Lecture 11 - BJT Common Emitter Amplifier

Review

1. Small Signal Model of the BJT
   1. Input Resistance r_pi = Vt / I_B
   2. Output resistance r_o = V_A / I_C
   3. Transconductance gm = I_C / Vt
2. Simple Amplifier circuit

General Common Emitter Amplifier

1. Collector Resistor tied to positive supply, R_C
2. Emitter bypass capacitor after emitter resistor, R_E, current supply bias
3. Direct coupled (no bypass capacitor) - base is at DC ground
4. DC solution first
   1. I_E = current supply value
   2. I_C = I_E
5. Reduce this circuit to a common amplifier configuration - Compute R_in, R_out, G_M
6. Input Resistance
   1. v_ss = i_ss * r_p + (ß+1) * R_E * i_ss
   2. R_in = v_ss / i_ss = r_p + (ß+1) * R_E
   3. Base resistance reflectance rule - looking into base you find (ß+1) * emitter impedance
7. Output Resistance (measure with input shorted out)
   1. R_out = R_{C || ro}
   2. No current in supply - look at emitter node
8. Voltage Gain
   1. v_o = -(R_{C || ro)} * gm * vp
   2. vp = v_ss * rp / (r_p + (ß+1) * R_E)
   3. A_vo = -gm * (RC || ro) * rp / (r_p + (ß+1) * R_E) = -(R_C||ro) * ß / (r_p + (ß+1) * R_E)
   4. When RE goes to zero, -gm * (RC || ro)
9. Function of RE
   1. Tradeoff input resistance and gain
   2. RE up, Rin up
   3. RE up, Avo down

Common Emitter Amplifier Example

1. R_E1 = 500, R_E2=3.5k, R_E2 bypassed, R_C = 5k, RB1=100k, RB2=50k, ß = 100, V_CC = 15V, VA = 100V
2. Solve the DC bias problem
   1. V_BB = 5V, R_BB = 33k
   2. 5V = I_B * R_BB + 0.7V + (ß+1) I_B R_E
   3. I_E = 1mA
3. Compute the small signal parameters for the transistor
   1. g_m = I_C / Vt = 1mA / 25mV = 40mS
   2. r_p = ß Vt / I_C = 2.5k
   3. ro = VA / IC = 100 / 1mA = 100k
4. Compute the equivalent amplifier model
   1. Input Resistance
      1. R_BB is in parallel with the resistance looking into the base
      2. R_in = R_BB || (r_p + (ß+1) * R_E) = 33k || (2.5k + 101 * 500) = 20k
   2. Output Resistance
      1. R_out = R_{C || ro} = 5k || 100k = 4.8k ~ RC
   3. Voltage Gain (open circuit gain)
      1. A_vo = -R_C / (r_pi / ß+ R_E) = -5k / (2.5k/100 + 500) = -9.5
5. What is the overall voltage gain when driven by a small signal source w/ R_S = 10k, R_L = 10k
   1. A_V = A_vo * R_in / (R_S + R_in) * R_L / (R_out + R_L) = -9.5 * 20 / (20+10) * 10 / (10+5) = -4.25

Conclusions

• Solved the general version of the common emitter amplifier
• Designed a simple common emitter amplifier

Next Time

• Design Example for Common Emitter Amplifier