Recall Last Lecture Introduction to BJT Amplifier Small signal or AC equivalent circuit parameters Have to calculate the DC collector current by performing DC analysis first Common Emitter-Emitter Grounded

STEPS OUTPUT SIDE Get the equivalent resistance at the output side, RO Get the vo equation where vo = - gm vbeRO INPUT SIDE Calculate Ri Get vbe in terms of vi

TYPE 2: Emitter terminal connected with RE – normally ro =  in this type New parameter: input resistance seen from the base, Rib = vb / ib VCC = 5 V RC = 5.6 k 250 k 75 k 0.5 k RE = 0.6 k β = 120 VBE = 0.7V VA =  Voltage Divider biasing: Change to Thevenin Equivalent RTH = 57.7 k VTH = 1.154 V

Perform DC analysis to obtain the value of IC BE loop: 57.7 IB + 0.7 + 0.6 IE – 1.154 = 0 IE = 121 IB 57.7 IB + 0.7 + 0.6 (121IB) – 1.154 = 0 IB = 0.454 / (72.6 + 57.7) = 0.00348 mA IC = βIB = 0.4176 mA Calculate the small-signal parameters r = 7.46 k , ro =  and gm = 16.06 mA/V

0.5 k RTH = 57.7 k RC = 6 k 7.46 k RE = 0.6 k + vi - gmvbe vbe

STEPS OUTPUT SIDE Get the equivalent resistance at the output side, Ro Get the vo equation where vo = - gmvbeRo INPUT SIDE Calculate Rib = r + (1+ ) RE Calculate Ri = Rib // RTH vbe in terms of vi

1. Rout = RC = 6 k 2. Equation of vo : vo = - gmvbeRC = - 96.36 vbe 3. Calculate Rib  Rib = = r + (1+ ) RE = [ 121(0.6) + 7.46 ] = 80.06 k 4. Calculate Ri  RTH // Rib = 33.53 k 5. vbe in terms of vi  KVL vi=vbe+ R E vbe rπ +gmvbe =vbe+0.6 16.194 vbe =10.72 vbe

Equation of vo : vo = - gmvbeRC = - 96.36 vbe vi=vbe+ R E vbe rπ +gmvbe =vbe+0.6 16.194 vbe =10.72 vbe 5. Av vi = vo  open circuit voltage Av vi = - 96.36 vbe = - 96.36 vi = - 8.99 Av = - 8.99  open circuit voltage gain 10.72

To find new voltage gain, vo/vs with input signal voltage source, vs RS = 0.5 kΩ vS vo Ri = 33.53 k 6 k To find new voltage gain, vo/vs with input signal voltage source, vs 6. vi in terms of vs  use voltage divider: vi = [ Ri / ( Ri + Rs )] * vs = 0.9853 vs 7. vo = Avvi  because there is no load resistor vo = -8.99 (0.9853 vs) vo/vs = -8.86

Voltage Divider biasing: Change to Thevenin Equivalent RTH = 10 k VCC = 10 V RC = 2.3 k 20 k 5 k β = 125 VBE = 0.7V VA =  Bypass capacitor Voltage Divider biasing: Change to Thevenin Equivalent RTH = 10 k VTH = 5 V

Perform DC analysis to obtain the value of IC β = 125 VBE = 0.7V VA =  Perform DC analysis to obtain the value of IC BE loop: 10 IB + 0.7 + 5 IE – 5 = 0 IE = 126 IB 10 IB + 0.7 + 5 (126 IB) – 5 = 0 IB = 4.3 / (10 + 630) = 0.00672mA IC = βIB = 0.84 mA Calculate the small-signal parameters r = 3.87 k , ro =  k and gm = 32.3 mA/V

r = 3.87 k , ro =  k and gm = 32.3 mA/V 1. Rout = RC = 2.3 k 2. Equation of vo : vo = - gmvbeRC = - 74.29 vbe 3. Calculate Rib  Rib = = r + (1+ ) RE = [ 126(5) + 3.87 ] = 633.87 k 4. Calculate Ri  RTH // Rib = 9.8 k 5. vbe in terms of vi  KVL vi=vbe+ R E vbe rπ +gmvbe =vbe+5 32.558 vbe =163.79 vbe

Equation of vo : vo = - gmvbeRC = - 74.29 vbe vi=163.79 vbe 5. Av vi = vo  open circuit voltage Av vi = - 74.29 vbe = - 74.29 vi = - 0.4536 Av = - 0.4536  open circuit voltage gain 163.79

To find new voltage gain, vo/vs with input signal voltage source, vs 2.3 k 9.8 k To find new voltage gain, vo/vs with input signal voltage source, vs 6. vi in terms of vs  in parallel vi = vs 7. vo = Avvi  because there is no load resistor vo = - 0.4536 (vs) vo/vs = -0.4536

TYPE 3: With Emitter Bypass Capacitor, CE Circuit with Emitter Bypass Capacitor There may be times when the emitter resistor must be large for the purpose of DC design, but degrades the small-signal gain too severely. An emitter bypass capacitor can be used to effectively create a short circuit path during AC analysis hence avoiding the effect RE

CE becomes a short circuit path – bypass RE; hence similar to Type 1 gmvbe RTH vS vO RC vbe CE becomes a short circuit path – bypass RE; hence similar to Type 1

Voltage Divider biasing: Change to Thevenin Equivalent RTH = 10 k VCC = 10 V RC = 2.3 k 20 k 5 k β = 125 VBE = 0.7V VA =  Bypass capacitor Voltage Divider biasing: Change to Thevenin Equivalent RTH = 10 k VTH = 5 V

Follow the steps 1. Ro = RC = 2.3 k gmvbe RTH = 10 k vS vO RC = 2.3 k 3.87 k vbe Follow the steps 1. Ro = RC = 2.3 k 2. Equation of vo : vo = - (RC ) gmvbe= - 74.29 vbe 3. Calculate Ri  RTH||r = 2.79 k 4. vbe = vi

Equation of vo : vo = - 74.29 vbe vbe = vi 5. Avvi = vo  open circuit voltage Av vi = -74.29 vbe = -73.58 vi Av = -74.29  open circuit voltage gain

To find new voltage gain, vo/vs with input signal voltage source, vs 2.278 k Ri = 2.79 k To find new voltage gain, vo/vs with input signal voltage source, vs 6. vi = vs  in parallel 7. vo = Avvi vo = - 74.29 (vs) vo/vs = -74.29 CONFIGURATION GAIN With RE - 0.4536 With bypass capacitor, CE - 74.29

Voltage Divider biasing: Change to Thevenin Equivalent RTH = 10 k EXAMPLE 2 – with ro VCC = 10 V RC = 2.3 k 20 k 5 k β = 125 VBE = 0.7V VA = 200 V Bypass capacitor Voltage Divider biasing: Change to Thevenin Equivalent RTH = 10 k VTH = 5 V

Perform DC analysis to obtain the value of IC β = 125 VBE = 0.7V VA = 200 V Perform DC analysis to obtain the value of IC BE loop: 10 IB + 0.7 + 5 IE – 5 = 0 IE = 126 IB 10 IB + 0.7 + 5 (126 IB) – 5 = 0 IB = 4.3 / (10 + 630) = 0.00672mA IC = βIB = 0.84 mA Calculate the small-signal parameters r = 3.87 k , ro = 238 k and gm = 32.3 mA/V

Follow the steps 1. Rout = ro || RC = 2.278 k gmvbe RTH = 10 k vS vO RC = 2.3 k 3.87 k vbe 238 k Follow the steps 1. Rout = ro || RC = 2.278 k 2. Equation of vo : vo = - ( ro || RC ) gmvbe= -73.58 vbe 3. Calculate Ri  RTH||r = 2.79 k 4. vbe = vi

Equation of vo : vo = - ( ro || RC ) gmvbe= -73.58 vbe vbe = vi 5. Avvi = vo  open circuit voltage Av vi = -73.58 vbe = -73.58 vi Av = -73.58  open circuit voltage gain

To find new voltage gain, vo/vs with input signal voltage source, vs 2.278 k 2.79 k To find new voltage gain, vo/vs with input signal voltage source, vs 6. vi = vs  in parallel 7. vo = Avvi vo = - 73.58 (vs) vo/vs = -73.58