# Solved Problems on Transistor Biasing

### Q1. An npn silicon transistor has V_{CC} = 6 V and the collector load R_{C }= 2.5 kΩ. Find : (i) The maximum collector current that can be allowed during the application of signal for faithful amplification. (ii) The minimum zero signal collector current required.

### Solution :

Fig.1

Collector supply voltage, V_{CC }= 6 V

Collector load, R_{C }= 2.5 kΩ

(i) We know that for faithful amplification, V_{CE} should not be less than 1V for silicon transistor.

∴ Max. voltage allowed across R_{C} = 6 − 1 = 5 V

∴ Max. allowed collector current = 5 V/R_{C} = 5 V/2.5 kΩ = 2 mA

Thus, the maximum collector current allowed during any part of the signal is 2 mA. If the collector current is allowed to rise above this value, V_{CE} will fall below 1 V. Consequently, value of β will fall, resulting in unfaithful amplification.

(ii) During the negative peak of the signal, collector current can at the most be allowed to become zero. As the negative and positive half cycles of the signal are equal, therefore, the change in collector current due to these will also be equal but in opposite direction.

∴ Minimum zero signal collector current required = 2 mA/2 = 1 mA

During the positive peak of the signal [point A in Fig. 1(ii)], i_{C} = 1 + 1 = 2mA

And during the negative peak (point B), i_{C} = 1 − 1 = 0 mA

### Q2. A transistor employs a 4 kΩ load and V_{CC }= 13V. What is the maximum input signal if β = 100 ? Given V_{knee }= 1V and a change of 1V in V_{BE }causes a change of 5mA in collector current.

### Solution :

Collector supply voltage, V_{CC } = 13 V

Knee voltage, V_{knee }= 1 V

Collector load, R_{C} = 4 kΩ

∴ Max. allowed voltage across R_{C} = 13 − 1 = 12 V

∴ Max. allowed collector current, i_{C} =12 V /R_{C} = 12 V/ 4 KΩ = 3 mA

Maximum base current, i_{B} = i_{C }/ β = 3 mA / 100 = 30 μA

Now Collector current / Base voltage (signal voltage) = 5 mA/V

∴ Base voltage (signal voltage) = Collector current / (5 mA/V )= 3 mA /( 5 mA/V) = 600 mV

### Q3. Fig. 2 (i) shows biasing with base resistor method. (i) Determine the collector current I_{C }and collector-emitter voltage V_{CE }. Neglect small base-emitter voltage. Given that β = 50. (ii) If R_{B }in this circuit is changed to 50 kΩ, find the new operating point.

### Solution :

Fig. 2

In the circuit shown in Fig. 2 (i), biasing is provided by a battery V_{BB} (= 2V) in the base circuit which is separate from the battery V_{CC} (= 9V) used in the output circuit.

The same circuit is shown in a simplified way in Fig. 2 (ii). Here, we need show only the supply voltages, + 2V and +9V. It may be noted that negative terminals of the power supplies are grounded to get a complete path of current.

(i) Referring to Fig.2 (ii) and applying Kirchhoff ’s voltage law to the circuit ABEN, we get,

(ii) When R_{B} is made equal to 50 kΩ, then it is easy to see that base current is doubled i.e. I_{B }= 40 μA.

### Q4. Fig. 3 (i) shows that a silicon transistor with β = 100 is biased by base resistor method. Draw the d.c. load line and determine the operating point. What is the stability factor ?

### Solution :

Fig. 3

Here, V_{CC} = 6 V, R_{B} = 530 kΩ, R_{C} = 2 kΩ

**D.C. load line**

Referring to Fig. 3 (i), V_{CE} = V_{CC} − I_{C}R_{C}

When I_{C} = 0, V_{CE} = V_{CC} = 6 V. This locates the first point B (OB = 6V) of the load line on collector-emitter voltage axis as shown in Fig. 3 (ii).

When V_{CE} = 0, I_{C} = V_{CC}/R_{C} = 6V/2 kΩ = 3 mA.

This locates the second point A (OA = 3mA) of the load line on the collector current axis. By joining points A and B, d.c. load line AB is constructed as shown in Fig. 3(ii).

**Operating point Q**

As it is a silicon transistor, therefore, V_{BE} = 0.7V. Referring to Fig. 3(i), it is clear that :

Fig. 3 (ii) shows the operating point Q on the d.c. load line. Its co-ordinates are I_{C} = 1mA and V_{CE }= 4V.

### Q5. (i) A germanium transistor is to be operated at zero signal I_{C} = 1mA. If the

collector supply V_{CC} = 12V, what is the value of R_{B} in the base resistor method ? Take β = 100.

(ii) If another transistor of the same batch with β = 50 is used, what will be the new value of zero signal I_{C} for the same R_{B }?

### Solution :

Given, V_{CC }= 12 V, β = 100

As it is a Ge transistor, therefore, V_{BE} = 0.3 V

**(i)** Zero signal I_{C }= 1 mA

### Q6. Calculate the values of three currents in the circuit shown in Fig. 4.

Fig. 4

### Solution :

Applying Kirchhoff ‘s voltage law to the base side and taking resistances in kΩ and currents in mA, we have,

### Q7. Design base resistor bias circuit for a CE amplifier such that operating point is V_{CE} = 8V and I_{C }= 2 mA. You are supplied with a fixed 15V d.c. supply and a silicon transistor with β = 100. Take base-emitter voltage V_{BE }= 0.6V. Calculate also the value of load resistance that would be employed.

### Solution :

Fig. 5 shows CE amplifier using base resistor method of biasing.

Fig. 5

### Q8. A base bias circuit in Fig. 6 is subjected to an increase in temperature from 25°C

to 75°C. If β = 100 at 25°C and 150 at 75°C, determine the percentage change in Q-point values ( V_{CE }and I_{C}) over this temperature range. Neglect any change in V_{BE }and the effects of any leakage current.

Fig 6

### Solution :

**At 25°C :**

**At 75 °C :**

### Q9. In base bias method, how Q-point is affected by changes in V_{BE} and I_{CBO}.

### Solution :

In addition to being affected by change in β, the Q-point is also affected by changes in V_{BE} and I_{CBO }in the base bias method.

**(i) Effect of V _{BE} :**

The base-emitter-voltage V_{BE} decreases with the increase in temperature (and vice-versa). The expression for I_{B} in base bias method is given by ;

It is clear that decrease in V_{BE} increases I_{B} . This will shift the Q-point (I_{C} = βI_{B} and V_{CE} = V_{CC} – I_{C} R_{C}). The effect of change in V_{BE} is negligible if V_{CC }>> V_{BE} (V_{CC} atleast 10 times greater than V_{BE}).

**(ii) Effect of I _{CBO }:**

The reverse leakage current I_{CBO} has the effect of decreasing the net base current and thus increasing the base voltage. It is because the flow of I_{CBO} creates a voltage drop across R_{B} that adds to the base voltage as shown in Fig. 7. Therefore, change in I_{CBO} shifts the Q-point of the base bias circuit.

Fig. 7

However, in modern transistors, I_{CBO} is usually less than 100 nA and its effect on the bias is negligible if V_{BB} >> I_{CBO} R_{B}.

### Q10. Fig. 8 (i) shows the base resistor transistor circuit. The device (i.e. transistor)

has the characteristics shown in Fig. 8 (ii). Determine V_{CC}, R_{C }and R_{B }.

Fig. 8

### Solution :

From the d.c load line, V_{CC} = 20V.

### Q11. What fault is indicated in (i) Fig. 9 (i) and (ii) Fig. 9 (ii) ?

Fig. 9

### Solution :

(i) The obvious fault in Fig. 9(i) is that the base is internally open. It is because 3V at the base and 9V at the collector mean that transistor is in cut-off state.

(ii) The obvious fault in Fig. 9 (ii) is that collector is internally open. The voltage at the base is correct. The voltage of 9V appears at the collector because the ‘open’ prevents collector current.

### Q12. For the emitter bias circuit shown in Fig. 10, find I_{E}, I_{C}, VC and V_{C }and V_{CE }for β= 85 and V_{BE} = 0.7V.

Fig. 10

### Solution :

### Q13. Determine how much the Q-point in Fig. 11 will change over a temperature range where β increases from 85 to 100 and V_{BE ,}decreases from 0.7V to 0.6V.

Fig. 11

### Solution :

For β = 85 and V_{BE} = 0.7V

As calculated in the above Question.12, IC = 1.73 mA and VCE = 14.6V.

For β = 100 and VBE = 0.6V

### Q14. Fig. 12 shows a silicon transistor biased by collector feedback resistor method.

Determine the operating point. Given that β = 100.

Fig. 12

### Solution :

V_{CC} = 20V, R_{B} = 100 kΩ, R_{C} = 1kΩ

Since it is a silicon transistor, V_{BE }= 0.7 V.

Assuming I_{B} to be in mA and using the relation,

### Q15. (i) It is required to set the operating point by biasing with collector feedback

resistor at I_{C} = 1mA, V_{CE } = 8V. If β = 100, V_{CC} = 12V, V_{BE }= 0.3V, how will you do it ?

(ii) What will be the new operating point if β = 50, all other circuit values remaining the same ?

### Solution :

Given, V_{CC } = 12V, V_{CE }= 8V, I_{C }= 1mA, β = 100, VBE = 0.3V

**(i)** To obtain the required operating point, we should find the value of R_{B}.

Now, collector load is

**(ii)** Now β = 50, and other circuit values remain the same.

### Q16. It is desired to set the operating point at 2V, 1mA by biasing a silicon transistor with collector feedback resistor R_{B}. If β = 100, find the value of R_{B}.

### Solution :

Fig. 13

### Q17. Find the Q-point values ( I_{C }and V_{CE}) for the collector feedback bias circuit shown in Fig. 14.

Fig. 14

### Solution :

Fig. 14 shows the currents in the three resistors (R_{C}, R_{B} and R_{E}) in the circuit. By following the path through V_{CC }, R_{C}, R_{B}, V_{BE }and R_{E} and applying Kirchhoff’s voltage law, we have,

### Q18. Find the d.c. bias values for the collector-feedback biasing circuit shown in Fig. 15. How does the circuit maintain a stable Q point against temperature variations ?

Fig.15

### Solution :

The collector current is

**Stability of Q-point :**

We know that β varies directly with temperature and V_{BE} varies inversely with temperature. As the temperature

goes up, β goes up and V_{BE} goes down. The increase in β increases I_{C} (= βI_{B}). The decrease in V_{BE} increases I_{B} which in turn increases I_{C} . As I_{C} tries to increase, the voltage drop across R_{C} (= I_{C} R_{C} ) also tries to increases. This tends to reduce collector voltage V_{C } and, therefore, the voltage across R_{B}. The reduced voltage across R_{B} reduces I_{B} and offsets the attempted increase in I_{C} and attempted decrease in V_{C} . The result is that the collector feedback circuit maintains a stable Q-point. The reverse action occurs when the temperature decreases.

### Q19. Fig. 16 shows the voltage divider bias method. Draw the d.c. load line and determine the operating point. Assume the transistor to be of silicon.

Fig. 16

### Solution :

**d.c. load line :**

The collector-emitter voltage V_{CE} is given by :

This locates the second point A (OA = 5 mA) of the load line on the collector current axis. By joining points A and B, the d.c. load line AB is constructed as shown in Fig. 17.

Fig. 17

**Operating point :**

For silicon transistor, V_{BE} = 0.7 V

Fig.17 shows the operating point Q on the load line. Its co-ordinates are I_{C }= 2.15 mA, V_{CE} = 8.55 V.

### Q20. Calculate the emitter current in the voltage divider circuit shown in Fig. 18.

Also find the value of V_{CE} and collector potential V_{C}.

Fig. 18

### Solution :

### Q21. For the circuit shown in Fig. 19, find the operating point. What is the stability factor of the circuit ? Given that β = 50 and V_{BE} = 0.7V.

Fig. 19

### Solution :

Fig. 19 shows the circuit of potential divider bias and Fig. 20 shows it with potential divider circuit replaced by Thevenin’s equivalent circuit.

Fig. 20

### Q22. The circuit shown in Fig. 21 uses silicon transistor having β = 100. Find the operating point and stability factor.

Fig. 21

### Solution :

Fig. 21 shows the circuit of potential divider bias and Fig. 22 shows it with potential divider circuit replaced by Thevenin’s equivalent circuit.

Fig. 22

### Q23. In the circuit shown in Fig. 23, the operating point is chosen such that I_{C} = 2mA, V_{CE} = 3V. If R_{C }= 2.2 kΩ, V_{CC} = 9V and β = 50, determine the values of R1, R2 and R_{E}. Take V_{BE }= 0.3V and I1 = 10I_{B} .

Fig. 23

### Solution :

Given, R_{C }= 2.2 kΩ, V_{CC} = 9V and β = 50, V_{BE }= 0.3V and I1 = 10I_{B} .

As I_{B} is very small as compared to I1, therefore, we can assume with reasonable accuracy that I1 flowing through R1 also flows through R2.

Applying Kirchhoff ‘s voltage law to the collector side of the circuit, we get,

### Q24. An npn transistor circuit as shown in Fig. 24 has α = 0.985 and V_{BE }= 0.3V. If V_{CC }=16V, calculate R1 and R_{C} to place Q point at I_{C} = 2mA,V_{CE} = 6 volts.

Fig. 24

### Solution :

**Given, α = 0.985 , V _{BE }= 0.3V and V_{CC }=16V**