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I'm hoping someone can help me remedy this problem (or at least come up with a temporary solution), as I am a novice in electronic circuits.

The below circuit shows my setup. Vin is my input signal which ranges from 1-5VDC, and it goes through a voltage buffer and then a 250Ω which converts the signal into a 4-20mA output Vout.

schematic

simulate this circuit – Schematic created using CircuitLab

This circuit output was working as expected for most of the range for 1 to <5V, but as soon as the signal starts to reach 5V, the voltage stays stuck at ~4.79V (~19mA) and never gets to 5V (20mA). From this, I learned that Op amps have an output impedance which I assume causes this voltage drop problem. I calculated an output impedance of about ~11Ω in the TSV612A which -added with the 250Ω- causes the voltage to max out at 4.79V for an input signal of 5V.

One solution that worked was to increase Vop (Supply voltage of the Op amp) to 5.5V, so that it had a little more voltage to work with in order to compensate this drop.

My question is this: is there another solution that will help to fix this voltage issue without increasing the 5V supply on the Op amp or changing the 250Ω resistor? Are there any characteristics of an Op amp that I could be forgetting about that might help me remedy this problem?

Thank you!

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    \$\begingroup\$ For a given load resistance, opamps can only go so much near to their supply rails, how much exactly depends on the opamp, but none can get there really at 100% \$\endgroup\$ – PlasmaHH Jan 16 '17 at 16:03
  • \$\begingroup\$ @PlasmaHH this opAmp seems to be rail-to-rail IO \$\endgroup\$ – Nazar Jan 16 '17 at 16:04
  • \$\begingroup\$ Although superficially you don't seem to be violating the datasheet specs, the bottom of page 10 reads "These products are micro-power, low-voltage operational amplifiers optimized to drive rather large resistive loads, above 10 kΩ"... \$\endgroup\$ – brhans Jan 16 '17 at 16:08
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    \$\begingroup\$ @Naz: rail2rail isn't a magic word to attach to an opamp datasheet to defy laws of physics. Every real device has an output impedance (or comparable characteristic). The difference between Vcc and Vout may be immeasurably low, but it exists. In case of the mentioned opamp, the datasheet specifies 35mV away from Vcc and GND for a 10kΩ load. \$\endgroup\$ – PlasmaHH Jan 16 '17 at 16:09
  • \$\begingroup\$ Also, mA is a current, not a "Vout". Your Vout is still at best cases 1V to 5V, mA is the current forced through the resistor and can only be 4mA at 1Vin and 20mA at 5Vin if nothing at all is attached to Vout. \$\endgroup\$ – Asmyldof Jan 16 '17 at 16:28
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I think you completely mis-understand what a 4-20 mA loop is.

In your circuit, even if you could get the current to 20 mA through the R1 resistor you could not drive an external loop Rx with it.
Most 4-20 mA loops use an external power supply for loop power and may have significant voltage drop in their Rx circuits and connecting wiring.

You should study devices such as this which produce an accurate loop current, though it's supply voltage for the op-amp is much higher than 5V:

enter image description here

Your circuit might work better like this, though the input voltage is 0.4 - 2.0 V for 4-20 mA:

schematic

simulate this circuit – Schematic created using CircuitLab

The FET has a very low VGS(th) and the output loop could only withstand a 12 V loop supply.

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  • \$\begingroup\$ You're right, I do have very limited knowledge when it comes to 4-20mA loops. Does using the external supply prevent that voltage drop I'm seeing? Since the Op amp will no longer be sourcing the 20mA. \$\endgroup\$ – AvantGarde116 Jan 16 '17 at 19:52
  • \$\begingroup\$ Your circuit has no loop, and no, the external supply would not help your circuit. All you are trying to do is define a current of 20 mA through a 250 Ohm resistor. If you lifted the ground end of the resistor and tried to make a loop, any load would appear in series and alter your current. You simply can't do it that way. \$\endgroup\$ – Jack Creasey Jan 16 '17 at 20:48
  • \$\begingroup\$ I think I'm beginning to understand now. My circuit will only work for the exact case of a 250 Ohm load, so this will not work for my design. In your suggested circuit: That MOSFET helps to keep the voltage supplies isolated to prevent damage, but is still able to source 4-20mA. Would your circuit be considered a 2-wire 4-20mA transmitter? \$\endgroup\$ – AvantGarde116 Jan 16 '17 at 21:21
  • \$\begingroup\$ Yes, the circuit I provided could be called a 4-20 mA Tx, but they typically withstand up to 24 VDC loop supply, the power limitation of the FET I used means it would only be viable at 12 V. This NI pater is a great introduction: ni.com/white-paper/6940/en \$\endgroup\$ – Jack Creasey Jan 16 '17 at 23:18
  • \$\begingroup\$ Thank you for your help, this has really helped save me a lot of time and frustration. \$\endgroup\$ – AvantGarde116 Jan 16 '17 at 23:37
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My question is this: is there another solution that will help to fix this voltage issue without increasing the 5V supply on the Op amp?

No, driving that 250 ohms at 5 V requires 20 mA. I have yet to see an opamp which can supply 20 mA while having 0 V voltage drop between positive supply and output.

You will need to account for at least about 0.1 V drop between an opamp's positive supply and output. So at Vop = 5 V the best you can expect with a very good opamp is 4.9 V.

At no current an opamp can make the full supply voltage at the output (so 5 V in your case) but you need some current so a voltage will be dropped, no way around that.

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  • \$\begingroup\$ Thank you, that's what I was afraid of. I was hoping there might be a trick to compensate the drop with what limited changes I can make. But it looks like that might not be the case. \$\endgroup\$ – AvantGarde116 Jan 16 '17 at 16:29
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schematic

simulate this circuit – Schematic created using CircuitLab

I'm not prepared to write it up fully, but I think your solution will be to generate 4-20 mA from a lower voltage. Perhaps divide your input so 5V becomes 4V, and use a 200 ohm resistor on the output - something like that. No doubt there are cleverer or more efficient ways to achieve the same end, but that should set you on the right direction to get to them. Likewise, there can be quibbling about the precise resistor values chosen, etc. - I'm just going for the general idea.

Depending on your sensor you might want another op-amp to buffer the signal (or buffer and "amplify" by 0.8, or invert and buffer and amplify by -0.8.)

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  • \$\begingroup\$ Thanks for your input, I should have mentioned that I am restricted and can't change the 250Ω resistor. Although you're right - it would definitely work if it were lower. Will edit my question. \$\endgroup\$ – AvantGarde116 Jan 16 '17 at 16:26
  • \$\begingroup\$ Well, I'm sure a boost converter will solve your supply problem, but you'll tell me you can't do that, either, so either apply magic, change the resistor, or add a 1K resistor in parallel with it. Two of the solutions are a lot less work than the other one. \$\endgroup\$ – Ecnerwal Jan 16 '17 at 16:37
  • \$\begingroup\$ For the 1K resistor in parallel, where would it go exactly? In parallel with the 250Ω resistor? I would like to try that suggestion. \$\endgroup\$ – AvantGarde116 Jan 16 '17 at 16:50
  • \$\begingroup\$ You'd still need the voltage divider on the input to scale the sensor down. The 1K would then go in parallel with the 250 to make a 200 equivalent. \$\endgroup\$ – Ecnerwal Jan 16 '17 at 23:59
  • \$\begingroup\$ Ah I see! Thank you very much! Your suggestion works perfect as a temporary solution for the current circuit - which I will have to do before designing Jack's solution. \$\endgroup\$ – AvantGarde116 Jan 17 '17 at 15:17

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