Hjdcti

Redundant comparison & "if" before assignment

Does a 'pending' US visa application constitute a denial?

Did arcade monitors have same pixel aspect ratio as TV sets?

Start making guitar arrangements

Electoral considerations aside, what are potential benefits, for the US, of policy changes proposed by the tweet recognizing Golan annexation?

Does an advisor owe his/her student anything? Will an advisor keep a PhD student only out of pity?

What does routing an IP address mean?

How should I respond when I lied about my education and the company finds out through background check?

How much character growth crosses the line into breaking the character

Delivering sarcasm

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

Store Credit Card Information in Password Manager?

Creepy dinosaur pc game identification

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

Count the occurrence of each unique word in the file

Why electric field inside a cavity of a non-conducting sphere not zero?

Removing files under particular conditions (number of files, file age)

What is Cash Advance APR?

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

250 Floor Tower

What was the exact wording from Ivanhoe of this advice on how to free yourself from slavery?

What are the purposes of autoencoders?

How to indicate a cut out for a product window

How could a planet have erratic days?

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
  5 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
  5 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

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

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 4 hours ago

answered 4 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

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

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

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 4 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

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

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 4 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

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

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 4 hours ago

answered 4 hours ago

Dave Tweed♦Dave Tweed

122k9152263

$endgroup$

share|improve this answer

edited 4 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