# Biased Clipper Circuit

I was studying for my end-term exam through past year papers(of my college). I came across this question in which a diode is in series with an AC voltage source and a DC voltage source.

Info given:

• VB = DC Source
• Vin = AC Source
• Consider diode forward voltage drop is 0.7 V (Si)

I am not sure if this is indeed a biased clipper.

I was only taught simple biased clipper but I tried Googling and found out that this is a Biased Series Clipper--it is the exact opposite of usual examples...

## My attempt

Since D1 is reverse biased, it becomes an open circuit for positive input waveform--that means the max Vout in the positive cycle is 0 Volts.

Am I correct?

Edit: Somebody has posted the exact question on Chegg--sadly I don't have an account on Chegg.

• Diode becomes forward biased during positive cycle when $V_{in}\ge V_{B} + V_{D1 fwd}$. Where $V_{D1 fwd}$ is the diode threshold voltage for conduction. With this info, can you solve the question?
– AJN
Commented Jun 4, 2021 at 15:45
• 5 is greater than 2.7--so this would mean a forward bias irrespective of the DC source? In that case, KVL gives Vout = 7.7 Volts... Would that be correct @AJN ? Commented Jun 4, 2021 at 16:16
• To use KVL, you must first draw a diagram with reference directions clearly marked. Please update the circuit diagram. Otherwise, there could be confusion, as to whether, voltages add or subtract.
– AJN
Commented Jun 4, 2021 at 16:32
• Yeah sorry @AJN. I could just use Vin - Vout = 2.7 (Is their a name for this "law"?) Thanks once again for telling that it would be forward biased. Vout = 2.3 V ,right? Commented Jun 4, 2021 at 16:39
• You are right that, It is KVL. I intended my comment to mean that a diagram is needed to check if the add/subtract operations of KVL are correct/consistent.
– AJN
Commented Jun 4, 2021 at 16:49

Note the following:

• $$\V_2\$$ is always $$\V_{out} + V_B\$$
• because of the diode current in the circuit can only flow in one direction
• therefore there are two cases:
1. no current flows and $$\V_{out}\$$ is zero
2. current flows in the circuit, $$\V_{out}\$$ is positive and $$\V_2 = V_{in} - 0.7\$$

simulate this circuit – Schematic created using CircuitLab

In the second case $$\V_{out} = V_{in} - 0.7 - V_B\$$ and the condition $$\V_{out} > 0\$$ means we must have $$\V_{in} > 0.7 + V_B\$$.

If $$\V_{in} < 0.7 + V_B\$$ then we are in the first case and $$\V_{out} = 0\$$.

• So, Max Vout = 2.3 V? I think I do understand it now after combining your and @AJN's post. Btw what is the law that gives Vin-Vout=2.7 called?(I know I got the wrong result by calling it KVL in my comment above--can't apply KVL on a wire) THANKS ONCE AGAIN!(Can't upvote due to low reputation) Commented Jun 4, 2021 at 16:36
• Yes, the max for Vout is 5-0.7-2 = 2.3. Not sure about the name of that law. Commented Jun 4, 2021 at 16:47