Thevenin voltage

[OrangeArav]

Hello, I need help with figuring out the open circuit voltage in the circuit below. Namely, the question is asking to plot Vz(Vs) and in order to do that we need a series circuit where the Zener diode is our load. However, when I go to verify my answer with the solutions, the Vth value is wrong. Please help me find my mistake.

How I approached the problem (labeled diagram below):

Using mesh currents I ended up with the following equations,

(1)

;

(2)

;
Adding (1) and (2) we get,

Knowing I_2 we can then find Vth by multiplying I_2 by R4 to get a value of 0.2Vs. However, in the solutions the answer is 0.125Vs... What did I do wrong? How should one approach this type of problem?

P.S. Rth I got correct. It is 600 ohms.

 

[Ratch]

Hello, I need help with figuring out the open circuit voltage in the circuit below. Namely, the question is asking to plot Vz(Vs) and in order to do that we need a series circuit where the Zener diode is our load. However, when I go to verify my answer with the solutions, the Vth value is wrong. Please help me find my mistake.

How I approached the problem (labeled diagram below):

Using mesh currents I ended up with the following equations,

(1)

;

(2)

;
Adding (1) and (2) we get,

Knowing I_2 we can then find Vth by multiplying I_2 by R4 to get a value of 0.2Vs. However, in the solutions the answer is 0.125Vs... What did I do wrong? How should one approach this type of problem?

P.S. Rth I got correct. It is 600 ohms.

Your calculations are correct. the Voc is 0.2 volts. So the source voltage has to increase to 25 volts for the zener to start conducting. After that, the voltage across the zener will be fairly constant at 5 volts. The voltage across the zener will change slightly when the 8 ohm resistor reacts with the current through the zener. Notice that the voltage source will always supplies current to the circuit even when the zener is not conducting, unless the source voltage falls to zero. That fact might not be obvious when looking at the source voltage, Thevenin resistance, and zener resistance in series.

Ratch

 

[Laplace]

Indeed, the given answer for Vth is incorrect. However, when one is looking for the voltage, why calculate the current? Why not calculate the node voltage instead of the mesh current? But since this problem is a simple cascaded voltage divider, one can use a voltage divider approach instead. That way you would have an intuitive feeling for the correctness of the solution, rather than wondering where is the mistake in the mesh equations. Note that for Vth < Vzo the Zener diode is essentially not part of the circuit.

 

[OrangeArav]

Thank you for your help!
This is what I came up with in simulation...

Also, please let me know if I have the polarity set up correctly for both the forward bias and breakdown regions:
1) In the upper set output voltage, Vz, is positive and the graph rises above 0
2) In the bottom set, Vz is negative and the graph falls below 0

 

[Laplace]

... if I have the polarity set up correctly for both the forward bias and breakdown regions:

Where does the problem indicate that the Zener diode should be forward biased?

 

[OrangeArav]

Where does the problem indicate that the Zener diode should be forward biased?

It doesn't say directly in the statement but it's assumed that the plot should include it. I'm just not sure about that transition from breakdown to forward, does the polarity on the Vd0 and Vz0 change so that current flows + to - in tandem with the signal polarity?

 

[Laplace]

In that case I can't see anything wrong. In the transition region from forward conduction to reverse breakdown there should be no current flowing in the Zener diode (ignoring leakage current) so the voltage polarity would be as though the diode were not present in the circuit.