Same approach as for diode circuits here. You have to do part of the work.
Transistors  Level 2
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Here is the circuit that you will analyze. You have to calculate the currents and voltages for all components.
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Question 1 of 8
1. Question
The first question is whether the transistor is in cutoff mode or not. Remember that a BJT behaves like a diode between base and emitter, so an equivalent circuit would look like this.
So what do you think?
Correct
Incorrect

Question 2 of 8
2. Question
The transistor can indeed not be in cutoff mode. We do not know yet how the transistor will behave between collector and emitter, but the base current can be calculated. Take 0.7 V for the knee voltage.
 The base current is (10) µA (microamps).
Correct
Incorrect

Question 3 of 8
3. Question
It is not always an obvious task to predict whether the transistor will be in saturation or in active mode. So, one way to proceed would be to start with an assumption. Let’s assume that the transistor is in active mode. In that case, the transistor will behave like a current source between collector and emitter, with a current equal to β times the base current.
Now calculate the following quantities.
 The collector current is (1) mA. The voltage across the 2kΩ resistor is (2) V. The voltage V_{CE} is (2) V.
Correct
Was the assumption of active mode correct? Well, apparently V_{CE} is still considerably larger than 0 V, so the result makes sense.
Incorrect
Was the assumption of active mode correct? Well, apparently V_{CE} is still considerably larger than 0 V, so the result makes sense.

Question 4 of 8
4. Question
The conclusion of the previous question was that the transistor is in active mode. But what would the result have been if we assumed that the transistor was in saturation? And how could we know that the obtained result was impossible? Let’s try.
Calculate the following quantities.
 The voltage V_{CE} is (0.2) V. The voltage across the 2 kΩ resistor is (3.8) V. The collector current is (1.9) mA.
Correct
A collector current of 1.9 mA is impossible. Why? Because the maximum current for the given base current (10 μA) is 1 mA (= β.I_{B}), which is the value for the active mode. In saturation, the collector current is always less than the current in the active mode.
Incorrect
A collector current of 1.9 mA is impossible. Why? Because the maximum current for the given base current (10 μA) is 1 mA (= β.I_{B}), which is the value for the active mode. In saturation, the collector current is always less than the current in the active mode.

Question 5 of 8
5. Question
Now suppose the resistor at the collector is changed to a 10 kΩ resistor.
What would the results be like? Again, start with the assumption of active mode, and calculate the following values.
 The collector current is (1) mA. The voltage across the 10 kΩ resistor is (10) V. The voltage V_{CE} is (6) V.
Correct
So, V_{CE} is negative! This is impossible. This voltage cannot drop below zero, so the assumption of active mode was incorrect.
Incorrect
So, V_{CE} is negative! This is impossible. This voltage cannot drop below zero, so the assumption of active mode was incorrect.

Question 6 of 8
6. Question
 The voltage V_{CE} is (0.2) V. The voltage across the 10 kΩ resistor is (3.8) V. The collector current is (0.38) mA.
Correct
This result makes sense. The obtained current is smaller than 1 mA, the current in active mode.
Incorrect
This result makes sense. The obtained current is smaller than 1 mA, the current in active mode.

Question 7 of 8
7. Question
Finally, let’s use the load line to visualize the results.
Calculate the quantities in the figure below.
 I_{C,max}, the current in the active mode, is (1) mA. V_{OC}, the open circuit voltage, is (4) V. I_{SC1}, the short circuit current for the 2 kΩ resistor, is (2) mA. I_{SC2}, the short circuit current for the 10 kΩ resistor, is (0.4) mA.

Question 8 of 8
8. Question
Perhaps one more…
Suppose the resistor at the collector is again 2 kΩ. How can the base resistor be changed, such that the transistor is driven into saturation? The new load line would look like this.
 The collector current you want to obtain is the short circuit current, equal to (2) mA. In the active mode, the base current then would have been (20) µA. But the base current has to be larger, in order to ensure saturation. The minimum value for the base resistor is then (165) kΩ.
Correct
Incorrect
Transistors  Level 3
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Question 1 of 6
1. Question
First calculate some voltages and currents. Do this in the order given below. Give one decimal after the decimal point, if needed.
 The voltage at the base (relative to ground) is (1.7) V. The voltage at the emitter (relative to ground) is (1) V. The emitter current is (10) mA. Take the same value for the collector current. The voltage across R_{L} is (4) V. The voltage across the transistor V_{CE} is (1) V. Assume β is 50. The base current I_{B} is then (200) μA. The voltage across R_{B} is (4.3) V. The current through R_{B} is (500) µA. The current through the Zener diode is (300) µA. The current the power supply has to deliver is (10.5) mA.
Correct
Perhaps this went easy but note that you have to learn to find the order of calculations yourself.
Incorrect
Perhaps this went easy but note that you have to learn to find the order of calculations yourself.
Hint
Three hints for the prize of one!
 You can assume that the Zener diode breaks down (why?).
 The voltage at the base (relative to ground) is also the voltage across the Zener diode.
 The voltage at the emitter (relative to ground) is also the voltage across the emitter resistor R_{E}.

Question 2 of 6
2. Question
Correct
There is an important conclusion to this result: The current through R_{L} does not depend on the value of R_{L}. That is the behavior of a current source!
Note that the transistor is used in the active mode or linear mode: Both voltage and current are nonzero.
Incorrect
There is an important conclusion to this result: The current through R_{L} does not depend on the value of R_{L}. That is the behavior of a current source!
Note that the transistor is used in the active mode or linear mode: Both voltage and current are nonzero.
Hint
First consider what will happen with the voltage across R_{E} if R_{L} decreases.

Question 3 of 6
3. Question
There is however an upper limit to R_{L}. Indeed, the larger it becomes, the larger the voltage across it will be (assuming that the current does not change – remember, this circuit is a current source of 10 mA). As a result, the lower the voltage V_{CE} across the transistor will be. The lower limit for this voltage is 0 V. Now, calculate the maximum value for R_{L}.
 The maximum value for R_{L} is (500) Ω.
Correct
If R_{L} would increase further, the transistor will go into saturation. The current I_{C} will drop, and the Zener diode will no longer break down.
Incorrect
If R_{L} would increase further, the transistor will go into saturation. The current I_{C} will drop, and the Zener diode will no longer break down.
Hint
First calculate the maximum voltage across R_{L}.

Question 4 of 6
4. Question
Now the question is what can be done to increase the current source’s current I_{C}. Which of the statements below is true? You may assume that the Zener diode will always break down, and that R_{L} is never too large.
Correct
V_{CC} indeed has barely an effect on I_{C}. What will change if V_{CC} increases, is the current through R_{B}, and the voltage across the transistor V_{CE}. As a result, if V_{CC} increases, the upper limit for R_{L} will increase (see previous question).
Incorrect
V_{CC} indeed has barely an effect on I_{C}. What will change if V_{CC} increases, is the current through R_{B}, and the voltage across the transistor V_{CE}. As a result, if V_{CC} increases, the upper limit for R_{L} will increase (see previous question).

Question 5 of 6
5. Question
Now the question is what will happen if the transistor’s β changes. Suppose β increases, which of the statements below is true?
Correct
This circuit is an example of a circuit in which I_{E} is controlled directly. I_{B} is then obtained by dividing I_{C} (≈ I_{E}) by β (assuming linear mode of course). This type of transistor biasing is a very common practice.
Incorrect
This circuit is an example of a circuit in which I_{E} is controlled directly. I_{B} is then obtained by dividing I_{C} (≈ I_{E}) by β (assuming linear mode of course). This type of transistor biasing is a very common practice.

Question 6 of 6
6. Question
A final one. Let us have a look at the components at the base of the transistor.
Suppose R_{B} increases, other components remain unaltered. Which of the statements below are true?
Correct
This result also means that R_{B} should not be too large. The current through it should be at least the desired base current. If this condition is not met, the Zener diode will not break down.
Incorrect
This result also means that R_{B} should not be too large. The current through it should be at least the desired base current. If this condition is not met, the Zener diode will not break down.
Hint
Lowering R_{B} has no effect on the voltage across it!
Transistors  Level 4
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The circuit below contains two transistors. The emitter current of the first transistor is also the base current of the second one. This configuration is called a Darlington pair, and allows for a higher current gain. At the start of this level you may assume that β is ∞. This means that you can ignore the base current, since I_{B} = I_{C} / β = 0 A. Note however, that this does not imply that the transistor is in cutoff. In practice, the base current is still nonzero, so the transistor is not in cutoff mode. It’s just small enough so that you can ignore it in your calculations.
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Question 1 of 4
1. Question
Calculate all currents, voltages and powers, assuming that β is ∞. Use one digit after the decimal point, if necessary.
 The voltage at the base of the first transistor (relative to ground) is (6) V. The voltage at the base of the second transistor is then (5.3) V. The voltage at the emitter of the second transistor is then (4.6) V. The emitter current of the second transistor is (9.2) mA, which is also equal to I_{C2}. I_{B2}, I_{C1} and I_{B1} are all (0) A. V_{CE2} is (7.4) V. The power of the second transistor is (68) mW (fill in an integer number). V_{CE1} is (6.7) V. The power of the first transistor is (0) mW (fill in an integer number).
Correct
Keep in mind that some results are 0, but in practice these values will be nonzero, though very small.
Incorrect
Keep in mind that some results are 0, but in practice these values will be nonzero, though very small.
Hint
Two hints:
 The left part of the circuit is a voltage divider with two equal resistors. Since you may ignore the base current, the voltage division formula can be used.
 Do you think the baseemitter diodes will be on?

Question 2 of 4
2. Question
Suppose now that β is 60. In that case the base current will not be ignored. In order to make a more accurate calculation, the circuit will be modified somewhat. The voltage divider at the left will be replaced by it’s Thévenin equivalent circuit (at the right). Calculate the Thévenin voltage V_{th} and the Thévenin resistance R_{th}.
 The Thévenin voltage is the voltage between nodes a and b, and equals (6) V. The Thévenin resistance is the resistance between nodes a and b, with the voltage source replaced by a short circuit, and equals (500) kΩ.
Correct
Incorrect
Hint
If the voltage source is replaced by a short circuit, the two resistors are in parallel between nodes a and b.

Question 3 of 4
3. Question
The circuit now looks like this. Note that this modification will have no influence on the transistor currents and voltages. In order to find the emitter current I_{E2}, you will need to solve an equation. Apply KVL in the loop at the left. The base current I_{B1} may be approximated by I_{E2} / β^{2}. Remember that β = 60.
 The emitter current I_{E2} equals (7.2) mA (give one digit after the decimal point).
Correct
The previous answer was 9.2 mA (for β = ∞). It is a different result, of course, but not that different. A difference of 20% is acceptable.
Incorrect
The previous answer was 9.2 mA (for β = ∞). It is a different result, of course, but not that different. A difference of 20% is acceptable.
Hint
Use 0.7 V for the baseemitter voltages.
Apply Ohm’s Law for the resistors. 
Question 4 of 4
4. Question
Now calculate the other currents, voltages and powers. Assume I_{E2} equal to 7.2 mA. Give one digit after the decimal point, if necessary.
 The voltage at the emitter of the second transistor is (3.6) V. V_{CE2} is then (8.4) V. Assuming I_{C2} equal to I_{E2}, the power in this transistor is (60) mW (integer number, ignore the contribution of the base current). V_{CE1} is then (7.7) V. I_{C1} is approximately equal to I_{E1}, which is (120) µA. The power in the first transistor is (1) mW (integer number). I_{B1} is (2) µA.
Correct
For β = ∞, the power in the second transistor was 68 mW. Now we found 60 mW. A ‘normal’ difference.
For the first transistor, we obtain 0 mW and 1 mW respectively.
Incorrect
For β = ∞, the power in the second transistor was 68 mW. Now we found 60 mW. A ‘normal’ difference.
For the first transistor, we obtain 0 mW and 1 mW respectively.
Hint
I_{E1} = I_{B2} = I_{C2} / β
Transistors  Level 5
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Another popular transistor configuration is the current mirror. In the circuit below, the two transistors are identical (typically implemented on the same chip), so they have identical values for β, and the relationship between base current and baseemitter voltage is also identical. Since both bases are connected, and also both emitters, the baseemitter voltages are also the same. As a result, both transistors have the same base current, and since they have the same β, they also have the same collector current. Hence the name “current mirror”.
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Question 1 of 1
1. Question
Calculate all relevant currents and voltages. Ignore the base currents (β = ∞). Note that the order given below, is NOT the order in which you can make the calculations. You have to find this order yourself.
 I_{C1} equals (1) mA. I_{C2} equals (1) mA. V_{CE1} equals (0.7) V. V_{CE2} equals (1.8) V. The baseemitter voltage of both transistors is (0.7) V. The voltage across R_{C1} is (3.3) V. The voltage across R_{C2} is (2.2) V.
Correct
So indeed, the collector current for both transistors is 1 mA, whatever the value of R_{C2}.
However, it may not be too large, because in that case transistor 2 will saturate (why?).
If the value of the base current is taken into account, the results will change slightly.
As illustrated below, the current through resistor R_{C1} will be equal to the collector current plus two times the base current.Incorrect
So indeed, the collector current for both transistors is 1 mA, whatever the value of R_{C2}.
However, it may not be too large, because in that case transistor 2 will saturate (why?).
If the value of the base current is taken into account, the results will change slightly.
As illustrated below, the current through resistor R_{C1} will be equal to the collector current plus two times the base current.Hint
Note that the collector of the first transistor is connected to the base, so V_{CE1} equals V_{BE1}.
Transistors  Level 6
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In the previous challenges, the transistor was used in the linear mode. As mentioned in the introduction, transistors are very often used as controllable switches, so in cutoff mode or saturation mode. The circuit below is an example.
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Question 1 of 5
1. Question
The input voltage v_{in} is either 0 V or 4.5 V. What can you say about the LEDs when v_{in} is 0 V?
Correct
The LEDs will be off, since the transistor is in cutoff mode. No collector current also means no current through the LEDs.
Incorrect
The LEDs will be off, since the transistor is in cutoff mode. No collector current also means no current through the LEDs.
Hint
What will be the transistor mode? Cutoff or saturation? And what will be the collector current?

Question 2 of 5
2. Question
Now suppose v_{in} is 4.5 V. The idea is that the transistor will then go into saturation mode, so take 0.2 V for the collectoremitter voltage.
If each LED has a knee voltage of 1.5 V, what is then the voltage across R_{C}? The voltage across R_{C} is (1.3) V.
Correct
Incorrect

Question 3 of 5
3. Question
 R_{C} is (6.5) Ω.
Correct
Incorrect
Hint
First calculate the desired current through R_{C}.
The voltage across R_{C} is 1.3 V (see previous question). 
Question 4 of 5
4. Question
In order to get the transistor in saturation, the base current should be large enough. Again, with a current of 100 mA through each LED, and the transistor’s β equal to 40, what is the minimum value for the base current?
 The minimum value for the base current is (5) mA.
Correct
The base current should be larger than 200 mA divided by 40 (I_{B} > I_{C }/ β).
Incorrect
The base current should be larger than 200 mA divided by 40 (I_{B} > I_{C }/ β).
Hint
The collector current is equal to the current through R_{C}.

Question 5 of 5
5. Question
Now we can calculate a limit for R_{B}. First find the voltage across the base resistor if v_{in} is 4.5 V, then use Ohm’s law.
 The voltage across R_{B} is (3.8) V. The minimum value for R_{B} is (760) Ω.
Correct
In practice, a value far below 760 Ω will be chosen, e.g. 390 Ω.
Incorrect
In practice, a value far below 760 Ω will be chosen, e.g. 390 Ω.
Hint
You have to take 5 mA for the base current (see previous question).
The website sure is useful, especially the quizzes. It would be nice to get some explanation if an exercise were incorrect though.
I just noticed that some questions do have an explanation, thanks!
Hi Jasper. In case the right answer does not help you further, post a comment explaining your problem. This might help me to add more explanations.
level 3, question 1: Why can we assume that the zenerdiode breaks down?
Because the zener diode is fed with a 6V power supply, and the breakdown voltage is lower than 6V.
Note that this is still an assumption that should be checked afterwards.
Suppose e.g. that the base current is very large (which is usually NOT the case), then the voltage across R_{B} might so large that the breakdown voltage is not reached.
Level 4, question 1: You asked 2x the power of the second transistor, i guess this is has to be 1x power of the first, and 1x of the second.
Indeed, this was an error.
Thanks for letting me know.
Deze website is geweldig en echt behulpzaam! Awesome gedaan, meneer Geurts!
Thanks, you’re welcome 😉
Transistor lvl 4:
On the exam, may you assume that Ib=0 or do you have to work with thevenin?
Never mind, it depends on the value of β
Indeed. You should be able to use both methods.
Transistors level 2, question 8, shouldn’t it be the resistor at the collector instead of transistor?
You’re right. I’ve changed it. Thanks.
level 4 question 1: bèta is infinite so Ib=0 in both transistors. This means they are both in cutoff. How do we know the voltage Vbe in this case? I assumed 0 but this turned out to be wrong.
No, no, no, no, no!
I understand your reasoning, but it is flawed.
“bèta infinity” means that you can ignore the base current in your calculations.
In practice, it is still nonzero, so the transistor is NOT in cutoff mode.
And so Vbe is 0.7 because it’s not in cutoff?
Yep.
level 5 question 1, how do we know the voltage over the 2.2k resistor?
By multiplying the resistor value with the current 😉
And how do we find the current then?
The hint for that is in the explanation of the circuit, when you start the quiz.
That paragraph was very helpfull, thank you!
Level 3 question 3: I don’t get why it is 500 instead of 430? isn’t the maximum value over RL 4.3V?
Probably you took 0.7 V for the voltage V_{CE} across the transistor.
This is a common mistake.
The voltage V_{BE} is 0.7 V, but V_{CE} can approach 0 V as indicated in the question.
No, i took 1.7V for the voltage across Re because of the zener diode. In the question it said that Vce =0V so i assumed that 6V0V1.7V=4.3V=Vrl
Ah, OK, but than you forgot to subtract the 0.7 V of the baseemitter voltage from the zener voltage.
LEVEL 3 – Q1
Why can we assume that the transistor is in active mode?
It is an assumption, so you are not sure yet.
If e.g. the voltage across the transistor would be negative, the assumption would be false, and the transistor would be in saturation (see level 2).
You do know that the transistor is not in cutoff though. You know why?
Level 2 : Why can the transistor not be in cut off mode ? How can you see that ?
Look at the equivalent circuit. The baseemitter diode is biased with a voltage higher than the knee voltage.
So, the diode conducts, and cutoff mode can excluded.
Ok, thanks!
you’re welcome.
Level 3 – Question 3:
The explanation says the Ic will drop when the transistor goes in saturation mode, which I understand, but then it says the Zener diode will no longer break down, why is that?
That is indeed a tricky one.
If Ic drops, the voltage across Re drops.
This voltage plus 0.7 V for Vbe will be less than the breakdown voltage, so the zener will no longer break down.
Maybe you could add the B for the transistors at question 3 of level 4, I began doubting…
How do I calculate the current Ie2? What I did is detract the diode voltages from Vth, so it looks like this:
60,70,7= 4,6V
and then you just have a voltage divider left between the 2 resistors of 500 and 500k Ohm, so I calculated the voltage for the 500 Ohm resistor:
(500/500500)*4,6= 0,0045954V
and divided by 500 Ohm that becomes 0,00000919 A (9,19 µA).
I feel like there’s another way to do it, using the equation with beta, but I can’t think of it.
Well, your way of calculating makes no sense. Sorry.
I have never seen a voltage division formula like yours (500/500500 ????).
When you use the Thévenin equivalent circuit, there is no voltage divider left.
What you can do, is use KVL in the loop consisting of Vth Rth Vbe1 Vbe2 and Re.
I used KVL in the loop: 6+500*Ib1+1.4+500Ie2=0 with Ib1=Ie2/60^2 and my Ie2 still is 9.197?
You multiply Ib1 with 500, but it should be 500000 (500 kohm).