Digital and Interfacing
Systems
Ceng 306
Operational Amplifier (Op-Amp)
Supplement
Prepared by Mike Crompton (Rev 28 May 2002)
Operational Amplifiers (Op-Amps)
The original big, bulky vacuum tube operational amplifiers were only used to perform
mathematical ‘operations’ in the early analog computers, hence the name Operational
Amplifiers. With the advent of semiconductors they were built as tiny integrated circuit
devices and became probably the most versatile and common linear analog devices in
electronics/computers today. Based on the two I/P differential amplifier principle, they
simply amplify the difference between their two inputs, and using this basic concept they
can be used to perform many and varied simple or complicated tasks.
The requirements for most amplifiers are:
 Gain – To get a larger signal (data, intelligence, information, audio) out of the
amplifier than what was put in.
 High input impedance – Provide a high resistance to ground that prevents the I/P
signal from ‘disappearing’ by flowing to ground.
 Low output impedance – Allows the O/P to be ‘matched’ to the low resistance of
wires, coax or other devices. (maximum power is transferred when source
resistance = device resistance).
 Good frequency response (Bandwidth) – The amplifier will amplify a broad range
of frequencies by the same amount giving a ‘flat’ response, and not unequal
amplification of different frequencies.
Op-Amps have these qualities plus several unique features that make them ideal for a
myriad applications.
Offset Null
+V
Differential
Amp.
Voltage
Amp.
Inverting I/P
Output
Amp.
O/P
Non–Inv. I/P
Offset Null
-V
All Op-Amps have the same basic construction (see the above diagram), consisting of a
two input differential amplifier feeding a voltage amplifier that is connected to an output
amplifier. The differential amplifier has nulling circuits that allow balancing to prevent
one I/P from producing greater amplification or offset voltages than the other. Provision
for dual power supply voltages of +V and –V (not to be confused with I/P voltages which
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are signals containing intelligence), and of course an output make up the external
connections to the chip. Very high gain figures of up to 1 million : 1 are achievable and,
depending on the configuration, very high I/P and very low O/P impedances can be
maintained.
Probably the most common Op-Amp in chip form is the ‘741’, so it is usually selected as
the learning model, as well as the chip of choice for many circuits. It is cheap, effective,
readily available and comes in many different packages, the 8 pin DIP being most
popular. The specification sheet for the 741 gives the following data:





Gain (Open loop) Minimum 25,000 : 1
Nominal 200,000 : 1
I/P Impedance 2M
O/P Impedance 75
Bandwidth (Gain dependant) at 1 : 1 bandwidth = 1 MHz
Common Mode Rejection Ratio (CMRR) 70dB. The gain ratio between two
identical I/Ps and two different I/Ps – remember that differential amplifiers
amplify the DIFFERENCE between their two inputs, if both are the same then
theoretically there will me NO output)
The diagram at right shows the symbol and pin numbers of the 741 Op-Amp. The 2
Offset Nulling pins are omitted as they are not used in the simple experiments conducted
as part of this course.
The 2 I/Ps are labeled as + and – and this
can sometimes lead to confusion when
first studying Op-Amps. The + I/P is in
fact the ‘non-inverting’ I/P. This means
that when this I/P alone is used, the O/P
will be the same polarity (for DC) or
phase (for AC) as the I/P. The – I/P is the
‘inverting’ I/P and when used the O/P
will be the opposite polarity (for DC) and
180 out of phase (for AC) as the I/P.
The very high (200,000 : 1) voltage gain denoted as AV of the ‘raw’ Op-Amp, also
known as the ‘Open Loop Gain’ (AVOL), makes it impractical to use as an amplifier.
Theoretically an I/P of 1V would produce an O/P of 200,000V. Since “nothing can be got
for nothing” in this world, the supply voltage for the Op-Amp would have to exceed
200,000V and it can be seen that this is impractical, not to mention very dangerous. Some
method of controlling the gain is necessary to produce reasonable gain with normal levels
of supply voltages. The method used is called ‘negative feedback’.
Negative feedback is taking a portion of the amplifier O/P and feeding it back to the I/P
in opposition to the I/P signal. Under normal conditions, as the I/P signal increases, so the
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O/P increases. If some of this increase is fed back to the inverting (-) I/P of the Op-Amp
that increase will be inverted causing the O/P to decrease thereby reducing the gain. This
sequence of changes will continue until a stable condition at a pre-determined gain is
achieved. Determining how much of the O/P to feedback could involve some heavy
calculations, but in the final analysis it turns out to be a simple ratio of 2 resistors that
will be connected to the Op-Amp. More about this later.
There are some facts and assumptions about Op-Amps that need to be discussed. First is
the fact that, due to the 2M or more very high I/P impedance (opposition to current
flow), almost no current can/will flow into the amplifier. e.g. If the I/P voltage was a
relatively high 1V, the current flowing into the amplifier through 2M of resistance
would, according to good old Ohm’s law, be 1V/2,000,000 = 0.0000005A or 0.5A.
Not an amount of current to seriously consider. This is significant because it means the
actual Op-Amp can basically be ignored when looking at the other components that will
be connected to it. This will make many of the calculations much easier.
Second is the fact that the 2 I/P pins will have so little difference in voltage between
them, that they can be considered to be at the same potential. This may seem strange, but
consider the conditions whereby the O/P of an Op-Amp is say 1Volt. With a gain of
200,000:1 this means that the I/P voltage must be 1/200,000 = 0.000005 or 5V. This in
turn means the difference between the 2 I/Ps is only 5V (remember the Op-Amp
amplifies the difference between it’s I/Ps). 5V is hardly a large amount of Voltage and
can, to all intents and purposes, be ignored leaving the I/Ps at basically the same
potential.
Third, the maximum voltage O/P from an Op-Amp, the saturation voltage or VSAT, is
determined by the power supply voltages and the value of the load resistance connected
to the amplifier O/P. As a rule of thumb, + VSAT and – VSAT are equal to +Vcc and – Vcc
less 2V when RLOAD is less than 10k, and +Vcc and –Vcc less 1V if RLOAD is greater
than 10k.
What does all this mean, and how does it relate to an Op-Amp as a useful device? Well,
let us look at one configuration of Op-Amp that is in common use today, the ‘Inverting
Op-Amp’. As the name implies the – or inverting I/P is where we will apply the signal to
be amplified.
The circuit at right depicts an Inverting OpAmp. As can be seen it is a very simple circuit
consisting of just 2 resistors and a 741
amplifier. The supply voltage is + and –
10Volts. First take note that the + or noninverting I/P (pin 3) is grounded. Since we
know that the other I/P (pin 2, inverting) is
never going to be more than a few Volts we
can say that it also is ‘virtually’ at ground. This
means the junction of R1 and Rf is ground too.
Considering that the actual Op-Amp can be
ignored because of it’s very high I/P
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impedance, this makes the circuit a simple 2
resistor voltage divider with Vin at one end,
ground in the middle and Vout at the other end.
See the equivalent circuit diagram at right.
Now it should be simple to work out the
voltage gain of the amplifier.
Gain is equal to Vout/Vin. Looking at the full circuit above we can see that Vout is from
pin 6 to ground and on the equivalent circuit we can see that Vout is actually the same as
the voltage across Rf (VRF). Vin on the full circuit above is from one end of R1 to ground
which on the equivalent circuit is the voltage across R1 (VR1). It should not take a genius
to figure out that if you want Vout (VRF) to be 10 times bigger than Vin (VR1) for a gain
of 10:1 then Rf must be 10 times bigger than R1. This is in fact the method of calculating
the ‘closed loop gain’ (AVCL) in an inverting Op-Amp (‘closed loop’ meaning the
amplifier has feedback from O/P to I/P creating a closed loop):
Gain (AVCL) = -Rf/R1
Where Rf is the feedback resistor
R1 is the input resistor
The minus sign means the O/P is inverted, it does not mean negative gain.
In the case of the example in the full circuit above, Rf is 33k and R1 is 2.2k. The gain
therefore is 33/2.2 = 15:1 and the O/P is inverted. Remember that gain is a ratio, not a
voltage. Using this ratio it means that if the I/P is a 1V Pk to Pk sine wave, the O/P is a
15V Pk to Pk sine wave 180 out of phase with the I/P. One other parameter of this
circuit is the I/P impedance of the whole circuit (not the actual chip itself). R1 is where
the I/P signal will be applied, and pin 2 is at ‘virtual’ ground so the circuit I/P impedance
(total resistance to ground the I/P signal will ‘see’) is the same as R1. This is usually
much less than the raw 2M of the Op-Amp alone.
The circuit at right is the Non-Inverting
Op-Amp where the I/P signal is fed to the
+ or non-inverting I/P. The end of R1,
which was the I/P in the previous example,
is grounded. The same voltage divider
principle applies to the feedback and it’s
effect on the basic gain, but applying the
I/P to the + I/P gives an additional gain of
1. Therefore the closed loop gain (AVCL) of
a non-inverting Op-Amp is:
AVCL = (Rf/R1) + 1.
In the case of the example:
AVCL = (33/2.2) + 1 = 15 + 1 = 16:1
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One major advantage of the non-inverting Op-Amp, besides the fact that the O/P is
always in phase with the I/P, is the extremely high circuit I/P impedance. Unlike the
inverting Op-Amp, there is no I/P resistor and no virtual ground. The negative feedback
will actually enhance the raw chip I/P impedance of 2M, giving a total circuit I/P
impedance of many M. This makes it suitable for working with very small I/P signals,
as they will not ‘leak’ to ground and disappear.
The Voltage Follower Op-Amp shown at right has
limited use as an amplifier due to the fact that it has a
gain of only 1 : 1 ( the O/P is identical to the I/P). The
I/P signal is applied to the non-inverting I/P and with Rf
being 0 (a straight piece of wire) the feedback is
actually 100%. R1 is infinity as there is no resistor. By
looking at the gain formula for the non-inverting
amplifier and substituting the values of Rf and R1 we
get the gain of 1 : 1.
AVCL = (Rf/R1) + 1
= 0/ + 1
= 1
What this circuit is extensively used as, is a ‘Buffer’ to isolate and prevent interaction
between one circuit and another, or to ‘Impedance match’ a signal to a low I/P impedance
device (75) or coaxial cable.
The ‘Differential Op-Amp’, is shown at right.
Notice that the Rf:R1 feedback resistor network is
duplicated with R2:R3. That is, R2 must be
identical to R1 and R3 identical to Rf. Any
differences in those values will cause an imbalance
resulting in the inverting/non-inverting gains being
different. Note also that it has 2 I/P signals, and
should the two I/P signals be identical each would
be amplified by the same amount, resulting in
identical but opposite outputs giving a net gain of
0, nothing, no output. Any difference between the
two I/Ps will result in that difference being
amplified with a gain equal to Rf/R1. This
configuration is very useful in eliminating ‘noise’.
(Unwanted random signals often generated by electrical machines and induced into the
actual wires feeding the wanted signals to their destination, in this case the I/Ps to the OpAmp). The noise induced in the wires will be identical in both I/Ps and will therefore
NOT be amplified. The wanted signals, that would be different from each other, will be
amplified resulting in an excellent signal : noise ratio.
As an example, say the two I/Ps were identical 5V Pk sine waves, and the resulting O/P
was only a 0.016V Pk sine wave. The dB gain is 20 * log Vout/Vin = 20 * log (0.016/5)
= -50dB. With the I/Ps different by 5V Pk the O/P is 50V and dB gain = 20 log (50/5) =
+20dB. Therefore the difference in dB between the two conditions is 70dB. What that
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ratio will be is determined by the CMRR of the Op-Amp (Common Mode Rejection
Ratio - see the specifications for the 741 Op-Amp on page 3).
Amplifiers are only one application of Op-Amps. Often, as amplifiers, their high gain
capabilities make them ideal for amplifying the relatively small signal voltages produced
by the sensors studied earlier in the course. There other uses are many and varied.
The Electronic Work Bench circuit shown at
right is an example of an Inverting Op-Amp
being used as a binary weighted D to A
converter. Vin will always be 12V, Rf will
always be 10k but R1 will vary dependant on
which switches are closed. If only switch A is
closed, corresponding to a 4 bit binary number
of 0001, then R1 is 400k and the gain of the
circuit is 10/400 = 0.025 : 1. Vout will
therefore be 12 * .025 = 0.3V. If just switch B
is closed (binary 0010) R1 is 200k and the
gain is 10/200 = 0.05 : 1 and Vout is 12 * .05 =
0.6V (twice as much as the 0.3V for 0001).
With both switches A and B closed (0011) R1 is 400k in parallel with 200k for a total
of 133.333k. The gain will be 10/133.333 = .075 : 1 giving an O/P of 0.9V or 3 times
the 0.3V for 0001. With all switches closed (1111) R1 is 400k//200k//100k//50k =
26.667k. The gain is 10/26.667 = 0.375 : 1 and an O/P of 4.5V or 15 times the 0.3V for
0001. Try constructing this circuit in EWB and testing it.
Another application for Op-Amps is as ‘comparators’, circuits that can compare two
parameters and indicate differences or activate other devices based on the result of the
comparison. These will be studied as a separate item.
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