Ideal Op-Amp Input impedance of op-amp is ∞ – No current flow in or out of input terminals Output impedance of op-amp (with respect to ground) is ‘0’ – Voltage V 0 does not change with load Open-loop gain A  ∞ – V d = V + - V - = V 0 / A – Since circuit is operated in linear stable mode, V 0 must be finite voltage (usually 13 V) – As A  ∞, Lim V d = Lim (V 0 /A) = 0 – V d  0; V + - V - = 0; V + = V -

Negative Feedback Gain might be too high for many applications Gain of op-amp can be reduced using negative feedback Negative feedback can also provide improvements in other amplifier characteristics Negative feedback – Noninverting configuration – Inverting configuration

Noninverting Configuration Open-loop voltage gain A OL Feedback factor: β (<1) V o = V id A OL V f = βV o V id = V in – V f = V in – βV o V o = A OL (V in – βV o ) V o = A OL V in /(1+βA OL ) A CL = V o /V in = A OL /(1+βA OL ) VoVo A OL VfVf β V in V id + - Σ

Noninverting Amplifier Under stable linear operation – A OL = ∞, R in = ∞; i in =0 and i 2 =0, – V id = 0; V in = V f – V f = V o β = V o [R 1 /(R 1 +R F )] – V in = V o [R 1 /(R 1 +R F )] – Closed loop voltage gain of circuit A CL = V o /V in = (R 1 +R F )/R 1 VoVo -+-+ RFRF VfVf R1R1 V in V id i2i2 i in

Problems Assume an ideal noninverting op-amp. Find A CL and V o if R F = 100 K-Ω, R 1 = 1 K- Ω, and V in = +20 mV – A CL = 1+(R F /R 1 ) = 1+(100/1)= 101 – V o = A CL V in = 101 x 20 mV = 2.02 V Determine voltage gain and V o if R F = 100 K-Ω, R 1 = 1 K-Ω, V in = +20 mV, and A OL = 100,000 – β = 1000 / ( ,000) = 9.9 x – A CL = A OL /(1+βA OL ) = 100,000/(1+ 990)= – V o = A CL V in = 20 mV x = V

Noninverting Op-amp: Input Resistance Typical input resistance: > 1 MΩ (can be increased using negative feedback) R in input resistance of op-amp without feedback R inF input resistance of op-amp with feedback R inF = V in /i in i in = V id /R in R inF = V in R in /V id R inF = R in (1+βA OL )

Inverting Configuration V in has opposite polarity from V o (180° phase reversal) A CL = - A OL /(1+β+βA OL ) β = R 1 /R F ; β << βA OL A CL = - A OL /(1+βA OL ) -1/A CL = (1+βA OL )/ A OL = (1/ A OL ) + (β/1) = (1/ A OL ) + (R 1 /R F ) A OL very high, -1/A CL = R 1 /R F A CL = -R F /R 1 VoVo A OL VfVf β V in V id - - Σ VoVo -+-+ RFRF R1R1 V in

Inverting Amplifier Under stable linear operation – A OL = ∞, R in = ∞ – V o = A OL (V in(+) – V in(-) ) – V id = (V in(+) – V in(-) ) = V o /A OL = 0 V – I 1 = V in /R 1 – I B(+) = I B(-) = 0 – I F = -I 1 – V o = I F R F = -I 1 R F = -V in R F /R 1 – Closed loop voltage gain of circuit A CL = V o /V in = -(R f /R i ) VoVo -+-+ RFRF R1R1 V in + - IFIF I1I1 V id I B(+) I B(-) Virtual ground

Inverting Op-amp: Input Resistance Input resistance in inverting mode is quite low R inF = R 1 Miller input resistance R’ F R’ F = R F /(A OL +1)

Voltage Follower Special case of noninverting amplifier – R f =0, R i =∞ – Closed loop voltage gain of circuit A CL = 1+(R f /R i ) =1 – Output voltage follows input voltage – Used where isolation between a source and a load is desired – Used where exact level of the original voltage is to be maintained – Even with series resistance A CL = 1 No drop across R -+-+ A CL =1 R -+-+ R