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 V
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
vb β = 100 VBE = 0.7 V VA = 80 V RS = 0.5 k 3.28 k RTH = 25 k vS vO 3.28 k 100.88 k vb RE = 2 k RS = 0.5 k Output at emitter terminal
Redraw the small signal equivalent circuit so that all signal grounds connected together. x vS vb RS = 0.5 k RTH = 25 k 3.28 k x x vO RE = 2 k x
RTH = 25 k vS vO 3.28 k 100.88 k vb RE = 2 k RS = 0.5 k x
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 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) 3.28 k RS = 0.5 k x vO io vb RTH = 25 k vS Rout 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
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
Current Gain Output side: io = vo / RE = vo / 2 RS = 0.5 k vs RE = 2 k Ri = 22.24 k Output side: io = vo / RE = vo / 2 Input side: iI = vs / (RS + Ri ) = vS / 22.74 Current gain = io / iI = vo (22.74) = 0.9621* 11.37 = 10.94 vs (2)
The voltage gain of common collector is less than 1. Hence, the common collector doesn’t do any voltage amplification. For that reason, this circuit is sometimes called a voltage follower. But, as you can see, this circuit does have great potential as a current amplifier.
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 3.28 k + Vx - + Vx - Independent voltage source turned off In parallel In parallel The output resistance, 1. vbe in terms of vx 3.28 k + Vx - + Vx - vbe = - 3.28 3.28 + 0.49 vx vbe = - 0.87 vx 0.49 k 1.96 k
r + 0.49 k + Vx - 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
The circuit above is a common collector configuration The circuit above is a common collector configuration. Given = 135 and VBEon = 0.66 V and VA = Calculate the DC collector current, IC Draw the AC equivalent circuit
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