In the given circuit, the three 741 operational amplifiers ①, ② and ③ are operated by power supplies of ± 15 V, ± 10 V, and ± 8 V, respectively. The approximate values of the output voltages V₁, V₂ and V₃ are respectively given by,

2018 A/L Physics Past Paper · Question 27 (English Medium)

Three 741 comparators — output voltages

In the given circuit, the three 741 operational amplifiers ①, ② and ③ are operated by power supplies of $\pm 15$ V, $\pm 10$ V, and $\pm 8$ V, respectively. The approximate values of the output voltages $V_1$ , $V_2$ and $V_3$ are respectively given by,

(1) $+2$ V, $-4$ V, $-4$ V
(2) $+15$ V, $-10$ V, $-8$ V
(3) $+2$ V, $+4$ V, $-4$ V
(4) $-15$ V, $+10$ V, $+8$ V
(5) $+15$ V, $+10$ V, $+8$ V

Answer and explanation

Answer: (2) $+15$ V, $-10$ V, $-8$ V

What the question asks

We want the outputs $V_1$ , $V_2$ and $V_3$ of the three op-amps.

None of them has feedback. So all three work as comparators.

Relevant theory

In this circuit, when V1 > V2 the difference input is positive and the output becomes nearly + VS (saturation at positive).

The magnitude of the output voltage (V0) does not exceed the magnitude of the supply voltage.

no current is flown into the input terminal.

Applying it to this question

Four equal resistors $R$ are in series from 8 V to earth. That is a potential divider.

Each 741 has its $+$ and $-$ inputs joined to points on the divider.

No current flows into the inputs.

So the op-amps take no current from the divider. They do not change its potentials.

Reasoning and calculation

$\tfrac{2}{R}$

$I = \dfrac{8}{4R}$

$2$

$IR = \dfrac{2}{R} \times R$

$8,\ 6,\ 4,\ 2,\ 0$

The comparator rule:

If $V_+ > V_-$ , the output is $+V_S$ .

If $V_+ < V_-$ , the output is $-V_S$ .

$V_S$ is about the supply voltage.

Each op-amp goes to its own supply.

①

$V_+ = 8,\ V_- = 6$ : $V_1 = +15$ V

②

$V_+ = 0,\ V_- = 4$ : $V_2 = -10$ V

③

$V_+ = 2,\ V_- = 6$ : $V_3 = -8$ V

Reading the options

Each option is a set $(V_1, V_2, V_3)$ .

Only option (2) is $(+15, -10, -8)$ V.

Why the other options fail

It takes the input difference as the output.

It takes the input difference as the output. The sign for ② is wrong too.

The $+$ and $-$ inputs are swapped. Every sign is reversed.

In ② and ③, $V_+ < V_-$ . So those outputs are negative.

Checks

No output is bigger than its own supply.

The inputs differ by 2 V or 4 V. That is far more than a 741 needs to reach its supply voltage.

Conclusion

$V_1 = +15$ V, $V_2 = -10$ V, $V_3 = -8$ V.

The answer is option (2).

For an op-amp with no feedback, write $V_+$ and $V_-$ first.

See which is bigger. Then write that supply voltage.

Exam tip

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