Question 3
The circuit of an
amplifier incorporating an ideal operational amplifier (op-amp) is shown in
Fig. 7.1.
Fig. 7.1
(a)
By reference to the properties of an ideal op-amp,
(i)
explain why point P is referred to as a virtual
earth, [4]
(ii)
derive an expression, in terms of the resistances R1 and R2,
for the gain of the amplifier circuit. [4]
(b) In the circuit of Fig. 7.1, the
ratio R2 / R1 is 4.5.
The
variation with time t of the input potential VIN is shown in Fig. 7.2.
Fig. 7.2
On Fig.
7.2, show the variation with time t of the output potential VOUT. [3]
[Total: 11]
Reference: Past Exam Paper – November 2017 Paper 41 & 43 Q7
Solution:
(a)
(i)
{For an ideal op-amp,} the gain of the amplifier
is very large
V+ is connected directly to earth (at zero potential)
For the amplifier not to saturate,
the difference between V- and V+ must be very small or V- must be equal to V+ {or else, the op-amp’s infinitely
large gain would lead to saturation}
(ii)
{For an ideal op-amp,} the input impedance is
infinite.
(So,) current in R1 = current in R2
{From Ohm’s law: I = V / R}
(VIN – 0) / R1 = (0 – VOUT) / R2
{VIN / R1 = – VOUT / R2 }
(gain =) VOUT / VIN =
– R2 / R1
(b)
graph: correct inverted shape (straight
diagonal line from (0,0) to a negative potential, then a horizontal line, then
a straight diagonal line back to the t-axis at the point where VIN =
0)
{Gain =
- R2 / R1 = - 4.5
Gain =
VOUT / VIN = - 4.5
VOUT
= - 4.5 VIN }
horizontal line at correct potential of (–)9.0
V
{When VIN = 3 V, VOUT = -
4.5 × 3 = - 13.5 V
But the supply voltage is ± 9V, so the VOUT saturates at - 9 V.
That is, when VIN = 3 V, VOUT
= - 9 V}
both ends of horizontal line occur at correct
times (coinciding with when VIN = 2.0 V)
{We need to
find at which value of VIN the output becomes – 9 V.
VOUT / VIN = - 4.5
- 9 / VIN = - 4.5
VIN = - 9 / -.45 = 2 V
So, when VIN = 2 V, VOUT
= - 9 V
But
as VIN increases from 0 V to 2 V, the value of VOUT also
increases (with a negative sign) from 0 V to – 9 V.}
