GATE E&C Engineering Analog Circuits Syllabus

Analog Circuits syllabus — full chapter and topic list

Expand any chapter to see its topics and sub-topics. This is the whole examinable outline for Analog Circuits in GATE E&C Engineering, not a summary of it.

1. 1

   Diode circuits

   3 topics

   • Clipping
   • Clamping
   • Rectifiers
2. 2

   BJT and MOSFET amplifiers

   4 topics

   • Biasing
   • AC coupling
   • Small signal analysis
   • Frequency response
3. 3

   Current mirrors and differential amplifiers

   2 topics

   • Current mirrors
   • Differential amplifiers
4. 4

   Op-amp circuits

   7 topics

   • Amplifiers
   • Summers
   • Differentiators
   • Integrators
   • Active filters
   • Schmitt triggers
   • Oscillators

Analog Circuits flashcards for GATE E&C Engineering

22 of 51 cards from the Analog Circuits deck — real questions with worked answers.

1. In a positive series clipper using an ideal diode in series with the load, which part of the input waveform appears at the output?

   A positive series clipper removes (clips) the positive half of the input. With an ideal series diode oriented to conduct on negative inputs, only the negative half-cycle (below $0\,\text{V}$) reaches the output; the positive portion is clipped to $0\,\text{V}$.

2. For a biased shunt clipper with a diode in parallel with the load and a reference battery $V_R$ (diode conducts when $v_i > V_R$), what is the clipped output level?

   When the diode conducts, the output is held at the reference level: $$v_o = V_R + V_\gamma$$ where $V_\gamma$ is the diode cut-in voltage (about $0.7\,\text{V}$ for silicon). Below this, $v_o = v_i$.

3. What is the fundamental difference between a clipper and a clamper circuit?

   A clipper removes a portion of the input waveform (limits amplitude) and does not shift the DC level. A clamper shifts the entire waveform up or down to a new DC reference without changing its shape or peak-to-peak amplitude; it uses a capacitor to add a DC offset.

4. In a negative clamper, to what value is the positive peak of the output clamped (ideal diode)?

   A negative clamper shifts the waveform down so its positive peak sits at $0\,\text{V}$. The capacitor charges to the peak input voltage $V_m$, giving output swinging from $0$ down to $-2V_m$.

5. State the design condition (time-constant requirement) for a clamper circuit to work properly.

   The RC time constant must be much larger than the period of the input: $$\tau = RC \gg T$$ Typically $RC \geq 10T$ (or $5T$ per half-period) so the capacitor holds its charge and the DC shift is maintained.

6. For a half-wave rectifier with sinusoidal input of peak $V_m$, give the DC (average) output voltage and the ripple factor.

   DC output: $$V_{dc} = \frac{V_m}{\pi}$$ RMS output: $V_{rms} = \frac{V_m}{2}$. Ripple factor: $$r = \sqrt{\left(\frac{V_{rms}}{V_{dc}}\right)^2 - 1} = 1.21$$

7. For a full-wave (or bridge) rectifier with sinusoidal input peak $V_m$, give $V_{dc}$, $V_{rms}$, ripple factor, and rectification efficiency.

   $$V_{dc} = \frac{2V_m}{\pi}, \quad V_{rms} = \frac{V_m}{\sqrt{2}}$$ Ripple factor $r = 0.482$; maximum efficiency $\eta = 81.2\%$.

8. Compare the Peak Inverse Voltage (PIV) of a center-tapped full-wave rectifier versus a bridge rectifier (each delivering peak $V_m$).

   Center-tapped FW rectifier: $\text{PIV} = 2V_m$. Bridge rectifier: $\text{PIV} = V_m$. The bridge needs cheaper, lower-rated diodes but uses four diodes versus two.

9. What is the ripple factor of a full-wave rectifier with a shunt capacitor filter (capacitor $C$, load $R_L$, supply frequency $f$)?

   $$r = \frac{1}{4\sqrt{3}\, f R_L C}$$ Ripple decreases as $C$ and $R_L$ increase; the ripple frequency for full-wave is $2f$.

10. State the four conditions a BJT amplifier must satisfy to operate in the active region for faithful amplification.

    For active region: emitter-base junction forward biased, collector-base junction reverse biased. The Q-point must lie within the active region (not in saturation or cutoff) and the signal swing must keep $V_{CE}$ and $I_C$ within active limits.

11. Define the stability factor $S$ for BJT bias and give its expression for the voltage-divider (self-bias) configuration.

    $$S = \frac{\partial I_C}{\partial I_{CO}}\Big|_{V_{BE},\beta}$$ For voltage-divider bias: $$S = \frac{(1+\beta)(1 + R_B/R_E)}{1 + \beta + R_B/R_E}$$ where $R_B = R_1 \parallel R_2$. Smaller $S$ (closer to 1) means better thermal stability.

12. Why is the voltage-divider (potential-divider) bias the most widely used BJT biasing scheme?

    It makes the Q-point essentially independent of $\beta$. With $R_E$ providing negative feedback and $V_B$ set by the divider, $I_C \approx (V_B - V_{BE})/R_E$, giving good thermal stability and a low stability factor $S$.

13. What is the purpose of the bypass capacitor across the emitter resistor $R_E$ in a CE amplifier?

    $R_E$ provides DC bias stabilization through negative feedback, but it also reduces AC gain. The bypass capacitor shorts $R_E$ at signal frequencies, restoring full AC gain $A_v \approx -g_m R_C$ while keeping DC stability intact.

14. What roles do coupling (AC-coupling) capacitors serve between amplifier stages?

    Coupling capacitors block DC so each stage's bias (Q-point) is undisturbed, while passing the AC signal between stages. They also set the lower cutoff frequency, behaving as a high-pass element.

15. How does a coupling capacitor influence the lower cutoff frequency of an amplifier?

    A coupling capacitor $C$ with the resistance $R$ it sees forms a high-pass filter with cutoff $$f_L = \frac{1}{2\pi R C}$$ Below $f_L$ the capacitor's reactance attenuates the signal at $-20\,\text{dB/decade}$.

16. In the hybrid-$\pi$ small-signal model of a BJT, give the transconductance $g_m$ and base input resistance $r_\pi$.

    $$g_m = \frac{I_C}{V_T}, \qquad r_\pi = \frac{\beta}{g_m} = \frac{\beta V_T}{I_C}$$ where $V_T \approx 26\,\text{mV}$ at room temperature.

17. What is the small-signal voltage gain of a common-emitter amplifier (with bypassed $R_E$), and what is its phase relationship?

    $$A_v = -g_m (R_C \parallel R_L)$$ The negative sign indicates a $180^\circ$ phase inversion between input and output.

18. Give the small-signal voltage gain and key characteristics of a common-collector (emitter follower) amplifier.

    $$A_v = \frac{g_m R_E}{1 + g_m R_E} \approx 1$$ It has high input impedance, low output impedance, no phase inversion, and is used as a buffer / impedance matcher; current and power gain are high though voltage gain $\approx 1$.

19. Summarize the comparison of CE, CB, and CC BJT amplifier configurations in terms of gains, impedances, and phase.

    CE: high voltage and current gain, medium $Z_{in}$/$Z_{out}$, $180^\circ$ phase shift. CB: high voltage gain, current gain $<1$, low $Z_{in}$, high $Z_{out}$, no phase shift. CC: voltage gain $\approx 1$, high current gain, high $Z_{in}$, low $Z_{out}$, no phase shift (buffer).

20. For a MOSFET in saturation, write the drain current equation and the small-signal transconductance $g_m$.

    $$I_D = \frac{1}{2}\mu_n C_{ox}\frac{W}{L}(V_{GS}-V_{th})^2$$ $$g_m = \mu_n C_{ox}\frac{W}{L}(V_{GS}-V_{th}) = \sqrt{2\mu_n C_{ox}\frac{W}{L} I_D} = \frac{2I_D}{V_{GS}-V_{th}}$$

21. What limits the high-frequency response of a CE amplifier, and which effect dominates?

    The internal junction capacitances $C_\pi$ and $C_\mu$ limit high-frequency gain. The Miller effect multiplies $C_\mu$ by $(1+|A_v|)$, giving an effective input capacitance $C_{in} = C_\pi + C_\mu(1+|A_v|)$, which dominates and sets the upper cutoff frequency.

22. State the Miller effect: how an impedance $Z$ bridging input and output of an inverting amplifier of gain $-A$ appears at the input.

    The bridging impedance appears at the input as $$Z_{in} = \frac{Z}{1+A}$$ For a capacitor $C$, the effective input capacitance becomes $C_{in} = C(1+A)$, greatly reducing bandwidth.

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