# Solving for Zener Diode Voltage

• d3thkn1ght
In summary: The Thevenin model would include the Rin and RL. The load line takes the open circuit voltage and short circuit current of the Thevenin equivalent as its end points. So work with...A schematic would be helpful.The cathode of the ideal zener pointing up or down? What would the voltage and resistance be?The cathode of the ideal zener would be pointing up. The voltage and resistance would be 9.5 V and 1 kΩ.
Thanks for clarifying. I was confused a little on the Vout part of it all. I'll try this.

Thanks again.

Is this formula correct:
$$\frac{ (E - Vz)} {Rin} = Iz + \frac {VL} {RL}$$

If so, is Vout = VL?

No, I don't see that as being correct. You haven't accounted for the zener resistance rz, and you've introduced a current variable Iz for no particular reason. An expression for Iz can be written directly using the given component values and the node voltage.

You have three voltages to work with: E, Vz, and Vout. You have three resistances to work with: Rin, RL, and rz. You shouldn't need any other variables or values.

Okay, thanks for clarifying.

Is this equation any better:
$$\frac{E-V_z}{R_{in}} = \frac{V_z}{rz} - \frac{V_L} {R_L}$$

No. You're taking Vz as the voltage at the node, but it isn't. Vz is the zener's threshold voltage, but the zener model in use comprises the voltage source that is Vz and the series resistance. The node voltage is across those two connected in series:

Your node equation should involve Vout, since that's the node voltage.

I was able to get the right answer using the following formula:
$$\frac{E-V_{out}}{R_{in}}=\frac{V_{out}-5}{rz} + \frac{V_{out}}{R_L}$$

d3thkn1ght said:
I was able to get the right answer using the following formula:
$$\frac{E-V_{out}}{R_{in}}=\frac{V_{out}-5}{rz} + \frac{V_{out}}{R_L}$$
Yes! That's a good node equation.

After plugging in the given resistor values, what expression did you get for Vout (in terms of E)?

The question wanted to know the contribution of VI which it called Vout and then the contribution from vi which was called vo.

For RL of 2K ohms:
E of 9.5 V gave a value of 5.001997004. I put in 5.1 since from my understanding the Zener diode keeps a constant voltage when open and it worked. This was before you had me work on the KCL equation.

For E of 0.05V, I got vo of .0000499251.
What I did was take the equation and get two answers, when E was 9.55 and when E was 9.45
Then I divided E(9.55)-E(9.45) by two. That was the correct answer. I don't know i I did that right but it worked. I am guessing, if I did it right, that the voltage source vi is a sine wave and sometimes has positive voltage and other times negative voltage. Thus you take the average value and that is what the vo voltage is.

Okay, that works. But you could also have substituted ##E = (V_I + V_i)## and expanded the equation for ##V_{out}##, grouping it into constant DC components and an AC component involving ##V_i##. Then you'd have your DC and AC terms separated.

Great to know. Thanks again for all the help.

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