Audio Amplifier-project Report 26516p

PROJECT REPORT Submitted to: Sir Nouman

Submitted by: Afraz Liaqat

2009-EE-13

Bilal Ahmed

2009-EE-18

Fuad Asim

2009-EE-26

Design Requirements: 1. Rin >= 15k 2. Vout (p-p) max before clipping =20V symmetrical with RL = 100 ohm. 3. Output DC level <=50mV 4. Output DC current <=10mA 5. Av variable from ≈ 0 to 2000. 6. Bandwidth from 10Hz to 20kHz.

Design Description: We have used 4 gain stages of BJT multistage amplifiers with each having a gain of 10. All stages are Common Emitter unbyed and identical. The DC quiescent point (VC) in each stage is set to 1.5V in order to achieve a max peak to peak output. The drop across emitter resistance is 5% of the total voltage and to compensate for the low negative , the Rth of each stage is kept higher than 10RE. For coupling different stages we have used RC coupling method. For deg, DC we have assumed IB = 0. For AC analysis β is taken to be 100.

1st Stage Design: VRE = 5 % of total voltage = 5/100 * 30 = 1.5V Av = 10 As we know, Av = RC / RE = 10 If we take RE = 560 Ω then RC comes out to be 5.6 kΩ As VRE = 1.5 V => VE = 1.5V - 15V = -13.5 V So, VB = VE + 0.7 = -12.8V Now we want Rin ≥ 15k ---------------- (a) We know that Rin = Rth || βRE Then Rth ≥ 24k to satisfy equation (a) Also, VB = R2 * 30 / (R1 + R2) - 15 = -12.8 ------- (1) And Rth = R1 || R2 ----------------------------------- (2) Solving equations (1) & (2) and replacing the resistors with standard values we get, R1 = 300kΩ R2 = 27kΩ RC = 5.6kΩ RE = 560Ω As all our stages are identical, deg each separately is unnecessary.

Coupling Capacitor Deg: We require the lower cutoff fL ≤ 10 Hz. The output impedance of each stage is RC = 5.6k and Rin = Rth || βRE = 17k We design the first stage to have a lower cutoff at 5Hz.

Using the formula fC = 1/(2πRC) where R is the sum of resistance appearing and the two terminals of the capacitor. For 1st stage: C1 = 1/(2π*(17k + 5.6k) * 5) = 1.4 µF Using standard value of 10µF. For 2nd stage we design for 0.5Hz , since R will be same the capacitor comes out to be C2 = 22µF (standard). Similarly, for 3rd stage we design for 0.05 Hz and C3 = 220µF.

Variable gain: Variable gain has been achieved in the circuit with a variable resistor in series with the RE of the last stage. The varistor varies from 0 Ω to 1MΩ, which in turn increases the RE of the last stage and reduces the gain. When the varistor is set to 0Ω, the gain of the last stage is 10 which produces the maximum overall gain. When it is increased to maximum i-e 1MΩ the gain of the last stage is 0.0056, which produces the overall gain ≈ 0.

Output Stage: We have used a CLASS AB AMPLIFIER circuit available on the internet. It consists of a complementary npn and pnp DARLINGTON PAIR. 4 diodes have been used to keep them forward biased in order to rectify the problem of Crossover Distortion.

Gain before output stage = AV1 * AV2 * AV3 * AV4 * Rin /(RS + Rin1) * Rin2/(Rin2 + Rout1) * Rin3 /(Rin3 + Rout2) * Rin4/ (Rin4 + Rout3) = 4019.74

Analysis: Since all our stages are identical we will only analyse 1st stage. We have designed R1 = 300k , R2 = 27k , RC = 5.6k , RE = 560Ω

VB = R2 /(R1 + R2) * 30 – 15 = -12.52 V VE = VB – 0.7 = -13.22V IE = (VE -15)/ IE = 2.36mA IC = IE = 2.36mA VC = 15 – ICRC = 1.784V VCE = VC – VE = 1.784 + 13.22 = 15.004 V

As evident above, the analysis is very close to the designed values.

The WinSpice Code and output are attached along with this report.