Electronics Class Notes — Tutorial 9

Teacher: Prof P. M. Sarun • NPHC206 • WINTER - 2025-2026 • Last updated:

Questions

  1. A differential amplifier has a differential voltage gain of \(2000\) and a common-mode gain of \(0.2\). Determine \(\text{CMRR}\) and express it in \(\text{dB}\).
  2. Find (i) the output signal, (ii) the noise on the output of a differential amplifier having a voltage gain of \(150\) and a \(\text{CMRR}\) of \(90 \,\text{dB}\). The input signals are \(50 \,\text{mV}\) and \(100 \,\text{mV}\) with \(1 \,\text{mV}\) of noise on each input.
  3. Find the bias voltages and currents for the differential amplifier circuit
  4. Tutorial 9
    Figure 1.
  5. Find the output voltage of the differential amplifier when the transistors are identical with \(\beta_{\text{dc}} = 100\).
  6. Tutorial 9
    Figure 2.
  7. Find the differential-mode voltage gain for the circuit.
  8. Tutorial 9
    Figure 3.

Answers

  1. \(\text{CMRR} = \frac{A_{\text{DM}}}{A_{\text{CM}}} = 10,000\) ; \(\text{CMRR}_{\text{dB}} = 80\,\text{dB}\)
  2. (i) \(v_{\text{out}} =A_{\text{DM}}(v_{1} – v_{2})\) = \(7.5 \,\text{V}\) (ii) \(A_{\text{CM}} = 4.7 \times 10^{–3}\); \(\therefore\) Noise on output \(= A_{\text{CM}} \times \text{noise}\) \(= 4.7 \times 10^{–6} \,\text{V}\)
  3. \(V_{\text{E}} = – 0.7 \,\text{V}\); \(I_{\text{E}} = \frac{V_{\text{EE}} - V_{\text{BE}}}{R_{\text{E}}}\) \(= 0.452 \,\text{mA}\); \(I_{\text{E1}} = I_{\text{E2}} = \frac{I_{\text{E}}}{2} = I_{\text{C1}} = I_{\text{C2}}\) \(=0.226 \,\text{mA}\); \(I_{\text{B1}} = I_{\text{B2}} = \frac{0.226 \,\text{mA}}{\beta}\) \(= 2.26 \,\mu\text{A}\); \(V_{\text{C1}} = V_{\text{CC}} = 12\,\text{V}\) ; \(V_{\text{C2}} = V_{\text{CC}} – I_{\text{C2}}R_{\text{C2}} = 9.7\,\text{V}\)
  4. \(I_{\text{E}} = 0.953 \,\text{mA}\) ; \(I_{\text{C}} \approx I_{\text{E}} = 0.477 \,\text{mA}\), \(V_{\text{out}} = V_{\text{CC}} – I_{\text{C}}R_{\text{C}} = 7.85 \,\text{V}\)
  5. \(I_{\text{E}} = 0.0565 \,\text{mA}\); \(I_{\text{E1}} = I_{\text{E2}} = \frac{I_{\text{E}}}{2} = 0.0283 \,\text{mA}\); \(r_{\text{e}} = \frac{25 \,\text{mV}}{I_{\text{E}}} = 883.4 \Omega\); \(A_{\text{DM}} = \frac{R_{\text{C}}}{2r_{\text{e}}}= 56.6\)