• MOSFET Configurations
• Op Amps
• 741, 356
• Imperfections
• Op‐amp applications
Acknowledgements:
Ron Roscoe,
Neamen, Donald: Microelectronics Circuit Analysis and Design, 3rd Edition
6.101 Spring 2015
Lecture 7
1
MOSFET Configurations
• Common source
• Common Drain ‐ Source follower
• Common gate
6.101 Spring 2015
Lecture 7
2
Basic FET Amplifiers
Common Source Common Drain Common Gate
Ai 
Av  Vo Vi   g m (ro RD )
Av 
RS ro
Ri
)
1
R
R

i
Si
 RS ro
gm
RO 
6.101 Spring 2015
(
IO
g R
RD
)( m Si )
(
Ii
R D  R L 1  g m RSi
Av 
g m ( RD RL )
1  g m RSi
1
RS ro
gm
Lecture 7
3
MOSFET Configuration Summary
Configuration
Common
Source
Source
Follower
Common
Gate
6.101 Spring 2015
Voltage
Gain
Av > 1
Av ≈ 1
Av > 1
Current
Gain
__
__
Ai ≈ 1
Lecture 7
Input
Output
Resistance Resistance
RTH
RTH
Low
Moderate
to high
Low
Moderate
to high
4
JFET Application Current Source
+15V
•
•
•
•
•
Household application: battery charger (car, laptop, mp3 players)
Differential amplifier current source
Ramp waveform generator
High Speed DA converter using capacitors
Simple circuit: 2N5459 Nchannel JFET
2N5459
iD
IDSS = current with VGS=0
VP = pinchoff voltage
 vGS
iD  I DSS 1 
 VP
6.101 Spring 2015



2
5
Op‐Amps
• Active device: V0 = a(V+-V-);
V+
note that it is the difference
V
Vof the input voltage!
• a=open loop gain ~ 105 – 106
• Most applications use negative feedback.
• Comparator: no feedback
• Active device requires power. No shown for
simplicity.
• Classics op-amps: 741, 357 ~ $0.20; one, two or
four in a package.
• Newer op-amps operate at <3.3V (OPA369 1.8V)
o
6.101 Spring 2015
Lecture 7
6
Op‐Amp Packaging
LM324
LF353
6.101 Spring 2015
Lecture 7
7
356 JFET Input Op‐amp
6.101 Spring 2015
Lecture 7
8
JFET Differential Pair
Small Signal Model
6.101 Spring 2015
9
741 Circuit
22 transistors,
11 resistors,
1 capacitor,
1 diode
6.101 Spring 2015
Lecture 7
10
Differential (Emitter Coupled) Pair
BJT Diff Pair
6.101 Spring 2015
Small Signal Model
11
Differential Pair – Common Mode Voltage
Small Signal Model
6.101 Spring 2015
12
Differential Pair – Differential Mode Voltage
6.101 Spring 2015
13
MOSFET Differential Pair
Small Signal Model
6.101 Spring 2015
14
Virtual Node Analysis
+
V2
Vout = a(V2-V1)
a
-
V1
Vx
+
a = gain
β = feedback or loop function

V1  Vx  Vout
•
If a>>1 and a>> β then
•
Current into input terminals zero by
design
•
Typical values:
a~100,000 & a β >>1
•
ok for a=a(s) and β = β(s) as long as
a(s) β(s) >> 1
•
β is the loop transfer function
V1  Vx  aV2  aV1
V1 1  a   Vx  aV2
(not to be confused of β of a BJT)
 1   a 
V1  Vx 

V2  V2

1
a

1
a

 

6.101 Spring 2015
v1 ~ v2
•
Lecture 7
a β is the loop gain
15
Op Amps – Virtual Node
•
•
With negative feedback, output will drive the input voltage difference to zero => V+ = V‐
Input current = 0
Benefits of Feedback
Stabilize gain against device variations, temperature, aging
Reduce distortion by the feedback factor [(1+aβ)]
Input and output impedances adjusted by (1+aβ)
Gain determined by passive components
Disadvantages of Feedback
Loss of gain; need more stages
Greater tendency for instability (oscillations)
 1   a 
  
V2  V2
V1  Vx 
1

a

1

a


 

6.101 Spring 2015
Lecture 7
16
741 Op Amp Max Ratings
common mode voltage
appears at both inputs
Need +Vcc, -Vee for operation
Idiot proof
6.101 Spring 2015
Lecture 7
17
741 Electrical Characteristics
Almost zero
6.101 Spring 2015
Lecture 7
18
LF356 not rail to rail
6.101 Spring 2015
Lecture 7
19
LM6152 – Rail to Rail Output
6.101 Spring 2015
Lecture 7
20
Decibel (dB)
 Vo 
dB  20 log 
 Vi 
log10(2)=.301
 Po 
dB  10 log 
 Pi 
100 dB = 100,000 = 105
80 dB = 10,000 = 104
3 dB point =
6.101 Spring 2015
half power point
Lecture 7
60 dB =
1,000 = 103
40 dB =
100 = 102
21
741 Open Loop Frequency Gain
6.101 Spring 2015
Lecture 7
22
Non‐Inverting Amplifer
Rs
+
_
vin
3
+15
+
7
2
4
-
v 
6
A
v+
Zero input current;
therefore v  vin
+
-15
v-
vout
R2
_
R1
R1
 vout but v  v
R 1 R 2
so vin 
R1
R1  R 2
vout R1  R 2

vin
R1
or Av  1 
β (not to be confused with β
of a BJT)

for finite A
6.101 Spring 2015
 vout ;
R2
R1
R1
R1  R2
vout
A
 Av 
1  A
vin
23
741 Open Loop Frequency Gain
Rs
+
_
vin
3
+
7
6
A
v+
2
-
4
+
-15
v-
vout
R2
R1
Examples at 1 Hz, 1000 Hz, and 10kHz
+15
Voltage gain Av=40dB = 100; R2= 100k,
R1= 1k; [101 = 40.1dB!] β=0.01
_
At 1 Hz, Avol = 100 dB = 1 x 105 = 100,000.
A
105
10 5
Av 


 100  40dB
1  A 1  10 5.01 10 3
At 1000 Hz, Avol = 60 dB = 103 = 1000.
A
103
103
1000
Av 



 90.9  39.2 dB
3
1  A 1  10 .01 1  10
11
At 10 kHz, Avol = 42 dB = 1.26 x 102 = 126.
Av 
A
126
126
126



 55.8  34.9dB
1  A 1  126.01 1  1.26 2.26
β is the loop transfer function
aβ is the loop gain
6.101 Spring 2015
Lecture 7
24
741 vs 356 Comparison
741
356
BJT
JFET
0.5uA
0.0001uA
Input resistance
0.3 MΩ
106 MΩ
Slew rate*
0.5 v/μs
7.5 v/μs
1 Mhz
5 Mhz
continuous
continuous
Input device
Input bias current
Gain Bandwidth product
Output short circuit duration
Identical pin out
* comparators have >50 v/μs slew rate
6.101 Spring 2015
Lecture 7
25
So Why BJT in Op‐amps?
BJTs have higher transconductance (gain), better
consistency in spec between pieces, and in some
applications, lower noise than FETs.
Like most JFET op amps, the LF356 has a relatively
high offset voltage, and relatively high drifts. BJT
op-amps tend to have much lower offset voltage and
drifts.
6.101 Spring 2015
26
Gain Bandwidth Product = Constant
(No free lunch)
Gain: 60dB = 103
Bandwidth = 5x103
Gain Bandwidth product = 5x106
6.101 Spring 2015
Lecture 7
27
Op‐Amp Imperfections – Real World
•
•
•
•
•
•
•
•
•
6.101 Spring 2015
Input offset voltage
Input Current Bias
Input Offset Current
Finite Output Voltage Swing
Finite Current
Finite Gain, gain bandwidth product
Voltage Noise – Johnson Noise
Phase Shifts
Slew Rate
Lecture 7
28
Input Offset Voltage *
741: 6000μv
357: 10,000μv
* Analog Devices MT-037 Tutorial
6.101 Spring 2015
Lecture 7
29
Offset Adjustments
6.101 Spring 2015
Lecture 7
30
Input Bias Current *
The input offset current, IOS,
is the difference between
IB– and IB+, or IOS = IB+ − IB–.
741: 200na
357: 0.05na
* Analog Devices MT-038 Tutorial
6.101 Spring 2015
Lecture 7
31
Inverting Amplifier Bias Current Compensation
I IN  I B  I F  0
RF
IIN
IB
RIN
+
_
VIN
IB
+
VOFF
IF
-2
V OFF   R1 I B
 V OUT
V IN  V OFF
V
 I B  OFF
0
R IN
RF
+15
7
A
6
3
VOUT
-15
R1
but with no input signal, V IN and we want V OUT  0, so :
+
4
+
_
_
 1
 1
1 
1 


- I B  V OFF 
 ; - I B   R1 I B 

RF 
RF 
 R IN
 R IN
 1
 1 
1 
thus : 

    as a condtion for no offset at V o
RF 
 R1 
 R IN


V OUT   I B R F // R IN  I B R1 AVOL
IB -
VDIFF
RF//RIN
R1
IB
+
+15
V OUT  0 if R F // R IN  R1
AVOL
-15
+
VOUT
_

V OUT   I B R F // R IN 

I B R1 AVOL
V OUT  0 if R F // R IN  R1
6.101 Spring 2015
Lecture 7
32
Common Mode Rejection Ratio CMRR
CMRR: ratio of the commonmode gain to differential-mode
gain.
Example, if a differential input
change of Y volts produces a
change of 1 V at the output, and
a common-mode change of X
volts produces a similar change
of 1 V, then the CMRR is X/Y.
CMRR often expressed in dB:
CMRR  20 log
6.101 Spring 2015
Lecture 7
AOL
ACM
33
Inverting Amplifer – Virtual Ground Analysis
Assumptions
Infinite input impedance: i  0;
v  0 because v is grounded .
A 
A  A
i  0
V

R
R
f
in
Rf
if
iin
2
Rin
+
_
vin
-
iin  i f  0
+15
A
3
+
vin  0 0  vout

0
Rin
Rf
7
6
4
-15
+
vout
vout  R f

 Av
vin
Rin
_
6.101 Spring 2015
Lecture 7
34
Schmitt Trigger
Vin
V-
Vo
•
Schmitt trigger have
different triggers
points for rising edge
and falling edge.
•
Can be used to
reduce false triggering
•
This is NOT a
negative feedback
circuit.
V+
R2
R1
6.101 Spring 2015
Lecture 7
35
Schmitt Trigger + RC Feedback = Oscillator
R3
10K
V-
0.33uf
+
•
741 op‐amp. •
R1=10k, R2=4.7k, R3=10K, C=.33uf
•
Display V‐ and Vout on the scope. Set R3=4.7k. Predict what happens to the frequency.
Vout
R1 10K
R2
6.101 Spring 2015
4.7K
Lecture 7
36
High Pass Filter HPF
AV (dB)
C
V1
R
V2
0
-3dB
slope = 6 dB / octave
slope = 20 dB / decade
log f
fLO or f-3dB
Av 
V2
R
j CR
s CR



V1 R  1
j CR  1 s CR  1
j C
Degrees
PHASE LEAD
90o
45o
0o
-45o
log f
fLO or f-3dB
6.101 Spring 2015
Lecture 7
37
Differentiator Insights
AV (dB)
0
-3dB
slope = 6 dB / octave
slope = 20 dB / decade
log f
fLO or f-3dB
Degrees
 R2
1
R1 
sC
Av 
at low frequency
Av


 sCR2
; s  j ;
sCR1  1
PHASE LEAD
90o
45o
0o
sCR1  1
-45o
log f
fLO or f-3dB
 sCR2
1
multiplying by s equals differentiation
6.101 Spring 2015
differentiation works only at f << flo
Lecture 7
38
Low Pass Filter LPF
AV (dB)
R
V1
C
V2
0
-3dB
slope = -6 dB / octave
slope = -20 dB / decade
log f
fHI or f-3dB
1
Av 
 j XC
V2


V1 R   j X C
1
Av 
sRC  1
j C
1

1
j RC  1
R
j C
Degrees
PHASE LAG
0o
-45o
-90o
fHI or f-3dB
6.101 Spring 2015
Lecture 7
log f
39
Integrator Insights
AV (dB)
0
-3dB
slope = -6 dB / octave
slope = -20 dB / decade
log f
fHI or f-3dB
Degrees
PHASE LAG
0o
-45o
R2
Av
 
R1
; s  j ;
1  sCR2
sCR2  1
at high frequency
R2
Av


R1
sCR2

-90o

1
sCR1
fHI or f-3dB
integration works only at f >> fHI
log f
sCR2  1
dividing by s equals integration
6.101 Spring 2015
Lecture 7
40
Why R2?
Without R2, any DC bias
current will saturate Vout
since the DC gain is the open
loop gain
6.101 Spring 2015
Lecture 7
41
Frequency Domain Insight
•
Integration and differentiation easy
to understand in time domain
•
In frequency domain, difference
between square wave and triangle
wave is amplitude and phase –
same harmonics.
•
Integrator (LPF) rolls off
harmonics and phase shift to
create a triangle wave
•
Differentiator (HPF) amplifies
harmonics and phase shift to
create a square wave.
6.101 Spring 2015
Lecture 7
42
Basic OpAmp Circuits
Non-inverting
Voltage Follower (buffer)
vin +
vin 

vout
-
vout 
Differential Input
vin 2
R1
R2
R1
R1  R2
R1
vin
Integrator
vout
vout 
R2
R1
vout  vin
vin1
vout
vin
C
R
vout
R2
R2
R1
v
in 2
 vin1 
vout  
1
RC

t

vin dt
43
Crossover Distortion (hole)
10k
+15
+15
10k
2N3904
+15
10k
2N3904
+15
0.1F
-
2
3
+
0.1F
vout
7
-
2
6
LF356
vin
10k
RL
0.1F
?k
vin
-15
2N3906
[a]
6
LF356
4
3
+
4
RL
0.1F
-15
?k
[b]
-15
vout
7
2N3906
-15
Why is [b] better?
6.101 Spring 2015
Lecture 7
44
Diode Biasing
+12 V
+12 V
RB1
2N2219
1N4001
RE=5.6
1/2 watt
D1 IN914
C
+
vin
_
[From
Preamplifier]
2
-
7
6
?opamp
3
+
D2 1N914
+12
RE=5.6
1/2 watt
2N2905
+
vout
RL
4
-12
_
RB2
1N4001
-12 V
-12 V
RF
6.101 Spring 2015
Lecture 7
45