# Circuit analysis of series diode clipper circuit

simulate this circuit – Schematic created using CircuitLab

I'm really unsure how to start with this. The instructions said to find Vin such that Vout will be 0.5Vin. The only given is Vd = 0.7 and Rd = 5 ohms. I tried to do a KVL equation but then I don't know what the current is since Vin is not given.

Additional info: Diode model that is used is the constant voltage drop model with resistance of Rd=5 ohms and forward voltage of 0.7

• Compared to your previous question, you've now clearly left the concept of the "ideal diode", and are now working with a diode that has a more complex mathematical model. But you forget to specify your diode's model – is it maybe the Shockley equation? No matter the model, you've also forgotten to specify the parameters of your diode (something like forward voltage in saturation, or saturation current). Commented Oct 8, 2020 at 13:06
• @MarcusMüller I should have mentioned that Vd = 0.7 meant that its the forward voltage of the diode. Hence, ill be using the constant voltage drop model here.
– user263783
Commented Oct 8, 2020 at 13:11
• Oh, Ok, that makes it... super easy? Then Vout = 0.7 V by definition, then 2Vout = Vin = 1.4 V by the problem statement? Commented Oct 8, 2020 at 13:14
• Thats what I was thinking as well. But then what im doubting about is the existence of the resistor Rd. Would it not affect Vout?
– user263783
Commented Oct 8, 2020 at 13:16
• that doesn't matter. Your problem statement says Vin = 2 Vout, and Vout=Vd is given. so, solved. (Unless your diode model is more complex, AND Vd is misleadingly named; but again, you're still not stating that model! We can't know what's in your head. Please define what a diode looks like for you, and what Rd is in case of your diode model.) Commented Oct 8, 2020 at 13:34

An explicit model to be used for the diode is missing in the original question, however the diode parameters provided ($$\V_d\$$ and $$\R_d\$$) point towards using a piece-wise linear model for the diode. Therefore, your circuit should be redrawn as follows.
Assuming $$\V_{out} > V_d\$$, the ideal diode has a forward bias and acts as a short-circuit. Voltage equations: $$\frac{V_{out}-V_i}{R_1}+\frac{V_{out}-V_d}{R_d}=0$$ With the additional constraint $$\V_{out} = 0.5V_i\$$: $$\frac{-0.5V_i}{R_1}+\frac{0.5V_i-V_d}{R_d}=0$$ $$0.5\left(\frac{1}{R_d}-\frac{1}{R_1}\right)V_i=\frac{V_d}{R_d}$$ $$V_i=2\frac{V_d}{R_d\left(\frac{1}{R_d}-\frac{1}{R_1}\right)}$$ Checking numerically if the forward bias assumption holds true: $$V_i=\frac{1.4}{5(0.2-0.1)}=\frac{1.4}{0.5}=2.8$$ $$V_{out}=0.5V_i=1.4 > V_d$$