7.1 Explain what an operational amplifier (op amp) is - NSC Electrical Technology Power Systems - Question 7 - 2017 - Paper 1
Question 7
7.1 Explain what an operational amplifier (op amp) is.
An operational amplifier (op amp) is an integrated circuit that functions as a high-gain voltage amplifier. I... show full transcript
Worked Solution & Example Answer:7.1 Explain what an operational amplifier (op amp) is - NSC Electrical Technology Power Systems - Question 7 - 2017 - Paper 1
Step 1
7.6.1 Output voltage of the amplifier.
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Answer
To calculate the output voltage of the amplifier, we can use the formula for an inverting amplifier:
VOUT=−VIN×RinRf
Given that:
VIN=0.7V
Rf=170kΩ
Rin=10kΩ
Substituting these values into the formula:
VOUT=−0.7V×10000170000=−11.9V
Therefore, the output voltage of the amplifier is approximately -11.9V.
Step 2
7.6.2 Voltage gain of the amplifier.
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Answer
The voltage gain for an inverting amplifier is given by the same formula used for the output voltage, but expressed in terms of gain:
Av=−RinRf
Substituting the given resistor values:
Av=−10000170000=−17
Thus, the gain of the amplifier is -17.
Step 3
7.12.1 Output voltage of the amplifier.
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Answer
Using the inverting amplifier formula again:
VOUT=−VIN×RinRf
Given:
VIN=5V
Rf=200kΩ
Rin=20kΩ
Calculating:
VOUT=−5V×20000200000=−50V
Therefore, the output voltage is -50V.
Step 4
7.12.2 Gain of the amplifier.
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Using the gain formula:
Av=−RinRf
Substituting the known values:
Av=−20000200000=−10
Thus, the gain of the amplifier is -10.
Step 5
7.13 State ONE application of a Schmidt trigger.
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A Schmidt trigger can be used in signal conditioning to clean noisy signals, providing a stable digital output.
Step 6
7.14 A Hartley oscillator calculation.
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Answer
To calculate the resonant frequency f of a Hartley oscillator, we use:
f=2πLTC1
Where:
LT=27mH
C=47μF
Calculating:
f=2π(27×10−3)×(47×10−6)1≈141.28Hz
So, the resonant frequency is approximately 141.28 Hz.
Step 7
7.15 RC phase-shift oscillator calculation.
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Answer
For the RC phase-shift oscillator:
f=2πRC1
Given:
R=25kΩ
C=45μF
Calculating:
f=2π(25×103)(45×10−6)1≈57.76kHz
Thus, the resonant frequency is approximately 57.76 kHz.