Transcript instrumentation

A-GROUP
 130330109036- Gansham Malavya
 130330109037- Prashant Malavya
 130330109038- Rajat Mandal
 130330109039-Dhiren Manga
 130330109040- Alhas Manya
Guided by- Prof. Reema
Operational Amplifier
 Instrumentation Amplifier
 Differential input & differential output amplifier
 V to I and I to V convertor
 Integrator
 Differentiator
 Comparator
 Non-Linear Amplifier
 Phase Shift Oscillator
Instrumentation Amplifier
 Defination:An instrumentation (or instrumentational) amp
lifieris a type of differential amplifier that has been outfitted
with input buffer amplifiers, which eliminate the need for
input impedance matching and thus make the amplifier
particularly suitable for use in measurement and test equipment.
 Additional characteristics include very lowDC offset, low drift,
low noise, very high open-loop gain, very high common-mode
rejection ratio, and very highinput impedances. Instrumentation
amplifiers are used where great accuracy and stability of
the circuit both short and long-term are required.
Instrumentation Amplifier
vo 1
v1
I 
I
v1  v2
R1
I
v2
vo 2
Although the
instrumentation amplifier is
usually shown schematically
identical to a
standard operational
amplifier (op-amp), the
electronic instrumentation
amp is almost always
internally composed of 3 opamps. These are arranged so
that there is one op-amp to
buffer each input (+,−), and
one to produce the desired
output with adequate
impedance matching for the
function.[1][2]
The most commonly used
instrumentation amplifier
circuit is shown in the figure.
The gain of the circuit is
The rightmost amplifier, along with the resistors labelled and is just the standard differential
amplifier circuit, with gain = and differential input resistance = 2· . The two amplifiers on the left
are the buffers. With removed (open circuited), they are simple unity gain buffers; the circuit
will work in that state, with gain simply equal to and high input impedance because of the
buffers. The buffer gain could be increased by putting resistors between the buffer inverting
inputs and ground to shunt away some of the negative feedback; however, the single
resistor between the two inverting inputs is a much more elegant method: it increases the
differential-mode gain of the buffer pair while leaving the common-mode gain equal to 1. This
increases the common-mode rejection ratio (CMRR) of the circuit and also enables the buffers to
handle much larger common-mode signals without clipping than would be the case if they were
separate and had the same gain. Another benefit of the method is that it boosts the gain using a
single resistor rather than a pair, thus avoiding a resistor-matching problem (although the two s
need to be matched), and very conveniently allowing the gain of the circuit to be changed by
changing the value of a single resistor. A set of switch-selectable resistors or even a potentiometer
can be used for , providing easy changes to the gain of the circuit, without the complexity of
having to switch matched pairs of resistors.
The ideal common-mode gain of an instrumentation amplifier is zero. In the circuit shown,
common-mode gain is caused by mismatches in the values of the equally numbered resistors and
by the mis-match in common mode gains of the two input op-amps. Obtaining very closely
matched resistors is a significant difficulty in fabricating these circuits, as is optimizing the
common mode performance of the input op-amps.[3]
An instrumentation amp can also be built with two op-amps to save on cost and increase CMRR,
but the gain must be higher than two (+6 dB).[4][5]
Single-ended output
If the differential output is not desired, then only one output can
be used (taken from just one of the collectors (or anodes or
drains), disregarding the other output without a collector
inductor; this configuration is referred to as single-ended output.
The gain is half that of the stage with differential output. To
avoid sacrificing gain, a differential to single-ended converter can
be utilized. This is often implemented as a current mirror
 Single-ended input
 The differential pair can be used as an amplifier with a single-
ended input if one of the inputs is grounded or fixed to a
reference voltage (usually, the other collector is used as a singleended output) This arrangement can be thought as of cascaded
common-collector and common-base stages or as a buffered
common-base stage.
 The emitter-coupled amplifier is compensated for temperature
drifts, VBE is cancelled, and the Miller effect and transistor
saturation are avoided. That is why it is used to form emittercoupled amplifiers (avoiding Miller effect), phase splitter circuits
(obtaining two inverse voltages), ECL gates and switches
(avoiding transistor saturation), etc.
Example #1: Voltage Comparator
is = 0
i1 = 0
i2 = 0
Note that the inverting input and non-inverting input
terminals have rotated in this schematic.
Example #1 (con’t)
 The internal circuitry in the op amp tries to force the
voltage at the inverting input to be equal to the noninverting input.
 As we will see shortly, a number of op amp circuits have
a resistor between the output terminal and the inverting
input terminals to allow the output voltage to influence
the value of the voltage at the inverting input terminal.
Inverting
Integrator
Now replace resistors R and R by complex
a
Zf
f
components Za and Zf, respectively,
therefore
Zf
Vo 
Vin
Za
Supposing
in
(i) The feedback component is a capacitor C,
i.e., Z  1
f
jC
(ii) The input component is a resistor R, Za =
R
1
Therefore, the closed-loop
gain (Vo/Vin)
vo (t ) 
v
(
t
)
dt
i
become:
RC 
Za
V ~
in
V
o
V ~
vi (t )  Vi e jt

+
C
R

V
o
+
where
What happens if Za = 1/jC whereas, Zf = R?
Inverting differentiator
Ref:080114HKN
Operational Amplifier
13
Op-Amp Integrator
Example:
(a) Determine the rate of change +5V
0
of the output voltage.
C
R
100s
(b) Draw the output waveform.
V
i
10 k
0.01F

V
o
+
Vo(max)=10 V
Solution:
(a) Rate of change of the output voltage
Vo
V
5V
 i 
t
RC (10 k)(0.01 F)
 50 mV/ s
+5V
0
V
i
0
(b) In 100 s, the voltage decrease
Vo  (50 mV/ s)(100μs)  5V
Ref:080114HKN
Operational Amplifier
-5V
-10V
V
o
14
Op-Amp Differentiator
R
C
0
to
t1
t2
V

i
V
o
0
+
to
t1
t2
 dV 
vo   i  RC
 dt 
Ref:080114HKN
Operational Amplifier
15