Description

TRANSIMPEDANCE AMPLIFIER Optical photodiode detectors and optical photoconductive detectors are current sources, with the current produced being proportional to the light intensity illuminating them. As a result, they are ideally suited to a transimpedance amplifier (TIA) configuration. In particular, we simply let the photodetector replace the input resistor R1:
R2
photodiode
Vout
In this configuration, the photodiode generates the input current directly. Since the – input is at ground poten

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TRANSIMPEDANCE AMPLIFIER
Optical photodiode detectors and optical photoconductive detectors are current sources, with thecurrent produced being proportional to the light intensity illuminating them.
R
2
V
outphotodiode
As a result, they are ideally suited to a transimpedance amplifier (TIA) configuration. Inparticular, we simply let the photodetector replace the input resistor
R
1
:In this configuration, the photodiode generates the input current directly. Since the – input is atground potential here (
V
–
= 0), the TIA’s input impedance is nearly zero, which is exactly what isneeded if we want to capture the maximum current from the photodiode.Notice also that we are operating the photodiode at points along the
V
D
= 0 axis of thephotodiode’s I-V curve, which means that the photodiode is operating in its linear region, just asit does when operating with a reverse bias.In fact, instead of connecting the photodiode cathode to ground, it is often connected to apositive voltage, which reverse biases the diode and provides higher speed response by loweringthe photodiode’s capacitance.Transimpedance amplifiers are given their name because they translate the output from a veryhigh impedance current source (note the slope of the I-V curve above) to a low impedance opamp output.1
R
2
V
outphotodiode
C
Transimpedance amplifiers are usually operated at very high gain. This produces a strongtendency for the amplifier to go into oscillation at some high frequency above the gainbandwidth cutoff. This problem can be eliminated by adding a small capacitor in the feedback loop, which lowers the gain at very high frequencies.The drawback of operating any op amp with very high gain is that its frequency response isgreatly reduced because an op amps gain and bandwidth are inversely proportional to each other,a characteristic that will be discussed in detail later.
R
f
V
out2
RV
2
V
1
R
1
V
3
R
3
SUMMING AMPLIFIER
If we have two or more input signals that we want to add together, we can make good use of thefact that the
V
–
input is a virtual ground. We just tie the signals together at the
V
–
input.In this arrangement, each signal thinks it’s looking at a grounded input, and the other signalshave no effect on it. However, the currents from the inputs all add together and flow through thefeedback resistor to produce an output voltage that is just
⎟⎟ ⎠ ⎞⎜⎜⎝ ⎛ ++−=
332211
RV RV RV RV
f out
2
DIFFERENTIAL AMPLIFIER
V
2
VV
1
1
RR
1
R
2
out
R
2
If we need to amplify a differential voltage, it can easily be done by using both the
V
+
and
V
–
inputs.Because the input-output relationships for signals at the two terminals are linear, we can analyzethis circuit by superposition.First, turn off
V
2
(i.e., short
V
2
to ground). Then the
V
+
input will be at ground, and the circuitlooks just like the standard inverting op-amp with a output signal
2211
(off)
out
RV V R
= −
V
Next, turn off
V
1
by shorting it to ground. The feedback circuit at the
V
–
input then becomes thatfor a noninverting op-amp, whose output signal is given by
1211
(off)
out
R RV V V R
+
⎛ ⎞+=⎜ ⎟⎝ ⎠
But the signal at
V
+
is just
V
2
reduced by the
R
1
:
R
2
voltage divider,
2212
V R R RV
+=
+
so
1222121121
(off)
out
R R R RV V V V R R R R
⎛ ⎞ ⎛ ⎞+= =⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠
2
Adding the two results for
V
out
with
V
1
turned off and for
V
out
with
V
2
turned off, we obtain thesimple result
21221
(off)(off)()
out out out
RV V V V V V V R
= + =
1
−
which is just the output one would like from a differential amplifier.3
The primary reason for using a differential amplifier is to have an amplifier that is immune fromthe effects of common-mode signals or noise, i.e., voltages that appear at both the
V
1
and the
V
2
signals. One common source of common-mode noise is pick-up of a 60 Hz power line radiationby the wire leads coming from some sensor.The common mode voltage is defined to be
½
(
V
1
+
V
2
). When a common mode voltage ispresent, the actual output voltage of a differential amplifier is given by
2121
12
()(
out diff com
V A V V A V V
= − + +
)
4
122121
()20log20log
diff com
AV V CMRR A V V
⎛ ⎞⎛ ⎞+= =⎜ ⎟ ⎜ ⎟−⎝ ⎠⎝ ⎠
where
A
diff
is the differential mode gain
,
and
A
com
is the common mode gain. To remove theunwanted common-mode signals we want the amplifier to amplify only the
difference
betweenthe voltages
V
1
and
V
2
and nothing else.
The key parameter for a differential amplifier thatquantifies how well an amplifier does this is its
common mode rejection ratio
(CMRR), which isdefined to be the ratio of the differential-mode gain to the common-mode gain, expressed indecibels. Thus we haveThis differential amplifier is less than ideal because it relies on having a perfectly matched pairof resistors
R
1
and a matched pair of resistors
R
2
. Any mismatch will create an amplifier that hasboth common mode gain and differential mode gain, greatly reducing the common-moderejection ratio. When a precision differential amplifier is needed, a better choice is to turn to adevice called an instrumentation amplifier.

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