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5/13/2017 769863324 1/7 Analysis of the Inverting Amplifier Consider an inverting amplifier: i2 R2 i1 vin R1 v1 v- i- =0 v+ v2 ideal oc vout + Note that we use here the new notation v v2 and v v1 . Now what is the open-circuit voltage gain of this inverting amplifier? Let’s start the analysis by writing down all that we know. First, the op-amp equation: 5/13/2017 769863324 2/7 oc vout Aop v v Since the non-inverting terminal is grounded (i.e., v+ =0): oc vout Aop v Now let’s apply our circuit knowledge to the remainder of the amplifier circuit. For example, we can use KCL to determine that: i1 i i2 However, we know that the input current i- of an ideal op-amp is zero, as the input resistance is infinitely large. Thus, we reach the conclusion that: i1 i2 Likewise, we know from Ohm’s Law: i1 v1 R1 i2 v2 R2 and also that: And so combining: v1 v 2 R1 R2 5/13/2017 769863324 3/7 Finally, from KCL we can conclude: vin v1 v v1 vin v In other “words”, we start at a potential of vin volts (with respect to ground), we drop a potential of v1 volts, and now we are at a potential of v volts (with respect to ground). i2 R2 i1 vin R1 v1 v- i- =0 v+ v2 oc vout ideal + Likewise, we start at a potential of of v volts (with respect to ground), we drop a potential of v 2 volts, and now we are at a oc potential of vout volts (with respect to ground). oc v v2 vout oc v2 v vout 5/13/2017 769863324 i2 R2 i1 vin R1 v1 v- i- =0 v+ 4/7 v2 ideal oc vout + Combining these last three equations, we find: oc vin v v vout R1 R2 Now rearranging, we get what is known as the feed-back equation: oc R2 vin R1 vout v R1 R2 oc Note the feed-back equation relates v in terms of output vout . We can combine this feed-back equation with the op-amp equation: oc vout v Aop 5/13/2017 769863324 5/7 This op-amp equation is likewise referred to as the feedoc forward equation. Note this equation relates the output vout in terms of v . We can combine the feed-back and feed-forward equations to determine an expression involving only input voltage vin and oc output voltage vout : oc oc R2 vin R1 vout vout R1 R2 Aop Rearranging this expression, we can determine the output oc voltage vout in terms of input voltage vin . Aop R2 oc vout v R1 R2 Aop R1 in and thus the open-circuit voltage gain of the inverting amplifier is: oc Aop R2 vout Avo vin R1 R2 Aop R1 Recall that the voltage gain A of an ideal op-amp is very large— approaching infinity. Thus the open-circuit voltage gain of the inverting amplifier is: 5/13/2017 769863324 6/7 Aop R2 Avo lim Aop R R A R 2 op 1 1 R 2 R1 Summarizing, we find that for the inverting amplifier: Avo R2 R1 R oc vout 2 vin R1 One last thing. Let’s use this final result to determine the value of v-, the voltage at the inverting terminal of the op-amp. Recall: oc R2 vin R1 vout v R1 R2 Inserting the equation: R oc vout 2 vin R1 we find: 5/13/2017 769863324 7/7 R R2 vi R1 2 vin R1 v R1 R2 R v R2 vin 2 in R1 R2 0 The voltage at the inverting terminal of the op-amp is zero! Thus, since the non-inverting terminal is grounded (v2 =0), we find that: v v and v v 0 Recall that this should not surprise us, as we know that if opoc amp gain Aop is infinitely large, its output vout will also be infinitely large (can you say saturation?), unless v+ - v- is infinitely small. We find that the actual value of v+ - v- to be: oc vout R2 v v vin Aop Aop R1 a number which approaches zero as Aop !