Calculate the output voltage for an integrator operational amplifier.

For DC, the capacitor acts as an open circuit, so the DC voltage is divided between $R$ and $R_L$.

$$V_{DC,out} = V_{DC,in} \times \frac{R_L}{R + R_L}$$ $$V_{DC,out} = 15 \text{ V} \times \frac{10 \times 10^3 \Omega}{120 \Omega + 10 \times 10^3 \Omega}$$ $$V_{DC,out} = 15 \text{ V} \times \frac{10000}{10120}$$ $$V_{DC,out} \approx 15 \text{ V} \times 0.98814$$ $$V_{DC,out} = \boxed{\text{14.82 V}}$$

2.1.2 The ripple voltage across the load

Given:
Input peak-to-peak ripple voltage $V_{r(pp),in} = 1.5 \text{ V}$
Line frequency $f_{line} = 50 \text{ Hz}$
Filter resistor $R = 120 \Omega$
Filter capacitor $C = 150 \mu F = 150 \times 10^{-6} F$
Load resistor $R_L = 10 \text{ k}\Omega = 10 \times 10^3 \Omega$

$$X_C = \frac{1}{2\pi f_r C} = \frac{1}{2\pi (100 \text{ Hz}) (150 \times 10^{-6} \text{ F})}$$ $$X_C = \frac{1}{0.03\pi} \approx 10.61 \Omega$$

Since $R_L \gg X_C$, the parallel impedance is approximately $X_C$.

$$|Z_{eq}| = \frac{R_L X_C}{\sqrt{R_L^2 + X_C^2}} = \frac{(10 \times 10^3 \Omega) \times 10.61 \Omega}{\sqrt{(10 \times 10^3 \Omega)^2 + (10.61 \Omega)^2}}$$ $$|Z_{eq}| \approx 10.61 \Omega$$

$$V_{r(pp),out} = V_{r(pp),in} \times \frac{|Z_{eq}|}{\sqrt{R^2 + |Z_{eq}|^2}}$$ $$V_{r(pp),out} = 1.5 \text{ V} \times \frac{10.61 \Omega}{\sqrt{(120 \Omega)^2 + (10.61 \Omega)^2}}$$ $$V_{r(pp),out} = 1.5 \text{ V} \times \frac{10.61}{\sqrt{14400 + 112.57}}$$ $$V_{r(pp),out} = 1.5 \text{ V} \times \frac{10.61}{120.47}$$ $$V_{r(pp),out} \approx \boxed{\text{0.132 V}}$$

2.1.3 The percentage ripple factor of the first capacitor

Given:
Peak-to-peak ripple voltage across the first capacitor $V_{r(pp)} = 1.5 \text{ V}$
DC voltage across the first capacitor $V_{DC} = 15 \text{ V}$

For a triangular ripple, $V_{r(rms)} = \frac{V_{r(pp)}}{2\sqrt{3}}$.

$$V_{r(rms)} = \frac{1.5 \text{ V}}{2\sqrt{3}} \approx \frac{1.5}{3.464} \approx 0.433 \text{ V}$$

$$\text{Percentage ripple factor} = \frac{V_{r(rms)}}{V_{DC}} \times 100\%$$ $$\text{Percentage ripple factor} = \frac{0.433 \text{ V}}{15 \text{ V}} \times 100\%$$ \text{Percentage ripple factor} = \boxed{\text{2.89%}}

2.2 Briefly describe the operational principle of a voltage tripler.

A voltage tripler is a type of voltage multiplier circuit that converts an AC input voltage into a DC output voltage that is approximately three times the peak value of the AC input. It achieves this by using a network of diodes and capacitors. During different half-cycles of the AC input, capacitors are charged in parallel to the peak input voltage. These charged capacitors are then effectively connected in series with each other and the input voltage, allowing their voltages to sum up and produce a higher DC output voltage.

2.3 Draw a neat, labelled circuit diagram of a Zener-diode reference source.

A Zener diode reference source typically consists of a series current-limiting resistor ($R_S$) connected to an unregulated DC input voltage ($V_{in}$). A Zener diode is then connected in parallel with the output (and the load $R_L$), with its cathode connected to the positive side of the output and its anode to the negative side. The Zener diode maintains a constant voltage ($V_Z$) across the load, provided the input voltage is sufficiently high and the current through the Zener diode is within its operating range.

(Diagram cannot be rendered in text. It would show a voltage source $V_{in}$, a series resistor $R_S$, and a Zener diode in parallel with a load resistor $R_L$, with the Zener diode's cathode connected to the positive output and anode to ground.)

QUESTION 3: AMPLIFIERS

3.1 Explain the term, stabilisation.

In electronics, stabilisation refers to the process of maintaining a circuit's desired operating conditions or output characteristics constant despite variations in external factors. For an amplifier, this typically means keeping its Q-point (quiescent operating point) stable against changes in temperature, supply voltage, or component tolerances, ensuring consistent gain and preventing issues like thermal runaway.

3.2 Calculate the value of the base resistor with the aid of precise method if the following information about the transistor is known:

Given:
Amplifier input impedance $Z_{in} = 2.5 \text{ k}\Omega = 2500 \Omega$
Transistor h-parameters:
$h_{ie} = 13 \text{ k}\Omega = 13000 \Omega$
$h_{re} = 2.8 \times 10^{-4}$
$h_{fe} = 300$
Collector resistor $R_C = 480 \Omega$
Load resistor $R_L = 6 \text{ k}\Omega = 6000 \Omega$
Source resistance $R_S = 0$

$$R_L' = R_C || R_L = \frac{R_C \times R_L}{R_C + R_L}$$ $$R_L' = \frac{480 \Omega \times 6000 \Omega}{480 \Omega + 6000 \Omega} = \frac{2880000}{6480} \approx 444.44 \Omega$$

For a common-emitter configuration, $Z_{in(base)} = h_{ie} + h_{re} h_{fe} R_L'$.

$$Z_{in(base)} = 13000 \Omega + (2.8 \times 10^{-4}) \times 300 \times 444.44 \Omega$$ $$Z_{in(base)} = 13000 \Omega + 0.084 \times 444.44 \Omega$$ $$Z_{in(base)} = 13000 \Omega + 37.33 \Omega = 13037.33 \Omega$$

For a fixed bias common-emitter amplifier, the overall input impedance of the amplifier stage is $Z_{in} = R_B || Z_{in(base)}$.

$$Z_{in} = \frac{R_B \times Z_{in(base)}}{R_B + Z_{in(base)}}$$

Rearranging for $R_B$:

$$R_B = \frac{Z_{in} \times Z_{in(base)}}{Z_{in(base)} - Z_{in}}$$ $$R_B = \frac{2500 \Omega \times 13037.33 \Omega}{13037.33 \Omega - 2500 \Omega}$$ $$R_B = \frac{32593325}{10537.33}$$ $$R_B \approx \boxed{\text{3093.1 }\Omega}$$

3.3 Calculate the output voltage for an integrator operational amplifier.

Given:
Input signal $V_{in}$ is a square waveform of $\pm 2 \text{ V}$.
Frequency $f = 800 \text{ Hz}$.
Input resistor $R_{in} = 10 \text{ M}\Omega = 10 \times 10^6 \Omega$.
Feedback capacitor $C_f = 0.01 \mu F = 0.01 \times 10^{-6} F = 1 \times 10^{-8} F$.
Capacitor is discharged originally.

$$T = \frac{1}{f} = \frac{1}{800 \text{ Hz}} = 0.00125 \text{ s} = 1.25 \text{ ms}$$ $$T/2 = 0.625 \text{ ms}$$

The output voltage of an integrator is $V_{out}(t) = -\frac{1}{R_{in} C_f} \int V_{in}(t) dt$.
Since the capacitor is initially discharged, $V_{out}(0) = 0$.

$$V_{out}(t) = -\frac{1}{0.1 \text{ s}} \int_0^t (+2 \text{ V}) dt = -10 \text{ s}^{-1} \times (2t)$$ $$V_{out}(t) = -20t$$

At $t = T/2 = 0.625 \times 10^{-3} \text{ s}$:

$$V_{out}(T/2) = -20 \times (0.625 \times 10^{-3}) = -0.0125 \text{ V}$$

This is the peak negative voltage of the triangular wave.

The integration starts from $V_{out}(T/2)$.

$$V_{out}(t) = V_{out}(T/2) - \frac{1}{R_{in} C_f} \int_{T/2}^t (-2 \text{ V}) dt$$ $$V_{out}(t) = -0.0125 \text{ V} - 10 \text{ s}^{-1} \times (-2(t - T/2))$$ $$V_{out}(t) = -0.0125 \text{ V} + 20(t - T/2)$$

At $t = T = 1.25 \times 10^{-3} \text{ s}$:

$$V_{out}(T) = -0.0125 \text{ V} + 20(1.25 \times 10^{-3} - 0.625 \times 10^{-3})$$ $$V_{out}(T) = -0.0125 \text{ V} + 20(0.625 \times 10^{-3})$$ $$V_{out}(T) = -0.0125 \text{ V} + 0.0125 \text{ V} = 0 \text{ V}$$

The output voltage is a triangular waveform that swings between $0 \text{ V}$ and $-0.0125 \text{ V}$. The peak output voltage is the magnitude of the maximum deviation from zero.
The peak output voltage is $\boxed{\text{0.0125 V}}$.