Consider an ideal op-amp differential amplifier shown in the figure below:

$then the value of \frac{V_{out}}{V_{{in}_{2}} - V_{{in}_{1}}} is equal to$

\frac{R_{G}}{R_{1} R_{2} R_{3}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{R_{1} R_{2}}{R_{3} R_{G}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{R_{1} R_{2}}{R_{G}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{R_{2} R_{G}}{R_{1} R_{3}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\frac{R_{2} R_{G}}{R_{1} R_{3}}$

$For input V_{{in}_{1}}, we can write,$

$I_{1} = \frac{V_{{in}_{1}} - V}{R_{1}} = \frac{V}{R_{2}} . . . (i)$

$\frac{V_{{in}_{1}}}{R_{1}} = V [\frac{1}{R_{1}} + \frac{1}{R_{2}}]$

$V = (\frac{R_{2}}{R_{1} + R_{2}}) V_{{in}_{1}} . . . (i i)$

$For input V_{{in}_{2}}, we can write$

$I_{2} = \frac{V_{{in}_{2}} - V}{R_{1}} = \frac{V - V_{x}}{R_{2}} . . . (i i i)$

$∴ V_{{in}_{2}} = \frac{V (R_{1} + R_{2})}{R_{2}} - \frac{R_{1}}{R_{2}} V_{x} . . . (i v)$

$V_{{in}_{2}} = V_{in} - \frac{R_{1}}{R_{2}} V_{x} (from ii) . . . (v)$

$Now, I = \frac{V_{out - V_{B}}}{R_{G}} = \frac{V_{in} - V_{x}}{R_{3}}$

$V_{x} = \frac{R_{3}}{R_{G}} V_{out}$

Substitution equation (vi) in equation (v), we get,

$V_{{in}_{2}} = V_{{in}_{1}} + \frac{R_{1} R_{3}}{R_{2} R_{G}} V_{out}$

$V_{out} = (\frac{R_{2} R_{G}}{R_{1} R_{3}}) (V_{{in}_{2}} - V_{{in}_{1}})$

Suggest Corrections

Similar questions

Then, the cut-off frequency f_{c} is equal to

Both the resistance R_{1} and R_{2} are constructed with a tolerance of 5%, then the range in which the gain of the amplifier can approximately vary is

The value of output voltage V_{0} is equal to

V_{0}

A_{C L}

A_{O L}

Join BYJU'S Learning Program