Calculating Wattage for Resistor in High Frequency Application?












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
    5 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
    5 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 5 hours ago









Transistor

87.1k785189




87.1k785189










asked 5 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
    5 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
    5 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
5 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
5 hours ago










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











$endgroup$













    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









    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











    $endgroup$


















      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











      $endgroup$
















        3












        3








        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











        $endgroup$



        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 4 hours ago

























        answered 4 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

            Liste der Baudenkmale in Friedland (Mecklenburg)

            Image
            In der Liste der Baudenkmale in Friedland sind alle denkmalgeschützten Bauten der Stadt Friedland (Mecklenburg-Vorpommern) und ihrer Ortsteile aufgelistet. Grundlage ist die Veröffentlichung der Denkmalliste des Landkreises Mecklenburgische Seenplatte (Auszug) vom 28. September 2016. Inhaltsverzeichnis 1 Baudenkmale nach Ortsteilen 1.1 Friedland 1.2 Bresewitz 1.3 Brohm 1.4 Cosa 1.5 Dishley 1.6 Dishley-Ausbau 1.7 Eichhorst 1.8 Glienke 1.9 Heinrichshöh 1.10 Heinrichswalde 1.11 Jatzke 1.12 Jatzke-Meierei 1.13 Liepen 1.14 Ramelow 1.15 Schwanbeck 2 Vorgesehener Denkmalbereich 3 Quelle 4 Weblinks Baudenkmale nach Ortsteilen | Friedland | ID Lage Bezeichnung Beschreibung Bild 281 An der Kleinbahn 2 Windmühle mit Umgang 282 Anklamer Straße alter Friedhof mit Portalwand 282 Anklamer Straße drei Erbbegräbnissen und Gräbern u. a. H.A. Riemann 284
            Read more

            Single-Malt-Whisky

            Image
            Fünf verschiedene schottische Single-Malt-Whiskys Als Single-Malt-Whisky oder kürzer Single Malt werden Whiskys bezeichnet, die zwei besondere Bedingungen erfüllen: Sie stammen zum ersten aus einer einzigen Brennerei, sind also kein Verschnitt aus mehreren Whiskysorten (daher Single , deutsch: einzeln). Zum zweiten wird als Getreide ausschließlich gemälzte Gerste verwendet (deswegen Malt , deutsch: Malz). Der Zusatz Single Malt wird als Prädikat verstanden, die so ausgezeichneten Whiskys sind in aller Regel besonders hochwertig. Hinzu kommt ihre besonders lange Lagerung, die Reifungszeit währt oft ein Jahrzehnt und länger. Inhaltsverzeichnis 1 Allgemeines 2 Herkunftsländer 2.1 Schottland 2.2 Irland 2.3 Japan 2.4 Kontinentaleuropa 3 Produktionsschritte 3.1 Mälzen 3.2 Brauen 3.3 Brennen 3.4 Reifung und Finish 3.5 Abfüllung 4 Altersangaben 5 Lagerung von Whiskyflaschen 6 Geschmacksklassifikation 7 Sonstig
            Read more

            Czorneboh

            Image
            Dieser Artikel erläutert den Berg Czorneboh, für die Gottheit siehe Czorneboh (Gottheit). Czorneboh Čornobóh Berggasthof mit Turm Höhe 555,7  m ü. NHN [1] Lage Freistaat Sachsen, Deutschland Gebirge Lausitzer Bergland Dominanz 15,4 km →  Kottmar Koordinaten 51° 7′ 11″  N , 14° 31′ 31″  O 51.119722222222 14.525194444444 555.7 Koordinaten: 51° 7′ 11″  N , 14° 31′ 31″  O Gestein Granodiorit Besonderheiten Aussichtsturm Czorneboh Der Czorneboh , obersorbisch Čornobóh , ist ein Berg zwischen Hochkirch und Cunewalde in der Oberlausitz. Er gehört zu der südöstlich von Bautzen gelegenen Czorneboh-Bergkette und ist mit einer Höhe von 555,7  m ü. NHN [1] der höchste Punkt dieses Ausläufers des Lausitzer Berglandes. Er leitet als Landschaftselement die wellige Granitlandschaft der Lausitzer Vorbergzone ein. Der Gipfel des Czorneboh liegt in der Gemarkung Meschwitz (Gemeinde Hochkirch), die Czornebohbaude jedoch
            Read more