One specific type of diode is the Zener diode. Essentially, it has the same characteristic as a regular diode, but the difference is that a Zener diode can withstand breakdown. Even more, a Zener diode is made to be used in the breakdown region. Breakdown voltages are typically lower than in a regular diode and range from 1 or 2 V to 100 V or more. The symbol is shown in Figure 2.29.
Figure 2.29 Symbol of a Zener diode. In forward bias, the shown current and voltage are positive.
The characteristic of the Zener diode is similar to that of a regular diode (see fig. 2.30). However, since breakdown is “allowed” in a Zener diode, the diode model is extended to include the breakdown region. In this region, the voltage barely changes, while currents can range from nearzero values to relatively large currents. As a result, the zener diode is modeled as a voltage source in this region. Note however that the polarity of the voltage source is reversed with respect to forward bias.
Figure 2.30 Characteristic of the Zener diode. The diode appears to have three operating regions: forward bias, nonconducting and breakdown, and behaves differently in each region.
Diodes  Level 9
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A few quizzes will help you to understand how zener diodes behave in circuits, and what their function might be.
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Question 1 of 5
1. Question
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Hint
Estimate in which direction v_{in} “pushes” the current. Next, determine how the Zener diode will respond to that.

Question 2 of 5
2. Question
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Hint
Estimate again the direction of the current flow.

Question 3 of 5
3. Question
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Hint
The two Zener diodes are in series, so their current is identical.

Question 4 of 5
4. Question
One or more of the statements below are true. Indicate which.
Assume that the Zener’s breakdown voltage is larger than the diode’s knee voltage.Correct
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Hint
What can you say about the diode if you know that the Zener breaks down? Does it conduct or not?

Question 5 of 5
5. Question
One or more of the statements below are true. Indicate which.
The voltages v_{1}, v_{2} and v_{3} are defined relative to ground.Correct
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Diodes  Level 10
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In the next level, you will have to predict the influence of changing parameters in the circuit, e.g. what will happen if a certain resistor value increases. Remember that, if the diode conducts forward, the voltage across the diode equals the knee voltage. If it breaks down, this voltage equals the breakdown voltage.
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Question 1 of 5
1. Question
Assume that the input voltage v_{in} is sufficiently large (and negative) to cause diode break down.
What happens to the currents and the voltages if v_{in} decreases (in absolute value)?Correct
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Hint
What do you know about the voltage across a diode in break down?
And what’s the relation between the current through the diode and the current through the resistor? 
Question 2 of 5
2. Question
Assume that the input voltage v_{in} is sufficiently large to cause diode break down.
What happens to the currents if R_{2} increases?Correct
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Hint
First find out what you can say about the voltage across R_{2}.

Question 3 of 5
3. Question
Assume that the input voltage v_{in} is sufficiently large, so that there is a nonzero current through both diodes.
What happens to the currents if v_{in} increases?Correct
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Hint
First find out what you can say about the voltages.

Question 4 of 5
4. Question
Assume that the input voltage v_{in} is sufficiently large and positive to cause all diodes to conduct a nonzero current.
What happens to the voltages and the currents if a Zener diode with a smaller breakdown would be used?Correct
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Hint
You can assume that the diode still conducts.

Question 5 of 5
5. Question
Assume that the input voltage v_{in} is sufficiently large to cause all diodes to break down.
What happens to the voltages if v_{in} increases?Correct
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Diodes  Level 11
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The next challenges contain numerical values. You have to calculate voltages or currents. In case of noninteger results, use the decimal dot (.).
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Question 1 of 5
1. Question
First find out what the Zener diode will do (conduct forward, break down, not conduct). Then calculate the current in the loop.
Assume a knee voltage of 1 V (in all challenges). The current in the loop is (25) mA. If the polarity of the voltage source is reversed, the current will run in the other direction and be equal to (40) mA.
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Hint
First apply KVL to find the voltage across the resistor.

Question 2 of 5
2. Question
First find out what the Zener diode will do (conduct forward, break down, not conduct). Then calculate all currents.
 The current through the 330 Ω resistor is (10) mA. The current through the 100 Ω resistor is (12) mA. The current through the Zener diode is (2) mA.
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Hint
The Zener diode and the 330 Ω resistor are in parallel, so they have the same voltage. Use KVL to find the voltage across the other resistor.

Question 3 of 5
3. Question
Almost the same challenge as the previous one, but now with a 200 Ω resistor instead of 100 Ω. Calculate the current through the Zener diode.
 The current through the Zener diode (0) mA.
Correct
Indeed, the Zener diode will not break down, so it’s current will be 0 A.
Incorrect
Always interpret your results. If you assumed that the Zener diode was breaking down, you would obtain a forward current. This would make no sense, so that assumption was incorrect.
Hint
Booby trap alert! You are seduced to make a mistake here.

Question 4 of 5
4. Question
First find out what each Zener diode will do (conduct forward, break down, not conduct). Then calculate all currents. Assume a knee voltage of 1 V.
 The current through the 100 Ω resistor is (40) mA. The current through the 20 Ω resistor is (100) mA. The current through both Zeners is (60) mA.
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Hint
Take care of the polarity of the voltage source. Also, if one Zener diode breaks down, the other one has to conduct forward.

Question 5 of 5
5. Question
 The current through the right resistor is (20) mA. The current though the upperleft resistor is (60) mA. The current through the Zener diode is (40) mA.
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Diodes  Level 12a
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In these challenges you have to find the I/O characteristic, i.e. the relation between the input voltage v_{in} and v_{out}.
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 break down.
 conduct forward.
 not conduct.

For large and negative input voltages, the Zener diode will

For large and positive input voltages, the Zener diode will

For small input voltages around 0 V, the Zener diode will
Correct
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Question 2 of 5
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Question 3 of 5
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Question 4 of 5
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Question 5 of 5
5. Question
So, this is the I/O characteristic. Find the values for V_{T1} and V_{T2}. Assume a knee voltage of 1 V.
 V_{T1} equals (2.5) V. V_{T2} equals (1) V.
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Diodes  Level 12b
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A bit more complex now…
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 conduct forward.
 break down.
 not conduct.

For large and negative input voltages, the Zener diode will

For large and positive input voltages, the Zener diode will

For small input voltages around 0 V, the Zener diode will
Correct
Incorrect

Question 2 of 5
2. Question
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Question 3 of 5
3. Question
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Question 4 of 5
4. Question
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Question 5 of 5
5. Question
So, this is the I/O characteristic. Find the values for V_{T1} and V_{T2}. Assume a knee voltage of 1 V.
 V_{T1} equals (1.25) V. V_{T2} equals (5) V.
Correct
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Diodes  Level 12c
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And a final one.
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Question 1 of 5
1. Question
Sort elements
 break down.
 conduct forward.
 not conduct.

For large and negative input voltages, the upper Zener diode will

For large and positive input voltages, the upper Zener diode will

For small input voltages around 0 V, both Zener diodes will
Correct
Incorrect

Question 2 of 5
2. Question
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Question 3 of 5
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Question 4 of 5
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Question 5 of 5
5. Question
So, this is the I/O characteristic. Find the values for V_{T1} and V_{T2}. Assume a knee voltage of 1 V.
 V_{T1} equals (4.4) V. V_{T2} equals (3.3) V.
Correct
Incorrect
Level 9, question 4. I can’t think of values where the diode conducts and the Zener does not break down…what am I missing?
Note the remark on the breakdown voltage: It is larger than the diode’s knee voltage.
Suppose the diode conducts a very small current.
Then there is a very small voltage across R_{2}, and the knee voltage across the diode.
The sum of both is the voltage across the Zener, and this sum might still be below the breakdown voltage.
level 12 a question 2: applying KVL I found Vin – 2.5 – V0 = 0. I thought so because the direction of the current in breakdown. This would hiwever mean that V0 = Vin – 2.5 instead of Vin + 2.5.
Note that the current goes counterclockwise if the Zener diode breaks down.
KVL should then be Vin + 2.5 – Vo = 0V, etc.
thanks 😉
you’re welcome.
12 c : shouldn’t the right resistor be 10 R1?
You’re right. I’ve changed it now.
Thanks for letting me know.
the figure on question 12.c.5 doesn’t work anymore. And it would be nice if you putted every circuit drawing with the graphs, i’m thinking here about lvl 6,7 or 8 (not sure about this).
Problem with question 12.C.5 is fixed.
Thanks for letting me know.
I’ve added the circuit drawings with the graphs in levels 7a, 7b and 7c.
Diodes – Level 9, Question 3: There is a grammatical error in option 3. Thought you should know, sir.