Rrerku

Can I sign legal documents with a smiley face?

Offered money to buy a house, seller is asking for more to cover gap between their listing and mortgage owed

Fear of getting stuck on one programming language / technology that is not used in my country

What does chmod -u do?

Loading commands from file

Is it possible to have a strip of cold climate in the middle of a planet?

On a tidally locked planet, would time be quantized?

Open a doc from terminal, but not by its name

What is this cable/device?

How to bake one texture for one mesh with multiple textures blender 2.8

Which one is correct as adjective “protruding” or “protruded”?

Is it better practice to read straight from sheet music rather than memorize it?

Why is it that I can sometimes guess the next note?

What are the purposes of autoencoders?

Multiplicative persistence

How to implement a feedback to keep the DC gain at zero for this conceptual passive filter?

How can "mimic phobia" be cured or prevented?

Is it safe to use olive oil to clean the ear wax?

What is the evidence for the "tyranny of the majority problem" in a direct democracy context?

Lowest total scrabble score

The IT department bottlenecks progress. How should I handle this?

Count the occurrence of each unique word in the file

Biological Blimps: Propulsion

Delivering sarcasm

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 4 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
  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 4 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
  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 4 hours ago

Transistor

87.1k785189

asked 4 hours ago

Israr SayedIsrar Sayed

204

$endgroup$

share|improve this question

edited 4 hours ago

Transistor

87.1k785189

asked 4 hours ago

Israr SayedIsrar Sayed

204

share|improve this question

edited 4 hours ago

Transistor

87.1k785189

asked 4 hours ago

Israr SayedIsrar Sayed

204

edited 4 hours ago

Transistor

87.1k785189

edited 4 hours ago

Transistor

87.1k785189

asked 4 hours ago

Israr SayedIsrar Sayed

204

asked 4 hours ago

Israr SayedIsrar Sayed

204

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

$$25 W cdotfrac2 cdot 100 ns2.5 mu s = 2 W$$

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

$$Energy = frac12cdot Charge cdot Voltage$$

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

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 3 hours ago

Dave Tweed♦Dave Tweed

122k9152263

$endgroup$

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
);
);
, "mathjax-editing");

StackExchange.ifUsing("editor", function ()
return StackExchange.using("schematics", function ()
StackExchange.schematics.init();
);
, "cicuitlab");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "135"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f428730%2fcalculating-wattage-for-resistor-in-high-frequency-application%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

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

$$25 W cdotfrac2 cdot 100 ns2.5 mu s = 2 W$$

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

$$Energy = frac12cdot Charge cdot Voltage$$

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

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 3 hours ago

Dave Tweed♦Dave Tweed

122k9152263

$endgroup$

add a comment | 

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

$$25 W cdotfrac2 cdot 100 ns2.5 mu s = 2 W$$

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

$$Energy = frac12cdot Charge cdot Voltage$$

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

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 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

$$25 W cdotfrac2 cdot 100 ns2.5 mu s = 2 W$$

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

$$Energy = frac12cdot Charge cdot Voltage$$

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

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 3 hours ago

Dave Tweed♦Dave Tweed

122k9152263

$endgroup$

share|improve this answer

edited 3 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