Questions on the ideal op-amp and the comparator

Question 1

State the four properties of an ideal operational amplifier.

Accepted Answer

All four of the following, one mark per two correct (2): infinite open-loop gain; infinite input resistance, so neither input draws current; zero output resistance, so the output does not sag under load; and infinite bandwidth. Real chips manage about 10⁵, a few megohms and tens of ohms. The question asks for four properties, so 'very high gain' alone is not a complete answer.

Question 2

Write down the open-loop relation for an operational amplifier, and explain what is meant by saturation.

Accepted Answer

V_out = A_OL(V₊ − V₋) (1). Saturation is what happens when that equation gives an output beyond the supply rails: the output cannot pass them, so it sticks a volt or so inside the rail and stays there, clipping the waveform flat (1). The amplifier acts on the difference between the inputs, so applying the gain to one input alone gives the wrong output.

Question 3

State what negative feedback gives away in an operational amplifier circuit, and give two things it buys.

Accepted Answer

It gives away most of the available gain (1). In return, any two of: the closed-loop gain is set by the feedback resistors rather than by the chip; the bandwidth is far wider; distortion falls; the circuit's behaviour survives a change of chip (1).

Question 4

Explain what is meant by the virtual earth in an inverting amplifier.

Accepted Answer

Negative feedback holds the two inputs within microvolts of each other, and the non-inverting input is connected to 0 V, so the inverting input is held at 0 V as well (1) without being connected to earth at all, which is why it is called virtual (1). It carries the input and feedback currents, which a real earth connection would divert.

Question 5

State two differences between an inverting and a non-inverting amplifier, other than the equations for their gains.

Accepted Answer

The inverting amplifier's output is in antiphase with its input, while the non-inverting amplifier's output is in phase (1). The non-inverting gain can never fall below 1, and the signal arrives straight at the op-amp's own input terminal, so the circuit draws almost nothing from the source and suits a high-resistance sensor (1).

Question 6

State the relationship between the closed-loop gain of an amplifier and its bandwidth, and state what happens to the bandwidth when the gain is increased by a factor of ten.

Accepted Answer

Gain × bandwidth = constant, the gain-bandwidth product of the chip (1). Ten times the gain therefore means a tenth of the bandwidth (1).

Question 7

An operational amplifier has an open-loop gain of 1.0 × 10⁵ and its output saturates at ±13 V. Calculate the difference between its input voltages that is just enough to drive the output to the positive rail, and state what happens for any larger difference.

Accepted Answer

V₊ − V₋ = V_out/A_OL = 13/(1.0 × 10⁵) (1) = 1.3 × 10⁻⁴ V, that is 130 μV (1). Any larger difference would require an output beyond the rail, which the amplifier cannot deliver, so it saturates at about +13 V and stays there (1). Divide by the open-loop gain rather than multiplying by it: multiplying gives 1.3 × 10⁶ V, which no supply rail permits.

Question 8

An inverting amplifier uses R_in = 4.7 kΩ and R_f = 82 kΩ. Calculate its closed-loop gain and its output for an input of 0.25 V.

Accepted Answer

Gain = −R_f/R_in = −82/4.7 (1) = −17.4 (1). V_out = −17.4 × 0.25 = −4.4 V (1), comfortably inside the rails. The output of an inverting amplifier has the opposite sign to its input, so +4.4 V is the wrong sign.

Question 9

A non-inverting amplifier uses R₁ = 2.2 kΩ with R_f = 33 kΩ. Calculate its closed-loop gain and its output for an input of 0.60 V.

Accepted Answer

Gain = 1 + R_f/R₁ = 1 + 33/2.2 = 1 + 15 = 16 (2). V_out = 16 × 0.60 = 9.6 V, in phase with the input (1). The non-inverting gain is 1 + R_f/R₁, so the bare resistor ratio of 15 omits the leading 1.

Question 10

An inverting amplifier has R_in = 22 kΩ and R_f = 56 kΩ, and its input is 0.44 V. Calculate the current into the inverting input node and hence the output voltage, working from the currents rather than from the gain equation.

Accepted Answer

The node is a virtual earth at 0 V, so the whole 0.44 V sits across R_in and I = 0.44/(22 × 10³) = 20 μA (1). No current enters the op-amp, so all 20 μA continues through R_f (1), and the output sits at −20 × 10⁻⁶ × 56 × 10³ = −1.12 V (1). The gain equation agrees: −(56/22) × 0.44 = −1.12 V.

Question 11

An operational amplifier has a gain-bandwidth product of 1.0 MHz. Calculate the bandwidth available at a closed-loop gain of 200, and the bandwidth at a closed-loop gain of 25.

Accepted Answer

Bandwidth = gain-bandwidth product/closed-loop gain (1). At a gain of 200: 1.0 × 10⁶/200 = 5.0 × 10³ Hz, that is 5 kHz (1). At a gain of 25: 1.0 × 10⁶/25 = 4.0 × 10⁴ Hz, that is 40 kHz (1).

Question 12

An inverting amplifier with R_in = 6.0 kΩ and R_f = 120 kΩ runs from supplies that let its output saturate at ±13 V. Determine the output for an input of 0.90 V, and the largest input the amplifier can handle linearly.

Accepted Answer

Gain = −120/6.0 = −20 (1). The ideal calculation gives −20 × 0.90 = −18 V, which exceeds the −13 V output limit, so the amplifier saturates at approximately −13 V and the waveform is clipped flat (1). For linear working |gain × V_in| ≤ 13, so V_in ≤ 13/20 = 0.65 V (1).

Question 13

An operational amplifier with an open-loop gain of 1.0 × 10⁵ and rails at ±13 V is used with no feedback as a comparator. Its non-inverting input is at 2.62 V and its inverting input is held at a reference of 2.50 V. State the output voltage, and justify it.

Accepted Answer

The output is at the positive rail, about +13 V (1). The difference is 2.62 − 2.50 = 0.12 V, roughly a thousand times the 130 μV needed to reach the rail, so the amplifier is driven hard into saturation (1).

Question 14

The ideal operational amplifier is assigned infinite input resistance, zero output resistance and infinite open-loop gain. Explain what each of these three properties licenses in circuit design.

Accepted Answer

Infinite input resistance means neither input draws any current, the assumption every derivation in this topic leans on (1). Zero output resistance means the output does not sag under load, so one stage can drive the next (1). Infinite open-loop gain means an input difference of only microvolts drives the output the whole way to a rail, and it is what forces the two inputs together the moment feedback is applied (1).

Question 15

A sensor delivers a 15 mV peak signal that must be amplified to 3.0 V peak and must have a bandwidth of at least 25 kHz. The available op-amps have a gain-bandwidth product of 1.0 MHz and saturate at ±13 V. Calculate the overall gain needed, show that a single stage cannot provide it at that bandwidth, and design a two-stage inverting chain that can, checking the bandwidth of each stage and the output against the rails.

Accepted Answer

Overall gain = 3.0/(15 × 10⁻³) = 200 (1). One stage of gain 200 would be flat only to 1.0 × 10⁶/200 = 5 kHz, far short of 25 kHz (1). The most any single stage may have is 1.0 × 10⁶/(25 × 10³) = 40 (1). Two inverting stages of gain −20 and −10 multiply to +200, and each sits under 40 (1). Their bandwidths are 1.0 × 10⁶/20 = 50 kHz and 1.0 × 10⁶/10 = 100 kHz, both clear of 25 kHz (1). Bandwidth here is the −3 dB point, which is not the same as flat: each stage is already falling before its own corner, and at 25 kHz the two together leave the output about 1.2 dB down. A response that must be flat rather than merely wide enough needs its corners further out still. Output = 15 × 10⁻³ × 200 = 3.0 V (1), well inside the ±13 V rails, and the two inversions cancel so the output is in phase with the sensor (1).

Question 16

An inverting amplifier is to have a gain of −25 and an input resistance of 4.7 kΩ. Calculate the feedback resistance required, state the gain obtained if the nearest preferred value of 120 kΩ is fitted instead, and calculate the percentage error in the gain.

Accepted Answer

The input resistance of an inverting amplifier is R_in itself, because the far end of that resistor is a virtual earth (1). R_f = 25 × 4.7 = 117.5 kΩ (1). With 120 kΩ the gain is −120/4.7 = −25.53 (1), an error of (25.53 − 25)/25 = 2.1% (1).

Question 17

A comparator runs from ±13 V rails and senses light from a 12 V sensing supply. Its inverting input is fed from the junction of two equal 10 kΩ resistors connected across that 12 V supply. Its non-inverting input is fed from the junction of a light-dependent resistor connected to +12 V and a 4.7 kΩ resistor connected to 0 V. Calculate the reference voltage, calculate the LDR resistance at which the output changes state, and state which way the output moves as the light level falls.

Accepted Answer

Reference = 12 × 10/(10 + 10) = 6.0 V (1). The sensing divider gives 12 × 4.7/(R_LDR + 4.7), and the output changes state when that equals 6.0 V (1), which needs R_LDR = 4.7 kΩ (1). As the light falls the LDR's resistance rises, so the non-inverting input falls below the reference and the output swings to the negative rail (1).

Question 18

An operational amplifier has an open-loop gain of 1.0 × 10⁵ and a gain-bandwidth product of 1.0 MHz. Calculate the frequency up to which the open-loop gain holds up, and explain why a designer who wants a flat response to 100 kHz cannot have a closed-loop gain of 1000.

Accepted Answer

Bandwidth at open loop = 1.0 × 10⁶/(1.0 × 10⁵) (1) = 10 Hz (1). Above that the gain falls in proportion to frequency, the product staying at 1.0 MHz all the way down to a gain of one at 1 MHz. A closed-loop gain of 1000 gives a bandwidth of 1.0 × 10⁶/1000 = 1 kHz (1), so the response is already falling at 100 kHz; 100 kHz demands a gain of no more than 10 (1).