(a) Two properties of an ideal operational amplifier (op-amp) are infinite bandwidth and infinite slew rate. Explain what is meant by

Q#12 (Past Exam Paper – March 2018 Paper 42 Q8)

(a) Two properties of an ideal operational amplifier (op-amp) are infinite bandwidth and infinite slew rate.

Explain what is meant by
(i) infinite bandwidth, [1]

(ii) infinite slew rate. [1]


(b) An ideal op-amp is incorporated into the circuit of Fig. 8.1.


Fig. 8.1

(i) Determine the resistance Rof the thermistor T at which the output potential difference VOUT is zero. [1]

(ii) The temperature of the thermistor is gradually increased so that its resistance decreases from 1.5Rto 0.5RT.

On Fig. 8.2, draw a line to show the variation of the output potential difference VOUT with
the thermistor resistance.


Fig. 8.2
[2]

(iii) On Fig. 8.1, draw the symbol for a light-emitting diode (LED), connected at the output of the circuit, such that it emits light when the resistance of the thermistor is less than RT. [2]
 [Total: 7]




Solution:
(a)
(i) An infinite bandwidth means that all frequencies have the same gain

(ii) The output changes at the same time as input changes


(b)
(i)
{The output p.d. is zero when V+ is equal to V-.
Using the potential divider equation,
(since V+ = V-, the ratio of resistance in one branch is equal to the ratio of resistance in the other branch.)}

R/ 800 = 1.8 / 1.2
R1200 Ω


(ii)

stepped from –9 V to +9 V or v.v.
Vout negative < Rand Vout positive > RT



{The circuit is a comparator since there are inputs at both V+ and V-.
If V+ is greater than V-, the output is positive and if V- is greater than V+, the output is negative.

The output will always be saturated at ±9 V as there is no feedback.

When RT = 1200 Ω, V+ = V- and so, Vout is zero.

When RT is less than 1200 Ω, V- > V+ and so, Vout is negative (= -9 V).

When RT is greater than 1200 Ω, V+ > V- and so, Vout is positive (= +9 V).}


(iii)
correct LED symbol with connection between VOUT and earth
diode pointing upwards

{As explained above, when the resistance is than RT (i.e., less than 1200 Ω), the output is negative. So, current flows from the earth towards the output (that is, upwards). The LED is point upwards.}

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