Calculating Wattage for Resistor in High Frequency Application?Resistor wattage?Resistor wattage for HDMI...
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Calculating Wattage for Resistor in High Frequency Application?
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$begingroup$
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
edited 3 hours ago
Transistor
87.1k785189
asked 4 hours ago
Israr SayedIsrar Sayed
204
$endgroup$
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
edited 3 hours ago
Transistor
87.1k785189
asked 4 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
3 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
edited 3 hours ago
Transistor
87.1k785189
asked 4 hours ago
Israr SayedIsrar Sayed
204
$endgroup$
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
edited 3 hours ago
Transistor
87.1k785189
asked 4 hours ago
Israr SayedIsrar Sayed
204
edited 3 hours ago
Transistor
87.1k785189
asked 4 hours ago
Israr SayedIsrar Sayed
204
edited 3 hours ago
Transistor
87.1k785189
edited 3 hours ago
Transistor
87.1k785189
edited 3 hours ago
Transistor
87.1k785189
87.1k785189
asked 4 hours ago
Israr SayedIsrar Sayed
204
asked 4 hours ago
Israr SayedIsrar Sayed
204
asked 4 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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered 3 hours ago
Dave Tweed♦Dave Tweed
122k9152263
$endgroup$
add a comment |
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1 Answer
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active
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votes
$begingroup$
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.
answered 3 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.
answered 3 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.
answered 3 hours ago
Dave Tweed♦Dave Tweed
122k9152263
$endgroup$
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.
answered 3 hours ago
Dave Tweed♦Dave Tweed
122k9152263
edited 2 hours ago
answered 3 hours ago
Dave Tweed♦Dave Tweed
122k9152263
answered 3 hours ago
Dave Tweed♦Dave Tweed
122k9152263
answered 3 hours ago
Dave Tweed♦Dave Tweed
122k9152263
122k9152263
add a comment |
add a comment |
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$begingroup$
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