1
\$\begingroup\$

I want to translate a Pt100 RTD to a 0-10V range, to be read by a PLC.

I designed a circuit (instrumentation amplifier) using the LM324 op-amp. It is powered using a single 12V supply.The resistor R11 is to be referred to as a Pt100 RTD. I am supposed to get 0 volts at RTD resistance 100Ω, but I get either 2.2V or fluctuating mV if I adjust the pot R4.

Also, I am supposed to get 10V at RTD resistance 250Ω, but I get only 6.5V.

As per the gain calculations I took all flexible possibilities. However R4 (the gain pot) doesn't change the gain much.

Where am I going wrong?

schematic

simulate this circuit – Schematic created using CircuitLab


Update from the comments: You can think of R11 as a Pt100 heated from 0 to 400 degree C, giving 100 to 250 ohm resistance.

\$\endgroup\$
  • \$\begingroup\$ "Pot R12"? I see POT R11. "floating mV" - do you mean that the mV signal at the output is rising up and down irregularly - hint - "floating" is a word reserved for a signal that does not (or appears not to) have a ground connection reference. \$\endgroup\$ – Andy aka Nov 5 '18 at 12:23
  • \$\begingroup\$ yes rising up and down (What I mean by floating). \$\endgroup\$ – Ohbhatt Nov 5 '18 at 12:30
  • 1
    \$\begingroup\$ You won't get the output of the LM324 able to get near 0 volts without loading it with something like a 1 kohm resistor to the 0 volt rail. \$\endgroup\$ – Andy aka Nov 5 '18 at 12:49
  • \$\begingroup\$ @Andyaka I tried the 1K load. no difference at all. What I am curious about is the gain set by R1 and R4 combination. Should I try making OA1 and OA2 unity gain buffers and then amplify in the next stage? or add one more stage all together? Please suggest. \$\endgroup\$ – Ohbhatt Nov 5 '18 at 13:04
  • \$\begingroup\$ Can you possibly find a more complicated way to do this? \$\endgroup\$ – Robert Endl Nov 5 '18 at 13:07
3
\$\begingroup\$

Your are trying to operate your op-amps beyond the rails.

The reference leg of your bridge is at, $$ V_R = 12 \dfrac{100}{4800} = 250 \text{ mV} $$

The RTD leg of your bridge will span from 250 mV to $$ V_{T,max} = 12\dfrac{250}{4950} \simeq 606 \text{ mV} $$

Say for arguments sake, the RTD leg is outputting 300 mV. You now have a differential input voltage of 50 mV.

Now if the op-amps were operating in their linear region of operation, both input terminals of OA1 would be at 250 mV. Similarly, both input terminals of OA2 would be at 300 mV.

Ignore potenimoter R4 and treat R1 as the gain setting resistor. The two op-amps are trying to drive their outputs to develop the 50 mV differential potential across R1. The top of R1 at 250 mV, the bottom of R1 at 300 mV.

Now OA1 is going to try to sink the 50 uA flowing through R1. OA1 must drive its output low enough to allow 50 uA to flow through R2. In this that would be -1.1 V across R1, placing OA1 output terminal at -850 mV. OA1 clips at the ground rail (assuming it had the drive strength pull all the way to ground).

Since the bride output is uni-polar, OA1 is always going to be trying to sink current below with its output at or below 250 mV (where the LM324 has no drive strength left).


Loosely speaking, getting an linear output of 0 - 10V with an LM324 on a 12 V supply is going to be a challenge. I suggest you start with split 15V supplies to gain an understanding how this circuit works (though spice is an even easier place to start).

You may also want to add a resistor below your bridge to place the common-mode voltage mid-supply.

\$\endgroup\$
  • \$\begingroup\$ I was thinking similar to you, in drive strength terms, I hence modified the circuit. I worked. Above all, I got linear gain from 153mV - 10V even on a single 12V supply. I Conclude that the general 3 op-amp in-amp Configuration with varying gain(Using a single resistor Rg) is ideal for split supply. \$\endgroup\$ – Ohbhatt Nov 5 '18 at 18:04
1
\$\begingroup\$

You just have too many variables floating around. Tune your in-amp with some known voltages or resistors to verify it's function, Use 1% resistors (or better).

You also have no way to balance your bridge. If you need zero output at 100\$\Omega\$, replace the RTD with an accurate 100\$\Omega\$ resistor, and use a trim pot on the bridge to tweak it out (keeping in mind that the LM324 is not guaranteed to go below 20mV with a single supply -- You'd have better luck working around midscale).

Lastly, RTDs are not subtle devices. There's no reason to use an in-amp to amplify a 250% change in a resistance.

\$\endgroup\$
  • \$\begingroup\$ @Scott_Seidman I Desire a gain of around 28, which if amplifies a 20mV noise, I will get an error offset of 0.56V I can still tolerate that at a 5.6% error. The bridge tuning, I did and kind of worked. But what about the upper 6.5V limit against a maximum gain of 90? I also changed the R7 resistances and the gain increased. Can we conclude that the gain due to R1 And R4 is problematic here? other than just the minor % errors in resistances. \$\endgroup\$ – Ohbhatt Nov 5 '18 at 14:28
  • \$\begingroup\$ I also hooked it up to my bench power supply and applied a range of voltages to check the function of the in-amp and it behaved the same way. \$\endgroup\$ – Ohbhatt Nov 5 '18 at 14:30
  • \$\begingroup\$ @Ohbhatt -- then why are we even discussing the bridge? \$\endgroup\$ – Scott Seidman Nov 5 '18 at 14:32
  • \$\begingroup\$ I just wanted to clarify that and also you indicated a way to balance the bridge. \$\endgroup\$ – Ohbhatt Nov 5 '18 at 14:35
  • \$\begingroup\$ @Scott_Seidman I finally had it working. I figured with some help that the input buffer gain Resistor was the problem and that I replaced it with unity gain configuration to make it work. Thanks for the help. \$\endgroup\$ – Ohbhatt Nov 5 '18 at 18:07
0
\$\begingroup\$

The problem here is that the input buffer gain is not ideal for a single supply and that the reference I took this circuit from, did not mention this. The problem is pointed out by @sstobbe (Limited Driving power of LM358). This answer is for readers who come across the same problem and are looking for a circuit that just works.

So, here it is:

schematic

simulate this circuit – Schematic created using CircuitLab

Advantage here is that you can control the gain using only one resistor(R6). It works fine with 150mV @ 0°C and 9.98V @ 400°C Input differential Voltage(V2 - V1) is 0 to 300mV.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.