Try this test and see if you get 100%.
We are using a simple NPN or PNP transistor such as BC547, BC557 or BC338.
Not all the answers have been covered in the lecture-notes.
The Test is to see how much you know.
After completing the test you can find the lecture-notes here:
Link 1
Link 2
Link 3

1. Transistors are separated into two types. Name them.

Positive and Negative
P and N
PNP and NPN
PNN and NNP

2. Name the three leads of a common transistor:

Collector Bias Omitter
Base Collector Case
Emitter Collector Bias
Collector Base Emitter

3. Which way do you normally draw an NPN transistor in a circuit:

a
b
c
d

4. When testing a transistor with a multimeter, it is set to:

Volts
Low Ohms
High Ohms
High Volts

5. Name the leads of a transistor:

a
b
c
d

6. The easiest transistor to test is:

PNP
NPN
Both equal

7. The lead marked with the arrow is:

The Collector
The Base
The Emitter
The case

8. If the voltage on the base of a transistor increases, does it:

Turn on
Turn off
Not enough information
Remain the same

9. Which diagram is correctly identified:

a
b
c

10. Which circuit is called a self-biased stage:

a
b
c
d

11. Which transistor is being turned ON in an amplifying stage:

a
b
c

12. In the diagram below, what will happen to the pointer when a finger is applied to the leads:

       

The pointer will move across the scale
The pointer does not move

13. In the diagram below, name the fault with the transistor:

       

The transistor has shorted between collector and base
The transistor has shorted between collector and emitter
The transistor is not faulty

14. Which circuit is correctly biased:

a
b
c

15. Which circuit contains an input capacitor:

a
b
c
d

16. When a transistor is turned ON, it is like:

a
b
c

17. The 1k resistor is changed to 2M2, the collector voltage will:

rise
fall
remain the same
 

18. The transistor stage is called:

H-Bridge
self biased
resistor bridge
 

19. The stage is called:

DC coupled
AC coupled
Direct Coupled

20. When the input signal is rising, the output is:

falling
rising
unknown

21. The input capacitor is 100n, you would design the output capacitor to be:

1n
10n
100n
1u

22. Putting another 2M2 across the base-bias resistor will:

increase the collector voltage
decrease the collector voltage
collector voltage will remain the same

23. A transistor has a gain of 350 in the circuit above. Replacing it with a transistor having a gain of 150 will:
increase the collector voltage
decrease the collector voltage
collector voltage will remain the same

24. The base-bias resistor falls off a module. The collector voltage of the transistor in the circuit above will:
increase
decrease
remain the same

25. The collector resistor falls off a module. The collector voltage of the transistor in the circuit above will:
increase
decrease
remain the same

26. The gain of the transistor is 250. What is the gain of this self-biased stage:

250
70
22

Note: The gain of a self-biased stage is about 70 for all types of transistors as the 2M2 reduces the gain.

If you want to reduce the gain, you can add a resistor to the input or decrease the input capacitance:

Changing the input components will reduce the input signal and thus reduce the gain of the stage.


27. Which stage will produce the highest gain:

A
B 
both the same

Note: The base-bias resistors for circuit 27B reduces the input signal considerably and that's why the output will be less than 27A.

28. What is the gain of this H-bridge stage:

10k
10
1k
1

Note: The ratio of the collector resistor and emitter resistor determines the gain of the stage.

29. Which stage has the highest input impedance (resistance):

H-bridge
self-biased
both the same

30. The transistor is equal to a resistor of:

1k
10k
6k

31. The voltage on the collector is 1v. To increase the voltage, the 1M is:

increased in resistance
decreased in resistance
the collector voltage cannot be changed

32. The output capacitor should be about 10 times larger than the input capacitor. The reason:
the output will have a higher current
the output will have a higher voltage
no technical reason

Note: A higher current will allow a capacitor to charge faster. We do not want a coupling capacitor to charge as this will reduce the amplitude of the signal.
The output will have at least 10 times more current and so a larger capacitor will take longer to charge.

33. Can you design a self-biased stage without knowing the characteristics of the input and output LOAD:
yes
no
not relevant

Note: You MUST know the characteristics of the input device and output LOAD as these will control the amplitude of the signal produced by the stage.

34. The signal on the collector of a self-biased stage is 1,200mV. When a LOAD is connected to the output, the amplitude of the signal will:
decrease
increase
stay the same

Note: A waveform (amplitude) produced by a stage will ALWAYS decrease when the output is connected to a LOAD.

35. The input capacitor is reduced to 10n. The signal on the collector will:

decrease
increase
stay the same

Note: The signal on the collector may stay the same but theory says it will decrease. That's why coupling capacitors are as large as possible. 

36. Using our theory with decoupling capacitors, what size capacitor will be used for C3:

100n
1u
10u

Note: Increasing the value is good design and shows good circuit knowledge. Maybe 10u is too high.

37. Using our theory on stage amplifying, for each stage, what do you think will be the gain of the circuit in Q36:
4,900
70
200

Note: The gain can be as high as 70 for each stage. But when the amplitude reaches full rail voltage the signal will not increase.

38. Can this circuit amplify a signal as small as one micro-volt:

yes
no

Note: The circuit will detect a signal as small as one micro volt because it is already "turned ON" and will detect "hum" in leads and all noise on the input.

39. Is it important to have the collector voltage equal to mid-rail voltage:

yes
no

Note: When the collector voltage is mid-rail, the stage can produce the maximum output without "clipping" or "bottoming" or distorting a signal.

40. In an audio amplifier, does it matter if the signal emerges INVERTED:

yes
no

Note: Your ear cannot detect an inverted signal.
 

41. What is the input impedance of a self-biased common emitter stage:

5,000 ohms
1,000 ohms
cannot be determined

Note: All the answers above are correct. The input impedance (resistance) changes enormously, according to the value of the base-bias resistor.
As the base gets turned ON harder with a low value base resistor, the input resistance decreases enormously.
Here is how to consider the value.
In the circuit above, the base has a 2M2 resistor and 3v supply. This means less than 1 microamp is entering the base and thus another microamp (or less) will be detected by the stage.
This is all you have to know. 

42. What happens to a 1v input signal because we know the base will not rise above 0.7v:

The signal gets converted to current
The signal gets converted to current
cannot be determined

Note: As the input signal increases in amplitude, it will deliver a higher current and this is what the transistor will detect as the voltage on the base will rise very very little. So it does not detect the voltage, ONLY the current.

43. How does the signal from the electret microphone enter the base of the transistor:

The signal gets converted to current
The signal gets converted to current
The signal enters as voltage

Note: The electret microphone detects a whistle and produces a wave very similar to a sinewave. When the FET transistor inside the microphone turns OFF, the voltage on the left lead of the 100n coupling capacitor rises. This voltage passes through the capacitor and adds a little more current into the base.  This because the voltage is higher than the 0.65v on the base.
The transistor reacts to this extra current and allows 200 times more current to flow in the collector-emitter leads. This reduces the voltage on the collector. Connected to the collector is a 2M2 resistor and now a slightly smaller voltage will appear across the 2M2 and thus it will deliver slightly less current into the base.
This means the base will see slightly less current and so the transistor will not perform to the full extent of 200 times amplification. Normally we allow the gain to be 70.
This means the output waveform will be 70 times larger than the input waveform and will be inverted.
You cannot perform any mathematical calculations because you don't know how much "degenerating effect" the 2M2 will have on the signal.
Degeneration is the name we give to an effect that improves stability. It can also be called NEGATIVE FEEDBACK.

44. Does the first transistor pass energy to the second stage:

yes
no
unknown

Note: NO.  When the first transistor turns OFF, the 22k pulls the 100n high and delivers energy to the second stage.
When the first transistor turns ON, it pulls the 100n low and turns OFF the second stage. It is the 22k resistor that does all the "passing."

45. What effect does a coupling capacitor have on the transfer of energy:

none
a lot
unknown

Note: The effect is unknown. Passing energy into the base of a transistor is not a linear transfer. As the transistor becomes more saturated, the gain of the transistor reduces.
It is amazing that we can get a reasonable linear and undistorted outcome from so many non-linear features.
However a capacitor does not transfer 100%. There are a lot of losses and the high-gain of each transistor overcomes these losses.

46. Why does a coupling capacitor have losses:

because it charges
because it discharges
unknown

Note: When a signal passes through a coupling capacitor, it charges.
This means the voltage across the capacitor reduces the amplitude of the signal.  When the amplitude reduces, the current provided by the signal reduces and since the transistor operates on CURRENT, the transistor sees the capacitor as producing losses.

47. Can you determine coupling capacitor losses:

yes
no

Note: The answer is Yes and NO.  You can try to work out the losses but they increase and decrease according to the frequency.
The losses are due to the capacitor charging during each cycle and the more it charges, the higher the losses.
You cannot choose a single capacitor value for the range 20Hz to 20,000Hz and so you have to compromise.
You need a high-value capacitor when high currents are involved and if a high voltage is involved, this adds to the size of the capacitor.
And if a high current is flowing in and out of the capacitor, it must have a high ripple capability so it does not get hot.

48. How much current is going to turn the stage ON fully and turn it OFF fully:

2uA  and about 1.5uA
2mA  and about 1.5mA
unknown

Note: We cannot work out the values accurately because the operation of the stage is not linear, but the base current is about one microamps flowing through the 2M2 resistor and if the transistor has a gain of 200, the collector current will be 200uA. This current will produce a voltage of 0.0002 x 20,000 = 4v = the theoretical voltage across the 22k resistor. This means the 22k will have to be reduced to 10k for the collector to sit at mid rail. Or the transistor has to have a gain of 100.
An additional 1uA will turn the transistor ON fully and about 1uA will turn it off. The values of "2mA  and about 1.5mA " are far too high.
 

49. Why is the H-Bridge stage a bad choice for a pre-amplifier::

it has a low input impedance
it will have a small amplification-factor
it uses more components than the self-biased stage

Note: All answers are correct. The input impedance is controlled by the 47k resistor on the base for rising input signals and this reduces the input impedance by an unknown amount - it could be as high as 50%. The rest of the circuit will have the same gain as the self-biased arrangement.
Both circuits have a degeneration factor to keep the collector at about mid-rail voltage if the supply rises or falls by a small amount.
In the H-bridge arrangement the degeneration is the created by the emitter resistor.

50. If the rail voltage is changed from 3v to 5v, will mid-rail voltage on the collector be maintained:

yes
no
unknown

Note: Mid-rail voltage will be maintained reasonably successfully by the action of the "feedback" 2M2 resistor on the base.

If we replace the transistor with a variable resistor we see the resistor must be 22k in both circuits for the voltage on the collector to be mid-rail. This is a simple voltage-divider analysis.
This is how it is maintained:
When the rail voltage increases by 2v, we will take things very slowly. We will assume the collector voltage rises 2v. This will put an additional 2v across the 2M2 and cause a higher current to flow through this resistor. This extra current will flow into the base of the transistor and make the transistor turn ON more and reduce the collector voltage.
This is all the transistor can do and the voltage on the collector will reduce a small amount, but not necessarily exactly 2v.  That's the best the transistor can do.
It cannot come down the full 2v because the transistor has to be turned ON more to allow extra current to flow in the collector-emitter circuit due to the higher rail voltage. And this extra voltage has to appear across the 2M2.

51. How do you design a self-biased stage.  What is the collector current:

1mA
0.1mA
10mA

Note: The collector current should be as small as possible because higher currents produce "transistor noise."

52. How do you select the collector load resistor

choose any value
22k
2k2

Note: For a self-biased pre-amplifier stage, select 0.1mA for collector current.
This is the current flowing through the collector LOAD resistor. 
The voltage across the resistor will be half-rail voltage.
For each one volt across 1k resistor, 1mA will flow.
This means 0.1mA will flow in 10k resistor and 1mA through 22k for 2.5v

53. How do you select the base-bias resistor:

choose any value
220k
2M2

Note: Build the circuit and fit 220k.  The collector voltage will be too low. Increase the base-bias resistor until the collector voltage is mid-rail. It will be about 2M2 to 3M3. No mathematics required. You cannot work it out mathematically as you don't know the gain of the transistor. 
I have never mentioned the 0.7v base-emitter voltage because it is not important.
It is a feature of a transistor that is turned on and conducting. But it is so small that you can neglect it in every decision.

54. What amplitude waveform will the self-biased pre-amplifier stage amplify:

any amplitude
1mV to 1v
0.1mv to 3v

Note: The circuit will definitely amplify a signal with an amplitude as small as 0.1mV and as a pre-amplifier, the signals are generally about 10mV, 100mV and up to 3v.
The stage will have a gain of more than 100, but when it is connected to another stage, this gain will reduce and we generally allow a value of 70.

55. What will happen when a 100 ohm magnetic sensor is connected directly to the base of a self-bias stage:

the circuit will not amplify the signal
the circuit will amplify the signal
the output will be 700mV

Note: The circuit will not amplify the signal because the voltage across the 100 ohm coil will be very small - much small than 650mV needed to turn the transistor ON.
The 2M2 normally allows the transistor to produce this voltage, but that is because nothing is connected to the base.
A capacitor connected to the base will get charged by the 2M2 and the transistor will see 650mV.
But 100R in series with 2M2 and a supply of 2.5v, will produce only a few millivolts across the 100R.
The magnetic sensor must be connected to the base via a capacitor as per the following circuit:

56. What will be the output from the circuit in Q55:
the circuit will not amplify the signal
unknown
the output will be 700mV

Note: The capacitor will allow 650mV to be developed on the base and the transistor will amplify the 10mV signal about 70 times to produce 700mV output.
As soon as the base has about 650mV developed on it, the transistor will amplify all signals from 1uV to 3v or higher.

57. What happens if the input signal is higher than 5v:

only about 0v to about 50mV of the signal will be processed
the stage will process the signal
only 0v to 5v of the signal will be processed

Note: This is a very complex question because this amplifier is designed for very small amplitude signals.  However it will process all amplitudes and as soon as the amplitude delivers more than 2uA, the transistor will then be saturated and it will be turned ON fully. If the input signal continues to rise, the output signal will not change. This means only a very small part of the signal will conform to the input, however the distance between each peak and each minimum will be seen on the output and this will provide the frequency of the signal.
The signal will be distorted as shown in the following graphs:

Since there is very little difference (to the ear) between a sinewave and a square wave, the signal will be considered to be processed.
The 100n coupling capacitor will absorb the signal and charge to 6.3v
That's because the base can never rise above 700mV and the rest of the amplitude has to charge the capacitor.

58. The electret microphone is connected to a separate 3v power supply and is connected to the self-biased pre-amplifier via two 100n capacitors. Will it work::

yes
no
only if the microphone has a 5v supply

Note: The left side of each 100n can be any voltage. Only the change (amplitude) of the signal is detected by the base of the transistor. and the capacitor connected to the 0v rail is connecting the two circuits together. The amplitude of the signal will be less due to the 0v rail capacitor but the amplifier will produce and accurate amplification.

59. Can a piezo diaphragm be connected directly base of a self-biased amplifier:

yes
no
a 100n coupling capacitor is needed

Note: No coupling capacitor is needed as the piezo diaphragm is really a 22n capacitor and the transistor gets its 700mV base voltage from the 2M2 resistor and it will amplify the waveform from the piezo when it is tapped. A 100n coupling capacitor is only needed to make sure the transistor gets its 700mV "turn ON" voltage.

60. Can a piezo and magnetic sensor be connect to the input at the same time:

yes
no

Note: The two transducers (detectors) can be paced on the input and the circuit will work with slightly less sensitivity as the signal will be sent to the other detector and some of the energy will be lost.

60A. When we talk about the "gain" of a transistor, is this

the voltage gain
the current gain

60B. Is the "gain" of a transistor constant

yes
no

Note: The gain of a transistor reduces when a higher current flows

60C. Can the "gain" of a transistor reduce to 1 (having no gain)

yes
no

Note: The gain of a transistor reduces to 1 when it is oscillating at its maximum frequency or when the current is higher than the transistor can handle.

When comparing the H-Bridge amplifier with the self-biased stage, the self-biased stage is simpler to design, has the higher gain, uses less components, consumes less current, has a higher input impedance and is less susceptible to noise from the supply because it has a higher-value load resistor and a feedback resistor.
Both designs have an ability to keep the collector at mid-voltage when the supply increases or decreases.

The self-biased stage is not covered in any University lectures or text books because none of the instructors have ever built an amplifier circuit and have no clue of the capability of any circuit other than a bridge amplifier. You need no mathematical gobbledygook for this stage and a test question simply revolves around building the circuit and fitting a base-bias resistor that suits the gain of the transistor. You cannot work out anything before-hand. And a test question is almost impossible to produce. So, you are missing out on the most important amplifier in you career.

What is the order of design:
1. Select the supply voltage.
2. Select a transistor.
3. Decide on the collector current
The collector will see mid-rail voltage
The current through the load resistor should be 0.1mA to 3mA
for a pre-amplifier
For each volt across the load resistor and 0.1mA flowing, the resistance is 10k
For each volt across the load resistor and 1mA, the resistance is 1k
4. Select a base-bias resistor that produces mid-rail voltage on the collector.
5. Select 100n for input capacitor and 100n or greater for output capacitor.
6. The input device can have almost any resistance or impedance and any signal from less than 1uV to 20mV.
In this range the output will not be distorted.

Here is the circuit:

23-3-2024