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

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 57.7 k RC = 6 k 7.46 k RE = 0.6 k + vb -

STEPS OUTPUT SIDE Get the equivalent resistance at the output side, ROUT Get the vo equation where vo = -  ib ROUT INPUT SIDE Calculate Rib = vb / ib : KVL at loop (extra step) Calculate Ri Get vb in terms of vs – eg: using voltage divider. Go back to vo equation and calculate the voltage gain

1. Rout = RC = 6 k 2. Equation of vo : vo = -  ib RC = - 720 ib 3. Calculate Rib  using KVL: ib r + ie RE - vb = 0 but ie = (1+ ) ib = 121 ib so: ib [ 121(0.6) + 7.46 ] = vb  Rib = 80.06 k 4. Calculate Ri  RTH||Rib = 33.53 k 5. vb in terms of vs  use voltage divider: vb = [ Ri / ( Ri + Rs )] * vs = 0.9853 vs vs = 1.0149 vb

AV = vo / vs = - 8.86 so: vs = 1.0149 vb 6. Go back to equation of vo vo = - 720 ib = - 720 [ vb / Rib ] = -720 vb / 80.06 = - 8.993 vb vo / vs = - 8.993 vb / 1.0149 vb vo / vs = - 8.86 AV = vo / vs = - 8.86

Current Gain Output side: io = vo / RC = vo / 6 RS = 0.5 k vs RC = 6k Ri = 33.53 k Output side: io = vo / RC = vo / 6 Input side: ii = vs / (RS + Ri ) = vS / 33.53 Current gain = io / ii = vo (33.53) = -8.86 * 5.588 = - 49.5 vs (6)

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 = 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 = 1.8 / (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 3.74 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 in terms of vs vbe = vs since connected in parallel

6. Go back to equation of vo vo / vs = -73.58 vbe / vbe gmvbe RTH = 10 k vS vO RC = 2.3 k 3.74 k vbe 238 k so: vbe = vs 6. Go back to equation of vo vo / vs = -73.58 vbe / vbe vo / vs = - 73.58 AV = vo / vs = - 73.58

Current Gain Output side: io = vo / RC = vo / 2.3 vs RC = 2.3 k Ri = 2.79 k Output side: io = vo / RC = vo / 2.3 Input side: ii = vs / Ri = vS / ( 2.79) Current gain = io / ii = vo (2.79) = -73.58 * 1.213 = - 89.25 vs (2.3)