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Calculating Wattage for Resistor in High Frequency Application?

Resistor wattage?Resistor wattage for HDMI hackTLC5940NT + 12v 5050 led stripImprove Rise Time on 1Hz SignalDetermining the surge duration of a double exponential transient?Resistor surge ratingZero Crossing Detection of ~ 400 kHz Signal with MCUpower supply remote sense protection resistor value?calculating maximum sense speed of amplified phototransistor circuitMay I use a smaller wattage resistor as mosfet's gate driver for a very short time?

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

edited 5 hours ago

Transistor

87.1k785189

asked 5 hours ago

Israr SayedIsrar Sayed

204

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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

edited 5 hours ago

Transistor

87.1k785189

asked 5 hours ago

Israr SayedIsrar Sayed

204

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

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

edited 5 hours ago

Transistor

87.1k785189

asked 5 hours ago

Israr SayedIsrar Sayed

204

$endgroup$

share|improve this question

edited 5 hours ago

Transistor

87.1k785189

asked 5 hours ago

Israr SayedIsrar Sayed

204

share|improve this question

edited 5 hours ago

Transistor

87.1k785189

asked 5 hours ago

Israr SayedIsrar Sayed

204

edited 5 hours ago

Transistor

87.1k785189

edited 5 hours ago

Transistor

87.1k785189

asked 5 hours ago

Israr SayedIsrar Sayed

204

asked 5 hours ago

Israr SayedIsrar Sayed

204

1 Answer
1

active

oldest

votes

3

$begingroup$

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

edited 3 hours ago

answered 4 hours ago

Dave Tweed♦Dave Tweed

122k9152263

$endgroup$

add a comment | 

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3

$begingroup$

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

edited 3 hours ago

answered 4 hours ago

Dave Tweed♦Dave Tweed

122k9152263

$endgroup$

add a comment | 

3

$begingroup$

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

edited 3 hours ago

answered 4 hours ago

Dave Tweed♦Dave Tweed

122k9152263

$endgroup$

add a comment | 

$begingroup$

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

edited 3 hours ago

answered 4 hours ago

Dave Tweed♦Dave Tweed

122k9152263

$endgroup$

share|improve this answer

edited 3 hours ago

answered 4 hours ago

Dave Tweed♦Dave Tweed

122k9152263

answered 4 hours ago

Dave Tweed♦Dave Tweed

122k9152263

answered 4 hours ago

Dave Tweed♦Dave Tweed

122k9152263