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how to control voltage with 4-20mA

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panfilero

Member
Hello,

I'm trying to figure out a way to use 4-20mA to control 0-40V.... basically, I have a current output module that I can control with my computer... but I need to use these 4-20mA to supply 0-40V across a resistive load (it's a heater), so the heater is around 80Ω and will draw around 0.75A....

I'm googling around the internet for a transistor circuit that may accomplish this... but I was also thinking of maybe using an op amp...

Right now I'm thinking about a BJT with the load between Vcc and the Collector, and sending my 4-20mA to the Base... which would basically be using 4-20mA to control a larger current.... ic = βib ....

i was wondering if someone could tell me if I'm on the right path with this solution or if there's a better way

much thanks!

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I just found the TI OPA548, which is an op amp that can do up to around 60V and 5A.... plenty for what I'm trying to do, so I tried to spice my circuit (with the OPA547 which is limited to 0.5A) but my simulation isn't giving me the results I expected... there was an E/S pin that I needed to pull high according to the datasheet, to make this work... but I was shooting for 40V out and I'm only getting around 7.75V.... I calculated my values for the gain according to Av = 1 + R3/R2... but it's not giving me that gain... can anyone tell me if I hooked up something wrong here, thanks!

opa547.JPG
 
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MikeMl

Well-Known Member
Most Helpful Member
AFAIK, high-power OpAmps like the 548 are not Rail-to-Rail, so the inputs cannot be referenced to V-.
 

Mr RB

Well-Known Member
You can do it with one transistor if you don't mind a few percent error, and maybe not getting full range.

Like you said use a NPN transistor, the 4-20mA current to its base resistor to ground. So it produces a DC voltage on the base resistor. Then use a value of emitter resistor so the base voltage causes an emitter voltage and hence sets the load current.

I'd try maybe emitter resistor 2v = 0.8A (2.7 ohm?) and a darlington NPN like TIP31C which needs 1v base to turn on. So input 20mA = 3v base = 2v emitter. You would get a range; 6.7mA = 0.0A load, 20mA = 0.8A load.
 

colin55

Well-Known Member
How on earth did you work these figures:

I'd try maybe emitter resistor 2v = 0.8A (2.7 ohm?) and a darlington NPN like TIP31C which needs 1v base to turn on. So input 20mA = 3v base = 2v emitter. You would get a range; 6.7mA = 0.0A load, 20mA = 0.8A load.
 

panfilero

Member
MikeMI -
Ummm... so basically I hooked up this op amp wrong? I can't reference my load from ground if I have ground going to my -V? I thought this was how you hooked up a standard non-inverting, neg feedback op amp....

Mr RB/colin55 -
I'll pursue the BJT solution then... I guess I would put my load between my supply and the common terminal and then try to bias it and control it with my 4-20mA... which I guess I can't just shoot straight into the base, I'll have it go across a resistor (250) and then control the voltage across that resistor via the BJT... so, is 40V and .75A not a problem for a BJT, and I can just control it linearly with my 4-20mA? Is this a better approach than with an op amp?

thanks all!

I still really don't see what's wrong with the op amp circuit.....
 
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Mr RB

Well-Known Member
You can control it "reasonably" linearly with the circuit i suggested, I used an almost identical current control system on the Linistepper project.

Your 4-20mA goes to the base and the top of the base resistor. So 20mA = 3v on the base resistor (Colin!) ie use a 150 ohm base resistor.

The TIP31C drops almost exactly 1volt Vbe at those currents and only requires a fraction of a milliamp base current.

With 3v on the base of the transistor the emitter resistor voltage is 2v, and produces load current of 0.8A (ie 2.5 ohms).

So it makes 6.7mA in = 1v in = 0A, then 20mA in = 2v out = 0.8A. With the picture now Colin? ;)

As for the opamp, well if you can use one transistor and 2 resistors, does it need the opamp? You did not mention any need for high accuracy?
 

colin55

Well-Known Member
So 20mA = 3v on the base resistor (Colin!) ie use a 150 ohm base resistor.

The TIP31C drops almost exactly 1volt Vbe at those currents and only requires a fraction of a milliamp base current.

With 3v on the base of the transistor the emitter resistor voltage is 2v, and produces load current of 0.8A (ie 2.5 ohms).

So it makes 6.7mA in = 1v in = 0A, then 20mA in = 2v out = 0.8A. With the picture now Colin? ;)
I have never heard so much hocus pocus in all my life. Absolutley nothing of what you say makes any sense.
 

panfilero

Member
"As for the opamp, well if you can use one transistor and 2 resistors, does it need the opamp? You did not mention any need for high accuracy? "

That's fine about an alternate solution and not using the op amp... but I'm just trying to understand... what I'm doing wrong with the op amp... just for my own knowledge regardless of whether I use it or not... from my books this op amp set up should of worked... but spice says otherwise... I'm just trying to figure out what I did wrong

much thanks!
 

Mr RB

Well-Known Member
I have never heard so much hocus pocus in all my life. Absolutley nothing of what you say makes any sense.
No, its right. With the exception of my failing memory, the darlington I mentioned is a TIP122 not a TIP31C. Sorry about that. It's been a few years since I designed that product.

The whole thing is only 2 resistors and 1 transistor, what part of it are you not getting?
 

colin55

Well-Known Member
Mr RB. Don't you realise that nothing of what you say is correct.

1. What is the point of including a 150R base resistor for a transistor with a very high input impedance?
2. How do you work out that 6.7mA base current will produce zero collector current?
3. How do you work out that 20mA base current will produce 800mA collector current?
4. What is the purpose of the 2R5 emitter resistor?

I have never heard anything like this from any of my 15,000 students.
It must come from some sort of new text-book that I have not heard about.

You may be taking values from circuit you designed, but expressing it the way you have done in your post leaves out a lot of data and the end result is a complete schmozzle.
 
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Mikebits

Well-Known Member
schmozzle ? Is that from the Lavern and Shirley theme song? :)
 

Mikebits

Well-Known Member
Nah, its from a book of 'Tricks and Magic', :rolleyes:
Lol, My humor is so unappreciated... :eek:

Theme song from Lavern and Shirley

Spoken: One, two, three, four, five, six, seven, eight. Schlemeel, schlemazel, hasenfeffer incorporated.

We’re gonna do it!
Give us any chance, we’ll take it.
Give us any rule, we’ll break it.
We’re gonna make our dreams come true.
Doin’ it our way.

 

Diver300

Well-Known Member
Most Helpful Member
The way that I would approach the problem is like this:

Design a 40 V power supply, with feedback, probably using a switch-mode IC. That will have a feedback point that is held at 1.25 V, as long as you use and adjustable voltage regulator such as LM2576HVT-ADJ http://www.electro-tech-online.com/custompdfs/2009/05/LM2576HV.pdf

Convert the 4 - 20 mA to 1.25 - 6.25 V and buffer so that there is little impedance. A 250 Ω resistor and a op amp that gives a gain of 1.25 will do.

Connect the buffered 1.25 - 6.25 V to the feedback point using a suitable resistor, using the value of the feedback resistors for you calculations. For instance, if you have 10k and 310k as your feedback resistors, you need 38.75 k (calculation below)

That will give you 0 - 40 V for 4 - 20 mA. It will be reversed, so 4mA will give you 40 V and 20 mA will give you 0V, so sort it out in software or the op amp.

Calculation, if anyone cares:-
1.25 / 40 = 1/32 so that is why 10k and 310k would work as feedback resistors.

When the op amp output is 1.25 V, there is no voltage difference between that and the feedback point, so resistance doesn't matter.

When the op amp output is 6.25 V, the voltage output is 0 V, but the feedback point is still 1.25 V. 125 µA will flow through the 10k to ground. 4.0322 µA will flow through the 310 k to the voltage output, which is also at ground. Total is 129.0322 µA.

Control voltage is at 6.25 V, 5V above the feedback point. So the 129.0322 µA is given by a 38.75k resistor.
 

Diver300

Well-Known Member
Most Helpful Member
On your original OPA547 circuit, there are 3 things wrong with it.

1) The load resistor should be 53.3 Ω for 0.75 A load. You are getting 7.75V because that is the current limit of the OPA547 times the 10 Ω shown.

2) The max output voltage of the OPA547 is 2.2 V less than the supply at 0.5A load, so you need a 43V supply at least.

3) You don't have an offset for the 4 mA. That will give 1 V across R1, so it will give you 8 V on the output, if it wasn't for point 1). You need a resistor to a regulated +ve supply from the -ve input of the OPA547 to give you 0 V output with 4 mA in.

Also, you should be aware that you will generate about 10 W of heat in the OPA547 so you need a good heatsink.
 

Mr RB

Well-Known Member
Mr RB. Don't you realise that nothing of what you say is correct.

1. What is the point of including a 150R base resistor for a transistor with a very high input impedance?
Because the resistor from base to ground turns the 4-20ma current into a reference voltage at the base of the NPN darlington.

2. How do you work out that 6.7mA base current will produce zero collector current?
Because 6.7mA will produce about 1v at the base of the darlington, which will be the point that it starts to turn on. Currents below 6.7mA will cause No collector current.

3. How do you work out that 20mA base current will produce 800mA collector current?
20mA at the base resistor will produce a base voltage of about 3v. The darlington will drop very close to 1v Vbe. So the voltage at the emitter is 2v, setting the collector current at 0.8A by choosing the right emitter resistor.

4. What is the purpose of the 2R5 emitter resistor?
See above.

Now let's PLEASE put this to bed. It's an incredibly simple 3-component emitter-follower current controller. I was right, I explained it right, repeatedly in ALL of my posts with the exception of forgetting the transistor part number.

I've been an electronics professional for 25 years and don't like your implication that I am dumber than any of your "15000 students".

It's quite scary that you have had so much trouble understanding this... And you teach?? :eek:
 

colin55

Well-Known Member
Without seeing the circuit, the way you have described it infers a base resistor. What you are now saying is the resistor is connected between base and 0v. A "base resistor" comes directly off the base and is used to limit the current into the base.
Secondly, you say the circuit is an emitter follower and yet the original request was for the control of a collector voltage in the order of 40v.
The circuit you have described is not an emitter-follower. It is actually classified as a constant-current circuit.
How can you accurately set the collector current by selecting an emitter resistor?
The base-emitter voltage of a darlington transistor varies from just over 1v when the collector current is very small, to over 1.4v when the collector currrent is maximum.
You will have to adjust the emitter resistor to suit the transistor but 2R5 will produce a collector current of about 600mA to 700mA.
 
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panfilero

Member
MR RB, Much thanks for your input... I didn't find your explanation hard or confusing to understand, and my quick simulation proves that you were spot on form the get go, very much appreciate the help!

TIP122.JPG
 

Sceadwian

Banned
colin, it's hard to argue with a working circuit.
 
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