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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.

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Presentation on theme: "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."— Presentation transcript:

1 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

2 Summing Amplifier Circuit yields the weighted sum of different input voltages Each input voltage is connected to the negative input terminal of the op-amp by an individual resistor -+-+ RFRF R4R4 + IFIF I4I4 VoVo R3R3 + I3I3 V3V3 V4V4 R2R2 + I2I2 V2V2 R1R1 + I1I1 V1V1

3 Summing Amplifier KCL current law: I F = I 1 + I 2 + I 3 + I 4 I 1 = V 1 /R 1, I 2 = V 2 /R 2, I 3 = V 3 /R 3, I 4 = V 4 /R 4 I F = -V 0 /R F -V 0 /R F = V 1 /R 1 + V 2 /R 2 + V 3 /R 3 + V 4 /R 4 V 0 = -R F (V 1 /R 1 + V 2 /R 2 + V 3 /R 3 + V 4 /R 4 )

4 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

5 Differential Amplifier Op-amp should amplify V 1 and V 2 equally Above is possible if R 1 = R 2 and R 1F = R 2F V 3 = V 2 {R 2F /(R 2 + R 2F )} I = (V 1 – V 3 )/R 1 = (V 3 – V 0 )/R 1F (V 3 – V 0 ) = (V 1 – V 3 ) R 1F /R 1 V 0 = V 3 (1+ R 1F /R 1 ) - V 1 R 1F /R 1 = V 2 {R 2F /(R 2 + R 2F )} {(R 1 +R 1F )/R 1 } – (V 1 R 1F /R 1 ) =V 2 (R 1F /R 1 ) – (V 1 R 1F /R 1 ) V 0 = (V 2 – V 1 ) (R 1F /R 1 ) VoVo -+-+ R 1F R1R1 I V1V1 V2V2 R2R2 R 2F V3V3 I

6 Comparators Circuit that compares the input voltage with a reference voltage If (V in + V ref ) > 0 – V 0 = -13 V Else V 0 = +13 V VoVo -+-+ R2R2 R1R1 V in V ref R1R1

7 Rectifiers Full-wave rectifier circuit using one op-amp If V in < 0, circuit behaves like an inverting amplifier rectifier with gain = 0.5 If V in > 0, the op-amp disconnects and passive resistor chain yields a gain = 0.5 VoVo -+-+ R F = 1K R 1 =2K V in R L = 3K

8 Effect of Negative Feedback on Output Resistance R oF = R o /(1+βA OL ) For inverting amplifier – β = R 1 /R F For noninverting amplifier – β = R 1 /(R 1 +R F )

9 Typical Bode Plots elexp.com, people.seas.harvard.edu/

10 Bandwidth Limitations f c (corner frequency or break frequency or critical frequency): frequency at which the gain of the op-amp deviates from the passband gain f c frequency at which gain of op-amp has dropped 3 dB from passband gain Midband: range of frequencies from 0 to f c Bandwidth: range of frequencies for which gain of op-amp is within 3 dB of maximum Unity gain: 0 dB gain (numerical gain is 1)

11 Bandwidth Limitations f unity gain = frequency at which A OL = 1 Gain-Bandwidth Product (GBW) = numerical closed loop gain value X frequency For both inverting and noninverting amplifiers – f c = f unity / {1+(R F /R 1 )}

12 Cascaded Amplifiers Total gain = A CL1 A CL2 Overall bandwidth (BW T ) = BW s √(2 1/n -1) – Where n is the number of stages cascaded -+-+ 14.14R R + -+-+ R V in VoVo A1A1 A2A2


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