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7.1 Explain what an operational amplifier (op amp) is - NSC Electrical Technology Electronics - Question 7 - 2017 - Paper 1

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7.1 Explain what an operational amplifier (op amp) is. 7.2 State TWO advantages of using integrated circuits (such as op amps) over discrete components (circuits b... show full transcript

Worked Solution & Example Answer:7.1 Explain what an operational amplifier (op amp) is - NSC Electrical Technology Electronics - Question 7 - 2017 - Paper 1

Step 1

Explain what an operational amplifier (op amp) is.

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Answer

An operational amplifier, or op amp, is an integrated circuit that amplifies the difference in voltage between its two inputs. It typically consists of a high-gain differential amplifier with a high input impedance and low output impedance, making it useful for various linear applications.

Step 2

State TWO advantages of using integrated circuits (such as op amps) over discrete components.

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Answer

  1. Compactness: Integrated circuits can incorporate many components into a single chip, minimizing space and reducing overall size in electronic circuits.
  2. Cost-Effectiveness: They are generally cheaper to manufacture due to reduced parts and labor, lowering production costs for electronic devices.

Step 3

Describe how a differential amplifier works.

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A differential amplifier compares two input voltages and outputs a voltage that is proportional to the difference between these two inputs. It amplifies this difference while rejecting any voltage common to both inputs, known as common-mode voltage, which enhances signal integrity in applications like audio and sensor signal processing.

Step 4

Name the type of feedback found in the following circuits: 7.4.1 Amplifier circuits

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Answer

Negative feedback is commonly found in amplifier circuits, as it stabilizes gain and reduces distortion.

Step 5

Name the type of feedback found in the following circuits: 7.4.2 Oscillator circuits

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Answer

Positive feedback is typically used in oscillator circuits to maintain oscillation and generate sustained waveforms.

Step 6

Explain the difference between positive feedback and negative feedback.

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Positive feedback enhances and increases input signal deviations, potentially leading to instability or oscillation. In contrast, negative feedback reduces output fluctuations by feeding a portion back to the input with an opposite phase, improving stability and bandwidth.

Step 7

Calculate the: 7.6.1 Output voltage of the amplifier.

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Answer

Using the formula for an inverting amplifier:
V_{out} = - rac{R_f}{R_{in}} imes V_{in}
Here, (R_f = 170 k\Omega) and (R_{in} = 10 k\Omega):
Vout=17000010000×0.7=11.9VV_{out} = - \frac{170000}{10000} \times 0.7 = - 11.9 V

Step 8

Calculate the: 7.6.2 Voltage gain of the amplifier.

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The voltage gain (A) is calculated as follows:
Av=RfRin=170kΩ10kΩ=17A_{v} = - \frac{R_f}{R_{in}} = - \frac{170 k\Omega}{10 k\Omega} = -17

Step 9

Give ONE application of a monostable multivibrator.

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Answer

A common application of a monostable multivibrator is in pulse generation, such as creating a time delay for signal processing or as a debounce circuit in switch controls.

Step 10

Explain the main difference between a monostable multivibrator and a bi-stable multivibrator.

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Answer

The main difference is that a monostable multivibrator has one stable state and changes its state upon triggering, while a bi-stable multivibrator has two stable states and switches between them based on triggering events.

Step 11

Redraw the input waveforms for 7.10.1 INTEGRATOR OP AMP.

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The input waveform consists of a step function, and the output waveform will resemble a ramp function, indicating that the integrator creates a continuous voltage ramp over time.

Step 12

Redraw the input waveforms for 7.10.2 INVERTING COMPARATOR OP AMP.

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The input waveform will appear as a triangular wave, while the output will produce a square wave, showing the switching behavior of the comparator based on input threshold levels.

Step 13

Redraw the input waveforms for 7.10.3 INVERTING SCHMITT TRIGGER OP AMP.

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The input will show a sinusoidal waveform, and the output will be a square wave triggered at defined upper and lower threshold values, demonstrating hysteresis.

Step 14

Redraw the input waveforms for 7.11.

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The input waveforms will consist of multiple sine waves that interact, and the output should reflect the summation of these inputs, likely displaying a combined waveform characteristic of an inverting summing amplifier.

Step 15

Calculate the: 7.12.1 Output voltage of the amplifier.

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Answer

The output voltage can be calculated using the inverting amplifier formula:
Vout=RfRin×VinV_{out} = - \frac{R_f}{R_{in}} \times V_{in}
Substituting in values:
Vout=200kΩ20kΩ×5V=50VV_{out} = -\frac{200 k\Omega}{20 k\Omega} \times 5 V = -50 V

Step 16

Calculate the: 7.12.2 Gain of the amplifier.

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Answer

The gain is calculated as follows:
Av=RfRin=200kΩ20kΩ=10A_{v} = - \frac{R_f}{R_{in}} = - \frac{200 k\Omega}{20 k\Omega} = -10

Step 17

State ONE application of a Schmidt trigger.

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Answer

A Schmidt trigger is often used to convert analog signals to digital signals, providing clean signal transitions and eliminating noise.

Step 18

Calculate the resonant frequency for 7.14.

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Answer

The resonant frequency is calculated using the formula:
f=12πLCf = \frac{1}{2\pi\sqrt{LC}}
Substituting values:
L=27mH,C=47μFL = 27 mH, C = 47 \mu F
f=12π(27×103)(47×106)141.28Hzf = \frac{1}{2\pi\sqrt{(27\times10^{-3})(47\times10^{-6})}} \approx 141.28 Hz

Step 19

Calculate the frequency of the oscillator for 7.15.

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Answer

The frequency can be calculated using the formula:
f=12πRCf = \frac{1}{2\pi RC}
Here, (R = 20 k\Omega, C = 45 pF):
f=12π(20×103)(45×1012)57.76kHzf = \frac{1}{2\pi(20 \times 10^3)(45 \times 10^{-12})} \approx 57.76 kHz

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