Simple clamper circuit - ambiguous math - What's the catch?

In the clamper circuit, consider $$\V_{i} = V_{p}cos(\omega t)\$$, and consider only the first quarter of $$\V_{i}\$$ i.e from $$\\omega t=0\$$ to $$\\pi/2\$$:
(Assumption: Initial charge on capacitor is zero and diode is ideal)

Assuming diode to be reverse biased, then $$\V_d\$$ must be negative, otherwise the assumption is false.

$$\V_d = -V_{i} + v_c = -V_{m}cost(\omega t) + 0 =\$$ -ve from $$\\omega t=0\$$ to $$\\pi /2\$$ so the diode is indeed reverse biased.

This time assuming the diode to be forward biased, then the current $$\I_d\$$ must be positive, otherwise the assumption is false.

Now $$\I_d = -C\frac{dv_c}{dt} = -C\frac{dV_{i}}{dt} = +C\omega sin(\omega t) =\$$ +ve from $$\\omega t=0\$$ to $$\\pi/2\$$ and the diode turns out to be forward biased.

However we know the diode is indeed reverse biased, but where did my math go wrong?

• Your math is unclear. I know where you went wrong but you should be clearer. For instance you say Vd must be negative but negative to what and, which diode terminal do you refer to. Then you say the current Id must be positive but which direction of current are you talking about. Nov 4 '20 at 11:25
• @Andyaka Care to tell where did I go wrong?
– Viki
Nov 4 '20 at 11:55
• "This time assuming the diode to be forward biased" - why are you assuming that? Nov 4 '20 at 15:04
• @BruceAbbott THe diode can only be in one state at a given time and assuming either should give a coherent answer. One of the assumptions must fail but thats apparently not the case here.
– Viki
Nov 5 '20 at 4:34
• You stipulate that the initial charge on the capacitor is zero, and the initial voltage of the waveform is Vp, so the diode must be reverse biased. Nov 5 '20 at 4:48

If I understand correctly what you are trying to do, you want to prove that the diode is reverse biased under these initial conditions by assuming it is forward biased initially, and showing a contradiction....except that you are not getting the contradiction you expect.

Your math went wrong by setting up initial conditions that violate Kirchoff's Voltage Law (KVL), and then using KVL as if it was not violated:

Now $$\I_d = -C\frac{dv_c}{dt} = -C\frac{dV_{i}}{dt} = +C\omega sin(\omega t) =\$$ from $$\\omega t=0\$$ to $$\\pi/2\$$ and the diode turns out to be forward biased.

Your initial condition of zero charge on the capacitor forces $$\v_c=0\$$, and assuming that 'ideal diode' means that the forward voltage drop is zero, then you have a violation of Kirchoff's Voltage Law (KVL) because your voltage source is the only element with non-zero voltage across it. Therefore $$\V_i \neq v_c\$$ by hypothesis on your initial conditions, but the second step in your equation where you apply $$\-Cdv_c/dt =-CdV_i/dt\$$ is derived from $$\V_i=v_c\$$ as if KVL actually was satisfied.

Your scenarios look to be both mathematically correct and, in practice they will both be correct (but at different periods of time). During some parts of the AC cycle the diode will loosely clamp the most negative portion of the input waveform to about -0.6 or 0.7 volts then, when the waveform is more positive the diode won't be clamping.

Hence, what you describe is not the full story about a clamper circuit. Try simulating it and see what happens.

• Im really only concerned about the first quarter of the cycle. And the diode appears to be in both the states IN THE SAME TIME PERIOD i.e from wt=0 to pi/2 ! HOW!?? And consider the diode to be ideal for simplification.
– Viki
Nov 4 '20 at 12:28
• It totally depends on what acquired DC voltage is across the capacitor and that is set be previous events that you haven't defined. Nov 4 '20 at 12:38
• Ah! Please assume initial rest conditions! :) Updated the question with the assumptions
– Viki
Nov 4 '20 at 12:40
• Try simulating it and see what happens. Nov 4 '20 at 12:56
• You'll thank me for advising you to go down the sim route (one day). Nov 4 '20 at 13:02