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SVETLANKA909090 [29]
3 years ago
14

An op-amp differential amplifier is built using four identical resistors, each having a tolerance of ±5%. Calculate the worst p

ossible CMRR.
Engineering
1 answer:
MatroZZZ [7]3 years ago
4 0

Answer:

80.9 db.

Explanation:

So, in this particular Question/problem we need to determine/calculate for the the worst possible CMRR. The data or information provided by the Question above;

=> ''amp differential amplifier is built using four identical resistors''

Therefore, R1 = R2 = R3 = R4. From the question above, we also know that the tolerance is ±5% for resistor that is;

A1 = A2 = A3 = A4 = A ±5%.

Using the kirchoff formula to each case, we have;

(1). Non Inverting terminal:

[ J - J2 / (A1 = A3) ] + J/ (A2 = A4) = 0.

J = 0.53 J2.

(2). Inverting terminal:

[ J_ - J2 / (A1 = A3) ] + J_ - Jb / (A2 = A4) = 0.

Thus, Jb = 2.11 J_ - 1.11 J2.

Note that J = J_ = 0.53 J2.

Thus, Jb = 1.11 J2 - 1.11 J1.

(Note that we will have that A2 = A4 = A + 0.05 = 1.05A.

Also, A1 = A3 = A - 0.05 = 0.95A).

The voltage is c1 and c2.

Therefore, Jb = 1.11( ccm + cv/2) - 1.11 ( Ccm - cv/2).

= - 10^-4 ccm - 1.11 cv.

Thus, CMMR = 20 log [| 1.11 ÷ (- 10^-4)|] = 80.9 db.

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tatyana61 [14]

Answer:

\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}

Explanation:

Previous concepts

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\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2".

If we analyze the staritning point we see that the initial velocity can be founded like this:

v_o =\omega r_{OIC}=\omega (0.15m)

And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:

H_Ai +\sum \int_{t_i}^{t_f} M_A dt =H_Af

0+\sum \int_{0}^{4} 20t (0.15m) dt =0.46875 \omega + 30kg[\omega(0.15m)](0.15m)

And if we integrate the left part and we simplify the right part we have

1.5(4^2)-1.5(0^2) = 0.46875\omega +0.675\omega=1.14375\omega

And if we solve for \omega we got:

\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}

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Explanation:

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