Physics 160 Lecture 13

Physics 160
Lecture 13
R. Johnson
May 11, 2015
JFET Amplifiers
Common-Source Amp
Source Follower
Note that Zout does not
depend on the output
impedance of the
driving voltage source.
Ci
Cin
VDD
Generally the performance is
poor compared with a BJT
amp, except for the input
impedance. For a JFET it is
spectacularly high! This can
often be an overwhelming
reason to use a FET.
VDD
JFET
Cin
RD
JFET
1 gm
RG
RS
RG can be a very
large resistance!
1
Z out 
RS
gm
May 6, 2015
RS
CS
RG
Gain   gm RD
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JFET Amplifiers
•
Typically, the place you want to use a JFET amplifier is where
you need very high input impedance, for example because your
signal source has a very high impedance (cannot deliver much
current).
– Probably best used in differential amps or source-followers, not
common-source amps.
•
Another example is where the base current of a bipolar
transistor will cause a significant
g
error.
– The LF411 Op-amp that you will soon use in several circuits uses
JFETs at its inputs. This is very nice, because the current flowing
into the inputs is negligible in all cases.
•
Otherwise, bipolar transistors will usually give much better
performance in terms of gain, predictability, etc.
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JFET Amplifiers
•
You need to keep in mind:
– The gate-source junction must be reverse biased at all times (or at
worst, zero volts).
– The gate does need some bias current, although it can be very
small.
– The drain-to-source voltage
g cannot be too small, especially
p
y if the
drain current is substantial. For a bipolar transistor the collectoremitter voltage can be a fraction of a volt, but for a JFET count on a
few volts.
– Unlike
U lik th
the case off a bi
bipolar
l ttransistor,
i t th
there iis no simple
i l fformula
l ffor
the transconductance. You must consult the data sheet, and it will
be small compared with the bipolar transistor transconductance.
• Or equivalently,
equivalently the effective dynamic resistance of the source
will be a few hundred ohms, not the bipolar transistor value of
25 divided by the current in mA.
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Very high
impedance
source is no
problem (except
at high
frequency)
JFET Source Follower
JFET has rather
high output
impedance, so a
BJT follower is
useful to boost
the output drive.
Driving source
0V
R2 1.382pA
V2
1.382pA
15Vdc
VS
6.401mA
10Meg
J1
J2N5485
1Vac
0Vdc
4.011mA
+0.5 V
Q1
2.390mA
V
13.89uA
Q2N3904
-2.404mA
0.1
01V
C2
12uF
V3
15Vdc
6.401mA
May 11, 2015
Voltage division
between Zout of
the JFET (1/gm)
and RS.results in
gain less than
unity
3.997mA
R1
RS
Physics 160
3.9k
V
2.404mA
R3
6.2k
RL
1Meg
0A
5
Setting the Bias Current
Essentially the same procedure
as used for the JFET current
source.
15.5V
R
 3 .9 k 
4 A
4mA
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JFET Follower Gain
G
May 11, 2015
1
 250 
gm
RS g m
3.9  4

 0.94
1  RS g m 1  3.9  4
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Improving Follower Performance
4.274mA
R2 1.380pA
0V
V2
1.380pA
15Vdc
VS
6 654mA
6.654mA
J1
10Meg
J2N5485
1Vac
0Vd
0Vdc
V
+0.5 V
13.83uA
Q1
2 380mA
2.380mA
Q2N3904
-2.394mA
0.1
0.1 V
12uF
4.260mA
15 V
V3
15Vdc
C2
J2
2.394mA
-1.464pA
J2N5485
R3
14
14.5
5V
6.654mA
R1
6.2k
RL
1Meg
0A
125
Replace the source
resistor by a JFET (or
BJT) current source.
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Reminder from Lecture 12:
15.00V
6.869mA
J3
V1
15Vdc
-1.053pA
1 053pA
J2N5485
IDSS
4.254mA
J3
V1
-1.381pA
1 381 A
J2N5485
15Vdc
531.8mV
6.869mA
4.254mA
R1
125
R
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Physics 160
0. 5V
 125 
4mA
9
Gain with current source = 0.99!
The lack of high frequency
performance is related to
the high driving source
impedance. We will see
how to improve this with a
“b t t ”
“bootstrap”.
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Output Impedance
Zout is independent of source resistance!!
JFET Zin>>10 Meg
4.274mA
R2 1.380pA
V2
15Vdc
1.380pA
VS
6.654mA
10Meg
J1
JFET Zout=1/gm=250 ohms
J2N5485
1Vac
0Vdc
V
Q1
13.83uA
BJT : re  25 / 2.38  10.5 
2.380mA
Q2N3904
BJT Zin=22.5170=3800 ohms
BJT Zout=250/170 + 25/2.4=12 
-2.394mA
C2
100uF
4.260mA
J2
V
2.394mA
V3
-1.464pA
J2N5485
R3
15Vdc
6.2k
6.654mA
R1
RL
12
0A
12 ohm load
( t
(extreme
case))
125
 2380/13 83 172
=2380/13.83=172
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Voltage Division of Impedances
3800
 0.95
3800  200
JFET Output
BJT Output
May 11, 2015
0.95 
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12
 0.48
11.7  12
12
Bootstrapping the Drain of the Follower
Make the drain follow the gate to
minimize the voltage changes
between Gate and Drain. This
almost eliminates the low
low-pass
pass
filter formed by the source
impedance and the G-S
capacitance.
High pass filter
CGS
Low pass filter
+7.8 V
0V
+0.5 V
4 mA
0.1 V
15 V
This trick can
greatly increase the
response at high
frequency.
14.5 V
Don’t make RD too much larger or
else the VDS will become too small.
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No Bootstrap on Drain
The low
low-pass
pass filter
formed by the
source impedance
and gate-drain
parasitic
iti
capacitance kills
the gain above
about 10 kHz
10 Meg-ohm
Source
Impedance
1 k-ohm load
Note that the gain is less than unity above because of charge division
between the output impedance of the BJT emitter
emitter-follower
follower and the 1 kk-ohm
ohm
load resistor.
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With Bootstrapped Drain
10 Meg-ohm
Source
Impedance
1 k-ohm load
Bandwidth is now
nearly 1 MHz
Note that the gain is less than unity above because of charge division
between the output impedance of the BJT emitter
emitter-follower
follower and the 1 kk-ohm
ohm
load resistor.
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Bootstrapped Drain with Heavy Load
Roll-off due to AC coupling to the load
10 Meg-ohm
Source
Impedance
100 ohm load
Voltage division
between Zout and
the 100 ohm load
has reduced the
gain by about 10%.
The frequency
response has moved
down from 1 MHz to
100 kHz
With the load greatly increased (100 ohm load resistor), then the gain is so much
less than unity that the bootstrap no longer works very well
well. We lose a decade of
frequency response.
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AC Coupled Input
The gate doesn’t need much
current, but about 1pA still
has to flow, so a bias
resistor is essential!
The capacitor can be very
small, since the bias resistor
(input impedance) is large.
Zin is now completely dominated by
the bias resistor ((well above the 3dB
point).
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Bias resistor lowered Zin to 100 M
Note the 10% loss of gain
g
due to voltage division
between the 10 M
source impedance and
the 100 M input
p
impedance.
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AC Coupling with Bias Bootstrap
We can bootstrap the bias
network to raise it to even
higher impedance at
signal frequencies
frequencies, just as
for a BJT follower.
Only a small,
inexpensive
capacitor is
needed, since the
bias network
impedance is so
high!
May 11, 2015
R10 at signal frequencies looks
like about 600 M.
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Input impedance with bootstrap
0.983 V
Voltage at JFET gate
Reffective 
May 11, 2015
10 M
 588 M
1 - 0.983
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DC-coupled follower with nearly zero offset
By using a matched JFET pair (must be on
the same IC chip) and matched resistors
R1 and R4, we can achieve nearly zero DC
offset between the input and output
output.
15.00V
4.254mA
R2 1.381pA
V2
15Vdc
10Meg
VS
1Vac
0Vdc
J1
J2N5485
13.81uV
0V
531 8mV
531.8mV
R4
0V
125
0V
13.69uV
0V
J2
-1.381pA
V3
V
J2N5485
13.69pA
RL
1Meg
15Vdc
-14.47V
4.254mA
We do pay the price
that now the JFET
output impedance of
1/gm is in series with
a 125 ohm resistor.
R1
0V
125
0V
-15.00V
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Output of matched follower
There is no bootstrap of the
drain, hence the poor
frequency response with
high source impedance. A
BJT emitter follower could
be added to drive a
bootstrap if higher
frequency response is
needed (the JFET output
impedance is too high to
drive a bootstrap by itself
effectively).
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JFET Common-Source Amplifier
Gain is gm times RD:
Looks like ~2.3 mmho from
graph, predicting G=2.37.5=17
1.015mA
RD
7.5k
Observe only ~15 from PSpice
A BJT would have gm~1/25 mho,
giving a gain of ~300!
C1
RVS
V2
10u
J1
J2N5485
15Vdc
VS
1Meg
-1.215pA
1Vac
0Vdc
V
1.015mA
RS
C2
1.5k
100u
RL
1Meg
0A
Without
t out bypassing
bypass g RS
S with
t capac
capacitor
to
C2 the gain will be less than unity!
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JFET Common-Source Amplifier
Only the high input
impedance is
impressive. Better to
use the FET as a
follower and follow it
b a BJT amp.
by
(Note: when FETs are used as
voltage amplifiers, usually the load
on the drain is a current source
rather
th th
than a resistor).
i t ) S
See HW 4
4.
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JFET Variable Resistor
R1
INPUT
Voltage
Divider
OUTPUT
1.000V
433.2mV
1k
V
R3 2.717uA
V
1Meg
V3
V1 = 0
V2 = .5
5
TD = 0
TR = 0.5m
TF = 0.5m
PW = 0.001m
1.000V
PER = 1.002m
V4
R2
-2.283V
2.283V
1Meg
564.0uA
J1
J2N3819
2.717uA
-1.082pA
p
-5.0Vdc
1Vdc
566.8uA
V1
A lower voltage in V1 increases the
current through the 1 Meg resistors
and lowers the gate voltage of J1,
causing its resistance to increase,
thus increasing the voltage at the
OUTPUT.
0V
A circuit that you will
study in the lab. V1
controls the gate bias
and hence the
resistance of the JFET
(at low drain current).
The JFET then forms a
voltage divider
together with R1.
It looks quite linear as
long as we stay well
below saturation and
don’t have too large a
voltage swing.
R3 provides some negative feedback that helps stabilize the output.
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JFET Non-Saturation Region
Resistance is inverse of the slope
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Divider output with V1=-5V
In= green
O t red
Out=
d
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Divider output with V1=-7V
The output is
slightly curved
(nonlinearity)
May 11, 2015
In= green
Out= red
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Op-Amps and Feedback
•
DC-coupled differential amp
– Amplify from DC up to some
limiting frequency
•
Very high differential gain
8
– To a first approximation, just
consider the gain to be infinite
3
+
+ power
V+
Non-inverting input
-
4
Inverting input
2
1
Output
V-
OUT
 power
Without an external feedback network, the
transistors do not have a stable bias point.
They are either in saturation or turned off,
and the op-amp output is at one limit or
the other. This can be useful as a
comparator.
The first operational amplifiers
were constructed with vacuum
tubes, and later with discrete
t
transistors.
i t
Expensive!
E
i !
To use the op-amp as a linear device, an
external feedback network, with negative
feedback, is essential!
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Ideal Op-Amp
Such an ideal op-amp of course does not exist, but a first
analysis of op
op-amp
amp circuits can be done to a good
approximation usually by ignoring the non-ideal behavior.
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3
V+
Vin
In
8
Non-Inverting Amplifier
+
2
-
4
Virtual Vin
Note: Zin is very large
(transistor base or gate).
R2
90k
R1
10k
Out
1
V-
OUT
Vin  Vout 
Identical to the Lecture
Lecture10 example, except that
the Op Amp is a highperformance IC.
R1
R1  R2
0
Simplified analysis of an op-amp with negative feedback:
- Assume infinite gain, so the negative feedback always has to keep the two inputs
equal in order to have a finite output.
- Assume zero current flow into the op-amp inputs.
- Then calculate the relationship between input and output from the feedback
network.
Vout
R2
G
 1
 1  9  10
Vin
R1
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