This is an example problem in my workbook. Please can someone explain me the working of the circuit and how is this formula derived.

I got 1 more solution to the same problem.


  • \$\begingroup\$ Where are you stuck? I would prefer to see the formula in terms of tcyc = 1/f. \$\endgroup\$ Dec 11, 2017 at 6:00
  • \$\begingroup\$ @Sephro Sir, how we get this formula ? And as RC >>T, diode current should be 0 then. \$\endgroup\$
    – user170930
    Dec 11, 2017 at 6:12
  • \$\begingroup\$ Consider the capacitor only charges to Vm instantaneously at the positive peaks of the input voltage and the diode does not conduct otherwise. That is an approximation. The purpose of the first part of the formula is to determine the average DC voltage. \$\endgroup\$ Dec 11, 2017 at 6:22
  • \$\begingroup\$ Try to draw the diode current i(t). Then a simple mathematical average over the cycle should yield the average diode current. \$\endgroup\$ Dec 11, 2017 at 7:04
  • \$\begingroup\$ @SpehroPefhany I got what you were trying to say. Please check my edited question and tell me which one is correct. \$\endgroup\$
    – user170930
    Dec 11, 2017 at 7:09

1 Answer 1


I think your workbook is wrong with that formula. They have used the full wave rectifier formula. For HWR, It has to be :

$$V_{dc} = V_m - I_{dc}/2fC$$

Your derivation is correct.

enter image description here From the above waveform,

$$V_{dc} = V_m - V_{rpp}/2$$ from ripple waveform, the amount of charge stored by the capacitor = The charge lost by it in time T seconds. i.e., $$C V_{rpp}= I_{dc}T$$

which gives, $$V_{rpp} = I_{dc}/fC$$ Therefore,

$$V_{dc} = V_m - I_{dc}/2fC$$

  • \$\begingroup\$ His derivation of average load current is correct, however his diode current is not. \$\endgroup\$ Dec 11, 2017 at 15:34
  • \$\begingroup\$ Thanks @MITU RAJ and Bruce Abbott for answering. I applied your formula and got Idc=0.0975mA. Will this also be the diode current? I am really confused with diode current calculation. \$\endgroup\$
    – user170930
    Dec 11, 2017 at 16:26
  • \$\begingroup\$ Not really cz there's a small current flow through capacitor. That's why the question asks "approximately". So you may say so. \$\endgroup\$
    – Mitu Raj
    Dec 11, 2017 at 16:31
  • \$\begingroup\$ But RC>>T. So in steady state, most of the time discharging will take place while only for a short duration charging will happen (when diode conducts). I am trying to say that diode current should have been negligible compared to capacitor current \$\endgroup\$
    – user170930
    Dec 11, 2017 at 18:37
  • \$\begingroup\$ Since voltage across the load = voltage across capacitor, and its not pure dc, Cdv/dt current always exist through cap. So it adds up with the current through load at the node, to get the total current coming through diode. Since dv/dt is very small here, you can neglect it. After all GATE questions are full of assumptions :D \$\endgroup\$
    – Mitu Raj
    Dec 11, 2017 at 18:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.