Voltage Divider Biasing MCQ Quiz - Objective Question with Answer for Voltage Divider Biasing - Download Free PDF

Concept:

The circuit diagram of the voltage divider bias circuit is:

Collecter current in voltage divider biasing is given by:

\(I_c = {V_2-V_{BE} \over R_E}\).............(i)

I_c = collector current

V₂ = voltage across R₂

R_{E =} emiiter resistance

V_BE = base to emitter voltage

Explanation:

From the equation (i), it is observed that:

• ​I_c doesn’t depend upon current gain (β).
• V_BE is very small that I_c doesn’t get affected by V_BE .
• Thus I_c in this circuit is almost independent of transistor parameters (β, V_BE) and hence good stabilization is achieved.
• In voltage divider biasing, the transistor always remains in the active region which is not achievable in other biasing methods.

Therefore, voltage divider biasing in BJT is the universal biasing method.

Additional Information

1. Emitter Bias: Emitter bias provides excellent bias stability in spite of changes in β. Hence, this biasing method depends on current gain (β).
2. Base Bias: Base bias ensures that the voltage fed to the base is enough which can supply enough base current to switch the transistor ON.
3. Collector Bias: This circuit helps in improving the stability considerably. If the value of I_c increases, the voltage across load increases, and hence V_CE  also increases. This in turn reduces the base current I_B.

Concept:

A BJT voltage divider configuration is as shown:

Usually, the resistance R₂ is small, resulting in small base current. This allows us to apply voltage division to obtain the base voltage, i.e.

\(V_B=V_{CC}\times \frac{R_2}{R_1+R_2}\)

Calculation:

With V_CC = 18 V, R1 = 4.7 kΩ and R2 = 1500 Ω, the base voltage will be:

\(V_B=18\times \frac{1500}{1500+4700}\)

V_B = 4.35 V

Given:

β = 100

V_C = 5 V

\(I_C = {V_C \over R_C}={5\over3.3\;k}=1.515\;mA \)

\(I_B = {I_C\over β}=0.01515\;mA\)

I_E = (1 + β) I_B = 1.53 mA

- 12 + 1.2 × 10³ × 1.53 × 10^-3 + 0.7 + V_B = 0

V_B = 9.464 V

\(I_X =\dfrac {{12-V_B}}{4.7\;k}=\dfrac{{12-9.464}}{4.7\;k}=0.54\;mA\)

I_Y = I_X + I_B = (0.54 + 0.01515) mA = 0.5547 mA

\(R_2=\dfrac{V_B}{I_Y}=\dfrac{9.464}{0.5547\times10^{-3}}=17.06\;kΩ\)

∴ The value of R₂ = 17.06 kΩ.

Concept:

A BJT voltage divider configuration is as shown:

Usually, the resistance R₂ is small, resulting in small base current. This allows us to apply voltage division to obtain the base voltage, i.e.

\(V_B=V_{CC}\times \frac{R_2}{R_1+R_2}\)

Calculation:

With V_CC = 18 V, R1 = 4.7 kΩ and R2 = 1500 Ω, the base voltage will be:

\(V_B=18\times \frac{1500}{1500+4700}\)

V_B = 4.35 V

Concept:

Potential divider bias circuit or self-bias circuit is as shown:

Where:

\({V_{Th}} = \left( {\frac{{{R_2}}}{{{R_1} + {R_2}}}} \right){V_{cc}}\)

\({R_{Th}} = \left( {{R_1}\parallel {R_2}} \right){\rm{\;\Omega }}\) 

Using KVL in the collector-emitter loop, we get:

V_cc - I_c R_c - V_CE - I_E R_E = 0

And using KVL in the Base-emitter loop, we get:

V_Th - I_B R_m - V_BE - I_E R_E = 0

In case if β value is not given in the problem, we always assume a high value of β.

Hence, I_c = β I_B

\({I_B} = \frac{{{I_c}}}{\beta }\) 

High value of β

I_B ≈ 0

And, I_E = I_B + I_c

I_E ≈ I_c

Calculation:

Given that, R₁ = 50 kΩ, R₂ = 10 kΩ, R_E = 1 kΩ

V_cc = 12 V, V_BE = 0.1 V, I_c = ?

The potential divider circuit is as shown:

\({V_{Th}} = \left( {\frac{{{R_2}}}{{{R_1} + {R_2}}}} \right){V_{cc}}\) 

\( = \left( {\frac{{10}}{{50 + 10}}} \right)12\) 

\( = \frac{{10}}{{60}} \times 12\) 

V_Th = 2 V

\({R_{Th}} = \left( {{R_1}\parallel {R_2}} \right) = \frac{{{R_1} \times {R_2}}}{{{R_1} + {R_2}}}\) 

\(= \frac{{50 \times 10}}{{50 + 10}} = \frac{{500}}{{60}}\) 

\(= \frac{{25}}{3}k{\rm{\Omega }}\) 

Since β value is not given in the question, we have to assume β value very high.

Hence I_B ≈ 0 and I_c ≈ I_E

Using KVL, in the base-emitter loop, we get:

V_Th - R_Th I_B - V_BE - I_E R_E = 0

\(2 - \frac{{25}}{3} \times 0 - 0.1 - {I_c}{R_E} = 0\) 

I_c R_E = 2 – 0.1

\({I_c} = \frac{{1.9}}{{{R_E}}}\) 

\({I_c} = \frac{{1.9}}{1}\) 

I_c = 1.9 mA

Important Points:

• BJT biasing is done for the following purpose:
• To operate BJT in the Active region so that it can be used as an amplifier.
• To maintain the stability of the operating point.
• To avoid thermal runaway during operation.

Some important biasing Techniques are

• Fixed biased circuit
• Collector to the base biased circuit.
• Self-bias or potential divider biased circuit.

Given:

β = 100

V_C = 5 V

\(I_C = {V_C \over R_C}={5\over3.3\;k}=1.515\;mA \)

\(I_B = {I_C\over β}=0.01515\;mA\)

I_E = (1 + β) I_B = 1.53 mA

- 12 + 1.2 × 10³ × 1.53 × 10^-3 + 0.7 + V_B = 0

V_B = 9.464 V

\(I_X =\dfrac {{12-V_B}}{4.7\;k}=\dfrac{{12-9.464}}{4.7\;k}=0.54\;mA\)

I_Y = I_X + I_B = (0.54 + 0.01515) mA = 0.5547 mA

\(R_2=\dfrac{V_B}{I_Y}=\dfrac{9.464}{0.5547\times10^{-3}}=17.06\;kΩ\)

∴ The value of R₂ = 17.06 kΩ.

Concept:

The circuit diagram of the voltage divider bias circuit is:

Collecter current in voltage divider biasing is given by:

\(I_c = {V_2-V_{BE} \over R_E}\).............(i)

I_c = collector current

V₂ = voltage across R₂

R_{E =} emiiter resistance

V_BE = base to emitter voltage

Explanation:

From the equation (i), it is observed that:

• ​I_c doesn’t depend upon current gain (β).
• V_BE is very small that I_c doesn’t get affected by V_BE .
• Thus I_c in this circuit is almost independent of transistor parameters (β, V_BE) and hence good stabilization is achieved.
• In voltage divider biasing, the transistor always remains in the active region which is not achievable in other biasing methods.

Therefore, voltage divider biasing in BJT is the universal biasing method.

Additional Information

1. Emitter Bias: Emitter bias provides excellent bias stability in spite of changes in β. Hence, this biasing method depends on current gain (β).
2. Base Bias: Base bias ensures that the voltage fed to the base is enough which can supply enough base current to switch the transistor ON.
3. Collector Bias: This circuit helps in improving the stability considerably. If the value of I_c increases, the voltage across load increases, and hence V_CE  also increases. This in turn reduces the base current I_B.

Concept:

A BJT voltage divider configuration is as shown:

Usually, the resistance R₂ is small, resulting in small base current. This allows us to apply voltage division to obtain the base voltage, i.e.

\(V_B=V_{CC}\times \frac{R_2}{R_1+R_2}\)

Calculation:

With V_CC = 18 V, R1 = 4.7 kΩ and R2 = 1500 Ω, the base voltage will be:

\(V_B=18\times \frac{1500}{1500+4700}\)

V_B = 4.35 V

Given:

β = 100

V_C = 5 V

\(I_C = {V_C \over R_C}={5\over3.3\;k}=1.515\;mA \)

\(I_B = {I_C\over β}=0.01515\;mA\)

I_E = (1 + β) I_B = 1.53 mA

- 12 + 1.2 × 10³ × 1.53 × 10^-3 + 0.7 + V_B = 0

V_B = 9.464 V

\(I_X =\dfrac {{12-V_B}}{4.7\;k}=\dfrac{{12-9.464}}{4.7\;k}=0.54\;mA\)

I_Y = I_X + I_B = (0.54 + 0.01515) mA = 0.5547 mA

\(R_2=\dfrac{V_B}{I_Y}=\dfrac{9.464}{0.5547\times10^{-3}}=17.06\;kΩ\)

∴ The value of R₂ = 17.06 kΩ.

Concept:

The circuit diagram of the voltage divider bias circuit is:

Collecter current in voltage divider biasing is given by:

\(I_c = {V_2-V_{BE} \over R_E}\).............(i)

I_c = collector current

V₂ = voltage across R₂

R_{E =} emiiter resistance

V_BE = base to emitter voltage

Explanation:

From the equation (i), it is observed that:

• ​I_c doesn’t depend upon current gain (β).
• V_BE is very small that I_c doesn’t get affected by V_BE .
• Thus I_c in this circuit is almost independent of transistor parameters (β, V_BE) and hence good stabilization is achieved.
• In voltage divider biasing, the transistor always remains in the active region which is not achievable in other biasing methods.

Therefore, voltage divider biasing in BJT is the universal biasing method.

Additional Information

1. Emitter Bias: Emitter bias provides excellent bias stability in spite of changes in β. Hence, this biasing method depends on current gain (β).
2. Base Bias: Base bias ensures that the voltage fed to the base is enough which can supply enough base current to switch the transistor ON.
3. Collector Bias: This circuit helps in improving the stability considerably. If the value of I_c increases, the voltage across load increases, and hence V_CE  also increases. This in turn reduces the base current I_B.

Concept:

NPN transistor structure is shown with all the junction voltages.

- VBE – VCB + VCE = 0

VCE = - VBE + VCB

Current equations are related as:

IC = β × IB

IC = α × IE

Gains are related as:

\(\alpha = \frac{\beta }{{1 + \beta }}\)

\(\beta = \frac{\alpha }{{1 - \alpha }}\)

Voltage Divider Bias or Self Bias

Thevenin’s equivalent will be

\({V_{th}} = \frac{{{V_{CC}}{R_2}}}{{{R_1} + {R_2}}}\)

\({R_{th}} = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}\)

Analysis:

When ∝ ≈ 1 (i.e. very close to unity)

β ≈ ∞ ( approaches ∞ )

which results in IC ≈ IE

and IB ≈ 0

\({V_{th}} = \frac{{10}}{{\left( {10 + 33} \right)}} \times 18\)

= 4.18 V

R_th = 33 ∥ 10 = 7.67 kΩ

Concept:

A BJT voltage divider configuration is as shown:

Usually, the resistance R₂ is small, resulting in small base current. This allows us to apply voltage division to obtain the base voltage, i.e.

\(V_B=V_{CC}\times \frac{R_2}{R_1+R_2}\)

Calculation:

With V_CC = 18 V, R1 = 4.7 kΩ and R2 = 1500 Ω, the base voltage will be:

\(V_B=18\times \frac{1500}{1500+4700}\)

V_B = 4.35 V

Concept:

A BJT voltage divider configuration is as shown:

Usually, the resistance R₂ is small, resulting in small base current. This allows us to apply voltage division to obtain the base voltage, i.e.

\(V_B=V_{CC}\times \frac{R_2}{R_1+R_2}\)

Calculation:

With V_CC = 18 V, R1 = 4.7 kΩ and R2 = 1500 Ω, the base voltage will be:

\(V_B=18\times \frac{1500}{1500+4700}\)

V_B = 4.35 V

Concept:

Potential divider bias circuit or self-bias circuit is as shown:

Where:

\({V_{Th}} = \left( {\frac{{{R_2}}}{{{R_1} + {R_2}}}} \right){V_{cc}}\)

\({R_{Th}} = \left( {{R_1}\parallel {R_2}} \right){\rm{\;\Omega }}\) 

Using KVL in the collector-emitter loop, we get:

V_cc - I_c R_c - V_CE - I_E R_E = 0

And using KVL in the Base-emitter loop, we get:

V_Th - I_B R_m - V_BE - I_E R_E = 0

In case if β value is not given in the problem, we always assume a high value of β.

Hence, I_c = β I_B

\({I_B} = \frac{{{I_c}}}{\beta }\) 

High value of β

I_B ≈ 0

And, I_E = I_B + I_c

I_E ≈ I_c

Calculation:

Given that, R₁ = 50 kΩ, R₂ = 10 kΩ, R_E = 1 kΩ

V_cc = 12 V, V_BE = 0.1 V, I_c = ?

The potential divider circuit is as shown:

\({V_{Th}} = \left( {\frac{{{R_2}}}{{{R_1} + {R_2}}}} \right){V_{cc}}\) 

\( = \left( {\frac{{10}}{{50 + 10}}} \right)12\) 

\( = \frac{{10}}{{60}} \times 12\) 

V_Th = 2 V

\({R_{Th}} = \left( {{R_1}\parallel {R_2}} \right) = \frac{{{R_1} \times {R_2}}}{{{R_1} + {R_2}}}\) 

\(= \frac{{50 \times 10}}{{50 + 10}} = \frac{{500}}{{60}}\) 

\(= \frac{{25}}{3}k{\rm{\Omega }}\) 

Since β value is not given in the question, we have to assume β value very high.

Hence I_B ≈ 0 and I_c ≈ I_E

Using KVL, in the base-emitter loop, we get:

V_Th - R_Th I_B - V_BE - I_E R_E = 0

\(2 - \frac{{25}}{3} \times 0 - 0.1 - {I_c}{R_E} = 0\) 

I_c R_E = 2 – 0.1

\({I_c} = \frac{{1.9}}{{{R_E}}}\) 

\({I_c} = \frac{{1.9}}{1}\) 

I_c = 1.9 mA

Important Points:

• BJT biasing is done for the following purpose:
• To operate BJT in the Active region so that it can be used as an amplifier.
• To maintain the stability of the operating point.
• To avoid thermal runaway during operation.

Some important biasing Techniques are

• Fixed biased circuit
• Collector to the base biased circuit.
• Self-bias or potential divider biased circuit.

Explanation:

Voltage Divider Bias Circuit in CE Configuration

Problem Statement: For a voltage divider bias circuit in CE configuration using NPN BJT transistor, with given values of V_CC = 10 V, R_C = 2 kΩ, and R_E = 500 Ω, we are tasked to determine the approximate maximum value of the collector current (I_C) for the circuit, assuming a very high DC current gain (β).

Solution:

The maximum collector current (I_C) can be calculated by considering the saturation condition of the transistor, where the transistor acts as a closed switch. In this state, the voltage across the collector-emitter junction (V_CE) is approximately zero.

Step 1: Voltage Distribution in the Circuit

In the saturation condition, the voltage across the collector resistor (R_C) and emitter resistor (R_E) will be:

V_RC + V_RE = V_CC

Where:

• V_RC = Voltage drop across the collector resistor (R_C)
• V_RE = Voltage drop across the emitter resistor (R_E)

Step 2: Current Flow in the Circuit

The current flowing through R_C and R_E is the collector current (I_C), which is equal to the emitter current (I_E) in the saturation condition because the base current (I_B) is negligible for high β values.

Step 3: Maximum Collector Current Calculation

To calculate the maximum collector current (I_C), we use Ohm's Law:

I_C = I_E = V_CC ÷ (R_C + R_E)

Substituting the given values:

V_CC = 10 V, R_C = 2 kΩ = 2000 Ω, R_E = 500 Ω

I_C = 10 ÷ (2000 + 500)

I_C = 10 ÷ 2500

I_C = 0.004 A

I_C = 4 mA

Thus, the maximum collector current for the voltage divider bias circuit is approximately 4 mA.

Important Information

Let’s analyze other options:

Option 1: 5 mA

This value is incorrect because the calculation shows that the maximum collector current (I_C) is 4 mA, based on the circuit parameters. The resistance values and supply voltage do not support a collector current of 5 mA.

Option 3: 2.5 mA

This value is incorrect because it underestimates the current. Using the formula I_C = V_CC ÷ (R_C + R_E), the calculated value is 4 mA, not 2.5 mA.

Option 4: 20 mA

This value is incorrect because it overestimates the current. The resistance values and supply voltage do not support such a high collector current. The calculated maximum value is 4 mA.

Conclusion:

The correct option is Option 2, which gives the maximum collector current (I_C) as approximately 4 mA. This value aligns with the calculations based on the given circuit parameters, considering the saturation condition of the transistor and assuming a very high DC current gain (β).