A) Draw the circuit symbol and the truth table of an OR gate. b) For a transistor with current amplification, hfe, of 100, calculate the base current and emitter current when the base-emitter voltage

Given: $V_i = 2.0 \text{ V}$, $V_{BE} = 0.7 \text{ V}$, $R_B = 5 \text{ k}\Omega = 5000 \Omega$.
Using Kirchhoff's Voltage Law in the base-emitter loop:

Given: $h_{fe} = 100$.
The collector current is $I_C = h_{fe} I_B$.
The emitter current is $I_E = I_B + I_C = I_B + h_{fe} I_B = (1 + h_{fe}) I_B$.

The base current is $\boxed{260 \mu\text{A}}$ and the emitter current is $\boxed{26.26 \text{ mA}}$.

ii) What is the output voltage, $V_o$?

Given: $V_{CC} = +9 \text{ V}$, $R_C = 100 \Omega$.
Using Kirchhoff's Voltage Law in the collector-emitter loop:

The output voltage is $\boxed{6.4 \text{ V}}$.

c) Figure 4(a) shows a transistor circuit in the CE arrangement. The input voltage ($V_P$) is varied using the potential divider, P. The corresponding output voltage ($V_o$) is shown graphically in figure 4(b).
For this circuit to be used as an alternating voltage amplifier, the input voltage must be fixed at a suitable value by adjusting P.

i) What is the most suitable value for this fixed input voltage, explain your answer.

The most suitable value for the fixed input voltage ($V_P$) is the quiescent point (Q-point) in the linear region of the transistor's operation. From Figure 4(b), the linear region where the output voltage changes significantly and linearly with the input voltage is approximately between $V_P = 0.8 \text{ V}$ and $V_P = 1.2 \text{ V}$. The midpoint of this region, which allows for maximum undistorted output swing, is $V_P = 1.0 \text{ V}$.

The most suitable value for the fixed input voltage is $\boxed{1.0 \text{ V}}$. This value places the transistor in the middle of its linear operating region, allowing for maximum undistorted amplification of an AC signal.

ii) What is the use of the capacitor C?

Capacitor C is a coupling capacitor. Its purpose is to block the DC component of the input signal, allowing only the AC signal to pass to the base of the transistor. This ensures that the DC bias point of the transistor, set for amplification, remains stable and is not affected by the input signal source.

d) A sinusoidal alternating voltage of amplitude $0.5 \text{ V}$ is superimposed on this fixed voltage.

i) What will the amplitude of the output voltage be?

The fixed input voltage is $1.0 \text{ V}$ (from part c(i)). A sinusoidal AC voltage of amplitude $0.5 \text{ V}$ is superimposed on it.
So, the input voltage $V_P$ will vary from $1.0 \text{ V} - 0.5 \text{ V} = 0.5 \text{ V}$ to $1.0 \text{ V} + 0.5 \text{ V} = 1.5 \text{ V}$.

When $V_P = 0.5 \text{ V}$, the output voltage $V_o \approx 1.8 \text{ V}$.
When $V_P = 1.5 \text{ V}$, the output voltage $V_o \approx 0.2 \text{ V}$.

The peak-to-peak output voltage is the difference between the maximum and minimum output voltages:

The amplitude of the output voltage is half of the peak-to-