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 Common Emitter – with RE Voltage gain

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

vb CE becomes a short circuit path – bypass RE; hence similar to Type 1

β = 125 VBE = 0.7V VA = 200 V IC = 0.84 mA Bypass capacitor VCC = 5 V RC = 2.3 k 20 k 0 k RE = 5k Bypass capacitor

Short-circuited (bypass) by the capacitor CE β = 125 VBE = 0.7V VA = 200 V IC = 0.84 mA 10 k RC = 2.3 k 238 k 3.87 k vbe Short-circuited (bypass) by the capacitor CE r =3.87 k , ro = 238 k and gm = 32.3 mA/V

Follow the steps 1. Rout = ro || RC = 2.278 k vbe 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 RC = 2.3 k 238 k 3.87 k vbe 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

Common-Collector (Emitter-Follower) Amplifier

Remember that for Common Collector Amplifier, the output is measured at the emitter terminal. the gain is a positive value

β = 100 VBE = 0.7V VA = 80

Perform DC analysis to obtain the value of IC BE loop: 25IB + 0.7 + 2IE – 2.5 = 0 25IB + 0.7 + 2(1+ β)IB = 2.5 IC = βIB = 0.793 mA Calculate the small-signal parameters r = 3.28 k and ro = 100.88 k

Output at emitter terminal β = 100 VBE = 0.7V VA = 80 Small-signal equivalent circuit of the emitter-follower amplifier Output at emitter terminal

Redraw the small signal equivalent circuit so that all signal grounds connected together. x Vb x RTH = 25 k RS = 0.5 k r = 3.28 k RE = 2 k ro = 100.88 k Vb

STEPS OUTPUT SIDE Get the equivalent resistance at the output side, ROUT At node x, use KCL and get io in terms of ib where io = ib +  ib Get the vo equation where vo = io ROUT INPUT SIDE Find vb in terms of ib using supermesh Calculate Rib – input resistance seen from base: Rib = vb / ib Calculate Ri Get vb in terms of vs. Go back to vo equation and get the voltage gain

Supermesh Supermesh is defined as the combination of two meshes which have current source on their boundary

2. At node x, use KCL and get io in terms of ib vb 1. Get the equivalent resistance at the output terminal, ROUT  ROUT = ro ||RE = 1.96 k 2. At node x, use KCL and get io in terms of ib  io = ib+ib = ( 1+ )ib = 101 ib 3. Get the vo equation where vo = io ROUT  vo = Rout ( 1+ )ib = 197.96 ib

4. Find vb in terms of ib using supermesh: vb = ibr + io(Rout) x vb 4. Find vb in terms of ib using supermesh: vb = ibr + io(Rout) vb = ib (r +101 (1.96)) = 201.24 ib 5. Calculate Rib Rib = vb / ib  201.24 k 6. Calculate Ri  Ri = RTH||Rib = 22.24 k

Small-Signal Voltage Gain 7. Get vb in terms of vs using voltage divider 8. Go back to vo equation and replace where necessary vb = 22.24 22.24 + 0.5 vs vb = 0.978vs Vb vs = 1.02249 vb vo = 197.96 ib but ib = vb / Rib vo = 197.96 (vb / Rib) = 197.96 ( vb) = 0.9837 vb 201.24 AV = vo / vs = 0.9621 vo / vs = 0.9837 vb / 1.02249 vb vo / vs = 0.9621

Output Resistance of a Common-Collector

Steps Turn off independent sources Place test voltage supply, VX having current IX at the output Hence, RO = VX / IX

vx The output resistance, 1. vbe in terms of vx + Vx - + Vx - 3.28 + 0.49 vx vbe = - 0.87 vx 0.49 k 1.96 k

r + RS + Vx - 2. Use nodal analysis 3.77 k 1.96 k 2. Use nodal analysis vbe = - 0.87 vx and gm = 30.5 mA/V - Vx + gmvbe - Vx + Ix = 0 3.77 1.96 - 0.2653 Vx – 26.535 Vx - 0.5102 Vx + Ix = 0 - 0.2653 Vx – 26.535 Vx - 0.5102 Vx + Ix = 0 - 27.3105 Vx + Ix = 0 Ix = 27.3105 Vx 1 = Vx 27.3105 Ix The output resistance, 0.0366 k

Output Resistance The input signal source is short circuited and assume it is an ideal source so RS = 0 The output resistance, 1. vbe in terms of vx + Vx - + Vx - vbe = - vx 1.96 k

+ Vx - r 2. Use nodal analysis - Vx + gmvbe - Vx + Ix = 0 3.28 1.96 1.96 k r 2. Use nodal analysis - Vx + gmvbe - Vx + Ix = 0 3.28 1.96 - 0.3049 Vx – 30.5 Vx - 0.5102 Vx + Ix = 0 - 0.3049 Vx – 30.5 Vx - 0.5102 Vx + Ix = 0 - 31.3151 Vx + Ix = 0 Ix = 31.3151 Vx 1 = Vx 31.3151 Ix The output resistance, 0.0319 k