Question 14
An operational amplifier (op-amp) may be used as part of
the processing unit in an electronic sensor.
(a) State three properties of an ideal op-amp. [3]
(b) A comparator circuit incorporating an ideal op-amp is
shown in Fig. 9.1.
Fig. 9.1
(i) In one application of the comparator, V2 is kept constant at +1.5 V.
The variation with time t of the potential V1 is shown in Fig. 9.2. The potential V2 is also shown.
Fig. 9.2
On Fig. 9.2, show the variation with time t of the
output potential VOUT
. [4]
(ii) Two light-emitting diodes (LEDs) R and
G are connected to the output of the op-amp in Fig. 9.1 such that R emits light
for a longer time than G.
On Fig. 9.1, draw the symbols for the two diodes
connected to the output of the op-amp and label the diodes R and G. [3]
Reference: Past Exam Paper – November 2012 Paper 41 & 42 Q9
Solution:
(a) Choose
any 3:
Zero output impedance /
resistance
Infinite input impedance /
resistance
Infinite (open loop) gain
Infinite bandwidth
Infinite slew rate
(b)
(i)
{The output voltage of the
comparator can be obtained by
Vout = Ao
× (V2 – V1) where Ao about 105
As the gain A0
is very large, the output of the op-amp will most of the time be saturated,
that is, the voltage is either be + 5V or – 5V unless the difference between
the two input voltages is very small.
V2 is kept at +
1.5 V.
When the input voltage V2
is greater than V1, the output voltage is positive and equal to + 5
V (as the op-amp is saturated).
Vout = +5V (+ve
supply line)
At time t = 0, V2
is greater than V1. So, the output voltage is initially positive.
When the input voltage V2
is less than V1, the output voltage is negative and equal to -5 V
(as the op-amp is saturated).
Vout = -5V (-ve
supply line)
Thus, the polarity of the output
depends on which of the two input voltages is larger.
The output voltage is zero
whenever V1 = V2 (from Vout = Ao × (V2 – V1), Vout = 0 V). This is
when the polarity of the output changes.}
Graph: square wave
Correct cross-over points
where V2 = V1
Amplitude 5V
Correct polarity (positive
at t=0)
(ii)
{LED R emits light for a
longer time than LED G – from the graph, this corresponds to positive as the
positive section is longer.
Thus, LED R connects when
the output is positive. Current flows from positive to negative. So, LED R
should point downwards (from the positive output, towards earth).
Also, LED G would point
upwards.}
Correct symbol for LED
Diodes connected correctly
between Vout and earth
Correct polarity consistent with graph (i)
