Solving for Zener Diode Voltage

[d3thkn1ght]

Homework Statement

I have a problem where I need to calculate vo (i.e. output noise) and VO (i.e. DC output voltage) different resistances in a circuit with a Zener diode. The diode is reversed biased but I don't know the Vz. I have the IV curve but can't seem to wrap my head around how to calculate the load line for the Zener diode. Could someone give me some basic steps on calculating the load line and operating point, so I can calculate Vz?

Homework Equations

Unsure

The Attempt at a Solution

I have tried calculating the Thevenin voltage and resistance but still cannot get the load line correct as all the equations I have seen use the current through the diode to calculate the voltage and it doesn't make sense to me.

I appreciate any help.
Thank you.

 

[cnh1995]

Welcome to PF!

Please post the circuit diagram and problem statement.

I have tried calculating the Thevenin voltage and resistance

You need to show your attempt. It is mandatory here while posting in the HH forums.

 

[d3thkn1ght]

Welcome to PF!

Please post the circuit diagram and problem statement.

You need to show your attempt. It is mandatory here while posting in the HH forums.

VI=9.5V, RIN=1 kΩ, and vi=50.0mV.

I tried using KVL at the top node above the diode, using VI + Id*Rin + Vd = 0. I set I to 0 and calculated that Vd was -9.5 V and then set V to 0 and got I was -9.5mA. I drew a load line and it appeared to cross the iv curve at v=-4.75 and i=-4.75mA but I am pretty sure that is wrong and not sure where to go from there.

 

[gneill]

Were you given a value for the load resistor R_L?

 

[d3thkn1ght]

Yes we have to solve for the Vout for a Load Resistance of 2K ohms and for a Load Resistor of 4K ohms. The want the Vo voltage and vo voltage at both.

 

[gneill]

Yes we have to solve for the Vout for a Load Resistance of 2K ohms and for a Load Resistor of 4K ohms. The want the Vo voltage and vo voltage at both.

Okay. You've noted that the zener operates with a reverse bias. That's where the "Zener" breakdown occurs. That also means that the area of interest for the zener's characteristic curve is in the third quadrant, where both the component's voltage and current are negative.

To make using the curve more convenient you can rotate it and use treat the current and voltage as positive values. So for the curve you've shown, where the breakdown occurs with a reverse bias of 5 V, the region of interest for the rotated curve will look something like this:

To tackle the DC operating point for the zener your idea of using a Thevenin equivalent is a good one. Start by removing the zener from the circuit and find the Thevenin equivalent for what remains. Start with the R_L = 2 kΩ case. Can you place its load-line on the curve plot?

 

[d3thkn1ght]

First off, thank you for the informative reply. I've been driving myself crazy trying to figure this out.

A couple of questions:
1. Would the thevenin include the 2K ohms resistor? Or is it only the 1K ohm resistor?
2. For the load line, would I use the thevenin voltage for zero current, or would I use a simple voltage divider circuit formula to get the voltage drop across the load resistor (2K ohms)?

Thank you again.

 

[gneill]

First off, thank you for the informative reply. I've been driving myself crazy trying to figure this out.

A couple of questions:
1. Would the thevenin include the 2K ohms resistor? Or is it only the 1K ohm resistor?
2. For the load line, would I use the thevenin voltage for zero current, or would I use a simple voltage divider circuit formula to get the voltage drop across the load resistor (2K ohms)?

Thank you again.

The Thevenin model would include the Rin and R_L. The load line takes the open circuit voltage and short circuit current of the Thevenin equivalent as its end points. So work with the Thevenin equivalent.

 

[rude man]

You can also model the zener as an ideal zener in series with a resistor ...

 

[d3thkn1ght]

You can also model the zener as an ideal zener in series with a resistor ...

Would you be so kind to draw me a simple schematic? Would you have the cathode of the ideal zener pointing up or down? What would the voltage and resistance be?

I understand if you don't have the time but would be helpful.

 

[d3thkn1ght]

The Thevenin model would include the Rin and R_L. The load line takes the open circuit voltage and short circuit current of the Thevenin equivalent as its end points. So work with the Thevenin equivalent.

Does the Diode factor into the Thevenin equivalent circuit?

 

[gneill]

Would you be so kind to draw me a simple schematic? Would you have the cathode of the ideal zener pointing up or down? What would the voltage and resistance be?

The ideal zener just provides the inflection points in the characteristic curve (the forward bias turn-on and reverse bias zener turn-on points). Look at the characteristic curve of your zener. Everything you need to know is there.

Also note that as long as the zener is operating in "zener mode" you can model it as a voltage source in series with a resistor. If it might drop out of zener mode and turn off you can include an ideal diode in the path to accomplish that:

If there's a risk of the zener becoming forward biased, include another diode in parallel to handle that eventuality.

Does the Diode factor into the Thevenin equivalent circuit?

No. That comes into play after you've reduced the source network to a Thevenin model (one voltage source, one resistor). You want to establish a load line for the source.

 

[d3thkn1ght]

So do I treat the diode as a 1 ohm resistor or remove it entirely? If I remove it, would it be an open circuit or short?

 

[gneill]

So do I treat the diode as a 1 ohm resistor or remove it entirely? If I remove it, would it be an open circuit or short?

Which diode are you referring to? The zener itself or the one in the equivalent model?

 

[d3thkn1ght]

I am wondering how to treat the diode to determine the thevenin voltage and resistance. So, I guess I would be referring to the zener diode in the original circuit?

 

[d3thkn1ght]

From my understanding of Thevenin/Norton equivalents, and I am still a novice on those, you need to calculate the Voc and Isc by removing the load. I am not sure what the load is in this case, since you said to include the 2K ohm resistor in the Thevenin and according to the schematic and problem that is the RL or load resistor.

If I am mistaken here, please correct me.

 

[gneill]

I am wondering how to treat the diode to determine the thevenin voltage and resistance. So, I guess I would be referring to the zener diode in the original circuit?

You remove the zener diode completely from the circuit. You want to determine the load-line of the power supply and resistor network, which is why you want its Thevenin equivalent. THEN you place that load line on the zener's characteristic curve to see where its operating point is.

 

[d3thkn1ght]

You remove the zener diode completely from the circuit. You want to determine the load-line of the power supply and resistor network, which is why you want its Thevenin equivalent. THEN you place that load line on the zener's characteristic curve to see where its operating point is.

So to be clear, when I remove it, do I replace it with an open circuit or just remove the path completely so that it would be a simple circuit with 2 resistors in series.

 

[gneill]

From my understanding of Thevenin/Norton equivalents, and I am still a novice on those, you need to calculate the Voc and Isc by removing the load. I am not sure what the load is in this case, since you said to include the 2K ohm resistor in the Thevenin and according to the schematic and problem that is the RL or load resistor.

If I am mistaken here, please correct me.

You want to find the operating point of the zener. As far as the zener is concerned, everything else, including the load resistor, is part of the power supply network. Remove only the zener and find the Thevenin equivalent of everything else.

 

[gneill]

So to be clear, when I remove it, do I replace it with an open circuit or just remove the path completely so that it would be a simple circuit with 2 resistors in series.

Seems to me that's the same thing. Just erase the zener from the diagram. Pretend it was never there.

 

[d3thkn1ght]

Okay, thank you for your patience. Unfortunately, some of these concepts are hard for me to pick up.

I did that and calculcate Vth as 9.5V from VI times 2000/3000 and got Vth = 6.33 (not sure i I should at it the 50 mV from Vi

For the resistance I shorted both the diode and voltage sources and it appears that the Rth would be the Rin in parallel with RL
So Rth would be (1000)(2000)/3000 = 667 ohms

Am I in the right ballpark?

 

[gneill]

Okay, thank you for your patience. Unfortunately, some of these concepts are hard for me to pick up.

I did that and calculcate Vth as 9.5V from VI times 2000/3000 and got Vth = 6.33 (not sure i I should at it the 50 mV from Vi

Hold off on dealing with Vi. Just deal with the DC operating point for now. Later you can take advantage of superposition and address Vi separately.

For the resistance I shorted both the diode and voltage sources and it appears that the Rth would be the Rin in parallel with RL
So Rth would be (1000)(2000)/3000 = 667 ohms

Am I in the right ballpark?

Yup. Looks good. You should able to draw the load line now, especially since you've already found the Voc and Isc.

 

[d3thkn1ght]

Hold off on dealing with Vi. Just deal with the DC operating point for now. Later you can take advantage of superposition and address Vi separately.

Yup. Looks good. You should able to draw the load line now, especially since you've already found the Voc and Isc.

Thanks again. You are making this very understandable. I will draw the load line and try to see if I can find the operating point.

 

[d3thkn1ght]

Okay, so the load line appears to be from (0,10) to (6.33,0). Thus it crosses the iv line of xener diode slightly past 5 Volts. I don't have any easy way to draw the line to scale. Looked online but didn't find any apps that let me do it.

 

[gneill]

How did you determine the (0,10) point? What are the units, and is it really exactly 10?

You should be able to sketch it out by hand, and I'll assume that you can do that since you've got the idea that the load line crosses the zener's curve where it is conducting, and so is just a shade over 5 V. The important thing about the graphical method is that you can verify which part of the zener's IV curve is involved.

If you're satisfied that you can find the operating point graphically, we can tackle a method for finding it mathematically, too.

 

[d3thkn1ght]

I divided the Voc by the Rth and figured that would be the Isc. I believe I can get the operating point graphically, and would be interested in how to do it mathematically.

Also, now that I have the operating point, how do I found Vo and vo?

 

[d3thkn1ght]

And the units I used were milliAmps

 

[gneill]

I divided the Voc by the Rth and figured that would be the Isc. I believe I can get the operating point graphically, and would be interested in how to do it mathematically.

Voc is the same as Vth which is (in this case) about 6.33 V. Rth is 666.6 Ω. That makes Vth/Rth 9.5 x 10^-3 A, or 9.5 mA, right?

Also, now that I have the operating point, how do I found Vo and vo?

Let's do the operating point mathematically first as it dovetails nicely into the small signal solution. You essentially get two for one in terms of effort required, and finds a more accurate value for the operating point than the graphical method.

Since the two voltage sources are in series you can see that by superposition the same analysis can apply separately to both. The first thing to do is form the appropriate equivalent model for the zener that matches where on its curve it is operating, which you found that out via your graphical method of finding the operating point. What's your suggestion for the zener equivalent model?

Next, replace the input voltage sources with a single source and give it a variable name, say "E". With your zener model in place, find the output voltage in terms of E.

 

[d3thkn1ght]

Yes, I see my error on the voltage. I did it incorrectly in my head.

So, the operating point is 5V from the graphical method?

In the two sentences, would I use a single source of 9.55 Volts (9.5 from VI and .05 from Vi) and then use KVL at the top node?

 

[gneill]

Yes, I see my error on the voltage. I did it incorrectly in my head.

So, the operating point is 5V from the graphical method?

As you said before, just a bit over 5 V. An accurate sketch would give you a better idea of what "a bit" is. It's surely less than 0.1 V though.

In the two sentences, would I use a single source of 9.55 Volts (9.5 from VI and .05 from Vi) and then use KVL at the top node?

No, leave the source as a variable and solve symbolically. You can plug in numbers for the rest of the known values where appropriate though; You just want the flexibility of being able to plug in any value you want for E later on. You'll want to use KCL at the node (not KVL).

What's going to comprise your zener model?

 

[d3thkn1ght]

For the zener model, I would guess a 5 V source and a 1 ohm resistor? With the + end of the source pointing up

 

[gneill]

For the zener model, I would guess a 5 V source and a 1 ohm resistor? With the + end of the source pointing up

Yes. Very good.

 

[d3thkn1ght]

Great, I will try the equations and post my results. Thanks again for all your patience.

 

[d3thkn1ght]

So far, I came up with the following equation for KCL:
$$\frac{ (E - Vz)} {Rin} = Iz + \frac {Vz} {RL}$$

 

[gneill]

Usually the variable name Vz is reserved for the zener threshold voltage (which is 5 V in this case). Maybe you want to name the node voltage something else. Vout perhaps? If you've replaced the zener with its equivalent model then you should just have resistors and voltage supplies to deal with. That is, you can write an expression for Iz in terms of E and Vz (given that Vz is the zener threshold voltage).

You can plug in the values for the fixed components R_in, R_L, r_z, V_z. You'll be left with Vout in terms of E.