4-20ma using 780x

[gooberpat]

I've seen a circuit using a 7805 to generate a constant current depending on the value of a resistor(R1) connected from the output of the 7805 to the common lead. The common lead to ground then is supposed to supply a constant current depending on the value of R1 and NOT affected by the load from the common to ground.

When I built it, with a specific value of R1 I get the expected current if I measure from the common to ground.
However, if I add some series resistance from the common to ground, my current changes.

I'm using a 9V battery to power the circuit. Do I need a higher input voltage?

 

[itsmoked]

Common? Ground? I'm confused. Can post your schematic?

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[benta]

I sometimes marvel at how people insist on using 30+ years old technology (such as 7805), when better devices are available.

Anyway, that setup won't fly. You have 5 V drop over the resistor, and the 7805 takes 2...3 V by itself. You will maximum have an output voltage swing of 1...2 V.

Benta.

 

[MacGyverS2000]

pat, that's not a common lead, it's a sense lead... don't connect it to ground.

Battery + connects to 7805 input. 7805 out connects to resistor. Other side of resistor connects to 7805 sense lead and load. Your current will depend upon the resistor used and not your load.

This is a student-style posting (not allowed here), but I'll let it go as it doesn't seem to be directly classroom related...


Dan - Owner

http://www.Hi-TecDesigns.com

 

[Skogsgurra]

Dan, I do not think that it is a student posting. But a typical "computer guy meets hardware" posting.

For gooberpat, these three pin regulators have a pin that is denoted Common in the symbol. And it is usually connected to ground. In the circuit you use, it turns into a "sense pin" (or rather the zero reference for the controller, the actual sensing is tied to the output of the 7805) and the controller tries to keep voltage between sense and zero reference equal to nominal output, five volts in your case.

I think that you should use something more modern. There are voltage regulators that have an internal 1.25 V band-gap reference and they will perform a lot better. Then you only drop 1.25 V over the resistor and, with the low current you have, the regulator needs less than 3 V to operate. So your 9 V battery will be OK - provided you don't need many volts across your load. Then add another battery.

If you like to work with ancient technology, there are the LM317 and the uA723. But there are modern ones as well. Just google "three pin regulator"

Gunnar Englund

http://www.gke.org

 

[benta]

LM317 won't fly either. The OP needs to source 4...20 mA, according to the headline.
But the LM317 as current regulator will only go down to 10 mA (worst case, 5 mA typ.), as it needs to be able to dump its operating current somehow.

Regards,

Benta.

 

[Skogsgurra]

Yes. I missed that he needed to go all the way from 4 mA. Thought he was after a 20 mA source. Ditch the 317. But there are modern ones. Linear and National - perhaps even Moto (now Freescale, why?) has something for you.

Gunnar Englund

http://www.gke.org

 

[gooberpat]

Thanks Skogsgurra, for some useful feedback.

You are correct, I am not a student, but a 55 year old embedded firmware engineer. (70%SW, 30%HW)
The circuit I asked about was one I found here on an earlier post.

No-one was amazed at the ancient technology suggested by one of the members on that post. I am aware the '317 has a lower voltage drop, but the other post referenced a 7805.

Yes, I knew I could get a Burr-Brown (TI) part designed to do the job, but all I needed was a box with a switch to source either 5ma (yes I meant 5 not 4) or 20ma for our factory people to do a go/nogo test.

To the rest, thanks for your constructive(?) replies.

 

[OperaHouse]

Let's save some power if you are running on a little 9V battery. A computer guy ought to have an old power supply with a TL494 switching regulator in it. Since this chip is probably older than than the poster, it shouldn't be too scary. I'm quite fond of these supplies. I have a cutoff wheel on my grinder and can usually cut the chip out with a section of board that connects half the pins and has the timing RC. 13 components that can all be found in the old computer supply except for the pot.

 

[MacGyverS2000]

Okay, so it's not a student posting (seem to be a rash of them the last week or so). Anyway, the connections I listed will get you what you want, though I too would suggest a more recent component (will work the same way on more recent stuff).


Dan - Owner

http://www.Hi-TecDesigns.com

 

[felixc]

If the drift over temperature isn't such of a problem, you could do it with a zener (or say a TL431) a npn transistor and two resistors, to achieve a drop out of about 3 volts total. Current at will depending of a resistor that you could parallel at the flick of your switch.

 

[2dye4]

LM334 from national semiconductor.

get the data sheet and set up the temp compensated
circuit. (one extra resistor and diode).

 

[OperaHouse]

Frankly the 9V supply is going to be a problem with any series type regulator scheme. The device you will be driving could have a signifignat voltage burden. Interesting that so many are stuck in analog.

 

[Skogsgurra]

Hey Opera!

Stuck in analogue? I don't see a digital solution here. What would it look like?

Gunnar Englund

http://www.gke.org

 

[benta]

I think that before this problem is addressed further, some more info is needed.
What happens at the load when 20 mA is sent to it? Will the voltage over the load be 2 V? 200 V? 1000 V?
And at 5 mA?

Until this is addressed, there is no sane answer.

Regards,

Benta.

 

[itsmoked]

Or.. Stop reinventing the wheel?

http://www.calright.com/pd_408.aspx

http://www.scan-data.com/loopcal.htm

http://www.calright.com/_coreModule...ategoryID=52&gclid=CI3n-e-LiIcCFRhFCwod6U_OYQ

http://www.omega.com/search/esearch...ort=rank&search=loop+calibrator&submit=Search

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[Skogsgurra]

This seems to be the latest, received it in the mail this morning. Voltage regulators have come a long way since the 78xx and 79xx made their debut.

http://focus.ti.com/docs/prod/folders/print/tps79901.html

is an LDO from TI that has 40 uA consumtion, can be had with down to 1.00 V output voltage and a few hundred mV Vin-Vout difference.

If you compensate for the 40 uA, you will have a 5/20 mA (or 4/20) current source that can work from one 9 V battery and still have about 6 V compliance.

Gunnar Englund

http://www.gke.org