Voltage/resistance to 4-20mA

Thread Starter

cts_casemod

Joined May 14, 2013
35
Hi,

I am trying to build a circuit to convert a voltage into a 4 - 20mA signal
Basically I have a POT in which the voltage divider provides between 0.8 and 4.6V from a 5V supply (or 1.2 to 2.2KOhm impedance) and I want to convert that into a 4 - 20mA signal.

Any advice?

PS: No my pot wont go all the way down to zero.
Range of values is 1230 to 2200Ohm.

Many Thanks
 
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moot

Joined Sep 20, 2009
46
A resistor is the simplest voltage-to-current device. You should be able to convert 5V into a 4-20mA current with a 250-1250Ω potentiometer (i.e. variable resistor), and nothing else. Do you have a guess at what such a circuit would look like? Remember, V = IR...
 

wayneh

Joined Sep 9, 2010
17,496
I built a voltmeter using a 4-20mA panel meter. It used one op-amp for the offset (4mA at 0V) and a second op-amp to operate as constant current loop driver, with the current proportional to an input voltage.
 

Thread Starter

cts_casemod

Joined May 14, 2013
35
A resistor is the simplest voltage-to-current device. You should be able to convert 5V into a 4-20mA current with a 250-1250Ω potentiometer (i.e. variable resistor), and nothing else. Do you have a guess at what such a circuit would look like? Remember, V = IR...

My particular pot has a range of values between 1250 and 2200Ohms.

Remember, V = IR...? It wont work without a converter of some kind
 

Thread Starter

cts_casemod

Joined May 14, 2013
35
I built a voltmeter using a 4-20mA panel meter. It used one op-amp for the offset (4mA at 0V) and a second op-amp to operate as constant current loop driver, with the current proportional to an input voltage.
That might be what I am looking for. Do you have a schematic?
 

moot

Joined Sep 20, 2009
46
My particular pot has a range of values between 1250 and 2200Ohms.

Remember, V = IR...? It wont work without a converter of some kind
Sorry I read 2200Ohms as 220Ohms. Had that been the case, just putting the pot in series with the 5V would give you the proper conversion. (There's a nice discussion of passive V-to-I concepts here. The problem with a passive converter, of course, is that the output depends on the load.)

Here is an app note from apex about V-to-I conversion with op amps, with circuit ideas.
 

Thread Starter

cts_casemod

Joined May 14, 2013
35
Sorry I read 2200Ohms as 220Ohms. Had that been the case, just putting the pot in series with the 5V would give you the proper conversion. (There's a nice discussion of passive V-to-I concepts here. The problem with a passive converter, of course, is that the output depends on the load.)

Here is an app note from apex about V-to-I conversion with op amps, with circuit ideas.
Even so, it wouldn't work. I need an active converter because any voltage drop over the line needs to be compensated to archive the same current. Thats one of the many advantages of using a 4-20mA signal.

I might do it with an op-amp, seems good enough to have a go :cool:
 

moot

Joined Sep 20, 2009
46
Even so, it wouldn't work. I need an active converter because any voltage drop over the line needs to be compensated to archive the same current. Thats one of the many advantages of using a 4-20mA signal.

I might do it with an op-amp, seems good enough to have a go :cool:
Yes, as I said, a passive V-to-I converter has several problems, including being load-dependent. Have you tried the circuits suggested by Alec_t or wayneh?

If you don't have a transistor (but if you do have lots of op amps), you could also try the attached circuit. It's a current source I found here, plus a current mirror (from the app note I linked to earlier) to amplify the current 50x. But there is a problem with it: it doesn't give you your whole range. You could get by this by finding/buying a 2k pot, or using a switch with a resistor in parallel to R1 to perhaps double the range. (Note, I haven't tested or simulated this circuit. It's just another option to try.)
 

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Thread Starter

cts_casemod

Joined May 14, 2013
35
Yes, as I said, a passive V-to-I converter has several problems, including being load-dependent. Have you tried the circuits suggested by Alec_t or wayneh?

If you don't have a transistor (but if you do have lots of op amps), you could also try the attached circuit. It's a current source I found here, plus a current mirror (from the app note I linked to earlier) to amplify the current 50x. But there is a problem with it: it doesn't give you your whole range. You could get by this by finding/buying a 2k pot, or using a switch with a resistor in parallel to R1 to perhaps double the range. (Note, I haven't tested or simulated this circuit. It's just another option to try.)

Yes. I tried alec circuit but it does not seem to be stable at 5V-6V.
Works fine at 12V.

I am trying something a bit different. Ill post results soon
 

Alec_t

Joined Sep 17, 2013
14,280
I tried alec circuit but it does not seem to be stable at 5V-6V. Works fine at 12V.
I wouldn't expect it to work well with a 5V supply using the LM324, as it's max output is ~ 1.5V less than the supply voltage, i.e. only 3.5V with the 5V supply. That's why I suggested a higher supply voltage. Or you could use a different opamp.
 

Thread Starter

cts_casemod

Joined May 14, 2013
35
Here's a very simple circuit which gives ~3.6mA-20.5mA for a 0.8-4.6V input. Close enough?
Yes, Thank you!

I used an LM2904 and an 2N3904 Transistor with a base resistor of 1.8KOhm.

I wouldn't expect it to work well with a 5V supply using the LM324, as it's max output is ~ 1.5V less than the supply voltage, i.e. only 3.5V with the 5V supply. That's why I suggested a higher supply voltage. Or you could use a different opamp.
Yes, 5V to have 20mA in the 220 Ohm resistor I used, plus the losses on the comparator internal transistor and the external NPN transistor, so about 7.5V.

The passive isolated transducer takes another 5V to supply the internal electronics @ 20mA output

All working now!!
 
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