Amplifier Assignment
Tutorial 1 – Sept. 13, 14, and 18, 2006 The following pertain to the 7 questions on amplifiers. 1. +/- 10V supply, linear VTC through (0,0); output saturates at +7V and –9V, 50V/V gain (= slope!) for largest output signal, output centered halfway between saturation points -1V ouput -1V = 50 x Input
Input (operating point) = -20mV
output magnitude = output range/2 = 8V
2. no-load voltage gain: AVo = 100V/V loaded gain: vout = AVi vin (RL / (RL + Rout)) = AV vin AV = 70V/V = 100 V/V (1k/ (1k + Rout)) 1k + Rout = 10/7 x 1k Rout = 3/7 x 1k = 428.6
3. 1667 V/V = AVo (Rin / (Rin + 10k)) Rin || Rin = Rin/2 is the new input impedance 909 V/V = AVo (Rin / (Rin + 2 x 10k)) divide the two : 1667/909 = (Rin + 20k)/ (Rin + 10k) Rin = 1.99k
4. Want to minimize loading (voltage dividers) – match loads Av = (10 V/V)2 x (1/2)3 = 100/8 = 12.5V/V
5. VOL = 0 + 0.5 = 0.5V VOH = 5 – 1.5 = 3.5V Width of transition region = 3V / (10V/V) = 0.3V VIL = 2.5 – 0.15V = 2.35V VIH = 2.5 + 0.15 = 2.65V NMH = VOH – VIH = 0.85V, NML = VIL – VOL = 1.85V
Double the width of the transition region:
VIL = 2.2V, VIH = 2.8V NMH = 0.7V, NML = 1.7V
6. Transconductance amplifier
G M = GMS ⋅
RIN ROUT ⋅ RIN + RS ROUT + RL
large RIN, large ROUT needed to maximize GM
Transresistance amplifier
RM = RMO ⋅
RS RL ⋅ RIN + RS ROUT + RL
small RIN, small ROUT needed to maximize RM
7. ROUT’ = vx/ix
3 equations: vin = v x
vin = (i x − i Rout ) ⋅ (RS R IN ) i Rout =
v x − AVO vin ROUT
Solve :
i Rout =
v x (1 − AVO ) from 1 and 3 ROUT
Into 2 :
vx = ix
1 RS R IN
1 A 1 + − VO ROUT ROUT
Since RIN is large, RS||RIN ~ RS ROUT is small ROUT’ = vx/ix
the term –AVO/ROUT dominates in the denominator -ROUT/AVO
Input (operating point) = -20mV
output magnitude = output range/2 = 8V
2. no-load voltage gain: AVo = 100V/V loaded gain: vout = AVi vin (RL / (RL + Rout)) = AV vin AV = 70V/V = 100 V/V (1k/ (1k + Rout)) 1k + Rout = 10/7 x 1k Rout = 3/7 x 1k = 428.6
3. 1667 V/V = AVo (Rin / (Rin + 10k)) Rin || Rin = Rin/2 is the new input impedance 909 V/V = AVo (Rin / (Rin + 2 x 10k)) divide the two : 1667/909 = (Rin + 20k)/ (Rin + 10k) Rin = 1.99k
4. Want to minimize loading (voltage dividers) – match loads Av = (10 V/V)2 x (1/2)3 = 100/8 = 12.5V/V
5. VOL = 0 + 0.5 = 0.5V VOH = 5 – 1.5 = 3.5V Width of transition region = 3V / (10V/V) = 0.3V VIL = 2.5 – 0.15V = 2.35V VIH = 2.5 + 0.15 = 2.65V NMH = VOH – VIH = 0.85V, NML = VIL – VOL = 1.85V
Double the width of the transition region:
VIL = 2.2V, VIH = 2.8V NMH = 0.7V, NML = 1.7V
6. Transconductance amplifier
G M = GMS ⋅
RIN ROUT ⋅ RIN + RS ROUT + RL
large RIN, large ROUT needed to maximize GM
Transresistance amplifier
RM = RMO ⋅
RS RL ⋅ RIN + RS ROUT + RL
small RIN, small ROUT needed to maximize RM
7. ROUT’ = vx/ix
3 equations: vin = v x
vin = (i x − i Rout ) ⋅ (RS R IN ) i Rout =
v x − AVO vin ROUT
Solve :
i Rout =
v x (1 − AVO ) from 1 and 3 ROUT
Into 2 :
vx = ix
1 RS R IN
1 A 1 + − VO ROUT ROUT
Since RIN is large, RS||RIN ~ RS ROUT is small ROUT’ = vx/ix
the term –AVO/ROUT dominates in the denominator -ROUT/AVO
