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0 to 5V to 4 to 20mA

kurtclark

Feb 2, 2012
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I am generating a 0-5V signal from a PWM and need to convert it to 4 to 20mA, I tried a few circuits that I found online but it does seem to work as I thought it would. These circuits generally give 0 to 20ma output.

What I was looking for is 4mA for 0V input and 20mA for 5V input.
 

kurtclark

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See if this thread gives you some clues to the workings & requirements of such circuits.

It seems that the op-amps in that circuit are powered by 20V + , I would like the op-amps to be powered by 5V and only the output transistor or mosfet to be powered by 24Vdc.
 

Resqueline

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That should make it even simpler, provided the load is allowed to hang from +24V instead of being connected to ground.
 

Resqueline

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Anything is possible but the 5V op-amp requirement makes that more of a challenge. What's the reason behind that requirement/restriction?
 

kurtclark

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Anything is possible but the 5V op-amp requirement makes that more of a challenge. What's the reason behind that requirement/restriction?

I want to power the device from the current loop, so I have to ensure that circuit current is within the 3.5mA range.
 

Resqueline

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Ok, so it's to be a loop-powered sensor/transmitter.
That implies that the circuit initially receives almost 24V, but has to consume less than 4mA as you say. What's the load resistance?
It also seems to to me that it'll be floating. Are you sure about the ground requirement? Have a look at this circuit example from Maxim.
 

kurtclark

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Ok, so it's to be a loop-powered sensor/transmitter.
That implies that the circuit initially receives almost 24V, but has to consume less than 4mA as you say. What's the load resistance?
It also seems to to me that it'll be floating. Are you sure about the ground requirement? Have a look at this circuit example from Maxim.

Generally speaking the load would be 250ohm so you would get 1V for 4mA and 5V for 20mA, but I would like to make the circuit to work suitably upto 1000ohm load.
The supply voltage is typically 28VDC.
I'm sorry I didn't understand your comment about the ground requirement, yes the ground would be have to be self created.
Have a look at this circuit from Analog Devices it is loop powered, but it works on an input of 0 to 1.6V.
http://www.analog.com/en/circuits-from-the-lab/CN0145/vc.html

I do not need galvanic isolation per se even a non isolated loop transmitter would be fine.
Thanks for the link it gives me a good insight of loop powered systems.
 

Harald Kapp

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If you have a working circuit that gives you 0 mA... 20 mA output, scale it down so it gives 0 mA...16 mA output.
Then add a fixed current source (4 mA) in parallel. The total current will be 4 mA + (0 mA ... 16 mA) = 4 mA ... 20 mA.

You don't even need a second current source. Just the 4mA jus generate the power for the measurement circuit from it.

Harald
 
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kurtclark

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If you have a working circuit that gives you 0 mA... 20 mA output, scale it down so it gives 0 mA...16 mA output.
Then add a fixed current source (4 mA) in parallel. The total current will be 4 mA + (0 mA ... 16 mA) = 4 mA ... 20 mA.

Harald

Wonderful suggestion.
I will try this out.
 

kurtclark

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You don't even need a second current source. Just the 4mA jus generate the power for the measurement circuit from it.

Harald

I'm sorry I didn't understand that why I didn't need a second current source?
 
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Harald Kapp

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I'm sorry, I didn't express that clearly.
I thought along the following lines:
You have a circuit (let's call it A) that measures some property and controls at it's output a current source 0...16 mA. Now, somehow this circuit needs to be powered. The beauty of a 4...20 mA intzerface is, that there is always at least 4 mA available. You can use this idle current to generate the supply voltage for your circuit.
Look at the schematic:
attachment.php


And of course, yes, you need a second current source for that, too. I hadn't thought that through thoroughly.


Regards,
Harald
 

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kurtclark

Feb 2, 2012
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I'm sorry, I didn't express that clearly.
I thought along the following lines:
You have a circuit (let's call it A) that measures some property and controls at it's output a current source 0...16 mA. Now, somehow this circuit needs to be powered. The beauty of a 4...20 mA intzerface is, that there is always at least 4 mA available. You can use this idle current to generate the supply voltage for your circuit.
Look at the schematic:
attachment.php


And of course, yes, you need a second current source for that, too. I hadn't thought that through thoroughly.


Regards,
Harald

Excellent explanation Harald, and thanks for the drawing that indeed was quite nice!
But I have a question, Lets says my circuit takes only 1.2mA for its operation what to do?
how do I manage the remaining 2.8mA?
 

Harald Kapp

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In answer to your question:
Don't be afraid, no current will be lost. Since the sum of all current in node is 0 - Kirchhoffs law - not yet overruled by any parliament of the world :) - all current that is not required to operate your circuit will flow through the lone resistor and develop some voltage drop across it (2.8 V at 1 kOhm). By adjusting the resistor you can adjust this voltage. Adding a capacitor and a zener diode parallel to the resistor will help you to stabilize and smooth out the voltage.

Harald
 
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