Feedback (continued)
From Chapter 12 – Section 12.6
K
Feedback Lecture
1
Feedback & Op-Amps Go Together
Ideal Op-Amp has
A   (infinite)
Rin   (infinite)
Rout  0
vout
R1
 1
vin
R2
Op-Amps make the use
of feedback easy!
Feedback Lecture
2
Analysis Methodology
1.Identify the forward network
2.Identify the feedback network
3.Break the feedback network
4.Calculate the open-loop parameters
5.Determine the feedback factor
6.Calculate the closed-loop parameters
Feedback Lecture
3
Breaking a Loop
Loading input &
output nodes must
be accounted for.
How do we
Terminate?
• The correct way to break a loop is to do it so that the loop does not
know it has been broken. The feedback network is duplicated at both
input and output – this allows the correct loading to be placed at both
the input and output.
Feedback Lecture
4
Rules for Breaking the Loop of Amplifier Types
Sense-Return
Voltage-Voltage
Voltage-Current
Current-Current
Current-Voltage
Feedback network K is duplicated at both the input and output.
Feedback Lecture
5
Example: Voltage-Voltage Feedback
Input of Network K
sees zero impedance
Output of Network K
sees large impedance
Return
Sense
• Ideally, the input of the feedback network sees zero impedance (Zout
of an ideal voltage source), the return replicate needs to be grounded.
Also, the output of the feedback network sees an infinite impedance
(Zin of an ideal voltage sensor), the sense replicate needs to be open.
• Analogous arguments apply to the three other types.
Feedback Lecture
6
Example: Voltage-Voltage Feedback
Input of Network K
sees zero impedance
Output of Network K
sees large impedance
Return
Sense
• As step 4 we can calculate the open-loop parameters with the forward
network being correctly loaded from the feedback network (but with
the feedback network broken and not functional).
• Next, we go to step 5 which is to calculate the feedback factor and
then on to step 6 to calculate closed-loop parameters with the
feedback network accounted for.
Feedback Lecture
7
Calculation of Feedback Factor
Voltage-Voltage
Current-Current
Feedback Lecture
Voltage-Current
Current-Voltage
8
Example
Feedback
Network
CG
stage
Av ,open  gm 1  RD
Rin ,open 
1
gm 1
Rout ,open  RD
 R1  R2  
Open-loop gain calculated
with correct loading
 R1  R2 
Feedback Lecture
9
Determining the Feedback factor
K
V2
 R2 /( R1  R2 )
V1
and
Av ,open  gm 1  RD ( R1  R2 ) 
 R2 
Therefore, KAv ,open  gm 1 
  RD ( R1  R2 ) 
 R1  R2 
To get closed-loop gain, Av ,closed  Av ,open /(1  KAv ,open )
Feedback Lecture
10
Closed-Loop Parameters
K
V2
 R2 /( R1  R2 )
V1
and
Av ,open  gm 1  RD ( R1  R2 ) 
gm 1  RD ( R1  R2 ) 
Av ,closed 
R2
1
gm 1  RD ( R1  R2 ) 
R1  R2
R2
(1  KAv ,open )  1 
gm 1  RD ( R1  R2 ) 
R1  R2


R2
1

g
(
R
(
R

R
))


m1
D
1
2
R

R

1
2

( RD ( R1  R2 ))
/(1  KAv ,open ) 


R2
1

g
(
R
(
R

R
))


m1
D
1
2
R1  R2


Rin ,closed  Rin ,open (1  KAv ,open ) 
Rout ,closed  Rout ,closed
1
gm 1
Feedback Lecture
11
Another Example: Transimpedance Amplifier
Voltage Sense – Current Return
Feedback Lecture
12
Transimpedance Amplifier (continued)
AZ , open
 Vout 
RF RD 1


   gm 2  RD 2 RF 

1
 I in  open  RF  gm 1 
1
K
,
RF
so
RZ ,in , open 
1
RF
gm 1
 1  KA
Z , open

 1

RF RD 1
 1

  gm 2  RD 2 RF  
1
 RF  RF  g M

1


RZ , out , open  RD 2 RF
Feedback Lecture
13
Transimpedance Amplifier (continued)
Vout 
 I 
 in  open
 Vout 
 AZ ,closed 




 I in  closed
 Vout 
1 K 



 I in  open 

 RF RD1    g ( R R
 m2 D2 F 
RF  gm11
AZ ,closed 

1  RF RD1
1


g
(
R
R
 m2 D2 F  

RF  RF  gm11

RZ , in ,closed
RZ , out ,closed
 Vout 
 RZ ,in , open /(1  K 
 )
 I in  open
 Vout 
 RZ ,out ,open /(1  K 
 )
 I in  open
Feedback Lecture
14
Another application of feedback:
Electronic Oscillator
Feedback Lecture
15
RC Phase Shift Feedback oscillator
BJT
(or FET)
Feedback Lecture
16
RC High Pass Filter Section
+60
At 60 each RC section has loss of 0.185 (or -14.7 dB)
Feedback Lecture
17
Feedback Network for Oscillator
N=3
f osc
1

2 RC 2 N
where N = number of RC sections
Feedback Lecture
18
Feedback Network & Gain Stage
Any inverting gain
stage will work here
A > 29.2
KA  1
 180 o
Closed-loop gain 
Aopen
1  KAopen
Oscillation criterion: Barkhausen condition
KA  1 and KA  2 n (n  0, 1, 2, 3 ... .)
Feedback Lecture
19
6.5 kHz RC Oscillator using an Op-Amp
Simple and practical implementation of an RC oscillator
Inverting
Op-Amp
Feedback Lecture
20
RC Oscillator Startup
Closed-loop gain 
Aopen
1  KAopen
Initially, loop gain is too large for a steady-state oscillation,
so amplitude builds up to until amplitude compression occurs;
then loop gain falls until the condition KA = 1 is met.
Question: What starts the oscillation to begin?
Feedback Lecture
21
Now what?
Feedback Lecture
22