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Select the largest possible value for a feedback resistor RF so that at least ±10 V of output signal swing remains available.
  • a)
    10 kΩ
  • b)
    100 kΩ
  • c)
    1 MΩ
  • d)
    10 MΩ
Correct answer is option 'D'. Can you explain this answer?
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Select the largest possible value for a feedback resistor RF so that a...
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Select the largest possible value for a feedback resistor RF so that a...
90% of the input signal is dropped across the feedback resistor in a non-inverting amplifier with a gain of 10.

Given that the gain of the non-inverting amplifier is 10, the output voltage is 10 times the input voltage. Therefore, the voltage drop across the feedback resistor is 9 times the input voltage (since 10-1=9).

To drop at least 90% of the input voltage across the feedback resistor, we need to have:

Vout × 0.9 = 9 × Vin

Simplifying this equation gives:

RF = 8 Vin / Vout

We know that the maximum value of the output voltage is limited by the power supply voltage. Let's assume that the power supply voltage is 15V. Then, the maximum output voltage is:

Vout = 15V - Vd

where Vd is the voltage drop across the diode in the feedback loop, which is typically around 0.7V.

Vout = 15V - 0.7V = 14.3V

Assuming a maximum input voltage of 1V (to avoid clipping), we can calculate the largest possible value of RF as:

RF = 8 × 1V / 14.3V = 0.56 kΩ

Therefore, the largest possible value for a feedback resistor RF so that at least 90% of the input signal is dropped across the feedback resistor in a non-inverting amplifier with a gain of 10 is 0.56 kΩ.
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