4-20mA to 0-20mA converter sips mere microamps

Circuits for converting 4 to 20mA analog current loop signals to 0 to 20mA may be hot topics, but hot implementations of those circuits are not.

Circuit designs for conversion of 4 to 20mA analog current loop signals, which are ubiquitous in process monitoring and control, to 0 to 20mA are a hot topic recently. “Hot topic” is a perfect description because typical examples of such converters can dissipate half a Watt. Some cook even hotter than that! This results in some very un-green complications like TO220 packaged power pass transistors sporting substantial heatsinks. Would Greta T. approve? I think not!

Wow the engineering world with your unique design: Design Ideas Submission Guide

The design in Figure 1 offers a cool (and maybe even useful) efficiency improvement. It thriftily recycles most of the 4 to 20mA input current to generate the 0 to 20mA output while needing only microamps from its own local power supply. It consumes merely 250uA x 24v = 6mW (typical) and dumps only similar single-digit milliwatts from Q2 (which is the closest thing it has to a pass transistor). That’s not even enough heat to make a TO92 tepid.

Here’s how it works:

Figure 1 This circuit’s operation is governed by the following equation: Iout = Iin – 4mA(1 – (Iin – 4mA)/16mA)) = 1.25(Iin – 4mA). The maximum current drawn from the local V+ supply is only 1/80^th of max Iout, translating to an 80:1 efficiency gain. Asterisked resistors are 0.5% precision or better.

The 4 to 20mA input current (99.95% of it, to be precise) passes through current sense resistor R1 and from there to the load, generating 200mv to 1v as Iin goes from 4 to 20mA. The voltage to current converter R1+A1+Q1 makes that into Ic1 = Iin R1/R2 = 2uA to 10uA as the input to current to voltage converter R5+A2+Q2+R4, generating 0.5 to 2.5v across R5. A2 compares this to A3’s 2.50v internal reference, forcing Q2 to conduct so that the sum Vr4 + Vr5 = 2.5v.

Thus Ic2 = (2.5 – Vr5)/R4 and it decreases linearly from 4mA to zero as Iin increases from 4 to 20mA. The net effect is to force Q2 to subtract a linearly decreasing 4 to 0mA from Iin, so that its 4 to 16mA span is corrected to 0 to 20mA in Iout. In other (mathspeak) words: Iout = Iin – 4mA(1 – (Iin – 4mA)/16mA)) = 1.25(Iin – 4mA)

The payoff is that 98.8% of Iout comes from recycled Iin instead being sucked anew from V+.

V+ isn’t critical and needs only be sufficient to provide the compliance required by the (grounded) load. R3 needs to provide 50uA bias for the A3pin3 shunt reference, so R3 = (V+ – 2.5)/50uA = 390k for V+ = 24v. Noncritical Z1 provides a few volts of headroom for the transistors. The total voltage drop from input to output is 5.6v.

Finally, there’s a caveat. In the event of complete loss of the nominal 4 to 20mA input current, A2 will drive Q2 into saturation. This won’t damage anything, but will force A2 to draw 5mA from V+ to supply the required Q2 base current.

Stephen Woodward‘s relationship with EDN’s DI column goes back quite a long way. Over 200 submissions have been accepted since his first contribution back in 1974. They have included best Design Idea of the year in 1974 and 2001.

Related Content

The post 4-20mA to 0-20mA converter sips mere microamps appeared first on EDN.

Become a member

Become a member

Become a subscriber

Become a sponsor