Calculating Wattage for Resistor in High Frequency Application?Resistor wattage?Resistor wattage for HDMI...

Is this toilet slogan correct usage of the English language?

Why can Carol Danvers change her suit colours in the first place?

Non-trope happy ending?

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

Pre-mixing cryogenic fuels and using only one fuel tank

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

Lowest total scrabble score

Not using 's' for he/she/it

What prevents the use of a multi-segment ILS for non-straight approaches?

Aragorn's "guise" in the Orthanc Stone

Calculating Wattage for Resistor in High Frequency Application?

How do you make your own symbol when Detexify fails?

GraphicsGrid with a Label for each Column and Row

Creepy dinosaur pc game identification

A social experiment. What is the worst that can happen?

Start making guitar arrangements

What was this official D&D 3.5e Lovecraft-flavored rulebook?

Why did the Mercure fail?

How could a planet have erratic days?

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

How do I color the graph in datavisualization?

What is Cash Advance APR?

Why should universal income be universal?

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



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











$endgroup$












  • $begingroup$
    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.






share|improve this answer











$endgroup$













    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


















    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















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    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.






    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






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Electrical Engineering Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            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















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            What is the “three and three hundred thousand syndrome”?Who wrote the book Arena?What five creatures were...

            Image
            What happens when the centripetal force is equal and opposite to the centrifugal force? Why is "la Gestapo" feminine? Nested Dynamic SOQL Query How are passwords stolen from companies if they only store hashes? Exposing a company lying about themselves in a tightly knit industry: Is my career at risk on the long run? Can "few" be used as a subject? If so, what is the rule? Why are there no stars visible in cislunar space? categorizing a variable turns it from insignificant to significant Do I need an EFI partition for each 18.04 ubuntu I have on my HD? What are the consequences of changing the number of hours in ...
            Read more

            Gersau Kjelder | Navigasjonsmeny46°59′0″N 8°31′0″E46°59′0″N...

            Image
            Tidlegare republikkarSkipingar i 1433Kommunar i SchwyzDistrikt i SchwyzNedleggingar i 1798Stader ved Vierwaldstättersee kommunekantonenSchwyzSveitskm²VierwaldstätterseeLauerz Gersau Kommunevåpenet til Gersau Plassering Styresmakter Land   Sveits Kanton kantonen Schwyz Distrikt Gersau Språk tysk Geografi Flatevidd  - kommune 23,70 km² Innbyggjarar  - kommune (desember 2010)    - folketettleik 2 094   88,4 /km² Koordinatar 46°59′0″N 8°31′0″E  Koordinatar: 46°59′0″N 8°31′0″E  Høgd over havet 435 moh Diverse annan informasjon Postnummer 6442 Nettstad: www.gersau.ch Freistaat und Republik Gersau Fristaten og republikken Gersau By-republikk og alliert protektorat i Det gamle sveitsiske eidsforundet ←   ← 1433–1798 1814–1818 →   → Hovudstad Gersau Styreform Republikk Historisk periode Tidleg moderne tid  - Kjøpte fridom frå ...
            Read more

            Hestehale Innhaldsliste Hestehale på kvinner | Hestehale på menn | Galleri | Sjå òg |...

            Image
            Frisyrar frisyrehårhårstrikkhårklemmerøyrethesthovudverkeuropeisk motejobbtreningfletterparykkarungdomsopprørmotkulturmusikararmotorsykkelentusiastarpolitietpolitiskforretningverksemdfiskeflette Hestehale sett bakfrå. Hestehale på ein mannleg indianarleiar. Hestehale er ein frisyre der ein samlar det meste eller alt av langt hår [1] i bakhovudet med ein hårstrikk eller liknande. Avhengig av mote kan det òg vere vanleg å sette opp hestehalen med hårklemmer eller å ha hestehalen hengande på den eine sida av hovudet ved å samle håret ved øyret. Hestehale er ein vanleg frisyre for jenter og for gutar med langt hår. Han har fått namnet sitt etter likskapen med halen på ein hest. For stramme hestehaler kan føre til hårtap, [2] i nokre høve også hovudverk. [3] Innhaldsliste 1 Hestehale på kvinner 2 Hestehale på menn 3 Galleri 4 Sjå òg 5 Bakgrunnsstoff 6 Kjelder Hestehale på kvinner | Ungjente med hestehale som held håret unna ansiktet...
            Read more