Calculating Wattage for Resistor in High Frequency Application?












1












$begingroup$


I am making a MOSFET driving circuit.

Frequency : 400 kHz [50% duty cycle]

Gate voltage: 12 V

Total gate charge : 210 nC as per datasheet IRFP460

Rise time: 100 ns

[Q=I*t]

Current: 2.1 A

Gate resistor: V/I > 12/2.1 > 5.7 ohm

Peak power: I * I * R > 2.1 * 2.1 * 5.7 > 25.1370 W

Average power: [Peak Power/Frequency]: 25.1370/400000 > 0.0000628425 [Ws]



1 watt resistor is OK ?






resistors high-frequency







share|improve this question











$endgroup$












  • $begingroup$
    Dividing peak power by frequency doesn't make sense to me. As you say, the units are watt-seconds, not watts.
    $endgroup$
    – Elliot Alderson
    3 hours ago
















1












$begingroup$


I am making a MOSFET driving circuit.

Frequency : 400 kHz [50% duty cycle]

Gate voltage: 12 V

Total gate charge : 210 nC as per datasheet IRFP460

Rise time: 100 ns

[Q=I*t]

Current: 2.1 A

Gate resistor: V/I > 12/2.1 > 5.7 ohm

Peak power: I * I * R > 2.1 * 2.1 * 5.7 > 25.1370 W

Average power: [Peak Power/Frequency]: 25.1370/400000 > 0.0000628425 [Ws]



1 watt resistor is OK ?






resistors high-frequency







share|improve this question











$endgroup$












  • $begingroup$
    Dividing peak power by frequency doesn't make sense to me. As you say, the units are watt-seconds, not watts.
    $endgroup$
    – Elliot Alderson
    3 hours ago














1












1








1





$begingroup$


I am making a MOSFET driving circuit.

Frequency : 400 kHz [50% duty cycle]

Gate voltage: 12 V

Total gate charge : 210 nC as per datasheet IRFP460

Rise time: 100 ns

[Q=I*t]

Current: 2.1 A

Gate resistor: V/I > 12/2.1 > 5.7 ohm

Peak power: I * I * R > 2.1 * 2.1 * 5.7 > 25.1370 W

Average power: [Peak Power/Frequency]: 25.1370/400000 > 0.0000628425 [Ws]



1 watt resistor is OK ?






resistors high-frequency







share|improve this question











$endgroup$




I am making a MOSFET driving circuit.

Frequency : 400 kHz [50% duty cycle]

Gate voltage: 12 V

Total gate charge : 210 nC as per datasheet IRFP460

Rise time: 100 ns

[Q=I*t]

Current: 2.1 A

Gate resistor: V/I > 12/2.1 > 5.7 ohm

Peak power: I * I * R > 2.1 * 2.1 * 5.7 > 25.1370 W

Average power: [Peak Power/Frequency]: 25.1370/400000 > 0.0000628425 [Ws]



1 watt resistor is OK ?






resistors high-frequency




resistors high-frequency






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 3 hours ago









Transistor

87.1k785189




87.1k785189










asked 3 hours ago









Israr SayedIsrar Sayed

204




204












  • $begingroup$
    Dividing peak power by frequency doesn't make sense to me. As you say, the units are watt-seconds, not watts.
    $endgroup$
    – Elliot Alderson
    3 hours ago


















  • $begingroup$
    Dividing peak power by frequency doesn't make sense to me. As you say, the units are watt-seconds, not watts.
    $endgroup$
    – Elliot Alderson
    3 hours ago
















$begingroup$
Dividing peak power by frequency doesn't make sense to me. As you say, the units are watt-seconds, not watts.
$endgroup$
– Elliot Alderson
3 hours ago




$begingroup$
Dividing peak power by frequency doesn't make sense to me. As you say, the units are watt-seconds, not watts.
$endgroup$
– Elliot Alderson
3 hours ago










1 Answer
1






active

oldest

votes


















2












$begingroup$

Dividing the peak power by the frequency is not useful.



Instead, you would multiply it by the duty cycle. If you're dumping 25 W of power into the resistor for 2 × 100 ns out of every 2.5 µs. This would be an average power of



$$25 W cdotfrac{2 cdot 100 ns}{2.5 mu s} = 2 W$$



Clearly, your 1W resistor is not going to cut it!



However, the peak instantaneous power is not really a good estimate of the average power during the switching transient. A better estimate can be arrived at by considering the energy flow into and out of the gate capacitance.



For an R-C circuit, the energy dissipated in the resistor is basically equal to the energy that ends up on the capacitor. If your gate charge is 210 nC and your gate voltage is 12V, this represents



$$Energy = frac{1}{2}cdot Charge cdot Voltage$$



$$0.5 cdot 210 nC cdot 12 V = 1.26 mu J$$



This is the energy you're dumping into the gate capacitance, and then dumping out again on every switching cycle. All of this energy gets dissipated in the gate resistor.



To get the average power, multiply the energy per cycle by the number of cycles per second, giving



$$1.26 mu J cdot 2 cdot 400 kHz = 1.088 W$$



Your 1W resistor would be running at its limit, with no margin. I would use a 2W resistor here.






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    1 Answer
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    $begingroup$

    Dividing the peak power by the frequency is not useful.



    Instead, you would multiply it by the duty cycle. If you're dumping 25 W of power into the resistor for 2 × 100 ns out of every 2.5 µs. This would be an average power of



    $$25 W cdotfrac{2 cdot 100 ns}{2.5 mu s} = 2 W$$



    Clearly, your 1W resistor is not going to cut it!



    However, the peak instantaneous power is not really a good estimate of the average power during the switching transient. A better estimate can be arrived at by considering the energy flow into and out of the gate capacitance.



    For an R-C circuit, the energy dissipated in the resistor is basically equal to the energy that ends up on the capacitor. If your gate charge is 210 nC and your gate voltage is 12V, this represents



    $$Energy = frac{1}{2}cdot Charge cdot Voltage$$



    $$0.5 cdot 210 nC cdot 12 V = 1.26 mu J$$



    This is the energy you're dumping into the gate capacitance, and then dumping out again on every switching cycle. All of this energy gets dissipated in the gate resistor.



    To get the average power, multiply the energy per cycle by the number of cycles per second, giving



    $$1.26 mu J cdot 2 cdot 400 kHz = 1.088 W$$



    Your 1W resistor would be running at its limit, with no margin. I would use a 2W resistor here.






    share|improve this answer











    $endgroup$


















      2












      $begingroup$

      Dividing the peak power by the frequency is not useful.



      Instead, you would multiply it by the duty cycle. If you're dumping 25 W of power into the resistor for 2 × 100 ns out of every 2.5 µs. This would be an average power of



      $$25 W cdotfrac{2 cdot 100 ns}{2.5 mu s} = 2 W$$



      Clearly, your 1W resistor is not going to cut it!



      However, the peak instantaneous power is not really a good estimate of the average power during the switching transient. A better estimate can be arrived at by considering the energy flow into and out of the gate capacitance.



      For an R-C circuit, the energy dissipated in the resistor is basically equal to the energy that ends up on the capacitor. If your gate charge is 210 nC and your gate voltage is 12V, this represents



      $$Energy = frac{1}{2}cdot Charge cdot Voltage$$



      $$0.5 cdot 210 nC cdot 12 V = 1.26 mu J$$



      This is the energy you're dumping into the gate capacitance, and then dumping out again on every switching cycle. All of this energy gets dissipated in the gate resistor.



      To get the average power, multiply the energy per cycle by the number of cycles per second, giving



      $$1.26 mu J cdot 2 cdot 400 kHz = 1.088 W$$



      Your 1W resistor would be running at its limit, with no margin. I would use a 2W resistor here.






      share|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        Dividing the peak power by the frequency is not useful.



        Instead, you would multiply it by the duty cycle. If you're dumping 25 W of power into the resistor for 2 × 100 ns out of every 2.5 µs. This would be an average power of



        $$25 W cdotfrac{2 cdot 100 ns}{2.5 mu s} = 2 W$$



        Clearly, your 1W resistor is not going to cut it!



        However, the peak instantaneous power is not really a good estimate of the average power during the switching transient. A better estimate can be arrived at by considering the energy flow into and out of the gate capacitance.



        For an R-C circuit, the energy dissipated in the resistor is basically equal to the energy that ends up on the capacitor. If your gate charge is 210 nC and your gate voltage is 12V, this represents



        $$Energy = frac{1}{2}cdot Charge cdot Voltage$$



        $$0.5 cdot 210 nC cdot 12 V = 1.26 mu J$$



        This is the energy you're dumping into the gate capacitance, and then dumping out again on every switching cycle. All of this energy gets dissipated in the gate resistor.



        To get the average power, multiply the energy per cycle by the number of cycles per second, giving



        $$1.26 mu J cdot 2 cdot 400 kHz = 1.088 W$$



        Your 1W resistor would be running at its limit, with no margin. I would use a 2W resistor here.






        share|improve this answer











        $endgroup$



        Dividing the peak power by the frequency is not useful.



        Instead, you would multiply it by the duty cycle. If you're dumping 25 W of power into the resistor for 2 × 100 ns out of every 2.5 µs. This would be an average power of



        $$25 W cdotfrac{2 cdot 100 ns}{2.5 mu s} = 2 W$$



        Clearly, your 1W resistor is not going to cut it!



        However, the peak instantaneous power is not really a good estimate of the average power during the switching transient. A better estimate can be arrived at by considering the energy flow into and out of the gate capacitance.



        For an R-C circuit, the energy dissipated in the resistor is basically equal to the energy that ends up on the capacitor. If your gate charge is 210 nC and your gate voltage is 12V, this represents



        $$Energy = frac{1}{2}cdot Charge cdot Voltage$$



        $$0.5 cdot 210 nC cdot 12 V = 1.26 mu J$$



        This is the energy you're dumping into the gate capacitance, and then dumping out again on every switching cycle. All of this energy gets dissipated in the gate resistor.



        To get the average power, multiply the energy per cycle by the number of cycles per second, giving



        $$1.26 mu J cdot 2 cdot 400 kHz = 1.088 W$$



        Your 1W resistor would be running at its limit, with no margin. I would use a 2W resistor here.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 2 hours ago

























        answered 3 hours ago









        Dave Tweed♦Dave Tweed

        122k9152263




        122k9152263






























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