# Need help analyzing 4-20mA signal conditioning circuit

Discussion in 'The Projects Forum' started by eng_it, Oct 25, 2010.

1. ### eng_it Thread Starter New Member

Oct 25, 2010
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I have a circuit and am trying to understand it better. It takes a 0-4V signal and translates it to a 4-20mA output. I believe the offset resistor is supposed to generate the 4mA current when Vsig is 0V and the transistor portion is supposed to supply the additional signal current (up to 16mA) when Vsig is at 4V. The feedback current in addition to the sense current gives me my output current of 4 to 20mA.

I've been told that node A is always going to be 0V due to the virtual short. I have 2 problems with this and I think this is where my confusion is coming from.
1) This seems to me to be an open loop configuration. Do the opamp inputs in this configuration act as a virtual short, therefore making node A always 0V?
2) If both opamp inputs are always 0V, shouldn't the opamp output always be 0V? Without any base current, wouldn't the opamp always be operating in shut off?

If you could give me some insight as to why this works, I would really appreciate it.

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2. ### wayneh Expert

Sep 9, 2010
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I'm not sure that circuit will work. That doesn't mean that it won't, just that I don't quite understand it.

If your goal is to drive a 4-20mA meter, and you have a dual op-amp, then there is another way. I know the attached circuit works because I made one. My apologies to the author - I've forgotten where I got the design.

You use a resistor divider to supply a slightly negative voltage to the inverting input of the first op-amp. The signal is applied to the non-inverting input. So effectively you add a small voltage to the signal and then apply the gain all in the first op-amp. The second op-amp then drives a transistor to control the main current, and watches the voltage at the low-ohms shunt. It keeps increasing current through the 4-20mA loop until the voltage matches the input coming from the first op-amp.

You can adjust the voltage range by the choice of the shunt resistor and the gain factor on the first op-amp.

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3. ### thatoneguy AAC Fanatic!

Feb 19, 2009
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The difference between the two inputs of an op-amp and its output is zero volts. Put roughly: if 1V is applied on the - side, 1V will appear on the output.

If both inputs were always tied to ground/0V, it wouldn't work very well as an active component.

This link has some info that may help you to better understand the operation of an op-amp.

Last edited: Oct 25, 2010
4. ### wayneh Expert

Sep 9, 2010
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The diagram would make more sense if the ... sense resistor wasn't tied to ground. Is it possible that it should be moved to the other end of the shunt, so that it measures current across that shunt?

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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The circuit should work as drawn.

The trick is to realize that the ground terminal is simply a reference point. The circuit is ideal for operation as a loop powered current transducer. The arrows that point up (to the supply) would be fed from the positive side of the loop power supply and the terminal shown as an arrow marked with the 'I' would connect to the negative side of the loop power supply via another measurement resistance. This other resistor would perhaps be a precision resistor which would convert the 4-20mA signal to a measurement voltage - say for a data acquisition task. A 250Ω resistor would provide a 1-5V signal.

Circuit analysis as a loop powered transducer.

In the linear range for the op-amp node A will be at ground potential.

Denote the current flowing in Rfb as Ibias. Also denote the current flowing in Rsense as Isense.

The total current I = Ibias+Isense

Note that the voltage drop across Rfb equals the voltage drop across Rsense.

So Rfb*Ibias=Rsense*Isense

Apply a condition that I=K*Ibias where K is a constant (say 100).

Hence I=K*Ibias=Ibias+Isense

Isense=(K-1)*Ibias

So

Rfb*Ibias=Rsense*Isense=Rsense*(K-1)*Ibias

Giving the overall requirement that

Rfb=Rsense*(K-1)

Say Rsense is chosen as 22Ω and K=100 is required

Then Rfb=22*(100-1)=2.178KΩ

Rather than going directly to the loop power positive rail, Roffset would more likely be fed from a local (on-board) fixed voltage regulator referenced to the ground terminal. Say you have a regulated 5V feed to Roffset. Following on with the values selected above would require a bias current of 4mA/100= 40uA for the I=4mA offset condition. With a 5V regulated source Roffset would be 5/40uA or 125KΩ.

The 36.5Ω is probably intended to be a current limiting resistor. A value of 36.5Ω is a little strange for that simple task. The OP may have confused this with the Rsense value.