It's not very clear for me which are the differences between Idc^2 * Rl and Irms^2*Rl in a simple circuit where I have an AC source, a load resistor, and a diode:

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If I have a 40Vpp Sinus voltage with 0V DC offset (Vmax = +20V) and a 0.25Ohm Load resistor connected in series with a diode (a half wave rectifier), Rd much smaller than Rload (Rd = diode equivalent resistance), I will have a peak current Imax of 80A, when the diode conducts. The RMS current in this circuit will be Imax/2 and the Idc will be Imax/pi.

As I see on this website, the DC power delivered to the load will be Idc^2*Rload and the power dissipated in the load will be Irms^2 * R.

So, using these two formulas I get:

  • DC power delivered to the load = 162.278 W
  • Power dissipated in the load = 400W

If I want to buy a resistor and use it in this circuit, I should buy a 500W resistor or a 200W resistor?

I think is 500W but I'm not sure. If yes, what is the DC power delivered to the load?

I hope someone is kind to help me understand a bit better the differences between this two powers.

  • 1
    \$\begingroup\$ Can you solve this case without the diode ? The diode will simply block half the waves so power is halved. Simple as that. You make things too complicated by also considering DC current. Since there is no smoothing capacitor the power is zero when the diode blocks. (all this assuming an ideal diode). Also stop thinking that all circuits have their own formulas. If you first think about what happens then simplify things, you only need very basic and simple formulas. \$\endgroup\$ Jan 26, 2017 at 10:38
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    \$\begingroup\$ I would pay serious attention to the equivalent forward resistance of the diode in this case as well. At the current levels being discussed the diode will be sharing a bit of the power from the source. \$\endgroup\$ Jan 26, 2017 at 11:48

1 Answer 1


A 40 Vp-p sine wave across a 0.25 ohm load is a power of 800 watts. A perfect diode will block half this power so the power in the load is 400 watts.

Reason: 40 Vp-p is 14.14 V RMS so power in the resistor is this voltage squared divided by 0.25 ohms = 200/0.25 = 800 watts.

  • \$\begingroup\$ So my resistor should be rated at leat 800W(ideal situation)? \$\endgroup\$
    – NumLock
    Jan 26, 2017 at 11:10
  • \$\begingroup\$ I would say it's OK to run it at two-thirds so, if it dissipated a maximum of 400 watts then you should choose a 600 watt resistor. \$\endgroup\$
    – Andy aka
    Jan 26, 2017 at 11:12
  • \$\begingroup\$ "Reason: 40 Vp-p is 14.14 V RMS so power in the resistor is this voltage squared divided by 0.25 ohms = 200/0.25 = 800 watts." This calculation you did for the resistor connected without the diode, right? \$\endgroup\$
    – NumLock
    Jan 26, 2017 at 11:45
  • \$\begingroup\$ That's what I believe I said, yes. \$\endgroup\$
    – Andy aka
    Jan 26, 2017 at 11:53

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