Voltage-to-current converters feeding to grounded loads often find their way into industrial measurements and control applications. The conventional textbook circuit needs both positive and negative-supply rails (Fig. 1). In this circuit:
ein - e1 = IL RS
Therefore, the load current is:
IL = ein/RS - e1/RS
The first term is proportional to the input voltage ein, and the second term is a constant. Here, e1 is derived from the negative power-supply rail through a potentiometer:
IL = ein/RS + (-e1)/RS
RS is selected so that the first term gives 16 mA for full-scale input voltage, and the potentiometer is adjusted so that the second term provides a constant 4 mA. In effect, the output ranges from 4 to 20 mA, corresponding with zero to full input voltage. But failure of the negative power supply causes erroneous output. Moreover, there may be equipment where the negative power supply is not available, requiring generation just for this application.
In such cases, there's a slightly different circuit that works on a single-supply rail (Fig. 2). This circuit uses one half of the quad operational amplifier LM324. The first amplifier is configured as a subtractor, while the second amplifier is configured as a current converter.
The output of the first amplifier at A equals e1 minus ein. Here, e1 is derived from the positive power supply by potentiometer P1. The voltage at B equals V minus IL RS.
Op amp inputs at A and B are the same, so:
e1 - ein = V - IL RS
IL = ein/RS + (V - e1)/RS
The first term is proportional to the input voltage, with the second term a constant. RS is chosen so that the first term gives 16 mA for full-scale input voltage, and the potentiometer is adjusted such that the second term supplies a constant 4 mA. In effect, the output is 4 to 20 mA, corresponding to zero to full input voltage. Thus, this circuit works without using a negative power-supply rail. For the circuit shown in Figure 2, the current varies from 4 to 20 mA with an input of 0 to 1 V.
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