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I need to amplify the PT100 (temperature sensor) for giving it to an Arduino, so I used this circuit with LM358 (V1 and V2 values are not correct):

enter image description here

The V2 AND V1 came from this circuit (based of Measuring Temperature From PT100 Using Arduino):

enter image description here

So as you can see the gain could be calculated like this:

enter image description here

My circuit is like the above circuit with these values:

R4=R3=10K and R2=3.3K and R1=10K potentiometer this type:

enter image description here

So when I have used the R1=10K I have gained G=(1+2*(3.3/10))=1.66 but when I reduce R1 until 10 I can not get a gain of more than G=3.62 it means that R1=2.5k which is not true and for example R1=10.

I have used the 5V port of Arduino or external 6V power supply but this result have achieved again.

This is the circuit:

enter image description here

Here the VIN+ - VIN- =0.060V (V1=0.147V V2=0.208V, which V2 is PT100 voltage), and the output is V_out=0.201V which the gain is G=(0.201/0.060)=3.35 but the R1=2.8 which with this value:

gain=(1+2×3300/2.8)=2358.14

Also, the output of V_A1 or V U1:B = 0.120V for input of V1=0.150V && V_A2 or V U1:A = 0.156V for input of V1=0.208V && V_U2:A or V_A3=0.201V .

I have double-checked the resistors and LM358 and changed them with another of that type but the result didn't change.

Also, the maximum gain of the LM358 is 100,000 so it must be achievable, so why did this happen?

Edited:

I have added the details as the @WhatRoughBeast Wanted here:

enter image description here

I have coworker that consist you must use the LM358 with both Positive and negative power supply (i have used only positive voltage for LM358 ICs) So I checked google and seen this question :

Does op amp need negative voltage?

As long as the voltage on the op-amp input leads does not become negative, the circuit can handle negative input voltages.

So is my coworker rigth?

So what do you think about why did this happen?

Update:

I have one mistake in my measurement, and that was during the measuring I have changed the power supply because of increasing the level of current control in my power supply(and the voltage of power supply changed for almost 1 volt, I don't know why?), so the table have this incorrectness in measurement in last rows. I checked the circuit with negative power supply too, and the circuit worked correctly.

Anyway thanks all for yours attention and yours times.😂😍😘

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  • \$\begingroup\$ went not just buy/use a fully integrated instrumentation amplifier? It's 2019 after all.... \$\endgroup\$ – vicatcu Oct 19 at 19:21
  • \$\begingroup\$ It's hard to see how you get your numbers. V1 should be 0.101 volts, and at 25C V2 should be .147. Until you get that settled, nothing is going to work the way you think. And what, exactly, do you mean by a maximum gain of 3.62? What voltage readings are you getting, and why do you think this produces your gain? \$\endgroup\$ – WhatRoughBeast Oct 19 at 20:15
  • \$\begingroup\$ This sentence: "Until you get that settled, nothing is going to work the way you think" what dose means? I need more gain maybe G=20 could be enough for me, but the maximum gain, which I could achieve by using R4=R3=10k using R2=3. 3k, using R1= 10k variable resistance is 3.6 and this cannot be correct, so I asked why did this happen? \$\endgroup\$ – moha_alpha-web.net Oct 19 at 21:05
  • \$\begingroup\$ moha_alpha-web.net - Please use the @username syntax to reply to comments. Otherwise comment writers will not be notified of your reply and so they might not re-visit your question to see it. This comment will notify you, because it is a comment on your question, and the "question asker" is notified of all comments, even though I did not use @username syntax. But if you write a comment without using that syntax, then (except in one situation that doesn't apply now) no-one gets notified. Please read this for details. \$\endgroup\$ – SamGibson Oct 19 at 21:14
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    \$\begingroup\$ Let me rephrase this. Please measure and provide the following values: V1, V2, the output voltage of U1A, U1B, and U2A. Do so for R1 equals 22k, 10k, and 1k. Use separate fixed resistors in each case, rather than a pot. Use a 100 ohm resistor in place of a PT100. In each case, tell us what you think the gain is, and why. Present this in an orderly way - let's say, one paragraph for each value of R1, and do it as an identified edit, rather than a comment. Start the section with the three paragraphs with the phrase "EDIT". As it stands, you are not providing the information we need. \$\endgroup\$ – WhatRoughBeast Oct 20 at 1:36
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I simulated your circuit in LTspice - setting R1 to 2.222k to produce a gain of 10x, V1 to 101mV (same as your divider powered by +5V) and V2 ramped up from there by +0.25V over a period of 1 second.

Instead of getting the expected 2.5V out for 0.25V in I only got 0.646V. The plot below shows why.

enter image description here

At first the outputs of U1-A (blue line) and U1-B (red line) did what was expected (ramping up and down equally to produce a differential output) so the output of U2-A (green line) rose with the expected slope. However U1-B's output quickly approached Ground where it leveled off because it could not go any lower. As a result the large signal gain was reduced from 10 to 2.6.

To give U1-B some 'headroom' below Ground I applied -4V to the LM358's negative power supply input. Now the plot looks like this:-

enter image description here

Much better!

According to LTspice the output of U2-A 'tops out' at about +3.9V when powered by 5V. This should provide enough dynamic range for reasonably accurate A/D conversion.

The maximum negative op amp supply current was 440uA, which could easily be produced by an ICL7660 or similar negative voltage generator.

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  • \$\begingroup\$ Thanks for your time and... @BruceAbbott, so are saying that the U1_B op amp which is related to none pt100 input voltage, received negative voltage differences as input and have problem in amplifying this and need to negative power supply? So it means that the above circuit is not proper for pt100 amplifying? \$\endgroup\$ – moha_alpha-web.net Oct 20 at 15:48
  • \$\begingroup\$ Thanks @BruceAbbott I have checked the circuit in lab and seen you are Right and the U1_B op amp which is related to none pt100 input voltage, received negative voltage differences as input, so I give -6V instead of ground to this op-amp and the circuit worked correctly. 😂😍😘 \$\endgroup\$ – moha_alpha-web.net Oct 20 at 20:09
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1) The input offset voltage of the LM358 is say, 1mV. The input offset voltage is multiplied by the user set closed loop gain and appears at the output as an output offset. With a gain of 2300 there's going to be a huge DC offset at the output.

2) With those input voltage values the output of U1:B will be trying to swing negative but is limited by the 0V negative supply rail.

3) Bear in mind that with a +5V positive supply rail for the op amps, the maximum output will be limited to about 3.4V.

I would suggest you use a AD623 integrated instrumentation amplifier and power it with both positive and negative supply rails.

You could use a ICL7660 to generate the -5V from +5V....

Inst. amp and -ve supply generator.

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  • \$\begingroup\$ You make some interesting statements, I am not sure that this is the cause of the case at hand. Proposing an AD623 is interesting but its pricing is over 4€ while one can get a LM358 for under 0.1€. Using 3xLM358LVIDDFR is still a lot cheaper than using an AD623 . \$\endgroup\$ – le_top Oct 20 at 0:25
  • \$\begingroup\$ The output swing is limited is due mainly to my point 2. If the output of U1:B is limited to 0V then the output of U1:A doesn't have to rise very far at all to put its inputs at the same voltage. \$\endgroup\$ – James Oct 20 at 8:28
  • \$\begingroup\$ Thanks @James my problem was in negative voltage which have been on U1:B and when I added the negative power supply to it, the problem solved. So I know here we have three acceptable answers and can I mark all of them as correct answer?!! \$\endgroup\$ – moha_alpha-web.net Oct 20 at 20:19
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This will be rather long, and does not actually answer your question, but I hope it gets you pointed in the right direction.

The most obvious issue is that your readings don't make sense.

First, you are not reporting your Vsource accurately. When you use 3.3k/68 ohms, you should get V1 = Vsource x (68/3368), or .113 volts for 5.6 volts, and .129 volts for 6.4 volts. That you get two different values for the same 5.6 suggests that your source is varying wildly and you don't know it. I suggest you measure the source. If that is stable and accurate (and why would it vary between 5.6 and 6.4, anyways?) then you have other issues as well.

Using your 100 ohm (not PT100) data, you should be seeing the results of a simple 3.3k/100 ohm voltage divider, and get Vsource x (100/3400), or .164 at 5.6 volts, and .188 at 6.4 volts.

Moving on to other matters, for large gains you do need a negative voltage for your supply. Consider what would happen if you use 5.6 volts Vsource, a 100 ohm in place of a PT100, and R1 = 1k. IF the circuit is working correctly, the + and - inputs of each op amp will be very nearly equal. Then U1A(-) will be at .164, and U1B(-) will be at .113. Since op amps are very high input impedance, the current in R3 will be the same as the current in R1. Since R3 is 10 times higher than that in R1, the voltage across it will 10 times greater. Then V_A2 will be .164 - (11 x .051) or -0.397 (!!!) Since this is impossible with the - supply tied to ground, the assumption that V+ equals V- cannot be true, and the rest of the analysis falls apart.

Even with +/- 5 volts on the LM358s, the circuit simply will not work properly for R1 of about 100 ohms or less.

How about R1 = 10k? Then V_A2 will be about 0.164 - (2 x .051), or .062. In theory, the LM358 will do this, so you're doing something else wrong.

What could that be?

The most obvious candidate is your rat's nest of a breadboard. All those long jumper wire are an invitation to trouble. Get rid of those things except for bringing signals and power onto your breadboard. For on-board connections, go to digikey.com and search on "jumper wire". Buy a bunch of different lengths, concentrating on short ones.

Now move your two op amps to within about 2 spaces of each other. Use a 0.3 inch jumper to connect a horizontal row between the two ICs. Designate this row as ground, and make all your connections to this row. Keep your jumpers short.

In addition, get some 0.1 uF ceramic capacitors (but don't bother with high-voltage ratings - 25 to 50 volts is fine). Without using jumpers, connect the capacitors between your supply voltages and your ground row at each IC. While you're at it, cut some of the capacitor leads fairly short so that you don't have much extra after you've plugged them in. I suspect that your long leads are causing one or more of your op amps to oscillate, and this is messing up your meter readings.

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  • \$\begingroup\$ Thanks @WhatRoughBeast you are Right, I have one mistake in my measurement, and that was during the measuring I have changed the power supply because of increasing the level of current control in my power supply(and the voltage of power supply changed for almost 1 volt, I don't know why?), so the table have this incorrectness in measurement is last rows. I checked the circuit with negative power supply too, and the circuit worked correctly. anywhay thanks for your attention and your time. \$\endgroup\$ – moha_alpha-web.net Oct 20 at 20:15

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