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enter image description hereThe datasheet of the chip UC3843 recommends to add an external resistor of about 10-20 ohm in series with the MOSFET. Can anyone explain how to calculate the power dissipation in the series resistor? I'm operating the circuit at 200Khz. Other parameters are as follows:

  1. Vin= 10V
  2. Vout=18V
  3. Ouput Current= 3A
  4. Mosfet: FDP5500
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  • \$\begingroup\$ The datasheet of the chip UC3843 recommends to add an external resistor of about 10-20 ohm in series with the MOSFET Uhm, No it doesn't but if I'm wrong please prove it by providing a schematic !!!! (You should have included it anyway) \$\endgroup\$ Mar 31, 2016 at 14:17
  • \$\begingroup\$ If v in is 10 volt and a 10 ohm is in series with the switcher, input current will be limited to 1 amp max, thus if its a boost converter then the output will be much much less then 1 amp, because power conservation nature of the smps, that is assuming your 10 ohm dissipate 0 watt, which is not...!! \$\endgroup\$ Mar 31, 2016 at 14:37
  • \$\begingroup\$ ti.com/lit/an/slua143/slua143.pdf. \$\endgroup\$
    – Akash
    Mar 31, 2016 at 15:02
  • \$\begingroup\$ That resistor is in series with the gate, it wouldn't normally be referred to as in series with "the MOSFET". The power dissipation will be pretty small, if you must calculate it use the total gate charge figure in the MOSFET datasheet (Qt) along with your switching frequency. You're charging and discharging that capacitance through that resistor every cycle. \$\endgroup\$
    – John D
    Mar 31, 2016 at 15:09
  • \$\begingroup\$ Yeah, that was my fault as I didn't mention the gate in the original post.... Thanx for your help.... \$\endgroup\$
    – Akash
    Mar 31, 2016 at 15:25

2 Answers 2

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Simple, the power is \$P=RI^2\$, where I is the gate current of the MOS. But how much current enter in the gate? ;-)

The Application Note say that R is for damping, for have a dissipation element of the oscillations that could be occurs in the circuit formed by gate capacitance and wiring incuctance. If your circuit is well-designed you could use a very small signal resistor (0.25W could be enough).

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  • \$\begingroup\$ I think I should better go out for some holidays. Ive calculated this thing thousands of time in my previous designs and got messed up... Thanx for your help... \$\endgroup\$
    – Akash
    Mar 31, 2016 at 15:29
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Estimate the total charge at the gate needed to switch the MOSFET using the operating condition and the data sheet (let that be Qg). The energy dissipated can be estimated to be (0.5 Qg Vg) per transition. If you like to estimate using the equivalent lumped gate capacitance instead, Qg = Cg Vg, then the same energy can be estimated as \$ 0.5 C_g V_g^2 \$.

The power dissipation of the resistor is energy times transitions per second, which would be: $$ P = 0.5 Q_g V_g \times 2 freq = Q_g V_g freq $$

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