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 , gm = 30.5 mA/V, ro = 100.88 k

vb β = 100 VBE = 0.7 V VA = 80 V RS = 0.5 k 3.28 k gmvbe RTH = 25 k vS vO 3.28 k 100.88 k vb RE = 2 k RS = 0.5 k Output at emitter terminal gmvbe

Redraw the small signal equivalent circuit so that all signal grounds connected together. x vS vi RS = 0.5 k RTH = 25 k 3.28 k x x vO RE = 2 k x gmvbe

RTH = 25 k vS vO 3.28 k 100.88 k RE = 2 k RS = 0.5 k x gmvbe vi

STEPS OUTPUT SIDE Get the equivalent resistance at the output side, Req At node x, use KCL and get io in terms of vbe where io = gmvbe+ 𝑣𝑏𝑒 𝑟𝜋 Get the vo equation where vo = io Req INPUT SIDE Calculate Rib = r + (1+) Req Calculate Ri = Rib // RTH vbe in terms of vi using supermesh, vi = vbe + vo

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 RTH = 25 k vS vO 3.28 k 100.88 k RE = 2 k RS = 0.5 k x gmvbe vi 1. Get the equivalent resistance at the output terminal, Req  Req = ro ||RE = 1.96 k 2. At node x, use KCL and get io in terms of ib  io = gmvbe+ 𝑣𝑏𝑒 𝑟𝜋 = 30.805 vbe 3. Get the vo equation where vo = io Req  vo = io Req = 60.38 vbe

- 4. Calculate Rib = r + (1+) Req Rib = 3.28+101 (1.96) = 201.24 k RS = 0.5 k x io + vO - vi RTH = 25 k vS Ro gmvbe 4. Calculate Rib = r + (1+) Req Rib = 3.28+101 (1.96) = 201.24 k 5. Calculate Ri  Ri = RTH // Rib = 22.24 k 6. vbe in terms of vi using supermesh vi = vbe + vo = vbe + 60.38 vbe vi = 61.38 vbe

5. Av vi = vo  open circuit voltage Av vi = 60.38 vbe = 60.38 vi = 0.9837 Av = 0.9837  open circuit voltage gain 61.38

To find new voltage gain, vo/vs with input signal voltage source, vs RS = 0.5 kΩ vS vo Ri = 22.24 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 = 0.978 vs 7. vo = Avvi  because there is no load resistor vo = 0.9837 (0.978 vs) vo/vs = 0.962

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.

RTH = 15 k VTH = 1.594 V 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

Perform DC analysis to obtain the value of IC BE loop: 15IB + 0.66 + 1IE – 1.594 = 0 15IB + 0.66 + 1(1+ β)IB = 1.594 15IB + 1(136)IB = 0.934 IB = 0.0062 mA IC = βIB = 0.837 mA Calculate the small-signal parameters r = 4.194 k , gm = 32.19 mA/V, ro = 

2. At node x, use KCL and get io in terms of ib RTH = 15 k 4.19 k RE = 1 k gmvbe r = 4.194 k , gm = 32.19 mA/V vi 1. Get the equivalent resistance at the output terminal, Req  Req = RE = 1 k 2. At node x, use KCL and get io in terms of ib  io = gm𝑣𝑏𝑒 + 𝑣𝑏𝑒 𝑟𝜋 = 32.43 𝑣𝑏𝑒 3. Get the vo equation where vo = io Req  vo = io Req = 32.43 𝑣𝑏𝑒

4. Calculate Rib = r + (1+) Req Rib = 4.19 +136 (1) = 140.19 k RTH = 15 k vi gmvbe RE = 1 k r = 4.194 k , gm = 32.19 mA/V 4. Calculate Rib = r + (1+) Req Rib = 4.19 +136 (1) = 140.19 k 5. Calculate Ri  Ri = RTH // Rib = 13.55 k 6. vbe in terms of vi using supermesh vi = 𝑣𝑏𝑒 + vo = 𝑣𝑏𝑒 + 32.43 𝑣𝑏𝑒 vi = 33.43 𝑣𝑏𝑒

5. Av vi = vo  open circuit voltage Av vi = 32.43 vbe = 32.43 vi = 0.97 Av = 0.97  open circuit voltage gain 33.43

Now, we need to calculate RO Ri = 13.55 k v1 RL = 4 k Now, we need to calculate RO The calculation of RO for common-collector is based on using test voltage, Vx with current Ix

𝑅𝑜= 𝑟𝜋 (1+𝛽) | 𝑅𝐸 |𝑟o + Vx - RTH 1. vbe in terms of vx vbe = - vx the input voltage vi turned off + Vx - RTH 1. vbe in terms of vx 𝑅𝑜= 𝑟𝜋 (1+𝛽) | 𝑅𝐸 |𝑟o vbe = - vx 2. Use nodal analysis Vx + Vx + Vx = gmvbe + Ix ro RE r Vx + Vx + Vx = - gm Vx + Ix

vo / v1 = 0.963 So back to our example, we can now calculate Ro r = 4.194 k , RE = 1 k, RTH = 15 k and ro =  So, Ro = 0.030 k 𝑅𝑜= 𝑟𝜋 (1+𝛽) | 𝑅𝐸 |𝑟o Ro = 0.030 k Ri = 13.55 k v1 RL = 4 k vi in terms of vs vi = v1  in parallel vo = [ RL / ( RL + Ro )] * Avvi  this is because we have load resistor RL vo = [ 4 / ( 4.030)] (0.97 v1) vo / v1 = 0.963

CURRENT GAIN ii io Output side: io = vo / 4 = vo / 4 Ro = 0.030 k ii Ri = 13.55 k RL = 4 k v1 io Output side: io = vo / 4 = vo / 4 Input side: ii = v1 / Ri = v1 / 13.55 Current gain = io / ii = vo (13.55) = 0.963 * 3.3875 = 3.262 v1 (4)

io 235 A ii 72 A Current gain = io / ii = 235/72 = 3.26

AMPLIFIER INPUT RESISTANCE, Ri OUTPUT RESISTANCE, RO Common Emitter Equivalent to Ri as in the steps to find voltage gain Equivalent to Ro as in the steps to find voltage gain Common Collector Use the formula or find Ro using test voltage, Vx with current Ix.

r = 1.12 k , gm = 156.54 mA/V, ro =  Given IC = 4.07 mA, = 175 VA =  vS vo r = 1.12 k , gm = 156.54 mA/V, ro = 

2. At node x, use KCL and get io in terms of ib RTH = 9 k 1.12 k RE = 1 k gmvbe r = 1.12 k , gm = 156.54 mA/V vi 1. Get the equivalent resistance at the output terminal, Req  Req = RE = 1 k 2. At node x, use KCL and get io in terms of ib  io = gmvbe+ 𝑣𝑏𝑒 𝑟𝜋 = 157.43 vbe 3. Get the vo equation where vo = io Req  vo = io Req = 157.43 vbe

4. Calculate Rib = r + (1+) Req Rib = 1.12 +176 (1) = 177.12 k RTH = 9 k 1.12 k RE = 1 k gmvbe r = 1.12 k , gm = 156.54 mA/V vi 4. Calculate Rib = r + (1+) Req Rib = 1.12 +176 (1) = 177.12 k 5. Calculate Ri  Ri = RTH // Rib = 8.565 k 6. vbe in terms of vi using supermesh vi = vbe + vo = vbe + 157.43 vbe vi = 158.43 vbe

vo = 157.43 vbe vi = 158.43 vbe Av vi = vo  open circuit voltage Av vi = 157.43 vbe = 157.43 vi = 0.9937 Av = 0.9937  open circuit voltage gain 158.43

vo / vS = 0.9875 Since we have RL we have to calculate Ro r = 1.12 k , RE = 1 k, RTH = 9 k and ro =  So, Ro = 0.0063 k 𝑅𝑜= 𝑟𝜋 (1+𝛽) | 𝑅𝐸 |𝑟o Ro = 0.0063k Ri = 8.565 k RL = 1 k vS vi in terms of vs vi = vS  in parallel vo = [ RL / ( RL + Ro )] * Avvi  this is because we have load resistor RL vo = [ 1 / ( 1.0063)] (0.9937 vS) vo / vS = 0.9875

CURRENT GAIN ii io Output side: io = vo / 1 = vo / 1 Ro = 0.054 k ii Ri = 8.565 k RL = 1 k vs io Output side: io = vo / 1 = vo / 1 Input side: ii = vS / Ri = vS / 8.565 Current gain = io / ii = vo (8.565) = 0.9875 * 8.565 = 8.46 vS (1)

0.95 mA 0.11 mA