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4-20mA from .5V to 3v

Discussion in 'General Electronics Discussion' started by sid2286, Nov 19, 2011.

  1. sid2286

    sid2286

    102
    1
    Aug 24, 2011
    Hi,

    I'm trying to design a circuit where in my input from DAC is .5V to maximum is 3V I want a current output which represent 4mA for 0.5v and 20mA for 3V. I've tried with some designs but still not successful!

    Please suggest how do I achieve it!

    Thank you!
    Sid
     
  2. Resqueline

    Resqueline

    2,848
    1
    Jul 31, 2009
    You don't say exactly what you've had issues with, there could be a number of reasons.
    One "challenge" is the offset required by the slope of the current/voltage relationship. At 3V you want 20mA which is 6.67mA/V, but at 0.5V you want 4mA which is 8mA/V.
    The real slope needed is thus 6.4mA/V with a zero-Volt current of 0.8mA (or the voltage needs to be -0.125V for zero current).
    Then there's the choice of semiconductors. I don't know what you've used/tried so far but here rail-to-rail capable op-amp's are needed.
    You can try this circuit. It's insensitive to supply voltage.
    The 20k offset pot needs to be a multi-turn type, or it can be replaced with a smaller value + fixed resistors at both ends.
    The P-channel MOSFET can be replaced with a PNP Darlington transistor, or even an ordinary PNP transistor with as high gain as possible.
    The supply voltage needs to be high enough to drive the required current through the load resistance (+ 3V for the current sense resistor + some to spare).
     

    Attached Files:

  3. shrtrnd

    shrtrnd

    3,676
    454
    Jan 15, 2010
    The industry standard in U.S. is 4-20ma 1-5VDC.
    I'm guessing you believe you need the 3V at 20ma, but you're bucking the entire industry
    for process control devices. EVERYTHING is made 4-20ma/1-5V, your system won't
    be compatible with anybody else's (unless this is your own homebrew project).
    Just my 2 cents worth.
    If you can get your circuit to 1-5V, 4-20ma; you'll be able to find thousands of devices
    tailored for that use.
     
  4. sid2286

    sid2286

    102
    1
    Aug 24, 2011
    Thanks Resqueline and shrtrnd,

    The circuit that i was working is attached below.
    I would like to ask Resqueline how did you calculate the the slope, also I'm clueless about how to implement this circuit, will the circuit that you have attached will work on my .5v to 3v?
    also i'm confused on how PNP type will work?

    waiting for your reply!

    Sid
     

    Attached Files:

  5. Resqueline

    Resqueline

    2,848
    1
    Jul 31, 2009
    The circuit you attached has an input voltage of approx. 0.4V for a 20mA output. You must add a 100 ohm pot in series with R6 to make it compatible with your spec's.
    It should work, but I see it as a drawback however that it mixes the signal and the offset directly via R7 & R8.

    A slope is simply a division of two related numbers:
    20mA / 3V = 6.67mA/V
    4mA / 0.5V = 8mA/V
    (20-4)mA / (3-0.5)V = 6.4mA/V
    The two first slope numbers are "imaginary" slopes; applicable only if the curve had been crossing zero at both axis simultaneously.
    The third slope number is a so-called delta-calculation (from two points on the curve)and represents the real slope (or transfer function).

    "My" circuit will work with 0.5-3V with the range pot adjusted approximately half-way. The offset needs to be adjusted appropriately.
    Both circuits needs repeated alternate adjustments of offset and range: apply 0.5V - adjust offset for 4mA, apply 3V - adjust range for 20mA, repeat as necessary.

    PNP transistors works exactly the same as NPN transistors do.
    It may be easier to think about it if you just flip the drawing upside-down.
    It's only the polarities (& thus current directions) that change.
     
  6. sid2286

    sid2286

    102
    1
    Aug 24, 2011
    thanks Resqueline

    I tried the circuit that you had attached.
    I can get the 16mA current change but the problem is, it is not starting from 10ma to 26ma instead of 4ma.

    How do i make it start from 4mA??

    Please reply.
    Sid
     
  7. Resqueline

    Resqueline

    2,848
    1
    Jul 31, 2009
    The offset pot should take care of that. Is there no response from adjusting that? What op-amp (& power supply) did you use?
     
  8. sid2286

    sid2286

    102
    1
    Aug 24, 2011
    I have used LM324 as an opamp and supply single supply of 24v.
    rest of the components are as specified.
     
  9. sid2286

    sid2286

    102
    1
    Aug 24, 2011
    No replies yet :(
     
  10. Resqueline

    Resqueline

    2,848
    1
    Jul 31, 2009
    I'll ask again; Is there no response from adjusting the offset pot? It should have a dramatic influence.
     
  11. sid2286

    sid2286

    102
    1
    Aug 24, 2011
    hi Resqueline,

    There is a good response also i get a good amount of change for my voltage change of .5v to 3v, however, the minimum amount of current that i can go is 11mA and above i'm not able to set 4mA and if i reduce it to 4mA then the change which was drastic disappears and for a voltage change of .5v to 3v i only get a change of 4mA to 4.72mA :(

    how should i proceed ahead??

    thanks once again, please reply!

    Sid
     
  12. Resqueline

    Resqueline

    2,848
    1
    Jul 31, 2009
    Hmm, ok, it sounds like the op-amp outputs are unable to go all the way up to V+..
    Use paint to add pin numbers to my drawing, and then refer to those when measuring voltages.
    Adjust the circuit for 11mA at 0.5V - and then apply 0V. Then measure all 6 voltages (+ the supply voltage).
     
  13. sid2286

    sid2286

    102
    1
    Aug 24, 2011
    no success yet

    hi Resqueline,

    I did exactly as u suggested and the results in the attachment.


    Please suggest, i'm clueless after looking after the voltages.

    The input pins seems to be at 10v and still there is a change in the output.
    I dont know how that happens.

    I have used LM 324 as an opamp and supply voltage is 23v.

    Please look into it.

    thanks
    Sid
     

    Attached Files:

    • RND.jpg
      RND.jpg
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  14. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,214
    2,695
    Jan 21, 2010
    Can you check the voltage on pin 2 when the input voltage is 0.5V again?

    It looks like you're banging up against a limitation of the op-amp, but best to be sure.
     
  15. Resqueline

    Resqueline

    2,848
    1
    Jul 31, 2009
    Ok, thanks. The clue is there, in the details. You need a rail-to-rail op-amp.
    Although your table doesn't include a 0V input it does show the issue.
    The small voltage difference between pin 2 & 3 with a 0.5V input shows that the LM324 struggles to get the output as high as 21.23V - with a 23V supply.
    Here is a page that mentions the issue with the LM324 - and a rail-to-rail replacement.
    Other rail-to-rail examples from the first Google page includes AD822, LT1677, MCP6241 series, OPA342 series, & TLV2731.
    But another issue is the P-FET. It conducts 11mA even with a "negative" gate voltage (21.34V - 21.23V). It seems to me to be a depletion-mode MOSFET, or a JFET.
    You would not be able to get it to conduct 0mA even with a rail-to-rail op-amp.
    So an option adressing both issues is to supply the LM324 with an even higher voltage than the existing V+, while still keeping the +23V as a supply to the resistors.
     
  16. sid2286

    sid2286

    102
    1
    Aug 24, 2011
    OK tired with decreasing the voltages at the resistor and it worked!!!

    Can you please explain me the circuit!

    i have no clue on how it works!

    Thanks once again!

    it was really helpful...

    Resqueline...you are too good!

    Regards,
    Sid
     
  17. Resqueline

    Resqueline

    2,848
    1
    Jul 31, 2009
    Ok, attached is approximately the circuit you have now, which will work with a 324 and a P-JFET.
    (I had originally intended the use of a rail-to-rail op-amp and an enhancement-mode P-MOSFET.)

    The first op-amp simply inverts the ground-related signal, making it appear upside-down and relating to/from the positive rail.
    This is necessary since the range (current setting) resistors are connected to the positive rail, enabling a ground related load.
    The offset pot shifts the "hinge" point of the (inverted) signal up & down as needed.
    The second op-amp adjusts the transistor bias as neccessary to keep the inverted signal voltage constant across the range resistors.
    If the load resistance changes then the transistor bias will change in order to keep a constant voltage across the range resistors. Hence a constant current flows.

    The clue to understanding op-amps is remembering that they will always try to adjust their outputs as needed to keep their inputs at exactly the same voltage.
    Then it's only a matter of thinking consequences (if this voltage is here then that voltage has to be there and so this current has to flow there).
    No current ever flows into or out of the op-amp inputs.
     

    Attached Files:

    Last edited: Dec 30, 2011
  18. sid2286

    sid2286

    102
    1
    Aug 24, 2011
    Thanks Resqueline !

    I was trying this experiment , I tried reducing the resistor voltage away from 23v, As i go on reducing the voltage(at resistor), i found that the accuracy goes on decreasing on different loads.
    I tried this experiment with different loads such as 100E,150E,220E,470E

    On 470E it struggles to go on 20mA. it get stuck with 17mA(max).

    can you please suggest me what may be the problem?

    Also tell me how do i get my basics right....just the way you got it :)

    Thanks Once again!

    Resqueline.
     
  19. Resqueline

    Resqueline

    2,848
    1
    Jul 31, 2009
    20mA in 470 ohms requires only 9.4V which should be well within the capabilites of the circuit - driven from 18-20V.
    The question is how you reduced the voltage to the resistors, and what transistor did you use on the output?
    Even the basics takes time to absorb, you just need to keep on fiddling with it. It took me ten years to get on top of it all.
     
  20. sid2286

    sid2286

    102
    1
    Aug 24, 2011
    Ok! how did u calculate 9.4v for 20 mA??

    I have a variable power supply, wherein i can check the voltages(that is how i varied).

    and over you comment "Even the basics takes time to absorb, you just need to keep on fiddling with it. It took me ten years to get on top of it all."

    I can only say Resqueline, Take a bow master!

    Regards,
    Sid
     
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