Transcript Non-inverting amplifier

Electronics in High Energy Physics
Introduction to electronics in HEP
Operational Amplifiers
(based on the lecture of P.Farthoaut at Cern)
1
Operational Amplifiers



Feedback
Ideal op-amp
Applications
–
–
–
–

Non-ideal amplifier
–
–
–
–
–
–


Voltage amplifier (inverting and non-inverting)
Summation and differentiation
Current amplifier
Charge amplifier
Offset
Bias current
Bandwidth
Slew rate
Stability
Drive of capacitive load
Data sheets
Current feedback amplifiers
2
Feedback

Y is a source linked to X
– Y=mx

Open loop
– x=de
– y=mx
– s=sy=sdmx

e
Closed loop
m
d
y
s
s
b
x  de  b y
y  mx  mde  mby
edm
1  bm
esdm
s  sy 
1  bm
s
sdm

e 1  bm
x
y


m is the open loop gain
bm is the loop gain
3
Interest of the feedback
e
x
m
d
s
s
b

In electronics
– m is an amplifier
– b is the feedback loop
– d and s are input and output impedances

If m is large enough the gain is independent of the amplifier
s
sdm sd


e 1  bm b
4
Operational amplifier
-
-A e
100
80
A (dB)
e
120
60
40
20
+
0
1.0E+00
-20
1.0E+01
1.0E+02
1.0E+03
1.0E+04
1.0E+05
1.0E+06
1.0E+07
-40
Frequency (Hz)


Gain A very large
Input impedance very high
– I.e input current = 0

A(p) as shown
5
How does it work?

R2
Direct gain calculation
Vin  e  I R1
Vout   Ae ; Vout  ( R1  R 2) I
-
Vout
A

Vin 1  A R1
R1  R 2

e
Feed-back equation
s
sdm

e 1  bm
R1
; d s 1
R1  R 2
Vout
A

Vin 1  A R1
R1  R 2
I
R1
+
-A e
Vout
m  A; b 

Vin
Ideal Op-Amp
Vout R1  R 2
A  ;

Vin
R1
6
Non-inverting amplifier
R2

Gain
Vin  I R1
Vout  ( R1  R 2) I
Vout R1  R 2

Vin
R1


I
R1
-
Called a follower if R2 = 0
+
Input impedance
Zin  
Vout
Vin
7
Inverting amplifier
R2

Gain
Vin  I R1
Vout   R 2 I
Vout
R2

Vin
R1
I
R1
Vin

Input impedance
Zin  R1

Gain error
+
Vout
Vout
R2

Vin
R1
G
R
2
G
R
G
8
Summation
R

Transfer function
Vi
Ii 
Ri
I   Ii
Vout   R I   R 
I
R1
V1
Vi
Ri
I1
-
Rn
Vn
In

If Ri = R
+
Vout
Vout    Vi
9
Differentiation
R2
I1
R1
Vout   R 2 I1  R 2 I 2  R 2 (I 2  I1)
V1
I1
V1  R1 I1  R1 I 2  V 2 V 2  V1  R1 (I 2  I1)
Vout 
R2
( V 2  V1)
R1
-
R1
V2
I2
+
Vout
R2
10
Current-to-Voltage converter (1)
C
R
Iin
+


Vout
Vout = - R Iin
For high gain and high bandwidth, one has to take into
account the parasitic capacitance
11
Current-to-Voltage converter (2)
R1
R2
r
Iin
+
Vout

High resistor value with small ones

Equivalent feedback resistor = R1 + R2 + R2 * (R1/r)
– ex. R1 = R2 = 100 k ; r = 1 k ; Req = 10.2 M

Allows the use of smaller resistor values with less problems of
parasitic capacitance
12
Charge amplifier (1)
R
1
Vout (p )  
I (p )
Cp
I ( t )  d( t ) ; I ( p )  1
1
1
Vout (p )  
; Vout (t )   (t )
Cp
C

C
I
Requires a device to discharge the capacitor
– Resistor in //
– Switch
+
1.5
Time
0
2
4
6
8
10
1
Input current
0.5
0
RC network
-0.5
-1.5
12
1
Input & Output
Input and Output
1.5
-1
Vout
Capacitor only
Input current
0.5
0
-0.5
0
2
4
6
8
10
12
14
16
-1
Output
-1.5
-2
Time
13
Charge amplifier (2)
R
C
I
+
Input Charge
In a few ns
V1
C1
R2
R1
Output of the charge amplifier
Very long time constant
C2
V2
Shaping
a few 10’s of ns
14
Miller effect

Charge amplifier
–
–
–
–

Vin = e
Vout = -A e
The capacitor sees a voltage (A+1) e
It behaves as if a capacitor (A+1)C was
seen by the input
C
Vin
–Two circuits are equivalent
X
Z
e
A e
Vout
+
Miller’s theorem
–Av = Vy / Vx
-
Y
Y
X
Z1
Z2
»Z1 = Z / (1 - Av)
»Z2 = Z / (1-Av-1)
15
Common mode

The amplifier looks at the difference of the two inputs
– Vout = G * (V2 - V1)

The common value is in theory ignored
– V1 = V0 + v1
– V2 = V0 + v2

In practice there are limitations
– linked to the power supplies
– changes in behaviour

Common mode rejection ratio CMRR
– Differential Gain / Common Gain (in dB)
16
Non-ideal amplifier

Input Offset voltage Vd

Input bias currents Ib+ and IbIb-

Limited gain

Input impedance




Zc
e
Zd
-A e
Output impedance
Common mode rejection
Noise
Bandwidth limitation &
Stability
Zout
+
Vd
Ib+
Zc
17
Input Offset Voltage


“Zero” at the input does not give “Zero”
at the output
In the inverting amplifier it acts as if an
input Vd was applied
R2
I
R1
-
– (Vout) = G Vd

Notes:
– Sign unknown
– Vd changes with temperature and time
(aging)
– Low offset = a few mV and
Vd = 0.1 mV / month
– Otherwise a few mV
Vd
+
Vout
18
Input bias current (1)



(Vout) = R2 Ib(Vout) = - R3 (1-G) Ib+
Error null for
R3 = (R1//R2) if Ib+ = Ib-
R2
Ib-
R1
+
R3
Ib+
Vout
19
Input bias current (2)



In the case of the charge
amplifier it has to be
compensated
Switch closed before the
measurement and to
discharge the capacitor
Values
– less than 1.0 pA for JFET
inputs
– 10’s of nA to mA bipolar
C
Ib-
+
R3
Ib+
Vout
20
Common mode rejection



Non-inverting amplifier
Input voltage Vc/Fr (Vc common mode voltage)
Same effect as the offset voltage
R2
I
R1
Vc/Fr
+
Vout
21
Gain limitation
R2
R2
R1 A
A
 Gi
R1 R1 A  R1  R 2
A  1  Gi
R2
Gi  
R1
1  Gi
G  G  Gi  Gi
if Gi  A
A
G
I
R1
Vin
e
+

-A e
Vout
A is of the order of 105
– Error is very small
22
Input Impedance
R2
Zc-
R1
Zd
+
Vin
Vout
Zc+

Non-inverting amplifier

Zin = Zc+ // (Zd A / G) ~ Zc+ G= (R1+R2)/R1
23
Output impedance

R2
Non-inverting amplifier
Vout
Zout 
when Vin  0
Iout
Vout  e G  - Aee - Zo Io;
Vout  (R1  R2) (Io  Iout)
G
Zout  Zo
A
R1  R2
G
R1
I0 + Iout
R1
e
-A e
I0
Iout
Z0
+
Vout
24
Current drive limitation
Maximum
Output
Swing
R2
R1
I
+
Vout
RL
Vin



RL*Imax
RL
Vout = R I = RL IL
The op-amp must deliver I + IL = Vout (1/R + 1/RL)
Limitation in current drive limits output swing
25
Bandwidth
f3db= fT/G
120
100
Gain [dB]
80
fT
60
40
20
0
-20
1.0E+00
-40
1.0E+01
1.0E+02
1.0E+03
1.0E+04
1.0E+05
1.0E+06
1.0E+07
Frequency

Gain amplifier of non-inverting G(p) = G A(p) / (G + A(p))
– A(p) with one pole at low frequency and -6dB/octave
» A(p) = A0 / (p+w0)
– G = (R1+R2)/R1 40 dB
– Asymptotic plot
» G < A G(p) = G
» G > A G(p) = A(p)
26
Slew Rate
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0




0.5
1
1.5
2
2.5
3
3.5
Limit of the rate at which the output can change
Typical values : a few V/ms
A sine wave of amplitude A and frequency f requires a slew rate of
2pAf
S (V/ms) = 0.3 fT (MHz); fT = frequency at which gain = 1
27
Settling Time
1.4
Amplitude
1.2
1
0.8
0.6
0.4
0.2
0
0
5
10
15
20
Time

Time necessary to have the output signal within accuracy
– ±x%

Depends on the bandwidth of the closed loop amplifier
– f3dB = fT / G

Rough estimate
– 5 t to 10 t with t = G / 2 p fT
28
Stability

Unstable amplifier
G(p) = A(p) G / (G + A(p))
– A(p) has several poles

If G = A(p) when the phase shift is 180o
then the denominator is null and the
circuit is unstable
Simple criteria
– On the Bode diagram G should cut A(p)
with a slope difference smaller than -12dB
/ octave
– The loop gain A(p)/G should cut the 0dB
axe with a slope smaller than -12dB /
octave

Phase margin
– (1800 - Phase at the two previous points)

The lower G the more problems
120
100
80
-12 dB/octave
60
Gain [dB]

40
20
0
-20
-12 dB/octave
-40
-60
-80
1.0E+00
1.0E+01
1.0E+02
1.0E+03
1.0E+04
1.0E+05
1.0E+06
Frequency
- Open loop gain A(p)
- Ideal gain G
- Loop gain A(p)/G
29
Stability improvement
120
120
100
100
80
80
Gain [dB]
Gain [dB]
60
40
20
0
-6 dB/octave
-20
60
40
20
0
-40
-20
-60
-40
-80
1.0E+00
1.0E+01
1.0E+02
1.0E+03
1.0E+04
1.0E+05
1.0E+06
Frequency
Compensation

-60
1.0E+00
-6 dB/octave
1.0E+01
1.0E+02
1.0E+03
1.0E+04
1.0E+05
1.0E+06
Frequency
Pole in the loop
Move the first pole of the amplifier
– Compensation

Add a pole in the feed-back

These actions reduce the bandwidth
30
Capacitive load

Buffering to drive lines
R2
R1
-
C = 20 pF
10
C Load = 0.5 mF
+

The output impedance of the amplifier and the capacitive contribute to
the formation of a second pole at low frequency
– A’(p) = k A(p) 1/(1+r C p) with r = R0//R2//R
– A(p) = A0 / (p+w0)

Capacitance in the feedback to compensate
– Feedback at high frequency from the op-amp
– Feedback at low frequency from the load
– Typical values a few pF and a few Ohms series resistor
31
Examples of data sheets (1)
32
Examples of data sheets (2)
33
Current feedback amplifiers
e
-
-A e
Zt ie
ie
+

Voltage feedback
+


Current feedback
Zt = Vout/Ie is called the transimpedance
gain of the amplifier
34
Applying Feedback
R2
R1
Vin  ( I  Ie ) R1
I
-
Zt ie
Vout  ( R1  R 2 ) I  R1 Ie
Vout  Zt Ie
Vout R1  R 2 1
R1  R 2


if Zt  
Vin
R1 1  R 2
R1
Zt
ie
+
Vout
Vin


Non-inverting amplifier
Same equations as the voltage feedback
35
Frequency response
R2
R1
Vout R1  R 2 1

Vin
R1 1  R 2
Zt
Z0
Zt 
pw
Vout R1  R 2
1

Vin
R1 1  R 2( p  w)
Z0

I
-
Zt ie
ie
+
Vout
Vin
The bandwidth is not affected by the gain but only by R2
– Gain and bandwidth can be defined independently

Different from the voltage feedback
– f3dB = fT / G
36
Data sheet of a current feedback amplifier
37
Data sheet of a current feedback amplifier (cont’)

Very small change of bandwidth with gain
38
Transmission Lines





Lossless Transmission Lines
Adaptation
Reflection
Transmission lines on PCB
Lossy Transmission Lines
39
Lossless transmission lines (1)


L,C per unit length x
Impedance of the line Z
Z  Lx p 
Z
ZCx p  1
L
0
C
L
x  0; Z2 
C
L
Z
C
Z 2  ZLx p 

Pure resistance
Lx
Cx
Lx
Cx
Z
40
Lossless transmission lines (2)

Propagation delay
V2  V1  Lx p I  V1  Lx p
V1
 V1 (1  LC xp)
Z
1
1
After unity length ( cells) V2  V1 (1  LC x p ) x
x
x  0 ; V2  V1 e  LCp
V2 (t )  V1 (t  t ) (t  t ) ; t  LC

Pure delay
Lx
V1
Cx
I
V2
Z
41
Lossless transmission lines (3)
Z

Characteristic impedance pure resistance

Pure delay

Capacitance and inductance per unit of length

Example 1: coaxial cable
L
C
t  LC
L  Zt
t
C
Z
– Z = 50 
– t = 5 ns/m
– L = 250 nH/m; C = 100 pF/m

Example 2: twisted pair
– Z = 100 
– t = 6 ns/m
– L = 600 nH/m ; C = 60 pF/m
42
Reflection (1)

Source generator
Zs
Zo
– V, Output impedance Zs

Line appears as Z0
Is  V


1
Z0
; Vs  V
ZS  Z 0
ZS  Z0
V
Vs
Is
ZL
All along the line Vs = Z0 Is
If the termination resistance is ZL a reflection wave is generated
to compensate the excess or lack of current in ZL
VL  Z L I L
VR  Z 0 I R
VL  Vs  VR
IL  Is  I R

The reflected wave has an amplitude VR  Vs
Z L   ; VR  Vs
ZL  Z0
ZL  Z0
Z L  0 ; VR   Vs
43
Reflection (2)
The reflected wave travels back to
source and will also generate a
reflected wave if the source
impedance is different from Z0
1.2
1
0.8
Volt

– During each travel some amplitude
is lost

V
ZS = 1/3 Z0
ZL = 3 Z 0
0.6
0.4
Vs
VL
0.2
0
The reflection process stops when
equilibrium is reached
0
5
10
15
20
25
Time
– VS = VL
1.2

Zs < Z0 & ZL > Z0
Dumped oscillation
Zs > Z0 & ZL > Z0
Integration like
1
ZS = 3 Z 0
ZL = 3 Z 0
0.8
Volt

0.6
V
Vs
VL
0.4
0.2
0
0
5
10
15
20
Time
44
Reflection (3)
1.2
Adaptation is always better
– At the destination: no
reflection at all
– At the source: 1 reflection
dumped
1
0.8
Volts

V
0.6
VS
VL
0.4
1 transit time
0.2
» Ex. ZL = 3 Z0
0
0
5
10
15
20
Time

Can be used to form signal
1.2
1
– Clamping
0.8
2 transit time
Zs
Zo
Volt
0.6
V
0.4
VS
0.2
VR
0
-0.2 0
V
Vs
5
10
15
20
-0.4
-0.6
Time
45
Transmission lines on PCB

Microstrip
Z0 
C0 
 5.98 H 
87
 
ln
e r  1.41  0.8 W  T 
0.67 e r  1.41
pF / inch
 5.98 H 
ln

 0 .8 W  T 
tpd  1.016 0.475e r  0.67 ns / feet
Example : e r  5 ; H  0.6mm ; W  0.5mm ; T  35m
Z 0  106  ; C 0  1.4 pf / inch ; t pd  1.77 ns / feet  5.80 ns / m

Stripline
Z0 
C0 
60  1.92H  T 
ln
 
e r  0 .8 W  T 
1.41e r
pF / inch
 3.81H 
ln

 0.8 W  T 
tpd  1.016 e r ns / feet
Example : e r  5 ; H  0.8mm ; W  0.5mm ; T  35m
Z 0  53  ; C 0  5 pf / inch ; t pd  2.27 ns / feet  7.45 ns / m
46
Lossy transmission lines

Idem with RsL instead of L, Rp//C instead of C
Z

R s  Lp
1
Cp 
Rp
Rs
L
C
Rp
Characteristic impedance depends on w
– Even Rs is a function of w because of the skin effect


Signal is distorted
Termination more complex to compensate cable
characteristic
47
Bibliography

The Art of Electronics, Horowitz and Hill, Cambridge
– Very large covering

An Analog Electronics Companion, S. Hamilton, Cambridge
– Includes a lot of Spice simulation exercises

Electronics manufacturers application notes
– Available on the web
» (e.g. http://www.national.com/apnotes/apnotes_all_1.html)

For feedback systems and their stability
– FEED-2002 from CERN Technical Training
48