Design formulas simplify classic V/I converter

Dudley Nye, Nye Engineering Co, Fort Lauderdale, FL -- EDN, 1/20/2000

Figure 1 shows a classic voltage-to-current (V/I) converter. You can select the resistor values such that the output current in the load, R_L, varies only with the input voltage, V_IN, and is independent of R_L. The circuit is widely used in industrial instruments for supplying a 4- to 20-mA signal. The circuit has its limitations, however, because the resistor values must be quite accurate to obtain a true current source. The literature describing the circuit provides design methods that are for special cases or are for approximate designs. This Design Idea gives two simple design formulas you can use to determine the component values that produce a true current source. It also provides a general formula for the output current, I_L, for any selection of resistor values, not just the constant-current selection.

For a true current output, I_L, as a function of the input voltage, V_IN, you must satisfy the following two equations:

In Equation 1, you can arbitrarily select any four of the terms and then determine the fifth term by solving the resulting equation. In Equation 2, you can arbitrarily select either R₃ or R₄ and then determine the unselected resistor after substituting the applicable terms from Equation 1. For example, you can solve Equation 1 for R₂ when I_L=20 mA, R₁=100 kW, R_X=0.1 kW, and V_IN=4V yields R₂=49.9 kW. Now, let R₄=100 kW and, with Equation 2, solve for R₃ as follows: R₃=(49.9 kW+0.1 kW)=50 kW. This example configures a design for the popular current source of 4 to 20 mA. In a second example, if R_X changes from 100 to 400W, the feedback changes fourfold, and you would expect that the output current would change fourfold, to 1 to 5 mA. You can check the result by substituting in the general formula for the output current:

When the complete coefficient (the terms inside the square brackets) of R_L equals zero, a true current source results, and equations 1 and 2 are valid. Note that substituting the values from the first example above forces the coefficient to zero. Substituting the values from the second example above results in the following expression:

With R_L=0.2 kW and V_IN=4V, I_L=5.019 mA. Then, with V_IN=0.8V, I_L=1.003 mA. Thus, after changing the feedback resistor by 4-to-1, you still have currents close to the 1- to 5-mA standard. Note also that I_L=5.02 mA when R_L=0W; thus, the circuit is still almost a perfect current source. This result is unique, as you can convert from 4 to 20 ma to 1 to 5 mA by changing only one resistor. You can configure the less used standard of 10 to 50 mA by making R_X=100/2.5=40W. (DI #2471)