Electronic Circuit Design: Voltage Regulator, Oscillator, Power Amplifier and RC Coupled Amplifier Calculations

Series Voltage Regulator Vmin = 8V, Vmax = 14V, Vz = 6.8V, ILmin = 40mA, ILmax = 70mA, Izmin = 5mA, Izmax = 100mA, β = 50, RS1 = (V_min − V_Z) / (I_Zmin + (I_Lmax / β)) = (8 − 6.8) / (5 mA + (70 mA / 50)) = 187 Ω, RS2 = (V_max − V_Z) / (I_Zmax + (I_Lmin / β)) = (14 − 6.8) / (100 mA + (40 mA / 50)) = 71 Ω, Select RS = average of RS1 and RS2 = 129 Ω ≈ 120 Ω (standard), Select RL as 1k pot.

I

RC Phase Shift Oscillator

Design of the amplifier: Select transistor BC107. It can provide a gain of more than 29 because its minimum h_FE is 100.

DC biasing conditions: V_CC = 12 V, I_C = 2 mA, V_RC = 40% of V_CC = 4.8 V, V_RE = 10% of V_CC = 1.2 V, V_CE = 50% of V_CC = 6 V.

Design of R_C: V_RC = I_C × R_C, R_C = V_RC / I_C = 4.8 / 2 = 2.4 kΩ.

Design of R_E: V_RE = I_E × R_E, R_E = V_RE / I_E = 1.2 / 2 mA = 600 Ω.

Design of R₁ and R₂: I_B = I_C / h_FE = 2 mA / 100 = 20 µA. Assume the current through R₁ = 10 I_B and through R₂ = 9 I_B to avoid loading the potential divider by the base current.

V_R2 = voltage across R₂ = V_BE + V_RE = 0.7 + 1.2 = 1.9 V, V_R2 = 9 I_B × R₂, R₂ = V_R2 / 9 I_B = 10.5 kΩ.

V_R1 = voltage across R₁ = V_CC − V_R2, V_R1 = 10 I_B × R₁, R₁ = V_R1 / 10 I_B = 50.5 kΩ.

Design of bypass capacitor C_E: X_CE ≤ R_E / 10 = 60 Ω, C_E = 1 / (2π f × R_E / 10) = 2.65 × 10^-3 F = 4.7 µF.

Design of frequency selective network: Required frequency of oscillation is 1 kHz, f = 1 / (2π R C √(6 + 4R_C/R)) = 1 kHz. The frequency determined by R and R_C must be selected to avoid loading of the amplifier, so R = 2R_C = 2 × 2.4 kΩ = 4.8 kΩ ≈ 4.7 kΩ. Then C = 0.0117 µF ≈ 0.01 µF.

Power Amplifier

Class B power amplifier, Q1 = SL100, Q2 = SK100.

Class AB power amplifier, Q1 = SL100, Q2 = SK100.

Class B power amplifier: Power output required P = 100 mW, V_CC = 6 V, V_CE1 = V_CE2 = 50% V_CC = 3 V. Let R_L = 8 Ω. Then P = I_C² R_L = 100 mW, so I_C = 112 mA. From the circuit: V_CC = I_C R_C + V_CE, i.e. R_C = (V_CC − V_CE) / I_C = 26.7 Ω ≈ 27 Ω with 1/2 W. Let C_C = 10 µF.

Class AB power amplifier: Power output required P = 100 mW, V_CC = 12 V, β = 50, V_CE1 = V_CE2 = 50% V_CC = 6 V. Let R_L = 8 Ω. Then P = I_C² R_L = 100 mW, so I_C = 112 mA. From the circuit: R_C = (V_CC − V_CE) / I_C = 53.5 Ω ≈ 56 Ω with 1/2 W. From the input side: V_CC = 2V_R + 2V_D, so V_R = (V_CC − 2V_D) / 2 = 5.4 V. I_B = I_C / β = 2.24 mA. Assuming 10 I_B flows through R: R = V_R / 10 I_B = 241 Ω ≈ 220 Ω. Let C_C = 10 µF.

RC Coupled Amplifier

Choose NPN transistor BC548. Let V_CC = 9 V, I_C = I_E = 2 mA, β = h_fe = 200, S = 5. To fix the operating point in the middle of the load line, assume 50% of V_CC is dropped across V_CE. For stability, assume 10% of V_CC is dropped across R_E.

V_CE = 50% of V_CC = 4.5 V, V_RE = 10% of V_CC = 0.9 V, R_E = V_RE / I_E = 0.9 / (2 × 10^-3) = 450 Ω ≈ 470 Ω. V_RC = V_CC − V_CE − V_RE = 3.6 V, R_C = V_RC / I_C = 3.6 / (2 × 10^-3) = 1.8 kΩ.

Assume stability factor S = 5: S = (1 + β)(1 + R_B/R_E) / (1 + β + R_B/R_E), giving R_B = 1.93 kΩ.

V_R2 = V_BE + V_RE, and V_R2 = V_CC (R₂ / (R₁ + R₂)) = V_CC (R_B / R₁). Thus 1.5 = 9 (1.93 / R₁), R₁ = 12 kΩ, and R₂ = 2.2 kΩ.

Coupling and bypass capacitors: Minimum operating frequency f = 50 Hz. X_CC = 10% R_C = R_C / 10 = 180 Ω, C_C = 1 / (2π f X_CC) = 1 / (2π × 50 × 180) = 17.6 µF ≈ 22 µF. X_CE = 10% R_E = R_E / 10 = 47 Ω, C_E = 1 / (2π × 50 × 47) = 67 µF ≈ 100 µF.