# Clamper Circuit Without A Resistor Analysis

simulate this circuit – Schematic created using CircuitLab

We were given this clamping circuit and were tasked to plot Vc and Vout for varying input. The only given quantities are the forward voltage of the diode which is 0.7 and the Vin which is:

My general KVL equation (clockwise direction) for Vc and Vout are: $$Vc = Vi + 2V + Vd$$ $$Vout = Vi - Vc + 2V + Vd$$

Using this equation, my analysis is:

• At Vi = 0, Vd would be off because of its opposite polarity to the voltage source parallel to it. Vc would then be 2V. Vout would be $$Vout = 0 - 0 + 2V = 2V$$
• At Vi = 10, Vd would still be off. Vc would be holding its initial charge. Hence, $$Vo = 10-2+2 = 10V$$
• At Vi = -10 V, Vd will be ON. Vc would still be holding its initial charge. Hence, Vout would just be equal to 2V + the forward voltage 0.7 but because of the opposite polarity, it would be negative. $$Vo = -2 - 0.7 = -2.7$$

There is no resistor in the circuit so I'm unsure of the behavior of the capacitor. I just assumed that the capacitor would be holding the charge for a very long time since there is no resistor it can discharge to. The configuration looks like a clamper circuit but then my calculations don't show similar behavior. I'm guessing it has something to do with the capacitor carrying its charge to the next time interval? Should I be instead calculating for initial charge + new charge? So for eg., from an initial charge of 2V at t(0), at t(0+), Vc should be like this? $$Vc(0+) = (Vi + 2V) + V(0-)$$

I suppose it could be simulated but where’s the fun in that?

I think this isn’t so much a clamper as a level shifter.

Since there’s no explicit ground in this circuit let’s arbitrarily call the bottom node ground, or at least our zero voltage reference point.

At t(0) the input is zero, and Vout is -2.7V (diode plus battery). C1 is now holding a 2.7V charge.

***CORRECTION (after Batt's comment): no, Vout will be zero because no current can flow through diode/battery.

When Vi first jumps to +10V, the left side of C1 jumps up 10V. Since the voltage across a capacitor can’t change instantaneously, the right side jumps up 10V too.

When Vi then jumps to -10V, the left side of C1 jumps down by 20V. The right side does too but is quickly clamped by the diode/battery to limit Vout to -2.7V.

When Vi goes back up to +10V, again the left side of C1 sees a 20V jump. This causes the right side of C1 to jump up 20V too, to 17.3V.

The circuit is now stable, with the output jumping between -2.7V and +17.3V.

Did I get it right?

• Why will the diode be ON at t(0)? Doesn't the anode see a higher potential than the cathode?
– user263783
Oct 7 '20 at 2:11
• Yeah, you're right. No current will be flowing through diode/battery so Vout is actually floating with respect to our "ground". Which means at t(0) we'd have to say Vout is zero (if we put a resistor from Vout to ground no current would flow so voltage must be zero). So the first Vi jump to +10V will cause Vout to jump all the way to +10V. After that it would oscillate between -2.7V and 17.3V. Final answer Oct 7 '20 at 3:14
• Wouldn't Vout be 2V at t(0)? even if the diode will not conduct, the voltage source of 2V is still there?
– user263783
Oct 8 '20 at 8:33
• No, because we’re only concerned with the voltage with respect to our zero volt reference, the anode of D1. So while there is certainly still 2V from Vout to the diode’s cathode, if we were to put voltmeter leads between Vout and the anode we’d see no voltage (I just confirmed it with a battery, diode and voltmeter!) This makes sense, because if there were a voltage from Vout to diode’s anode, then if we were to install a resistor between those points we’d get current flow – but that clearly can’t happen because the diode is preventing any current flow through the battery. Oct 8 '20 at 16:43
• If that is the case, the first time Vi jumps to 10V, Vout is also equal to 0V? I apologize, ive only gone back to this problem today.
– user263783
Oct 11 '20 at 8:24