Question

Sometimes, we need an amplifier with an accurate gain. Even using 1% tolerance resistors may not be sufficient to provide the required gain tolerance. In these designs, trimmer resistors must be used that can be adjusted after the circuit is built. Each resistor in the circuit can be replaced by a series combination of a fixed resistor and an adjustable resistor. Consider the inverting amplifier below with R1=47kΩ and R2=450kΩ. The source signal internal resistance varies from 0 to Rsmax=1kΩ. Only one trimmer resistor can be added, and it is going to be put in series with the feedback resistor. What are the lowest and highest values R must be able to have so that the gain can be adusted to Av=−10±0.5%? Assume that all resistors have 1% tolerance.

Answer 1

Answer:

The value of R_low is 13.1 kΩ while that of R_high is 36.8 kΩ

Explanation:

The diagram is as given as

The circuit is an inverting op amp for which the gain is given as

$A_v=\dfrac{R_2+R}{R_1+R_z}$

Here R2 is given as 450 kΩ.

R1 is given as 47 kΩ.

The gain is given as

$A_v=-10\pm 0.5*\dfrac{10}{100}\\A_v=-10\pm 0.05$

For the Lowest value of R the value of the internal resistance rz is 0 and the gain is -10+0.05 so

$A_v=-\dfrac{R_2+R}{R_1+R_z}\\-10+0.05=-\dfrac{450+R_l_o_w}{47+0}\\-9.95=-\dfrac{450+R_l_o_w}{47+0}\\R_l_o_w=13.1$

So the value of R_low is 13.1 kΩ

For the Highest value of R the value of the internal resistance rz is 1 kΩ and the gain is -10-0.05 so

$A_v=-\dfrac{R_2+R}{R_1+R_z}\\-10-0.05=-\dfrac{450+R_{high}}{47+1}\\-10.05=-\dfrac{450+R_{high}}{47+1}\\R_{high}=36.8$

So the value of R_high is 36.8 kΩ