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 V0 does not change with load

• Open-loop gain A ∞– Vd = V+ - V- = V0 / A

– Since circuit is operated in linear stable mode, V0 must be finite voltage (usually 13 V)

– As A ∞, Lim Vd = Lim (V0/A) = 0

– Vd0; 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 AOL

• Feedback factor: β (<1)• Vo = VidAOL

• Vf = βVo

• Vid = Vin – Vf = Vin – βVo

• Vo = AOL (Vin – βVo)

• Vo = AOLVin/(1+βAOL)

• ACL = Vo /Vin = AOL/(1+βAOL)

VoAOL

Vf

β

Vin Vid+

-

Σ

Noninverting AmplifierUnder stable linear operation– AOL= ∞, Rin= ∞; iin =0 and i2=0,

– Vid = 0; Vin = Vf

– Vf = Voβ = Vo [R1/(R1+RF)]

– Vin = Vo [R1/(R1+RF)] – Closed loop voltage gain of

circuit ACL = Vo/Vin = (R1+RF)/R1

Vo

-

+

RF

Vf

R1

Vin

Vid

i2

iin

Problems

• Assume an ideal noninverting op-amp. Find ACL and Vo if RF = 100 K-Ω, R1 = 1 K- Ω, and Vin = +20 mV– ACL = 1+(RF/R1) = 1+(100/1)= 101– Vo = ACL Vin = 101 x 20 mV = 2.02 V

• Determine voltage gain and Vo if RF = 100 K-Ω, R1 = 1 K-Ω, Vin = +20 mV, and AOL = 100,000– β = 1000 / (1000+100,000) = 9.9 x 10-3

– ACL = AOL/(1+βAOL) = 100,000/(1+ 990)= 100.9– Vo = ACL Vin = 20 mV x 100.9 = 2.018 V

Noninverting Op-amp: Input Resistance

• Typical input resistance: > 1 MΩ (can be increased using negative feedback)

• Rin input resistance of op-amp without feedback

• RinF input resistance of op-amp with feedback

• RinF = Vin/iin

• iin = Vid/Rin

• RinF = VinRin/Vid

• RinF = Rin (1+βAOL)

Inverting Configuration• Vin has opposite polarity from Vo

(180° phase reversal)• ACL = - AOL/(1+β+βAOL)• β = R1/RF; β << βAOL

• ACL = - AOL/(1+βAOL)• -1/ACL = (1+βAOL)/ AOL

= (1/AOL) + (β/1) = (1/AOL) + (R1/RF)

• AOLvery high, -1/ACL = R1/RF

• ACL = -RF/R1

VoAOL

Vf

β

Vin Vid-

-

Σ

Vo

-

+

RF

R1

Vin

Inverting AmplifierUnder stable linear operation– AOL= ∞, Rin= ∞– Vo= AOL (Vin(+) – Vin(-))

– Vid = (Vin(+) – Vin(-)) = Vo/AOL = 0 V

– I1 = Vin/R1

– IB(+) = IB(-) = 0

– IF = -I1

– Vo= IFRF = -I1RF = -VinRF/R1

– Closed loop voltage gain of circuit ACL = Vo/Vin = -(Rf/Ri)

Vo

-

+

RF

R1

Vin

+ -IFI1

Vid

IB(+)

IB(-)

Virtual ground

Inverting Op-amp: Input Resistance

• Input resistance in inverting mode is quite low• RinF = R1

• Miller input resistance R’F

• R’F = RF/(AOL+1)

Voltage Follower• Special case of noninverting amplifier– Rf=0, Ri=∞

– Closed loop voltage gain of circuit ACL = 1+(Rf/Ri) =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 ACL = 1• No drop across R

-

+ACL =1

R

-

+ACL =1

R