Chapter 5, Problem 93CP

A voltage-to-current converter is shown in Fig. 5.110, which means that i_L= Av_i if R₁R₂ = R₃R₄. Find the constant term A.

To determine

Calculate the constant term A in the voltage-to-current converter in Figure 5.110.

Answer to Problem 93CP

The constant term A is $\underset{\_}{\frac{1}{\left(1+\frac{R_1}{R_3}\right)R_L-R_1\left(\frac{R_2+R_L}{R_2{R}_3}\right)\left(R_4+\frac{R_2{R}_L}{R_2+R_L}\right)}}$.

Explanation of Solution

Given data:

Refer Figure 5.110 in the textbook for the voltage-to-current converter circuit.

Calculation:

Modify the Figure 5.110 by indicating the node voltages. The modified circuit as shown in Figure 1.

In Figure 1, the $v_a$ and $v_b$ are the voltages at nodes a and b respectively.

Write the node voltage expression at node $a$ by the use of Kirchhoff's current law.

$\begin{array}{l}\frac{v_i-v_a}{R_1}=\frac{v_a-v_o}{R_3}\\ v_i-v_a=\frac{R_1}{R_3}\left(v_a-v_o\right)\\ v_i+\frac{R_1}{R_3}{v}_o=v_a+\frac{R_1}{R_3}{v}_a\end{array}$

$v_i+\frac{R_1}{R_3}{v}_o=\left(1+\frac{R_1}{R_3}\right)v_a$ (1)

From the general property of ideal op amp, consider that the voltage across the two input terminals of op amp is equal to zero.

$v_a-v_b=0$

Re-arrange the equation.

$v_a=v_b$ (2)

Consider the expression by using Figure 1.

$v_b=v_L$

Substitute $v_L$ for $v_b$ in Equation (2).

$v_a=v_L$

Substitute $v_L$ for $v_a$ in Equation (1).

$v_i+\frac{R_1}{R_3}{v}_o=\left(1+\frac{R_1}{R_3}\right)v_L$

$v_i=\left(1+\frac{R_1}{R_3}\right)v_L-\frac{R_1}{R_3}{v}_o$ (3)

Write the expression for the current $i_o$.

$i_o=\frac{v_o}{\left(R_4+R_2\parallel R_L\right)}$ (4)

Write the expression for the current $i_L$ by the use of Figure 1.

$i_L=\frac{R_2}{\left(R_2+R_L\right)}{i}_o$ (5)

Substitute Equation (4) in (5).

$i_L=\frac{R_2}{\left(R_2+R_L\right)}\frac{v_o}{\left(R_4+R_2\parallel R_L\right)}$

$v_o=i_L\left(R_2+R_L\right)\frac{\left(R_4+R_2\parallel R_L\right)}{R_2}$ (6)

Consider the expression for the voltage $v_L$.

$v_L=i_L{R}_L$ (7)

Substitute Equations (6) and (7) in (3).

$v_i=\left(1+\frac{R_1}{R_3}\right)\left(i_L{R}_L\right)-\frac{R_1}{R_3}{i}_L\left(R_2+R_L\right)\frac{\left(R_4+R_2\parallel R_L\right)}{R_2}$

$v_i=\left[\left(\frac{R_3+R_1}{R_3}\right)R_L-R_1\frac{\left(R_2+R_L\right)}{R_2{R}_3}\left(R_4+\frac{R_2\times R_L}{R_2+R_L}\right)\right]i_L$ ( 8)

Consider the given expression.

$i_L=A{v}_i$

Re-arrange the Equation.

$v_i=\left(\frac{1}{A}\right)i_L$ ( 9)

Compare Equations (8) and (9).

$\begin{array}{c}\frac{1}{A}=\left(\frac{R_3+R_1}{R_3}\right)R_L-R_1\frac{\left(R_2+R_L\right)}{R_2{R}_3}\left(R_4+\frac{R_2\times R_L}{R_2+R_L}\right)\\ =\left(1+\frac{R_1}{R_3}\right)R_L-R_1\left(\frac{R_2+R_L}{R_2{R}_3}\right)\left(R_4+\frac{R_2{R}_L}{R_2+R_L}\right)\end{array}$

Re-arrange the Equation.

$A=\frac{1}{\left(1+\frac{R_1}{R_3}\right)R_L-R_1\left(\frac{R_2+R_L}{R_2{R}_3}\right)\left(R_4+\frac{R_2{R}_L}{R_2+R_L}\right)}$

Conclusion:

Thus, the constant term A is $\underset{\_}{\frac{1}{\left(1+\frac{R_1}{R_3}\right)R_L-R_1\left(\frac{R_2+R_L}{R_2{R}_3}\right)\left(R_4+\frac{R_2{R}_L}{R_2+R_L}\right)}}$.