1. Esercizio n o 2 Si realizzino i seguenti montaggi basati sull’amplificatore operazione 741 e se ne descriva il funzionamento: –amplificatore –amplificatore invertente –integratore o derivatore Si caratterizzino in modo quantitativo gli ampificatori al variare della frequenza del segnale in ingresso Si verifichi la banda passante di uno dei due amplificatori al variare del guadagno

2. Winter 2012 UCSD: Physics 121; 2012 2 Inverting amplifier example Applying the rules:  terminal at “virtual ground”Applying the rules:  terminal at “virtual ground” –so current through R 1 is I f = V in /R 1 Current does not flow into op-amp (one of our rules)Current does not flow into op-amp (one of our rules) –so the current through R 1 must go through R 2 –voltage drop across R 2 is then I f R 2 = V in  (R 2 /R 1 ) So V out = 0  V in  (R 2 /R 1 ) =  V in  (R 2 /R 1 )So V out = 0  V in  (R 2 /R 1 ) =  V in  (R 2 /R 1 ) Thus we amplify V in by factor  R 2 /R 1Thus we amplify V in by factor  R 2 /R 1 –negative sign earns title “inverting” amplifier Current is drawn into op-amp output terminalCurrent is drawn into op-amp output terminal  + V in V out R1R1 R2R2

3. Winter 2012 UCSD: Physics 121; 2012 3 Non-inverting Amplifier Now neg. terminal held at V inNow neg. terminal held at V in –so current through R 1 is I f = V in /R 1 (to left, into ground) This current cannot come from op-amp inputThis current cannot come from op-amp input –so comes through R 2 (delivered from op-amp output) –voltage drop across R 2 is I f R 2 = V in  (R 2 /R 1 ) –so that output is higher than neg. input terminal by V in  (R 2 /R 1 ) –V out = V in + V in  (R 2 /R 1 ) = V in  (1 + R 2 /R 1 ) –thus gain is (1 + R 2 /R 1 ), and is positive Current is sourced from op-amp output in this exampleCurrent is sourced from op-amp output in this example  + V in V out R1R1 R2R2

4. Ref:080114HKNOperational Amplifier4 Noninverting amplifier Noninverting input with voltage divider Voltage follower Less than unity gain

5. Winter 2012 UCSD: Physics 121; 2012 5 Summing Amplifier Much like the inverting amplifier, but with two input voltagesMuch like the inverting amplifier, but with two input voltages –inverting input still held at virtual ground –I 1 and I 2 are added together to run through R f –so we get the (inverted) sum: V out =  R f  (V 1 /R 1 + V 2 /R 2 ) if R 2 = R 1, we get a sum proportional to (V 1 + V 2 ) Can have any number of summing inputsCan have any number of summing inputs –we’ll make our D/A converter this way  + V1V1 V out R1R1 RfRf V2V2 R2R2

6. Winter 2012 UCSD: Physics 121; 2012 6 Differencing Amplifier The non-inverting input is a simple voltage divider:The non-inverting input is a simple voltage divider: –V node = V + R 2 /(R 1 + R 2 ) So I f = (V   V node )/R 1So I f = (V   V node )/R 1 –V out = V node  I f R 2 = V + (1 + R 2 /R 1 )(R 2 /(R 1 + R 2 ))  V  (R 2 /R 1 ) –so V out = (R 2 /R 1 )(V   V  )  + VV V out R1R1 R2R2 V+V+ R1R1 R2R2

7. 7 Differentiator (high-pass) V in V out C R

8. Winter 2012 UCSD: Physics 121; 2012 8 Differentiator (high-pass) For a capacitorFor a capacitor So we have a differentiator, or high-pass filterSo we have a differentiator, or high-pass filter  + V in V out C R

9. Winter 2012 UCSD: Physics 121; 2012 9 Low-pass filter (integrator) I f = V in /R, so C·dV cap /dt = V in /RI f = V in /R, so C·dV cap /dt = V in /R –and since left side of capacitor is at virtual ground: –and therefore we have an integrator (low pass)  + V in V out R C