COMPS-Question 2Consider the asymmetric bipolar st

Sure, let's break down the questions one by one and provide the necessary answers. Question 2(a) Derive an expression for the two collector voltages (Vc1 and Vc2) at the equilibrium point arising when both inverters are operating in their linear regions. Given: - Forward voltage drop on a p-n junction (Vf) - Identical gain for Q1 and Q2 (β = 100) - Rc1 = Rc2 = 1kΩ - Rb1 = Rb2 = 10kΩ To derive the expression, we need to consider the voltage drops and the current through the transistors and resistors. Let's denote: -

I_{C1}

and

I_{C2}

as the collector currents of Q1 and Q2 respectively. -

V_{BE1}

and

V_{BE2}

as the base-emitter voltages of Q1 and Q2 respectively. At equilibrium, the base-emitter voltage

V_{BE}

for both transistors will be approximately equal to the forward voltage drop

V_f

. The collector voltage

V_{C1}

and

V_{C2}

can be expressed as:

V_{C1} = V_{cc} - I_{C1} \cdot R_{C1}

V_{C2} = V_{cc} - I_{C2} \cdot R_{C2}

Since

I_{C1} = I_{C2}

(due to symmetry and identical transistors):

V_{C1} = V_{C2} = V_{cc} - I_C \cdot R_C

Where

I_C

is the collector current and

R_C

is the collector resistor. Question 2(b)(i) For what value of V is Vc1 = Vc2 at this equilibrium point? At equilibrium,

V_{C1} = V_{C2}

implies that the currents through both transistors are equal. This happens when the base voltages are equal, which means:

V_{B1} = V_{B2}

Given that the base-emitter voltage drop is

V_f

, the base voltage

V_B

can be expressed as:

V_B = V_{C1} + V_f

Thus, the value of

V

for which

V_{C1} = V_{C2}

is:

V = V_{cc} - I_C \cdot R_C + V_f

Question 2(b)(ii) What is this value of Vc1? Using the derived expression:

V_{C1} = V_{cc} - I_C \cdot R_C

Given

R_C = 1kΩ

and assuming

I_C

is determined by the circuit parameters, we can substitute the values to find

V_{C1}

. Question 2(c)(i) If Rc1 = Rc2 = 1kΩ, Rb1 = Rb2 = 10kΩ and β = 100, state whether the equilibrium point you derived in part (a) above is stable or unstable. To determine stability, we need to analyze the small-signal model of the circuit. If the feedback is negative, the equilibrium point is stable; if the feedback is positive, it is unstable. Question 2(c)(ii) Explain your reasoning. The stability can be analyzed by considering the small-signal parameters and the feedback loop. If the loop gain is less than 1, the system is stable. Given the high gain (β = 100), the feedback is likely to be positive, making the equilibrium point unstable. Summary of Answers 1. Derive an expression for the two collector voltages (Vc1 and Vc2) at the equilibrium point:

A

Key Concept

Equilibrium point in linear region

Explanation

The collector voltages are derived based on the current through the transistors and resistors.

2. For what value of V is Vc1 = Vc2 at this equilibrium point?

B

Key Concept

Equal base voltages

Explanation

The base voltages must be equal for the collector voltages to be equal.

3. What is this value of Vc1?

C

Key Concept

Collector voltage calculation

Explanation

The collector voltage is calculated based on the supply voltage and the voltage drop across the collector resistor.

4. State whether the equilibrium point you derived in part (a) above is stable or unstable.

D

Key Concept

Stability analysis

Explanation

The stability is determined by the feedback loop and the gain of the transistors.

If Rc1 = Rc2 = 1kΩ, Rb1 = Rb2 = 10kΩ and β = 100, state whether the equilibrium point you derived in part (a) above is stable or unstable.

The equilibrium point derived in part (a) is stable.

Here is the reasoning:

▪

The equilibrium point is determined by the balance of currents and voltages in the circuit

▪

Given that

R_{c1} = R_{c2} = 1k\Omega

,

R_{b1} = R_{b2} = 10k\Omega

, and

\beta = 100

, the circuit is symmetric

▪

In a symmetric circuit with identical components, the equilibrium point where

V_{c1} = V_{c2}

is typically stable because any small perturbation will be counteracted by the feedback mechanisms in the circuit

▪

The feedback in the circuit ensures that if

V_{c1}

or

V_{c2}

deviates slightly, the circuit will adjust to bring the voltages back to equilibrium

▪

The stability is also supported by the fact that the transistors have a high gain (

\beta = 100

), which means they can quickly respond to changes and stabilize the voltages

Therefore, the equilibrium point is stable.

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