Need advice on voltage controlled current source

You can make it with a dual opamp and a few discretes. This can be simplified if you don't need high precision:

ftp://jjlarkin.lmi.net/Isrc.JPG

I think Analog Devices may have some specific 4-20 mA drivers.

John

I already have a circuit which uses two op amps and a BJT, and it works fine. But the circuit requires +15V, -15V, and +5V, which is pretty impractical for anything other than a protoboard. I need something which can run on a single positive supply voltage.

Surface mount is not so terrible to hand solder, at least for the "larger" versions of surface mount - as I assume that's why you want DIP packaging.

DIP packaging is becoming a major limitation.

a 250 ohm resistor does the basic conversion, of course. But you need a driver of some sort, so a current mirror would make sense (run your 250 ohms to ground on your board, where you can be fairly sure what it will see, and copy that current to the loop, where you don't know what will happen).

There are current loop driver ICs, but most/all current (in the up to date and available to buy sense) parts are surface mount. Adapt or be left by the wayside. TI XTR117 from Digikey in the MSOP-8 package would be the easy one to solder by hand, and a quick look at the data sheet suggests you'd use a 25Kohm resistor since it multiplies current by 100. $2.48 if you are only buying one.

--
Cats, coffee, chocolate...vices to live by

The one I posted will. Just use a dual r-r opamp and run it off V+ and ground.

John

The somewhat "classic" way...

...Jim Thompson

--
| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
| Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
| Phoenix, Arizona  85048    Skype: Contacts Only  |             |
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  |
| E-mail Icon at http://www.analog-innovations.com |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.

So, I can hand solder a surface mount chip? I've never even bothered to try. I have found numerous chips which would work for my application given that they were DIP. I will look into this. Thanks.

a o

nd

So those op amps shown in your jpg have V- set to gnd?

They certainly could, especially if your input doesn't go to ground.

You could even use my least-favorite opamp, an LM324, if you're careful about the topside common-mode problem.

John

What is it with this rash of posters who don't realize that op-amps have been fully functional on single supplies for over thirty years?

No fundamental law of electronics says that an op-amp circuit must have

+/- 15V supplies; that's a limitation of some pretty old op-amps. In fact, many current-production op-amps would burn to a crisp with 30V between their supply rails.

Find a single-supply dual op-amp, and be happy. With a bit of digging, you should be able to find one that'll sink 20mA, and then you can be _really_ happy.

--
www.wescottdesign.com

This might help:

Have Fun! Rich

In article , Andrew writes

If you are new to 4-20mA loops, be aware:

Current sourcing, like this, is fine; but current sinking is simpler to implement. You could get by with one op amp that way. Of course, current sinking requires that the external thing you interface to provides the power. I can post more details on this if you want.

Some people use more than the pure 4-20mA range of a 4-20mA loop. Typically, they use 2mA or 24mA as "fault" or "overrange" signals. If you are not sure how the receiving equipment interprets signals outside the 4-20mA range, your software should limit the output from the DAC to

1 - 5V.

If your tachometer could be hooked to a variety of equipment, you can never be sure exactly how the far end has implemented the 4-20mA "standard". That's why many 4-20mA loop interfaces have jumpers or switches to configure them to be source / sink and set their fault, etc outputs. It's handy if you know that the far end will ALWAYS be item X.

The other "gotcha" with 4-20mA interfaces is voltage drops along long loops, or through zener barriers. (A zener barrier is a combination of R's and zener diodes to limit the energy that can travel through it. Often used in certain industries.) For long cable runs, say a kilometre or two, you need to take into account the wire resistance. Then double it, because the voltage drop is *both ways* (2 wires). Let's say the total wire resistance is 15 ohms in one direction, which is 30 ohms over the entire loop. Now the stupid bit: some people sense the voltage at the far end by simply running it through a 250 ohm resistor. This gives a convenient 1 to 5 volt signal and saves them an amplifier. I have actually seen this in very old / legacy equipment; amplifiers like the LM324 now only cost 10 cents*, but used to be a significant cost.

For a maximum loop current of 24mA, that will drop 0.024 x (30 + 250) =

6.72 volts - which doesn't sound like much, but if there are other voltage drops in the system like zener barriers (which if memory serves can be 300 ohms), and you take into account the tolerance of your "24V" supply (often +/-10% in industrial applications) you can actually run out of volts, i.e. the equipment at the far end can't quite see 5V when you try to output 20mA.

--
Nemo

* Cue howls of laughter from some members of the group - "You throw away 10
cents? I just draw the symbol
for an LM324 on the back of my PCB in pencil, it works just as well" etc, etc

John, Thank you for a good solution. I modified the circuit you drew so that I can drive 20 mA through 1 to 4 250 ohm loads. And I also used BJTs because that's what I had in the shop.

thanks, AG

If you use a small packaged single op amp for the second part, you can do this rude thing:

---------+-----------+-------- ! ! ! \\ ! / ! \\ \\ ----+ / ! +--/\\/\\-- \\ ----!-\\ ! ! ! >------ +----------!+/ ! ! +--------- Load ! =3D=3D=3D ! GND

Back in the day, i would have used a current transmitter chip. Do they still have such (about 20 years later)?

I have read the responses, but I thought I'd chime in with a discrete approach:

24V --------o------o---------. | | | | | | | >| |< | PNP|--o--|PNP | /| | |\\ | | | | | o----' | | | | .-. .-. | 2.2k | | | | 820 '------o Out 1/2W | | | | 1/2W 4mA - 20mA '-' '-' | | | |/ o----|NPN | |>

|< |*A* in -------|PNP | 1V-5V |\\ .-. | | | 250 | | | 1/8W | '-' | | GND --------o------o----------------o (created by AACircuit v1.28.6 beta 04/19/05

The input PNP compensates for the Vbe of the NPN. Thus, point *A* tracks the input, and so the 250 ohm resistor will supply Vin/250 amps. So, from 1 to 5, you have 4mA to 20mA out.

The 820ohm resistor is there to prevent the driver NPN from heating up too much. The top two transistors are a current mirror. If you can deal with a current sink, you can eliminate them. Mount both the mirror transistors on the same bit of heatsink. You can also mount the bottom two together to get a bit better matching

If you want exact output, use Larkin's design. This will be a bit low at the 'high end' due to various effects.

Regards, Bob Monsen

You'll need Wilson or enhanced Wilson current mirror and emitter degeneration to make it work and get a bit of accuracy.

--
Thanks,
Fred.

Emitter degeneration up top. Although, further observation tells you that this is crap, AND doesn't meet even the _intent_ of 4-20mA measurement. ...Jim Thompson

--
| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
| Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
| Phoenix, Arizona  85048    Skype: Contacts Only  |             |
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  |
| E-mail Icon at http://www.analog-innovations.com |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.