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lozanna [386]
2 years ago
7

We intend to measure the open-loop gain (LaTeX: A_{open}A o p e n ) of an actual operational amplifier. The magnitude of LaTeX:

A_{open}A o p e n is in the range of 106 V/V. However, the signal generator in measurement setup can supply minimal voltage of 1 mV, and the oscilloscope used at amplifier output can measure maximal voltage level of 10 V. Can you design a simple measurement setup using this signal generator and oscilloscope, and accurately measure the LaTeX: A_{open}A o p e n
Physics
1 answer:
Korvikt [17]2 years ago
8 0

Answer:

voltage divider,  R₂ = 1000 R₁

measuring the output in the resistance R₁

Explanation:

Let's analyze the situation, in an op amp in open gain loop, the gain is maximum G = 10⁶ V / V

in this case the signal generator gives a minimum wave of 1 10⁻³ V, after passing through the amplified it becomes 10³ V which saturates the oscilloscope.

To solve this problem we must use a simple voltage divider, for this we use the fact that in a series circuit the voltage is the sum of the voltages of each element.

If we use two resistors whose relationship is

            R₂ / R₁ = 10³

            R₂ = 1000 R₁

When measuring the output in the resistance R₁ we have the desired divider, with a tolerance range, for the minimum output of the generator (1 10⁻³V) we have a reading of V = 1 V in the oscilloscope, for which we can use voltage up to 10V on the generator

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A spring with a spring constant of 25.1 N/m is attached to different masses, and the system is set in motion. What is its period
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T=2 \pi \sqrt{\frac{m}{k}}

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3) 0.13 s

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T=2 \pi \sqrt{\frac{m}{k}}

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4) 7.69 Hz

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5) 1.59 s

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T=2 \pi \sqrt{\frac{m}{k}}

In this part of the problem, we have

k = 25.1 N/m

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T=2 \pi \sqrt{\frac{1.6 kg}{25.1 N/m}}=1.59 s

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