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I have to make an isolated voltage sensor using an isolated amplifier (AMC1300, link to its datasheet). It produces a differential output centred around its common mode voltage, \$1.44V\$ approximately. I am simulating the following circuit which is mentioned on page 25 of the above mentioned datasheet:

Circuit to convert a differential signal into its single-ended equivalent:

enter image description here

A similar suggestion for conversion has been provided by TI in figures 3 and 5 of this application note as well.

I simulated the same concept (without \$C_1\$ and \$C_2\$) in PSpice for TI (because IC models were only available in this particular software) using the ICs:

  1. AMC1300 (link is mentioned above)
  2. TL084 (personally went for something with a higher slew rate instead of TLV6001)

Simulation schematic:

enter image description here

Simulation output:

enter image description here

As you can see, the output is saturated at \$V_{REF} = 1.5V\$ during the negative half-cycle of the input. However, when I use a dual bias \$±5 V\$ the output goes below the reference.

Further, if you look at figure 19 on page 12 of this document by BROADCOM, even though the \$V_{REF} = 0V\$, the op-amp is biased using a bipolar supply.

Am I doing something wrong here? Or is there an error in the documents of TI?

I would love to learn more about this.

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  • \$\begingroup\$ Whosoever has down voted my question, can you tell me why my question is irrelevant? \$\endgroup\$ Commented Oct 22 at 16:14

2 Answers 2

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In the datasheet for TL084, look for the common mode input range and you will see the explanation why it works with dual supply.

The common mode input is given below enter image description here

You can notice that the minimum input level is Vcc- +1.5V or 0V + 1.5V = 1.5V in your case.

So what happens when you input a differential voltage of Vin+=0.5V and Vin-=2.5V, is that it is multiplied by your gain of Gain=0.5 plus the 1.5V, which should give an output of 0.5V. But try and calculate what the voltage would be at the TL084 inputs, the inverting and non-inverting inputs. Non-inverting input is 1.11V and inverting input can be calculated from you simulation output. But if you would have gotten the 0.5V out, then it would have been 1.16V, again less than the common mode input range.

Because the inputs are out of common mode input range, the output is of course not correct. When you have this scenario the output can latch up to one of the rails or it can do phase reversals at the output. Basically you cannot count on the output to be correct anymore.

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  • \$\begingroup\$ Hello @Tyassin, Thank you for opening my eyes. I understand your point about the common-mode input range. I have been using op-amps as a black-box ever since I have begun to do hardware. To be honest I never fully understood the datasheet of an op-amp. All I would look at were its slew rate and if I can give it a bipolar supply or not. This is totally my mistake. So does it mean the op-amp will produce an output between \$V_{CC-} + 1.5\$ and \$V_{CC+}\$ in this case? Could you answer this as well, please? \$\endgroup\$ Commented Oct 21 at 18:44
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    \$\begingroup\$ The opamp will give and output between the supply you give it :-) So between 0V and 5V. \$\endgroup\$ Commented Oct 21 at 18:46
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    \$\begingroup\$ For sanity, try and calculate the input voltages when the inputs are VIN+=2.5V and VIN-=0.5V: The output should be 2.5V and both inputs are higher than 1.5V \$\endgroup\$ Commented Oct 21 at 18:49
  • \$\begingroup\$ So, if TL084 (hypothetically) had a common-mode input range between 0 and 5V then i wouldn't need a bipolar supply. Am I correct? Also, in simulation, because the supply is between 0 and 5V, the common mode input range becomes 1.5V to 5V and because the input is less than 1.5V during a particular time period, the output must be zero but because of the presence of the 1.5V reference, the output gets locked to 1.5V. Is this correct? \$\endgroup\$ Commented Oct 21 at 18:52
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    \$\begingroup\$ Yes correct about the common-mode input range. If you can find one that works down to GND you could potentially avoid a dual supply. About the output, no when you are outside the CM range, you cannot say how the output will behave. In this case in a simulation it is clamped to 1.5V. And notice that an opamp will try to keep its inputs equal, and it is not in this case, with 1.5V out. I have not read your graphs perfectly, but you can see I calculate the inputs to be fairly close (1.11V and 1.16V), when the output is 0.5V which it should have been. \$\endgroup\$ Commented Oct 21 at 18:57
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I also was presented this issue a few years back, my predecessor conjured up an complex single-supply differential to single ended converter circuit introducing offsets, noise and thermal drift.

I used two ADC channels and did some basic DSP.
Way more stable end result, dozens of parts removed from the BOM per channel.

The only thing to lookout for is skew error if you're not sampling simultaneous enough.

Perhaps this is also a feasible way out for you.

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  • \$\begingroup\$ Thank you @Jeroen3, you have suggested an efficient workaround. I sample all the feedback signals at the same time every ISR, however, I am only a beginner (with a bit of adult autism too?) who lacks confidence. I am not sure how I would calibrate the sensor in such a scenario. The way I calibrate my sensor is by passing a fixed voltage or a fixed current through the sensor which I read in my firmware for a fixed amount of time and then average the output of the code. I repeat the process for different values. \$\endgroup\$ Commented Oct 21 at 19:14
  • \$\begingroup\$ (continued) Then I obtain a linear equation between the actual sensor input and the code output using the linear regression tool of MS-EXCEL and thus I have my final calibrated sensor gain equation with respect to my DSP. \$\endgroup\$ Commented Oct 21 at 19:17
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    \$\begingroup\$ @ankitkumar that is a suitable way to calibrate. I do that as well, two point calibration with linear interpolation. For DSP doing the conversion to single end, a subtract should suffice after you've converted it to a signed number, then you can run it though your filters and scaling etc... \$\endgroup\$ Commented Oct 22 at 5:48
  • \$\begingroup\$ thank you for the comment. So are you saying to subtract the ADC results of the differential pair and then implement the calibration? \$\endgroup\$ Commented Oct 22 at 8:12

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