Download Op-Amp comparators

Digital and Interfacing
Systems
Ceng 306
Op-Amp Comparator
Supplement
Prepared by Mike Crompton (Rev 31 March 2005)
Op-Amp comparators.
The Op-Amp is often used in the open loop configuration as a device to compare two
voltage levels (Comparator). Since it amplifies the difference between it’s + and – inputs,
it will go to saturation in the direction (+ve or –ve) of the input that is the larger by only a
very small amount. Remember that with a gain of 200000:1 a 5V input will give a 1V
O/P, and theoretically, a 1mV input would give a 200V O/P. So it should be obvious that
only a very small difference between the two inputs will drive the O/P to saturation.
Recall that saturation is the amplifier’s maximum O/P voltage level (+ve and/or –ve), and
is determined primarily by the positive and negative levels of the supply voltage (+VCC
and -VCC ).
If a fixed voltage is applied to one of the inputs, it would become the reference voltage.
Any voltage applied to the other input must be positive or negative with respect to that
reference voltage, and would drive the amplifier to saturation in one direction or the
other. e.g. Say that a reference voltage of +2.5V is applied to the inverting (-) input of the
amplifier. See Fig.1. With no voltage on the other (+ or non-inverting) input, the
amplifier would go to –ve saturation and stay there, VOUT would be approximately –9V.
If we now apply a voltage (VIN) to the non-inverting input, the amplifier would go to +ve
saturation (VOUT = +9V) the instant that voltage exceeded +2.5V by just a few microvolts.
Using the same logic, if the +2.5V reference
was applied to the + (non-inverting) input the
amplifier would go to +ve saturation (+9V)
and stay there until a voltage on the –
(inverting) input exceeded +2.5V by a few
micro-volts when it would go to –ve
saturation (-9V). The voltage on the inverting
I/P being greater than the non-inverting I/P,
the inversion would take place giving a –ve
O/P not a +ve.
A practical application for this type of circuit
might be an over/under temperature
indicator. VIN could be the changing voltage
across a thermistor resulting from the
thermistor’s changing resistance with
temperature. At a pre-determined voltage
(resistance-temperature) the amplifier would
go from –ve saturation to +ve saturation.
Two different coloured LEDs back to back
with a current limiting resistor in series and
connected to the amplifier O/P would
provide a visual indication of under/over
temperature conditions. See Fig. 2.
2
Fig. 1
Fig. 2
In some cases it is desirable to have
Fig. 3
the change from positive saturation
to negative saturation occur at a
different voltage level. With the
previous comparator this change
was within a few micro volts + or of the fixed reference voltage level.
A considerable difference in
switching levels can be achieved by
the use of a Comparator with
Hysteresis. See Fig. 3 at right. In
this configuration the voltage at
point ‘A’ (the non-inverting I/P) is
determined by a fixed reference
voltage at one end of a two resistor
voltage divider and VSAT at the other end. Since VSAT can vary from a positive voltage to
a negative voltage the mid point will also vary by a considerable amount. In the circuit of
Fig. 3, two 10k resistors across a 5V supply put one end of the voltage divider (also a
10k resistor) at +2.5V. The other end (through a 100k ) point ‘B’ will be at VSAT.
Presume that point ‘B’ is at +10V (+ve saturation), we can look at the voltage divider in
complete isolation to calculate the voltage at point ‘A’. See Fig. 4.
The voltage divider formula states that:
+10V (+VSAT)
VRX = (VT/RT) * RX
To calculate the voltage drop across the 100k resistor this
translates to:
VTOTAL is 10V – 2.5V = 7.5V and RTOTAL is 110k
Fig. 4
VR100k = (7.5/110) * 100 = 6.82V
Therefore point ‘A’ is at +10V – 6.82V = +3.18V. This means VIN,
on the inverting (-) I/P, must go a few micro volts above +3.18V to
drive the Op-Amp to negative saturation, producing the voltage
divider circuit shown in Fig. 5.
-10V (-VSAT)
Now VTOTAL is –10V – (+2.5V) = 12.5V
And VR100k = (12.5/110) * 100 = 11.36V
Point ‘A’ therefore is –10V + 11.36V = +1.36V. This means that
VIN must drop below +1.36V to drive the Op-Amp to positive
saturation. Therefore the ‘hysterisis window’ between +ve and –ve
saturation is 3.18V – 1.36V = 1.82V.
3
Fig. 5