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Rin and Rout of the
Inverting Amplifier
Recall that the input resistance of an amplifier is:
Rin 
vin
iin
For the inverting amplifier, it is evident that the input current iin is equal to i1 :
R2
i2
R1
vin
v-
oc
vout
iin  i1
v+
Jim Stiles
+
The Univ. of Kansas
Dept. of EECS
5/1/2017
769821458
2/11
Its input resistance
From Ohm’s Law, we know that this current is:
iin  i1 
The non-inverting terminal is
“connected” to virtual ground:
vin  v1
R1
R2
v  0
i2
and thus the input current is:
iin  i1 
vin
R1
R1
vin
vin
 vin
iin
oc
vout
iin  i1
v+
We now can determine the input
resistance:
Rin 
v-
+
 R1 
   R1
 vin 
The input resistance of this inverting amplifier is therefore Rin  R1 !
Jim Stiles
The Univ. of Kansas
Dept. of EECS
5/1/2017
769821458
3/11
Output resistance is harder
Now, let’s attempt to determine the output resistance Rout.
Recall that we need to determine two values: the short-circuit output current
ioutsc  and the open-circuit output voltage voutoc  .
To accomplish this, we must replace the op-amp in the circuit with its linear
circuit model:
R2
i2
vin
R1
-
v-
op
Rout
ii  i1
+ v
+
Jim Stiles
+
-
vout  0
op
iout
Aop (v   v  )
The Univ. of Kansas
sc
iout
Dept. of EECS
5/1/2017
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First, the short circuit output current
R2
From KCL, we find that:
i2
op
sc
iout
 i2  iout
where:
op
out
i
and:
vin

oc
Aop v   vout
R
op
o

Aop v 
R1
-
v-
op
Rout
ii  i1
+ v
+
R
op
o
+
-
vout  0
op
iout
Aop (v   v  )
sc
iout
oc
v   vout
v
i2 
 
R2
R2
Therefore, the short-circuit output current is:
i
sc
out
Jim Stiles
op
v  Aop v   Rout  R2 Aop

 op  
op
 R2 Rout
R2
Rout


Aop
v    op v 

Rout

The Univ. of Kansas
Dept. of EECS
5/1/2017
769821458
5/11
Now, the open circuit output voltage
R2
The open-circuit output voltage can
likewise be determined in terms of Aop
and v- .
i2
R1
vin
-
v-
ii  i1
+ v
+
iout =0
op
Rout
+
-
op
iout
Aop (v   v  )
+
oc
vout
-
Here, it is evident that since iout = 0:
op
i2  iout
where we find from Ohm’s Law:
i2 
Jim Stiles
v   (  Aopv  )
op
R2  Rout
 1  Aop

op
 R2  Rout
The Univ. of Kansas

v 

Dept. of EECS
5/1/2017
769821458
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The open-circuit output voltage
R2
Now from KVL:
i2
oc
vout
 v   R2 i2
R1
vin
Inserting the expression for i2 :
 1  Aop 
oc
vout
 v   R2 
v
op  
 R2  Rout 
 R  R op R2 1  Aop
out
 2

op
op
 R2  Rout
R2  Rout

 Roop  R2 Aop 

 v
op
 R2  Rout



R2 Aop

v
op
R2  Rout

Jim Stiles
-
v-
ii  i1
+ v
+
  v


iout =0
op
Rout
+
-
op
iout
Aop (v   v  )
+
oc
vout
-

The Univ. of Kansas
Dept. of EECS
5/1/2017
769821458
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Now we find the output resistance
Now, we can find the output resistance of this amplifier:
Rout
oc
vout
 sc
iout
 R2 Aop    Aop 

op  
op 
R

R
R
2
o
o



op
R R
 2 o op
R2  Ro
1
 R2 Roop
In other words, the inverting amplifier output resistance is simply equal to the
op
value of the feedback resistor R2 in parallel with op-amp output resistance Rout
.
Jim Stiles
The Univ. of Kansas
Dept. of EECS
5/1/2017
769821458
8/11
This is zero if the op-amp is ideal
Ideally, of course, the op-amp output resistance is zero, so that the output
resistance of the inverting amplifier is likewise zero:
op
Rout  R2 Rout
 R2 0
0
Note for this case—where the output resistance is zero—the output voltage will
be the same, regardless of what load is attached at the output (e.g., regardless
of iout )!
R2
i2
vin
R1
-
v-
iout  0
ii  i1
+ v
+
+
-
op
iout
Aop (v   v  )
+
R
vout   R2 vin
1
-
Jim Stiles
The Univ. of Kansas
Dept. of EECS
5/1/2017
769821458
9/11
For real op-amps the
output resistance is small
Thus, if Rout  0 , then the output voltage is equal to the open-circuit output
voltage—even when the output is not open circuited:
vout  
R2
v
R1 in
for all iout !!
Recall that it is this property that made Rout  0 an “ideal” amplifier
characteristic.
We will find that real (i.e., non-ideal!) op-amps typically have an output
op
R2 ), so that the inverting amplifier output
resistance that is very small ( Rout
resistance is approximately equal to the op-amp output resistance:
op
Rout  R2 Rout
op
 Rout
Jim Stiles
The Univ. of Kansas
Dept. of EECS
5/1/2017
769821458
10/11
A summary
Summarizing, we have found that for the inverting amplifier:
Rin  R1
op
Rout  Rout
(ideally zero)
Thus, this inverting amplifier…
iin t 
R2
R1
iout t 
-

vin t 

+

vout t 

Jim Stiles
The Univ. of Kansas
Dept. of EECS
5/1/2017
769821458
11/11
The inverting amp equivalent circuit
…has the equivalent circuit:
iout t 
iin t 
op
Rout

vin t 

R1



  v t 
R2
R1

vout t 

in
Note the input resistance and open-circuit voltage gain of the inverting
amplifier is VERY different from that of the op-amp itself!
Jim Stiles
The Univ. of Kansas
Dept. of EECS