240VA or 2 amps off the 120V mains. The branch circuit wiring back to the panel is dropping 0.4 volts. Conductance is 5 Siemens == 1/5 ohm.
Now you fire up your most demanding game. Pulling 200A at 5V, the resistive losses alone inside your PC's wiring jumps to 0.1 volts. So the cards get 4.90 volts. That's a drop.
Meanwhile, the power supply draws 10A (1200VA) from the AC mains. Wiring voltage drop predictably increases to 2.0 volts, so voltage at the power suppy is 118V. Most likely a switching power supply pulls a skitch more current to compensate, otherwise its output voltage would sag also.
No current is being drawn on safety ground, so it's not dropping. Measured from ground, neutral is 1 volt and hot is 119 volts. And we can use this to affirm correct wiring. It's like the pointer bar on a torque wrench, it doesn't bend.
Of course, similar drops are happening all the way back to the power plant. There, increased load (in amps) sags the voltage because of the internal resistance of the generator, but also due to turbine horsepower. VA=W. If A increases beyond spec, V must decrease in proportion so W can remain within the turbine's ability. Having the turbine bog and slow down is not an option, because it's AC power and must stay in sync.
answered Dec 15 '18 at 2:12
HarperHarper
6,299826
$endgroup$
add a comment |
$begingroup$
This is true of all power supplies
Indeed, batteries sag their voltage on being loaded. So does everything else.
The main culprit is Ohm's Law, E=IR, where voltage drop across any conductor is proportional to its amperage drawn.
Part of a battery's sag is chemical, but part is simply the Ohm's Law resistance of its internal components.
Let's suppose you have a mad gaming rig with 4 paralleled video cards, the combo pulls 1000 watts when gaming. But it's just sitting at the Windows Homescreen and only pulling 100 watts. The power cables are carrrying 20A@5V, and dropping 0.01 volts, so the cards get 4.99 volts. (The wires are 2000 Siemens == 1/2000 Ohms.)
At this light load, the AC power supply is inefficient and poor power factor, so it is drawing 240VA or 2 amps off the 120V mains. The branch circuit wiring back to the panel is dropping 0.4 volts. Conductance is 5 Siemens == 1/5 ohm.
Now you fire up your most demanding game. Pulling 200A at 5V, the resistive losses alone inside your PC's wiring jumps to 0.1 volts. So the cards get 4.90 volts. That's a drop.
Meanwhile, the power supply draws 10A (1200VA) from the AC mains. Wiring voltage drop predictably increases to 2.0 volts, so voltage at the power suppy is 118V. Most likely a switching power supply pulls a skitch more current to compensate, otherwise its output voltage would sag also.
No current is being drawn on safety ground, so it's not dropping. Measured from ground, neutral is 1 volt and hot is 119 volts. And we can use this to affirm correct wiring. It's like the pointer bar on a torque wrench, it doesn't bend.
Of course, similar drops are happening all the way back to the power plant. There, increased load (in amps) sags the voltage because of the internal resistance of the generator, but also due to turbine horsepower. VA=W. If A increases beyond spec, V must decrease in proportion so W can remain within the turbine's ability. Having the turbine bog and slow down is not an option, because it's AC power and must stay in sync.
answered Dec 15 '18 at 2:12
HarperHarper
6,299826
$endgroup$
This is true of all power supplies
Indeed, batteries sag their voltage on being loaded. So does everything else.
The main culprit is Ohm's Law, E=IR, where voltage drop across any conductor is proportional to its amperage drawn.
Part of a battery's sag is chemical, but part is simply the Ohm's Law resistance of its internal components.
Let's suppose you have a mad gaming rig with 4 paralleled video cards, the combo pulls 1000 watts when gaming. But it's just sitting at the Windows Homescreen and only pulling 100 watts. The power cables are carrrying 20A@5V, and dropping 0.01 volts, so the cards get 4.99 volts. (The wires are 2000 Siemens == 1/2000 Ohms.)
At this light load, the AC power supply is inefficient and poor power factor, so it is drawing 240VA or 2 amps off the 120V mains. The branch circuit wiring back to the panel is dropping 0.4 volts. Conductance is 5 Siemens == 1/5 ohm.
Now you fire up your most demanding game. Pulling 200A at 5V, the resistive losses alone inside your PC's wiring jumps to 0.1 volts. So the cards get 4.90 volts. That's a drop.
Meanwhile, the power supply draws 10A (1200VA) from the AC mains. Wiring voltage drop predictably increases to 2.0 volts, so voltage at the power suppy is 118V. Most likely a switching power supply pulls a skitch more current to compensate, otherwise its output voltage would sag also.
No current is being drawn on safety ground, so it's not dropping. Measured from ground, neutral is 1 volt and hot is 119 volts. And we can use this to affirm correct wiring. It's like the pointer bar on a torque wrench, it doesn't bend.
Of course, similar drops are happening all the way back to the power plant. There, increased load (in amps) sags the voltage because of the internal resistance of the generator, but also due to turbine horsepower. VA=W. If A increases beyond spec, V must decrease in proportion so W can remain within the turbine's ability. Having the turbine bog and slow down is not an option, because it's AC power and must stay in sync.
answered Dec 15 '18 at 2:12
HarperHarper
6,299826
answered Dec 15 '18 at 2:12
HarperHarper
6,299826
answered Dec 15 '18 at 2:12
HarperHarper
6,299826
answered Dec 15 '18 at 2:12
HarperHarper
6,299826
6,299826
add a comment |
add a comment |
$begingroup$
All batteries have a memory effect when unloaded such that they return slowly to near the previous voltage after a short burst load. There is also momentary quick drop in voltage due to a load of ESR*I = Δ V.
So both measurements must be taken at the same time to check calibration for errors and consider the amount of hysteresis thresholds required to prevent oscillation of sleep, wake-up cycles.
The memory effect time constant can be several to many minutes depending on the "no-load" leakage current after a load.
Because of these combined effects which might be computed for a given cell (ΔV= ESR * V/Rload + t/ESR * C2) the cut-off voltage is often lowered to capture the charge stored in memory capacitance C2 as long you know it returns to the safe Vmin threshold. Battery rapid aging occurs for the amount of time below its Vmin threshold.
Review the battery datasheet for details.
answered Dec 13 '18 at 19:27
Sunnyskyguy EE75Sunnyskyguy EE75
67.4k22397
$endgroup$
add a comment |
$begingroup$
All batteries have a memory effect when unloaded such that they return slowly to near the previous voltage after a short burst load. There is also momentary quick drop in voltage due to a load of ESR*I = Δ V.
So both measurements must be taken at the same time to check calibration for errors and consider the amount of hysteresis thresholds required to prevent oscillation of sleep, wake-up cycles.
The memory effect time constant can be several to many minutes depending on the "no-load" leakage current after a load.
Because of these combined effects which might be computed for a given cell (ΔV= ESR * V/Rload + t/ESR * C2) the cut-off voltage is often lowered to capture the charge stored in memory capacitance C2 as long you know it returns to the safe Vmin threshold. Battery rapid aging occurs for the amount of time below its Vmin threshold.
Review the battery datasheet for details.
answered Dec 13 '18 at 19:27
Sunnyskyguy EE75Sunnyskyguy EE75
67.4k22397
$endgroup$
add a comment |
$begingroup$
All batteries have a memory effect when unloaded such that they return slowly to near the previous voltage after a short burst load. There is also momentary quick drop in voltage due to a load of ESR*I = Δ V.
So both measurements must be taken at the same time to check calibration for errors and consider the amount of hysteresis thresholds required to prevent oscillation of sleep, wake-up cycles.
The memory effect time constant can be several to many minutes depending on the "no-load" leakage current after a load.
Because of these combined effects which might be computed for a given cell (ΔV= ESR * V/Rload + t/ESR * C2) the cut-off voltage is often lowered to capture the charge stored in memory capacitance C2 as long you know it returns to the safe Vmin threshold. Battery rapid aging occurs for the amount of time below its Vmin threshold.
Review the battery datasheet for details.
answered Dec 13 '18 at 19:27
Sunnyskyguy EE75Sunnyskyguy EE75
67.4k22397
$endgroup$
All batteries have a memory effect when unloaded such that they return slowly to near the previous voltage after a short burst load. There is also momentary quick drop in voltage due to a load of ESR*I = Δ V.
So both measurements must be taken at the same time to check calibration for errors and consider the amount of hysteresis thresholds required to prevent oscillation of sleep, wake-up cycles.
The memory effect time constant can be several to many minutes depending on the "no-load" leakage current after a load.
Because of these combined effects which might be computed for a given cell (ΔV= ESR * V/Rload + t/ESR * C2) the cut-off voltage is often lowered to capture the charge stored in memory capacitance C2 as long you know it returns to the safe Vmin threshold. Battery rapid aging occurs for the amount of time below its Vmin threshold.
Review the battery datasheet for details.
answered Dec 13 '18 at 19:27
Sunnyskyguy EE75Sunnyskyguy EE75
67.4k22397
edited Dec 13 '18 at 19:36
answered Dec 13 '18 at 19:27
Sunnyskyguy EE75Sunnyskyguy EE75
67.4k22397
answered Dec 13 '18 at 19:27
Sunnyskyguy EE75Sunnyskyguy EE75
67.4k22397
answered Dec 13 '18 at 19:27
Sunnyskyguy EE75Sunnyskyguy EE75
67.4k22397
67.4k22397
add a comment |
add a comment |
$begingroup$
There is drop in voltage due internal resistance of the battery coming into play so you will see the voltage drop by a value of i * r ( where i is the current flowing and r is the internal resistance of the battery)
answered Dec 16 '18 at 5:12
Vinu PhilipVinu Philip
111
$endgroup$
add a comment |
$begingroup$
There is drop in voltage due internal resistance of the battery coming into play so you will see the voltage drop by a value of i * r ( where i is the current flowing and r is the internal resistance of the battery)
answered Dec 16 '18 at 5:12
Vinu PhilipVinu Philip
111
$endgroup$
add a comment |
$begingroup$
There is drop in voltage due internal resistance of the battery coming into play so you will see the voltage drop by a value of i * r ( where i is the current flowing and r is the internal resistance of the battery)
answered Dec 16 '18 at 5:12
Vinu PhilipVinu Philip
111
$endgroup$
There is drop in voltage due internal resistance of the battery coming into play so you will see the voltage drop by a value of i * r ( where i is the current flowing and r is the internal resistance of the battery)
answered Dec 16 '18 at 5:12
Vinu PhilipVinu Philip
111
answered Dec 16 '18 at 5:12
Vinu PhilipVinu Philip
111
answered Dec 16 '18 at 5:12
Vinu PhilipVinu Philip
111
answered Dec 16 '18 at 5:12
Vinu PhilipVinu Philip
111
111
add a comment |
add a comment |
$begingroup$
A new battery will have much less loaded voltage drop than you have.
An old, worn out, or damaged Lithium battery has a much higher internal resistance than a new battery. It is damaged if it has been fully charged for longer than a few months, if it has been discharged too low or if it has had too many charge-discharge cycles.
answered Dec 14 '18 at 0:28
AudioguruAudioguru
46213
$endgroup$
add a comment |
$begingroup$
A new battery will have much less loaded voltage drop than you have.
An old, worn out, or damaged Lithium battery has a much higher internal resistance than a new battery. It is damaged if it has been fully charged for longer than a few months, if it has been discharged too low or if it has had too many charge-discharge cycles.
answered Dec 14 '18 at 0:28
AudioguruAudioguru
46213
$endgroup$
add a comment |
$begingroup$
A new battery will have much less loaded voltage drop than you have.
An old, worn out, or damaged Lithium battery has a much higher internal resistance than a new battery. It is damaged if it has been fully charged for longer than a few months, if it has been discharged too low or if it has had too many charge-discharge cycles.
answered Dec 14 '18 at 0:28
AudioguruAudioguru
46213
$endgroup$
A new battery will have much less loaded voltage drop than you have.
An old, worn out, or damaged Lithium battery has a much higher internal resistance than a new battery. It is damaged if it has been fully charged for longer than a few months, if it has been discharged too low or if it has had too many charge-discharge cycles.
answered Dec 14 '18 at 0:28
AudioguruAudioguru
46213
answered Dec 14 '18 at 0:28
AudioguruAudioguru
46213
answered Dec 14 '18 at 0:28
AudioguruAudioguru
46213
answered Dec 14 '18 at 0:28
AudioguruAudioguru
46213
46213
add a comment |
add a comment |
$begingroup$
While all the other answers are great and tell what I would tell you too (the battery voltage actually get's lower when there is a load), I'd like to add something:
The reason which is presented that the voltage drops is the "internal resistance". I want to mention that the model of an internal resistance is JUST A MODEL that works very well in modelling the properties of a voltage source, while at the same time being simple and easy to calculate.
In reality, it's more complicated. The resistance of the internal components of the battery that current has to pass through (I intentionally don't call those "internal resistance", because this is a term from the above mentioned model) plays a role, but it's not the only role.
In most batteries there is a chemical reaction going on that separates charges at some border layer. This chemical reaction follows the laws of statistical physics. It comes to halt when the (chemical equilibrium is reached. The separation of charges generates the voltage that you can measure, and this voltage is a factor in the chemical equilibrium (the higher the voltage is, the less separation occurs to create a new pair of separated charges).
When you attach a load now, you take away charges at constant intervals (because there is an electrical current). If the system reaches the equilibrium situation now, the amount of separated charges and voltage will be less (because more charges have to be created).
answered Dec 16 '18 at 9:52
QuantumwhispQuantumwhisp
1335
$endgroup$
add a comment |
$begingroup$
While all the other answers are great and tell what I would tell you too (the battery voltage actually get's lower when there is a load), I'd like to add something:
The reason which is presented that the voltage drops is the "internal resistance". I want to mention that the model of an internal resistance is JUST A MODEL that works very well in modelling the properties of a voltage source, while at the same time being simple and easy to calculate.
In reality, it's more complicated. The resistance of the internal components of the battery that current has to pass through (I intentionally don't call those "internal resistance", because this is a term from the above mentioned model) plays a role, but it's not the only role.
In most batteries there is a chemical reaction going on that separates charges at some border layer. This chemical reaction follows the laws of statistical physics. It comes to halt when the (chemical equilibrium is reached. The separation of charges generates the voltage that you can measure, and this voltage is a factor in the chemical equilibrium (the higher the voltage is, the less separation occurs to create a new pair of separated charges).
When you attach a load now, you take away charges at constant intervals (because there is an electrical current). If the system reaches the equilibrium situation now, the amount of separated charges and voltage will be less (because more charges have to be created).
answered Dec 16 '18 at 9:52
QuantumwhispQuantumwhisp
1335
$endgroup$
add a comment |
$begingroup$
While all the other answers are great and tell what I would tell you too (the battery voltage actually get's lower when there is a load), I'd like to add something:
The reason which is presented that the voltage drops is the "internal resistance". I want to mention that the model of an internal resistance is JUST A MODEL that works very well in modelling the properties of a voltage source, while at the same time being simple and easy to calculate.
In reality, it's more complicated. The resistance of the internal components of the battery that current has to pass through (I intentionally don't call those "internal resistance", because this is a term from the above mentioned model) plays a role, but it's not the only role.
In most batteries there is a chemical reaction going on that separates charges at some border layer. This chemical reaction follows the laws of statistical physics. It comes to halt when the (chemical equilibrium is reached. The separation of charges generates the voltage that you can measure, and this voltage is a factor in the chemical equilibrium (the higher the voltage is, the less separation occurs to create a new pair of separated charges).
When you attach a load now, you take away charges at constant intervals (because there is an electrical current). If the system reaches the equilibrium situation now, the amount of separated charges and voltage will be less (because more charges have to be created).
answered Dec 16 '18 at 9:52
QuantumwhispQuantumwhisp
1335
$endgroup$
While all the other answers are great and tell what I would tell you too (the battery voltage actually get's lower when there is a load), I'd like to add something:
The reason which is presented that the voltage drops is the "internal resistance". I want to mention that the model of an internal resistance is JUST A MODEL that works very well in modelling the properties of a voltage source, while at the same time being simple and easy to calculate.
In reality, it's more complicated. The resistance of the internal components of the battery that current has to pass through (I intentionally don't call those "internal resistance", because this is a term from the above mentioned model) plays a role, but it's not the only role.
In most batteries there is a chemical reaction going on that separates charges at some border layer. This chemical reaction follows the laws of statistical physics. It comes to halt when the (chemical equilibrium is reached. The separation of charges generates the voltage that you can measure, and this voltage is a factor in the chemical equilibrium (the higher the voltage is, the less separation occurs to create a new pair of separated charges).
When you attach a load now, you take away charges at constant intervals (because there is an electrical current). If the system reaches the equilibrium situation now, the amount of separated charges and voltage will be less (because more charges have to be created).
answered Dec 16 '18 at 9:52
QuantumwhispQuantumwhisp
1335
answered Dec 16 '18 at 9:52
QuantumwhispQuantumwhisp
1335
answered Dec 16 '18 at 9:52
QuantumwhispQuantumwhisp
1335
answered Dec 16 '18 at 9:52
QuantumwhispQuantumwhisp
1335
1335
add a comment |
add a comment |
$begingroup$
Battery voltage doesn't usually drop just because there's a load connected. But the measured voltage does tend to drop
Here's what you've got to know about voltage measurements
A voltmeter uses a resistor with a very high resistance. Ideally, it's infinite. The voltmeter measures the voltage across this resistor.
So when you connect a battery to your voltmeter, the internal resistance of the battery is insignificant compared to the voltmeter's resistance. So most of the voltage drop occurs across the voltmeter's resistance and not the battery's internal resistor. Hence, you measure the right voltage.
However, your microcontroller may have a resistance that's not too high.
If the battery had an internal resistance of say, 1 milli ohm, and the voltmeter was using a 24000 ohm resistor, this error is expected.
answered Dec 14 '18 at 8:09
Pradyoth ShandilyaPradyoth Shandilya
486
$endgroup$
5
$begingroup$
There's no need to make things more confusing by talking about multimeter design. An ideal (infinite-resistance) voltmeter would experience the same effect. The issue is not comparing voltmeter resistance to the battery's internal resistance, but comparing the battery's internal resistance to the load resistance.
$endgroup$
– Sneftel
Dec 14 '18 at 8:37
$begingroup$
The multimeter's design explains why there's a difference in the readings. That's something anyone dealing with basic electronics must know
$endgroup$
– Pradyoth Shandilya
Dec 14 '18 at 8:41
7
$begingroup$
No, it doesn't explain that. There's a difference in the readings because the voltage is different. Even if there were no multimeter being used, the actual voltage across the battery is actually different when a load is connected.
$endgroup$
– Sneftel
Dec 14 '18 at 10:21
add a comment |
$begingroup$
Battery voltage doesn't usually drop just because there's a load connected. But the measured voltage does tend to drop
Here's what you've got to know about voltage measurements
A voltmeter uses a resistor with a very high resistance. Ideally, it's infinite. The voltmeter measures the voltage across this resistor.
So when you connect a battery to your voltmeter, the internal resistance of the battery is insignificant compared to the voltmeter's resistance. So most of the voltage drop occurs across the voltmeter's resistance and not the battery's internal resistor. Hence, you measure the right voltage.
However, your microcontroller may have a resistance that's not too high.
If the battery had an internal resistance of say, 1 milli ohm, and the voltmeter was using a 24000 ohm resistor, this error is expected.
answered Dec 14 '18 at 8:09
Pradyoth ShandilyaPradyoth Shandilya
486
$endgroup$
5
$begingroup$
There's no need to make things more confusing by talking about multimeter design. An ideal (infinite-resistance) voltmeter would experience the same effect. The issue is not comparing voltmeter resistance to the battery's internal resistance, but comparing the battery's internal resistance to the load resistance.
$endgroup$
– Sneftel
Dec 14 '18 at 8:37
$begingroup$
The multimeter's design explains why there's a difference in the readings. That's something anyone dealing with basic electronics must know
$endgroup$
– Pradyoth Shandilya
Dec 14 '18 at 8:41
7
$begingroup$
No, it doesn't explain that. There's a difference in the readings because the voltage is different. Even if there were no multimeter being used, the actual voltage across the battery is actually different when a load is connected.
$endgroup$
– Sneftel
Dec 14 '18 at 10:21
add a comment |
$begingroup$
Battery voltage doesn't usually drop just because there's a load connected. But the measured voltage does tend to drop
Here's what you've got to know about voltage measurements
A voltmeter uses a resistor with a very high resistance. Ideally, it's infinite. The voltmeter measures the voltage across this resistor.
So when you connect a battery to your voltmeter, the internal resistance of the battery is insignificant compared to the voltmeter's resistance. So most of the voltage drop occurs across the voltmeter's resistance and not the battery's internal resistor. Hence, you measure the right voltage.
However, your microcontroller may have a resistance that's not too high.
If the battery had an internal resistance of say, 1 milli ohm, and the voltmeter was using a 24000 ohm resistor, this error is expected.
answered Dec 14 '18 at 8:09
Pradyoth ShandilyaPradyoth Shandilya
486
$endgroup$
Battery voltage doesn't usually drop just because there's a load connected. But the measured voltage does tend to drop
Here's what you've got to know about voltage measurements
A voltmeter uses a resistor with a very high resistance. Ideally, it's infinite. The voltmeter measures the voltage across this resistor.
So when you connect a battery to your voltmeter, the internal resistance of the battery is insignificant compared to the voltmeter's resistance. So most of the voltage drop occurs across the voltmeter's resistance and not the battery's internal resistor. Hence, you measure the right voltage.
However, your microcontroller may have a resistance that's not too high.
If the battery had an internal resistance of say, 1 milli ohm, and the voltmeter was using a 24000 ohm resistor, this error is expected.
answered Dec 14 '18 at 8:09
Pradyoth ShandilyaPradyoth Shandilya
486
edited Dec 14 '18 at 9:33
answered Dec 14 '18 at 8:09
Pradyoth ShandilyaPradyoth Shandilya
486
answered Dec 14 '18 at 8:09
Pradyoth ShandilyaPradyoth Shandilya
486
answered Dec 14 '18 at 8:09
Pradyoth ShandilyaPradyoth Shandilya
486
486
5
$begingroup$
There's no need to make things more confusing by talking about multimeter design. An ideal (infinite-resistance) voltmeter would experience the same effect. The issue is not comparing voltmeter resistance to the battery's internal resistance, but comparing the battery's internal resistance to the load resistance.
$endgroup$
– Sneftel
Dec 14 '18 at 8:37
$begingroup$
The multimeter's design explains why there's a difference in the readings. That's something anyone dealing with basic electronics must know
$endgroup$
– Pradyoth Shandilya
Dec 14 '18 at 8:41
7
$begingroup$
No, it doesn't explain that. There's a difference in the readings because the voltage is different. Even if there were no multimeter being used, the actual voltage across the battery is actually different when a load is connected.
$endgroup$
– Sneftel
Dec 14 '18 at 10:21
add a comment |
5
$begingroup$
There's no need to make things more confusing by talking about multimeter design. An ideal (infinite-resistance) voltmeter would experience the same effect. The issue is not comparing voltmeter resistance to the battery's internal resistance, but comparing the battery's internal resistance to the load resistance.
$endgroup$
– Sneftel
Dec 14 '18 at 8:37
$begingroup$
The multimeter's design explains why there's a difference in the readings. That's something anyone dealing with basic electronics must know
$endgroup$
– Pradyoth Shandilya
Dec 14 '18 at 8:41
7
$begingroup$
No, it doesn't explain that. There's a difference in the readings because the voltage is different. Even if there were no multimeter being used, the actual voltage across the battery is actually different when a load is connected.
$endgroup$
– Sneftel
Dec 14 '18 at 10:21
5
5
$begingroup$
There's no need to make things more confusing by talking about multimeter design. An ideal (infinite-resistance) voltmeter would experience the same effect. The issue is not comparing voltmeter resistance to the battery's internal resistance, but comparing the battery's internal resistance to the load resistance.
$endgroup$
– Sneftel
Dec 14 '18 at 8:37
$begingroup$
There's no need to make things more confusing by talking about multimeter design. An ideal (infinite-resistance) voltmeter would experience the same effect. The issue is not comparing voltmeter resistance to the battery's internal resistance, but comparing the battery's internal resistance to the load resistance.
$endgroup$
– Sneftel
Dec 14 '18 at 8:37
$begingroup$
The multimeter's design explains why there's a difference in the readings. That's something anyone dealing with basic electronics must know
$endgroup$
– Pradyoth Shandilya
Dec 14 '18 at 8:41
$begingroup$
The multimeter's design explains why there's a difference in the readings. That's something anyone dealing with basic electronics must know
$endgroup$
– Pradyoth Shandilya
Dec 14 '18 at 8:41
7
7
$begingroup$
No, it doesn't explain that. There's a difference in the readings because the voltage is different. Even if there were no multimeter being used, the actual voltage across the battery is actually different when a load is connected.
$endgroup$
– Sneftel
Dec 14 '18 at 10:21
$begingroup$
No, it doesn't explain that. There's a difference in the readings because the voltage is different. Even if there were no multimeter being used, the actual voltage across the battery is actually different when a load is connected.
$endgroup$
– Sneftel
Dec 14 '18 at 10:21
add a comment |
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$begingroup$
Are you measuring the battery (connected vs disconnected) with the multimeter both times, or are you comparing something internal from the microcontroller? Use the multimeter to make the measurement while the controller is connected if you can.
$endgroup$
– mbrig
Dec 13 '18 at 20:06
7
$begingroup$
A car battery has over 13V when not connected, yet drops to 10.5V while starting the engine. Which voltage is correct? Both.
$endgroup$
– Mast
Dec 14 '18 at 4:19
$begingroup$
Just going to add a note. Some batteries, such as lithium ion, are pretty well modeled by the series resistance concept. There is a true cell voltage which is inaccessible, and it is equal to Vout + Iout * ESR. When you remove the load, the voltage recovers quickly. But with lead acid or alkaline batteries, it may take a lot longer to recover to the final open-circuit voltage after removing the load. In other words, it is more complicated than a voltage source in series with a resistor. There is some kind of long time-constant voltage recovery that occurs.
$endgroup$
– mkeith
Dec 16 '18 at 10:05