LM324 instrumentation amplifier gain and offset problem

Question

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?

^{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.

Answer 1

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.

Answer 2

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.

Answer 3

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:

^{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.