1. 1.6 Op-Amp Basics High input impedance Low output impedance Made using difference amplifiers having 2 inputs and at least 1 output 1 Note: Terminals for dc supply are omitted

2. 1.6 Op-Amp Basics AC equivalent circuit 2 Ideal Practical Input impedance, R i is typically very high Output voltage, V o is amplifier gain times input signal taken through output impedance, R o which is typically low. Ideal op-amp would have infinite input impedance, zero output impedance and infinite voltage gain.

3. 1.6 Op-Amp Basics Basic Op-amp – constant-gain multiplier 3 The input voltage V 1 is applied to the inverting terminal through R 1. The output voltage V o is fed back to the inverting terminal through R f. The output voltage V o is in antiphase with the input voltae V i The non-inverting terminal is grounded

4. Basic Op-amp – constant-gain multiplier AC equivalent circuit 1.6 Op-Amp Basics 4 Practical op-amp

5.  Basic Op-amp – constant-gain multiplier  If the practical op-amp is replaced by an ideal op-amp, the equivalent circuit becomes; 1.6 Op-Amp Basics 5

6. Basic Op-amp – constant-gain multiplier Redrawn equivalent circuit 6

7. 1.6 Op-Amp Basics 7 Basic Op-amp – constant-gain multiplier By superposition theorem i) V 1 only (set –A v V i = 0) ii) –A v V i only (set V 1 = 0)

8.  Basic Op-amp – constant-gain multiplier 1.6 Op-Amp Basics 8 Solving for V i, gives us;

9.  Basic Op-amp – constant-gain multiplier 1.6 Op-Amp Basics 9 Usually and Then; The ratio V o /V 1 is dependent on the external components R f and R 1 only.

10.  Basic Op-amp – unity gain 1.6 Op-Amp Basics 10 If R f = R 1 ; Shown that the output voltage equal to input voltage with 180° phase inversion.

11.  Basic Op-amp – constant magnitude gain 1.6 Op-Amp Basics 11 If R f = 10R 1 ; Shown that the output voltage is equal to 10X the input voltage with 180° phase inversion.