An ideal operational amplifier (op-amp) has infinite voltage gain and infinite slew rate.

Question 2

An ideal operational amplifier (op-amp) has infinite voltage gain and infinite slew rate.

(a) State what is meant by

(i) the voltage gain, [1]

(ii) infinite slew rate. [2]

(b) A non-inverting amplifier circuit incorporating an ideal op-amp is shown in Fig. 8.1.

Fig. 8.1

The supply to the op-amp is +9 V / −9 V.

The voltage gain of the amplifier circuit is 12.

Determine the resistance of resistor R. [2]

(c) For the circuit of Fig. 8.1, the variation with time t of the input potential VIN to the amplifier is shown in Fig. 8.2.

Fig. 8.2

On Fig. 8.3, show the variation with time t of the output potential VOUT for time t = 0 to time

Fig. 8.3

[4]

[Total: 9]

Reference: Past Exam Paper – June 2016 Paper 42 Q8

Solution:

(a)

(i)

The voltage gain is the ratio of the voltage output to the voltage input.

(ii)

An infinite slew rate means that the changes in the output voltage occur immediately when the input voltage changes.

(b)

{This is a non-inverting amplifier circuit.

Gain = 1 + R_F / R_IN }

12 = 1 + R / (1.5×10³)

R = 16.5 kΩ

(c)

straight line from (0,0) to (0.75t1, 9.0 V)

horizontal line from endpoint of straight line to t1

+9 V to –9 V (or v.v.) at t1

correct line to t2

{Gain = V_OUT / V_IN       giving V_OUT = Gain × V_IN = 12 × V_IN

The output voltage cannot be greater than the supply voltage of 9 V.

When V_OUT = 9 V, V_IN = V_OUT / Gain = 9 / 12 = 0.75 V

So, when V_IN = 0.75 V (in Fig 8.2), V_OUT = 9 V (in Fig 8.3).

V_IN = 0.75 V at t = 0.75t₁. So, in Fig 8.3, the line passes through (0.75t₁, 9.0 V)}

{The output voltage cannot be greater than 9.0 V. So, even if V_IN increases, V_OUT remains 9.0 V. So, from 0.75t₁ to t₁ (this is where V_IN is maximum), V_OUT is still 9.0 V. This is a horizontal line in the graph to t₁.}

{At t₁, V_IN changes from positive to negative. So, V_OUT also changes from positive to negative. V_IN is maximum (negative), causing V_OUT to be – 9.0 V.}

{Just as for the positive V_IN, V_OUT = – 9.0 V when V_IN = – 0.75 V. This occurs at time t = 0.25t₂ in Fig 8.2. Hence, in Fig 8.3, we have a horizontal line from (t₁, – 9.0 V) to (0.25t₂, – 9.0 V).}

{From t = 0.25t₂ to t₂, the graph is a straight line joining the points (0.25t₂, – 9.0 V) to (t₂, 0 V).}