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:

e_in − e₁ = I_L R_S

Therefore, the load current is:

I_L = e_in/R_S − e₁/R_S

The first term is proportional to the input voltage e_in, and the second term is a constant. Here, e₁ is derived from the negative power-supply rail through a potentiometer:

I_L = e_in/R_S + (−e₁)/R_S

R_S 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 e_in. Here, e₁ is derived from the positive power supply by potentiometer P₁. The voltage at B equals V minus I_L R_S.

Op amp inputs at A and B are the same, so:

e1 − e_in = V − I_L R_S

I_L = e_in/R_S + (V − e₁)/R_S

The first term is proportional to the input voltage, with the second term a constant. R_S 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.

• Microcontroller Interface Delivers Standard 4- To 20-mA Output
• Inside The VTM
• Digital Potentiometers

Reprints

Printer-Friendly

Email this Article

RSS

Submit PlanetEE del.icio.us Digg Reddit Newsvine StumbleUpon Netscape Yahoo MyWeb Live Bookmarks Fark  Submit

Font Size

What's This?

Network-On-Chip Tools Arrive for The Masses Tackling System Design Challenges Through Early Verification ESL Tools Take Center Stage As Designers Move Up Parasitic Extraction Tool Targets Next-Generation Custom ICs Synopsys Jumps Into ESL-Synthesis Pool Verify Control Systems Before Committing To Hardware You're Using How Many FPGAs? Tool Up For The FPGA Blitz

1) Build A Smart Battery Charger Using A Single-Transistor Circuit (181 views today) 2) Hot Hands For Some Cool Rock: Motion Sensing Meets Audio Engineering (169 views today) 3) Science Fiction Meets Science Fact In Today's Robot Research (101 views today) 4) What's All This Transimpedance Amplifier Stuff, Anyhow? (Part 1) (92 views today) 5) Adjustment-Free Fan Controller For Under $1 (85 views today) ALL TOP 20

Reader Comments Hi, I was wondering if you could assist with a query. If I have for example one source of 500volts and convert it to current. And another source of 500volts. Can I join the outputs of the two together to get single voltage/current output? Any points to take note of? Thank you. Nick -November 19, 2009 nice site for electronics khan -March 24, 2009
POST YOUR COMMENTS HERE Name: Email:
Your Comments: Enter the text from the image below Please refresh the page if you have trouble reading this text.