TRANSISTOR

TEST



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

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.

 

REVISION

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:

Here is a totally impractical circuit from YouTube:

It might work but the voltage on the collector will depend on the gain of the transistor (and the supply voltage) and if you use a different transistor with a slightly different gain, the mid-rail voltage will change.
But the most important reason NOT to use this design is the noise it produces. Any noise on the power rail will be amplified 200 times and passed to the LOAD resistor. A small portion of this noise will be passed to the power rail where it will be amplified another 200 times.
Our self-bias arrangement has a feedback resistor that helps compensate for different transistors and different voltages and acts against amplifying noise. Instead of amplifying noise, the feedback resistor reduces it by about 10% to 30% and so it is effectively removed. You will never see this design in any practical application . . .  because it so ineffective.  So don't study it or use it.

A question from a YouTube video:
How does a transistor amplify current?
A transistor is really a variable resistor and when it reduces resistance, (due to more current being supplied to the base lead) a higher current will flow in the LOAD resistor (or the device you have connected between the supply and collector lead).
When we say the transistor reduces resistance, it actually allows more current to flow in the collect-emitter circuit.
The transistor can also be placed in a configuration where it allows more current to flow in the emitter-collector junction and through the device connected to it. So you have to understand both concepts.
We have only covered the common-emitter stage . . . . only a fraction of the total field.

There is an enormous amount of instruction and circuitry on talkingelectronics.com website with entirely new ways to see how circuits work so you can "see" them working in your mind and be able to design and test them like an expert (like me). 
Here is just a few links to start you in the right direction:
Link 1
Link 2
Link 3


THE TRANSISTOR AS A SWITCH
This another field of operation for a transistor.
It is still an amplifier but in a different mode called DIGITAL MODE.
This mode has two states.  One state has the transistor fully turned OFF and the other state has the transistor fully turned ON.
These two states have an amazing feature. In either state the transistor stays the coolest or even coldest.
When a transistor is fully turned ON, the voltage across the collector-emitter terminals is the lowest and the transistor dissipates the least heat.
Not only that, but each state is called a DEFINED STATE or an OBVIOUS STATE or a dedicated state or a state that cannot be challenged or a state that is "known."
There are a number of names for these two modes (states).
ON and OFF
Digital Mode
CUT-OFF and BOTTOMING
"0" and "1"
HIGH and LOW
Conducting and Not-conducting
and the transistor stage can be called an Inverter
It can also be called a SWITCH.

This stage or state is used to turn things ON and OFF.  Not half-ON but fully ON.
To put a transistor into this mode, you have to follow some rules.
The first rule is to allow 100 gain for LEDs and resistors as the LOAD.
For globes and motors the gain is 20.  That's because a globe or motor takes 5 times more current to get it to start to glow or start to turn.
You can now work out the base current, by dividing the collector current by 20 or 100.
You can deliver extra current but it will not achieve anything and it will be wasted. 

Q61.  Will the globe illuminate:

yes
no

Note: Using the theory above, the 100mA globe requires about 500mA to make sure it starts to glow.
The next factor to take into account is the gain of the transistor is only 50 when a high current flows.
This means the base current has to be 10mA.  (10 x 50 = 500)
With a control voltage of 5v, the resistor to the base has to be 5/0.01= 500 ohms.  Select 470R resistor. The 1k in the circuit above should be 470R

Q62  Name the function of the following circuit:

amplifier
inverter

Note: In the context of this discussion, the circuit is an Inverter

Q63 In the circuit above, the current into the base will be about 5mA.  The transistor will have a gain of 100.  What is the collector current:
500mA
12mA

Note: The collector current is limited by the value of the LOAD resistor. Even though the transistor will pass up to 500mA, the 1k resistor in the collector limits this to 12mA.

Q64 Will the circuit operate the motor:

yes
no

Note: The motor will require 500mA to start revolving.  The base will see 5/220 = 22mA.  The transistor will amplify this to 22 x 50 = 1,100mA but the motor only requires 500mA, so the circuit will work.
 

Q65 Will the circuit illuminate the two white LEDs:

yes
no

Q66 What is the collector current:
20mA
12mA
5mA

Note: LEDs have a characteristic voltage across them when they are illuminated. This voltage depends on the colour and does not alter for the brightness or size.  For a white LED, the voltage is 3.2v   This means 6.4v is dropped "lost" across the LEDs and only 12 - 6.4 = 5.6v is available as the supply voltage for working out the current through the circuit.
We now have 5.6v across a 1k resistor and this means 5.6mA will flow in the collector circuit.  The base will see 5/10,000 = 0.5mA and the collector will deliver up to 100 x 0.5 = 50mA.  This means the circuit will work.

The transistor can be used to control the flow of signals and it can stop and start the flow. This is called GATING and we have covered gating on talkingelectronics website:

In the first circuit, both inputs have to be LOW and if either input goes HIGH, the gate changes from HIGH to LOW.
For the second circuit, both inputs must be LOW and both must go HIGH for the output to change from HIGH to LOW.
The three 10k resistors have a purpose.
When the two inputs are LOW, the voltage on the voltage on the join of the first and second 10k resistors will be 0.7v This would normally be high enough to change the state of the gate. But the second and third 10k form a voltage divider to produce mid-voltage of 0.35v and this is not enough to change the gate.
This allows the input to rise to 1v and the gate will still not change.
The 0.3v allowance allows the inputs to be connected to the collector of a stage and when the transistor in this stage is "turned ON" the collector-emitter voltage will be 0.2v (we have a 0.3v allowance).

Q67  The motor requires 250mA to start and 50mA to run. Will this circuit work:

yes
no

Note:  No.  The circuit will not work.
The BC547 transistor is only able to pass 100mA.  The motor requires 250mA. Secondly, the transistor will only have a gain of 50 when trying to pass a high current and the circuit will only produce 0.5 x 50 = 25mA.
The circuit will not activate the motor, BUT . . . what will happen?
The transistor will pass about 50mA to 100mA and because the motor is not rotating, it will not produce a back EMF and thus the stalled resistance of the armature will pass the current. This means a very small voltage will appear across the motor and thus more than 10v will appear across the transistor. When you multiply the voltage and current you get nearly: 10x 0.1 = 1watt of heat dissipation in the transistor and it will be destroyed.

Q68  The motor requires 250mA to start and 50mA to run. Will this circuit work:

yes
no

Note:  The circuit will work. The base current is 5mA  and the BC338 will have a gain of 50, so it will deliver a maximum of 5 x 50 = 250mA.
You have to take the capability of the transistor into account when designing a circuit.
When a transistor is fully turned ON, the voltage across it will be 0.3v to 0.5v and this represent a wattage of about 25milliwatts.
If the motor is loaded and takes 150mA, the wattage will be 75milliwatts.
During "start-up" this loss will be much higher, but it is only for a very short period of time.

Q69  If we increase the resistance of the LOAD resistor, will the collector voltage:

rise
fall
remain the same
unknown

Note:  The collector voltage will fall.  This is the fact that we have been covering in the above questions. When a motor is operating, its resistance is effectively increased and the voltage across the transistor falls to the lowest value. That's why the transistor remains cold.

Q70  Which stage do you think is the best design for a pre-amplifier:

H-Bridge
self-bias

You will be using the transistor very little as an amplifier in your electronics designs and unless you can visualise it as a variable resistor, you will not be able to do any real designing.  
Universities, text books and YouTube videos have never gone further than the common-emitter amplifier because they have not realised the variable resistor concept and have never designed a circuit in their life.  Go to Link 3 for a discussion on the transistor as a variable resistor.

What is the "voltage gain" of a common-emitter stage?

Note:  In 50 years of designing circuits, I have never been asked or needed to know the voltage gain of a stage. A transistor works on amplifying CURRENT and any voltage delivered to the base is converted to a current by the transistor.

If you want to know how much current is required into the base of a transistor to deliver 12v across the LOAD, a simple equation can be used.
Suppose the transistor has a gain of 100. For each 1mA into the base, the collector will deliver 100mA. Each 100mA through the 100R LOAD will produce
[V=IR] 0.1 x 100R = 10v
So you need just over 1mA base current to produce almost rail voltage across the LOAD.

Remember this:  One amp flowing through one ohm produces one volt across the resistor.
So, if you pass 100mA (that is one-tenth of an amp) across 10 ohms, the result is one volt (you have divided one value by 10 and multiplied the other value by 10).
If you pass 100mA across 100 ohms, you produce 10 volts. No mathematics needed.  Just multiplication and division.

Q71  The supply voltage is gradually increased. At what voltage will the transistor turn-ON:

 

 

 

 


5v
21.8v

Note: Forget everything you have read about the input impedance of a transistor being less than 10k. The input impedance is very high and it will detect current as small as one microamp or less.
As the supply is increased, the small leakage current through the two zener diodes will pass current to the base and as soon as 0.7v is developed across the base-emitter junction, the transistor will start to turn ON. The supply can be as low as a few volts and the zener voltages of 9v1 and 12v DO NOT have to be developed across the zener diodes. This is called a ZENER LEAKAGE situation. As the supply is increased, the leakage current will increase at a faster rate because it is "diode leakage" but you can consider the two zeners as high-resistance resistors.

Q72  A 680R resistor has been placed between base and 0v rail. The supply voltage is gradually increased. At what voltage will the transistor turn-ON:


5v
21.8v

Note: The 680R between base and 0v rail has changed the operation of the circuit completely. The 680R is a very low impedance compared to the input of the transistor and a voltage will only develop across this resistor when a noticeable current is flowing. When the supply reaches 9v1 + 12v = 21.1v the voltage across these two zener will not be allowed to get any higher and so any increase in supply voltage will be passed directly to the 680R and every 0.1v increase will appear across this resistor. When the supply increases by about 0.6v or 0.65v or 0.7v, this voltage will appear on the base of the transistor and turn it ON fully. It must be noted that a further increase of 0.1v will blow up the transistor because the supply voltage is being passed DIRECTLY to the base as the zener diodes are RIGID ITEMS and do not allow a voltage higher than their marking to appear across the leads.
The 680R has been added to make sure the circuit ONLY turns ON when the supply reaches 21.8v
You can increase the 680R to 1k or 10k and the circuit will operate exactly the same. That's because 10k is considerably smaller than the very high input impedance of the transistor.
Almost no voltage will be developed across the 10k with "leakage current." It takes "real current" to develop a recognised voltage.
And only "real current" starts to flow when 9v1 + 12v is developed across each zener and they start to "break-down." In other words they become like a fully extended shock absorber and any further extension will pull on the other end by a direct amount as it is fully extended.  That's the way the zener works. As soon as it gets 9v1 across it, any increase will be passed directly to the component below (that's not really correct as voltage develops across each zener by about the same amount and when the total combined voltage is reached, the next increase is directly passed to the component below)
This is a very good lesson on understanding "impedance." Impedance is used when an actual resistance-value is not known and it changes when the current increases. It's a value you cannot measure because test equipment will not supply the exact same current  (since resistance is determined with a specified voltage and a certain current is flowing).

Q73  What is the function of this circuit:
 

amplifier
constant current

Note: The circuit delivers a constant current of 10mA to the LED when the supply is between 5v and any other higher voltage.
The 10mA current is produced by the circuit by the 72R resistor and here's how it works:
Place two ordinary diodes between the base and 0v rail. Use a "turn-ON" between base and 5v rail. This can be any value from 1k to 10k or higher and only has to pass one hundredth of 10mA.
Place any colour LED in the collector circuit and NO current-limiting resistor is needed.

When the circuit is turned ON, 0.7v will be developed across each diode.
The characteristic voltage across the base-emitter junction is 0.7v and this cancels one of the diodes.
This means 0.7v remains for the emitter resistor.
If we want 10mA to flow through the LED, the same current must flow through the emitter resistor.
From Ohm's Law:  I=V/R
0.01 = 0.7/R
R = 0.7/0.01  = 70/1  = 70    use 72R resistor.
If you use 140R resistor the current will be 5mA
If you use 36R resistor the current will be 20mA
The circuit limits the current because the transistor is turned ON by the 4k7 and current flows through the 72R resistor and produces a voltage drop across it. This voltage increases and increases until it reaches 0.7v.
The voltage on the base is 1.4v
The voltage across the resistor cannot increase to 0.8v because this will mean there is only 0.6v available for the base-emitter junction and the transistor will be "turned OFF."
So, only 10mA will flow as this will produce 0.7v across the 72R.
If the emitter resistor is removed, a very large, uncontrollable, current will flow and the LED may be destroyed.
This is because the two diodes are equal to a 1.4v zener diode and when any voltage below 1.4v is across it, the zener (the two diodes)  do not function and they are effectively OUT OF CIRCUIT.
This mean the 4k7 is connected to the base and the transistor will amplify the base current 100 times and pass this current through the LED.
With no emitter resistor, the base will never rise above 0.7v and the two diodes will have no effect.
You will notice all these calculations do not involve the supply voltage and it can be almost any value as only a very small current is flowing.

Q74  What is the gain of this circuit:

100
1,000
10,000

Note: This circuit produces a very high gain and uses very few components. It is difficult to get high gain with direct coupling and this circuit very reliable. The only problem is the quiescent current is 5mA or 50mA, depending on the input voltage.
The circuit is very difficult to test because the voltage on the collector of the first transistor is either 0.3v or 0.7v and you will have to be able to detect the difference.
Here's how to work out the gain.
The first transistor needs to be turned ON fully to bring the collector below 0.7v so the second transistor changes state.
This means the first transistor will pass 5mA through the 1k resistor.
The base will have to be supplied with 1/100th of 5mA.
To pass 5/100th mA through 100k, the input voltage must be: 1/00 x 100
 x 5/100 x 1/1,000 x 100,000 =5v
If the gain of the first transistor is 200, the input voltage needs to be 2.5v
When the first transistor is OFF, the base of the 2nd transistor will have 1k.
This will allow 5mA into the base. 
When the second transistor is turned ON, the current through the 100R will be:  5/100 = 5/100 x 1,000 mA = 50mA.
When 5mA flows into the base the transistor will be capable of multiplying this current by 100 = 500mA.
But the load only accepts 50mA so the second stage has a gain of 10.
This means the overall gain of the circuit is 100 x 10 = 1,000
These are the sort of calculations you can carry-out in your head  . . . AND YOU NEED TO  . . . .to be able to see if a circuit will function.
Normally this type of circuit uses high-value load resistors so the quiescent current is very small.

Q75  When the collector voltage is too high, does the transistor:

break-down
zener

Note: It is called "breakdown" but the transistor actually "zeners" and a fixed maximum voltage appears across the collector-emitter terminals.  If only a small current flows, the transistor will not be damaged and this zener voltage can be used in a circuit as a reference voltage. Some transistors will "zener" at 10v and they cannot be used in a 12v circuit as they will never "turn-off" and their operation may mess up the operation of the circuit.
The "breakdown" voltage is sometimes accurately specified in the data sheet and sometimes it is much higher than specified.

Q76  Will this circuit work?

no
yes

Note: The circuit will not work because the "piezo speaker"  "piezo diaphragm" is a 22n capacitor and when the transistor turns ON, the diaphragm "clicks" but when the transistor turns OFF, there is no way for the capacitor to discharge.   The circuit needs a resistor, such as 1k, across the piezo diaphragm to discharge the voltage across the plates.

Q77  Here is a Hall Effect circuit:


How do you connect the LOAD:


A
B
neither

Note: Not circuit A or B.  Circuit A needs a current limiting resistor.  Circuit B will not work AT ALL.

Q78  Will this circuit detect faint sounds:


Yes
No

Note: A piezo diaphragm will detect sounds as well as produce a very loud output, depending on the size and quality of the diaphragm.
When the two diaphragms are placed near each other the circuit will produce a squeal (feedback).
The piezo can be connected directly to the base because the piezo is a very high impedance device and the base of the transistor in this circuit also has a very high impedance and so the two match each other perfectly.
When the piezo is placed on a window, it has the effect of "listening through the glass." 

Q79  What will happen to this circuit when the multimeter measures the base-emitter voltage:


No effect
The transistor will turn OFF

Note: The multimeter has a resistance and in addition, it needs current to move the needle. That's why it upsets some voltage readings.
The base voltage will be initially 0.65v and the transistor will be "turned-ON."
The 1M base-bias resistor will form a voltage-divider with the multimeter and the voltage on the base will reduce.
The transistor will turn OFF and it will effectively disappear from the circuit and the final voltage on the mid-point will be determined by the value of the 1M and the internal resistance of the multimeter (when it is set to say 2,000mV setting).
That's why analogue multimeters cannot be used in high-impedance sections of a circuit as they will produce false readings and change the operation of the circuit.

Q80  Why does this circuit only have a gain (amplification) of 70 when the transistor has an Hfe of 250:


Note:  When the transistor is sitting in "idle" (quiescent) conditions, the 2M2 is chosen so the collector voltage is mid-rail.
This means the 2M2 is supplying base current. Say it is 2 microamps.
It will only take say 2 more microamps to turn the transistor ON fully.
But when you supply this extra current, the collector voltage will drop to nearly zero and the 2M2 will stop delivering 2 microamps. So you have to deliver 2 microamps plus 2 microamps = 4 microamps to make the transistor turn ON fully.
This means you have to deliver more current than what you expect.
In other words, the transistor is not amplifying the input signal 250 times because you are losing the help of the 2M2.
Already you are down to a gain of 125.  The gain drops even further to 70 because the transistor needs added current to turn ON fully and it may need 5 or 6 microamps.

These are facts you will never find in any text book or video and you will wonder why your circuit is not producing the expected gain. 

Q81  If the base-bias resistors pass 10 times the base current, what percentage of the input amplitude will be lost:


zero
10%
90%

Note:  Do not use the H-Bridge circuit above to amplify small signals from a source that has a high impedance as 90% of the amplitude will be lost (attenuated) by the base-bias resistors.

Q82  Which circuit will have the higher gain:

A
B
Note:  We have explained the H-Bridge arrangement attenuates 90% of the input signal via the base-bias resistors so circuit A will have a higher gain of about 70.

Q83. What is the gain of this H-bridge stage:

100
10
1,000
Unknown

Note:  The gain of any stage is unknown if you don't know the characteristics of the input signal and what the output signal is "driving into."
It is pointless saying the stage has a gain of 100 when the input signal will be attenuated by 90% via the base-bias resistors and further reduced if the output signal drives into a low resistance.
Furthermore, the gain changes according to the frequency and the quiescent current and the gain of the transistor.
The end result is you will have no idea of the result.
In 50 years I have never tried to work out any values of gain because I know they are completely erroneous, a waste of time and achieve NOTHING.
It takes 10 minutes to make the circuit on bread-board and test it, make adjustments and get everything working perfectly before using a CAD program to make a Printed Circuit Board.
But no University can test you on these skills, so they charge $70,000 and "lead you down the garden path" with equations, mathematics and accuracy that doesn't hold-up in electronics.
Analogue electronics is only 5%, 10% or 50% accurate, if you take some of the sensors into account.
Thermal sensors, Hall effect devices, IR sensors and magnetic detectors all have a wide range of detection capabilities and to give an answer to 1% accuracy shows you have no idea of the problem.
 
Q84  What is the advantage of circuit B over circuit A:

lower quiescent current
higher gain
more stable
None

Note:  Circuit B has no advantages over circuit A. 
Circuit B consumes more current and produces a lower gain and requires more components.
In circuit B, 90% of the input signal is lost in the base-bias resistors.
No lecture has compared the two stages or detailed circuit A - the self-biased stage.  So you have been left in the "dark."
Circuit A is most suited as the first stage in an amplifier as it has the highest input impedance and generally provides a gain of 70, when you take into account the matching of the input signal to the stage and the output of the stage to a high-impedance following-stage.

Q85  Should you learn how to work out the gain of the self-based and H-bridge stages:

Yes
No

Note:  The answer is NO. You cannot work out any of the operating conditions of either of these stages as the characteristics of the input signal and the impedance of the load on the output of the stage will change the values enormously. You cannot accurately asses how the output of one stage will match the input impedance of a following stage. And the frequency of operation will also change the gain.
I would not have a clue about the gain of either circuit and I have been designing circuits for 50 years.
That's why I simply bread-board the stage(s) and change the value of the components until the desired result is obtained.
None of the videos, text books or University lectures have ever provided the slightest information on determining the values of the biasing components because they simply cannot be worked out without building a circuit and taking measurements.
I hope this test has steered you in the right direction.
The only way to learn electronics is to build circuits. All the lectures in the world are worthless if you don't build things.
I have repaired over 37,000 TV's and electronic appliance and I can remember how I fixed every item. You need to remember so the next fault will take less time to diagnose.
And because you are fixing faulty designs, you learn how to design circuits correctly.
It's called REVERSE LEARNING. It's the fastest way to learn electronics.

How does the input capacitor work:


The input capacitor is just like a shock absorber. It stretches and contracts.
The base of the transistor cannot rise above 0.75v.
The input signal has an amplitude of 5v.
As the signal rises from 0v to 0.75v, it is passed to the base ONLY during the portion of 0.55v to 0.75v. As it rises above 0.75v, the shock absorber stretches and the transistor does not accept any more signal.
It stretches to 4.25v.
When the signal reduces, the first 0.2v of the signal is passed to the base to turn the transistor OFF.
As the signal reduced further, the base sees a negative voltage of       -4.25v. The transistor can do nothing with this this negative voltage and the shock absorber remains in a stretched state.
However the 2M2 base-bias resistor has the effect of being able to compress the shock absorber, but only very slowly because it is a very high resistance.
It reduces it a small amount and when the signal produces the second cycle, it will be delivered to the base when the amplitude is very close to its maximum.
You can see, 5v does not enter the transistor as it only accepts 0.2v (0.75v - 0.55v = 0.2v).    The capacitor does not linearly pass the signal to the base, that's why there is a lot of distortion. 
If you want more of the signal to be detected by the transistor, it is not a matter of increasing the value of the capacitor but DECREASING it so that it slowly charges over the whole time the signal is rising. And this smaller-value capacitor will discharge via the 2M2 base-bias resistor, ready for the next cycle. In other words, the shock absorber must be of a type that can be extended and compressed VERY EASILY.
You can see the capacitor in this application is nothing like a capacitor in an AC circuit where the input and out is a sinewave. In the transistor circuit, the transfer is for only a very short portion of the cycle and yet no lecture has explained the realities of the transfer.
You need to know what is really happening if you want to find out where the distortion is coming from.
And because the transfer is so short you cannot assign any impedance values to the capacitor. You are jut being led "down the garden path" with all the gobbledygook mathematics, trying to assign a value of gain for a stage when none of the values can be assessed.

Q87  What is the approximate gain of the second transistor:

70
4

Note:  We have already covered the fact that the first transistor has a gain of about 70, when all the losses are taken into account. This applies to a transistor with a gain as low as 175 or as high as 450. So it is pointless doing any mathematical calculations.
The component that produces the output capability is not the transistor but the 22k resistor. The first transistor pulls the output "down" and the resistor pulls the output "up."
The 4k7 will pull the output up with 4 times more "strength" because its resistance is one-quarter of 22k and so the second stage will have a gain of 4.
I keep stressing, no mathematics is needed to instantly see how to calculate values in a circuit and this allows you to easily select values when designing.
As soon as you get away from mathematical calculations, you will start to understand how a stage works.
I have already explained the input capacitor works for only a fraction of the cycle, so you can't use any mathematics that refers to a capacitor in a normal AC circuit.

                                     ooooo000000oooooo

Learning to build and design the common-emitter stage is only a tiny fraction of the transistor circuits you need to understand.
None of the courses cover the 99% of transistor arrangements that you need to work with.
That's because they are too hard to understand and explain (from any of the lecturers). See the links below for hundreds of circuits using transistors and build them to get an understanding of how they work and how to alter their operation.
 


You need to go to these links:
Link 1  &  Link for Capacitor understanding
Link 2
Link 3
and then 200 Transistor Circuits, to get some idea on how to use the transistor.
And then go to: TRANSISTOR AS A DIGITAL DEVICE.
 


24-3-2024 23-3-2024