Analog Electronics Test 1

The ripple factor measures AC fluctuations in a rectifier’s output. A full-wave rectifier has a lower ripple factor than a half-wave rectifier because it rectifies both halves of the AC input, producing an output with twice the frequency (e.g., 120 Hz for a 60 Hz input). This higher frequency results in smaller voltage fluctuations between peaks, reducing the ripple factor.

Option A: Correct, as higher frequency reduces ripple magnitude.

The ripple factor for an inductor filter is

Diodes are essential components in electronic circuits, primarily used to manage voltage levels. They perform this function by behaving like a voltage source under certain conditions. Here are the key points regarding their operation:

• Clipping Voltage: Diodes can limit the maximum voltage in a circuit, preventing damage to other components.
• Forward Bias: When a diode is forward-biased, it allows current to flow, effectively acting as a voltage source.
• Reverse Bias: In reverse bias, the diode blocks current, protecting the circuit from excessive voltage.
• Applications: They are commonly used in power supply circuits and signal processing.

In summary, diodes play a crucial role in controlling voltage levels by functioning as a voltage source under specific conditions, making them indispensable in many electronic applications.

When a capacitor is connected across the output terminals of a rectifier, the output voltage:

• Essentially becomes a DC voltage.
• Undergoes minimal fluctuations, smoothing out the signal.
• Helps to store energy, providing a stable output.

This process is critical in converting the alternating current (AC) into a usable direct current (DC), which is essential for most electronic devices. The capacitor acts as a reservoir, filling in the gaps between the peaks of the rectified voltage and reducing the ripple effect.

• Diode is off for v_i  < 5 V. Hence v_o = 5 V.
• For v_i > 5 V, v_o = v_i .Therefore (D) is correct option.

• For v_i  < 4 V the diode is ON and output v_o = 4 V.
• For v_i  >4 V diode is off and output v_o = v_i.
• Thus (D) is correct option.

• During positive cycle when v_s < 8V both diode are OFF v_o = v_s . For v_s > 8V , v_o = 8V , D₁ is ON.
• During negative cycle when |v_s| < 6 V, both diode are OFF, v_o = v_s .
• For |v_s| > 6V , D₂ is on v₀ = -6V .Therefore (C) is correct.

The output voltage cannot exceed the positive power supply voltage and cannot be lower than the negative power supply voltage.

Concept:

The BJT configuration for a pnp and npn transistors are as shown:

The emitter-collector voltage can be calculated using KVL

Analysis:

The configuration given is of a PNP transistor.

Given, V_E = 3 V

V_EB = 600 × 10^-3 V = 0.6 V

V_E – V_B ⇒ 0.6 V

3 – V_B ⇒ 0.6 V

So, V_B = 2.4 Volts

Applying Ohm’s law across 60 kΩ resistor,

\({I_B} = \frac{{{V_B}}}{{60 \times {{10}^3}}} = \frac{{2.4}}{{60 \times {{10}^3}}} \)

\(I_B= 0.04 \times {10^{ - 3}}A\)

I_C = β I_B (where β is the current gain)

Given, β = 50,

I_C = 50 × 0.04 × 10^-3 A = 2 mA

I_C = 2 × 10^-3 A

Applying Ohms law across 500 Ω resistor, we get:

\({I_C} = \frac{{{V_C}}}{{500}} = \frac{{{V_C}}}{{500}} \)

\(I_C= 2 \times {10^{ - 3}}\)

\({V_C} = 1\;Volts\)

Now,

V_EC = V_E­ – V_C = 3 – 1 = 2 Volts

So, V_EC = 2 Volts

Common Mistakes:

Mark emitter & collector terminals accordingly. It is common to notice that students always apply npn-transistor concepts even for a given pnp transistor. 

For an n-channel enhancement mode MOSFET, transition point is given by:

V_DS(sat) = V_GS - V_th

= V_GS – 2

From the circuit,

V_DS = V_GS

⇒ V_Ds(sat) = V_DS – 2

⇒ V_DS = V_DS(sat) + 2

⇒ V_­DS > V_DS(sat)

So, the transistor will always be in the Saturation region.

I_D = K (V_GS - V_th)²

⇒ 4 = K (V_GS - 2)²

V_GS = V_DS = 10 – I_D R_­D = 10 – 4(1) = 6 V

⇒ 4 = K (6 -2)²

⇒ K = 0.25

Now R_D is increased to 4 kΩ, Let current \(I{'}{_d}\) and voltages are \(V{'}{_{DS}} = V{'}{_{GS}}\)

Applying current equation

\(I_D' = K{\left( {V_{GS}' - {V_{TH}}} \right)^2}\)

\(I_D' = \frac{1}{4}{\left( {V_{GS}' - 2} \right)^2}\)

\(V_{GS}' = V_{DS}' = 10 - I_D'R_D' = 10 - 4I_D'\)

\(4I_{DS}' = {\left( {10 - 4I_D' - 2} \right)^2} = {\left( {8 - 4I_D'} \right)^2}\)

\(= 16{\left( {2 - I_D'} \right)^2}\)

\(I_D' = 4\left( {4 + I_D' - 4I_D'} \right)\)

\(\Rightarrow 4{\left( {I_D'} \right)^2} - 17I_D' + 16 = 0\)

\(\Rightarrow I_D' = 2.84\;mA~or~1.4~mA\)

It cannot be 2.84 mA because:

For I_D = 2.84 mA, VGS will be negative.

So, I_D' = 1.4 mA