Test: FET Biasing - Electrical Engineering (EE) MCQ

Bias stabilization is a process of making Q-point independent of changes in temperature and changes in the transistor parameter.

Current stability factor
 

Fixed bias circuit:

Base current IB is fixed. Hence it is called a fixed biased circuit.

S = 1 + β

Since S is very large, IC will be unstable

Concept:

Amplifier

It is an electronic device which is used to increase the strength of the signal. Strength means power.

Consider the below CE amplifier as shown:

The advantage of the CE amplifier is its power gain will be high.

Coupling capacitors

These are used in an electronic circuit to pass the AC signal and blocks the unwanted DC components.

• The unwanted DC signals comes form the electronic devices or preceding stage of the electronic circuit.
• The reactive nature of the capacitor is useful for different behavior to different frequencies.
• In audio systems, DC sources are used to power audio circuits.

Coupling capacitors are essential components in amplifier circuits. They are used to prevent interference of a transistor’s bias voltage by AC signals.

In most amplifier circuits, this is achieved by driving the signal to the base terminal of a transistor through a coupling capacitor.

Conclusion:

Option C is correct.

The purpose of Biasing in BJT amplifier:

i) To Stabilize Q-point in its linear region of operation (Q-point is an operating point).

[Statement 2 correct]

ii) To stabilize Q-point in against variation of temperature because

I_c = βI_B + (1 + β) I_co

I_co → leakage current.

Where I_co, β_1, and I_B are temperature dependent.

i.e. Temp ↑ → I_co ↑ → β ↑ → α ↑     

(∵β=α1−α)

[Statement 1 correct], [Statement 3 wrong]

iii) To reduce distortion and increase dynamic range.

Since biasing fixed the Q-point so that there is no variation in the output so distortion is reduced hence dynamic range i.e. operating range of the amplifier is improved.

[Statement 4 correct]

So, statement 1, 2, 4 are correct.

iv) Biasing must provide the operating point independent of β of the transistor. So that variation in β due to temperature change does not affect the Q-point of an amplifier.

A field-effect transistor works as a voltage variable resistor in the ohmic region.

It works as an amplifier in the saturation region.

Pinch off voltage:

The voltage at which channel will be pinched off i.e. no further movement of carriers from source to drain and the current will be practically constant is called the ‘Pinch off voltage’.

V_P = V_{GS (Cutoff)} = V_{GS (OFF})

Condition for pinch off saturation:

V_DS = V_GS - V_P

To operate FET as an amplifier, it is operated in the saturation region.

Cutoff: I_D = 0

Linear: I_D increases linearly

Saturation: I_D is constant

Note:

1) I_D ≤ I_DSS always

2) This also can be used as a ‘Voltage variable resistor. 

The FET physical structure which contains silicon dioxide provides infinite resistance. Hence no current will flow through the gate terminal.

We know that for an FET same current flows through the gate and source terminal,
Hence source current=1mA.

V_Rs = I_D R_s
But V_Rs + V_gs = 0
V_gs = -I_D R_s.

V_G = R₂ × V_DD/R₁ + R₂
⇒ V_G = 1×5/2
⇒ V_G =  2.5V.

Increasing values of R_s result in lower quiescent values of I_D and more negative values of V_gs.

The above equation called as Shockley’s equation depicts the relation between I_D and V_gs. When V_gs becomes equal toV_p, the current will become zero, which clearly satisfies the physical nature of FET.

 V_DS = V_DD - I_D R_D
⇒10 = 12-R_D × 1mA
⇒ R_D = 2/1mA
= 2 KΩ.

V_DS = V_DD - I_D (R_D + R_s)
⇒ V_DS = 12 - 1mA(1KΩ + 1KΩ)
⇒ V_DS = 10V.

V_G = R₂ × V_DD/R₁ + R₂
⇒ V_G = 2 × 12/4
⇒ V_G = 6V
⇒ V_GS = V_G - I_D R_s
⇒ V_GS = 2V.

I_R1 = I_R2 = V_DD/R₁ + R₂
⇒ I_R1 = 10/2KΩ
⇒ I_R1 = 5mA.

Concept:

For Voltage divider bias amplifier:

Input Resistance:

Output Resistance: 

Voltage Gain:

Calculation:

Voltage gain = A_V