# How can I get this large value of peak current in half wave rectifier?

Given values are:

C = 1.4F, V_D = 0.8V, V_P = 4.5V, and 1/T is 60Hz, 0.436ohm load.

I have to find the peak diode current.

I understand when diode is on and delta T starts, the diode current get peak value.

I get ripple voltage V_R = 0.1V and By following equation I get i_DMAX with approximation

$$i_{DMAX} = \frac{V_p-V_d}{R} + C\cdot V_p\cdot \frac{2\pi}{T}\sqrt{\frac{2V_R}{V_p}} = 483.49A$$

How can this large value can come out from these small given values? Is this instantaneous value? I'm not sure I'm doing right.

From first principles:

I get Vr = 123mV for approximate ripple. (Edit: 101mV is closer to the approx value)

So peak current is approximated by the load current ((Vp-Vd)/0.436 = 8.49A) plus the current due to dv/dt when the source is at 4.5 - Vr.

v(t) = Vp sin($$\\omega t\$$) so when v(t)/Vp = ((4.5-0.123)/4.5) $$\\omega t\$$ is 1.3364

dv/dt = $$\V_P\cdot \omega \cos(\omega t)\$$ = 394v/s at $$\\omega t\$$ = 1.3364

The current into the capacitor is thus 551A, and the peak diode current is about 560A

In actual practice the peak current will be less than that because the source and wires will have some impedance and the diode Vf increases with current - typically 25mV for a factor of 10 increase in current.

• I understood this with your detailed values. So I can find (wt) point by guessing, turning- diode-on-point is when ripple voltage is completely effected, right? So in that point source voltage is Vd-Vr, and I can find wt. Did I understand right? I'm not perfectly sure. Commented May 14, 2020 at 20:03
• The voltage source is Vp-Vr (the capacitor voltage is Vp-Vr-Vd). Here, I'm considering the repetitive peak current that occurs every cycle, not the start-up peak current. Commented May 14, 2020 at 20:30
• I get slightly different numbers. Did you by any chance use 50 Hz instead of 60 Hz when calculating the ripple? Commented May 14, 2020 at 21:35
• @DaveTweed I used 60Hz but the wrong current because I ignored the diode drop. Commented May 14, 2020 at 21:58

Given values are : C = 1.4F

For a capacitor $$\i = C\dfrac{dv}{dt}\$$.
So, if $$\V_{peak}\$$ is 4.5 volts, the maximum $$\\dfrac{dv}{dt} = 4.5\cdot 2\cdot\pi\cdot 60\$$ = 1,696 volts per second.