4
\$\begingroup\$

I've got 16 current sense transformers that are being used to monitor AC loads. The SCTs are current output type, producing 50mA on a 100A load. The maximum burden resistance is 10Ω, thus the voltage signal generated from the voltage drop over that resistor is ±50mV. I'm feeding these outputs into an ADC whose input range is 0V to 5V.

While I could technically DC offset this ±50mV signal and feed it directly into a 12-bit ADC, this wastes a ton of the measurement range and leaves most of the signal behind under the noise floor of the ADC. I'm trying to figure out a way to bring the 50mV bipolar sense signal up to somewhere around ±2.0V centered around a 2.5V DC offset, without losing accuracy or running into offset issues.

I've seen other designs buffer the signal and send it straight to the ADC, but those designs also make weird decisions like not using rail-to-rail amplifiers, and adding 20Ω or even 30Ω of burden resistance to the SCT, presumably to increase the voltage range, but this is way in excess of the rated maximum in the SCT datasheet. In short, I don't trust them all too much when it comes to accurate results.

I currently do not have negative supply rails on my board, so my first thought is something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

This is just for reference - the exact opamp would be something other than the TL081, and the D1/R2 part of the circuit would be implemented via a voltage reference IC. The idea is that the 0.0625V reference voltage biases the burden resistor (R1), such that the signal going into the opamp has a range of 0.0125V to 0.1125V. The opamp then amplifies this by 40x, resulting in a range of 0.5V to 4.5V.

My primary concern with this approach is that any input offset voltage in the opamp would cause problems when multiplied by 40. Same goes for any variance in the DC bias reference voltage.

My other thought would be to use an instrumentation amplifier and feed its output and a DC offset voltage to an adder circuit. It means biting the bullet and adding a negative rail to the board, but it's probably a more robust solution. Here's a rough schematic:

schematic

simulate this circuit

The instrumentation amplifier is likely to have a much smaller overall offset voltage, and since we're only amplifying the input signal by 40x, instead of both the signal and the DC bias, my guess is it'll perform better. OA1's offset voltage is less of a concern, partly because it is relatively low by comparison to the voltage range of the signal, and also because it is a constant that can be measured and accounted for in software. The TLC2274A seems like it should work ok - common mode voltage specs seem ok, and its max input offset voltage is 950µV.

The problem with the instrumentation amplifier approach is that I need 16 of them, plus the additional opamp for the voltage adder, so that increases costs somewhat. Using a 4-channel opamp does help somewhat here, since it's then just one IC per channel, but it's still a whole bunch of them.

Am I thinking this through correctly? Am I missing a cheaper solution that gets the same level of accuracy?

\$\endgroup\$
3
  • \$\begingroup\$ In your first circuit why have you picked a bias of 62.5 mV? Why not Vcc/2? You probably need to add a big cap in parallel with D1 to hold it steady no matter what's happening on the CT. \$\endgroup\$
    – Transistor
    Commented Jan 30, 2021 at 19:14
  • \$\begingroup\$ @Transistor I answered that in the paragraph below the circuit. The offset gets multiplied by 40, so 62.5mV * 40 = 2.5V, and the zener/resistor circuit is just there as an example - I'd use a voltage reference IC. (in practice this is wrong, because it's 500mV not 50mV as Andy pointed out, so it'd be 2.5V/4 = 625mV) \$\endgroup\$
    – Polynomial
    Commented Jan 30, 2021 at 19:18
  • 1
    \$\begingroup\$ You're right. I was thinking of an inverting configuration where the IN+ is also referenced from mid-supply. Keep going ... \$\endgroup\$
    – Transistor
    Commented Jan 30, 2021 at 19:22

5 Answers 5

5
\$\begingroup\$

Am I thinking this through correctly?

And

The SCTs are current output type, producing 50mA on a 100A load. The maximum burden resistance is 10Ω, thus the voltage signal generated from the voltage drop over that resistor is ±50mV.

Here are a few points that might influence the choice: -

  • 50 mA through ten ohms produces 500 mV. Hence the gain required is only 4 and the DC offset problem you originally alluded to might not really be a problem.

  • The TL081 will not be a happy bunny with the inputs at 62.5 mV above the most negative rail - the lowest the input can be from that negative rail and still work is 1.5 volts.

  • The InAmp circuit will need input bias current bleed resistors to 0 volts/mid-rail for it to work correctly. You have to provide a DC current path for InAmp inputs to work correctly. Given the circuit you show, there is no DC path. It's a rule.

Personally, I'd go for op-amps and a decent and low output impedance mid-rail generator that can be shared amongst all your circuits: -

enter image description here

\$\endgroup\$
11
  • \$\begingroup\$ Whoops! Yes, of course it's 500mV! Could you elaborate on the second bullet point, though? Not sure what you mean by "input bias current bleed resistors". \$\endgroup\$
    – Polynomial
    Commented Jan 30, 2021 at 19:13
  • \$\begingroup\$ Regarding your edit, I don't think the issue with the TL081 is a problem. I'd be using a 625mV DC offset rather than 62.5mV, because it'd be 2.5V/4 not 2.5/40 due to my miscalculation, and I'd be using something more like the TLC2274 than the TL081. \$\endgroup\$
    – Polynomial
    Commented Jan 30, 2021 at 19:20
  • \$\begingroup\$ If you are not using a TL081 why did you show the schematic with a TL081? Granted on the 0.625 volts but that isn't a great circuit; you should use a mid-rail generator and connect R3 to the output of that mid-rail generator. Maybe with the gain being only needing to be 4 it's not so bad regarding the error budget. \$\endgroup\$
    – Andy aka
    Commented Jan 30, 2021 at 19:23
  • 1
    \$\begingroup\$ No, create a mid-rail generator of 2.5 volts and connect that to the CT and connect R3 also to that same point. I'll add a picture. \$\endgroup\$
    – Andy aka
    Commented Jan 30, 2021 at 19:31
  • 3
    \$\begingroup\$ You certainly can share that reference point as I said in my answer. You might want to pick a decent reference chip that has low output impedance to suit multiple connections hence, the low output impedance prevents cross talk from one channel to another. I've done the same with 16 channels of analogue processing but I toiled hard on the choices because my equivalent to R3 was about 1 kohm and now yours can be 10 kohm. \$\endgroup\$
    – Andy aka
    Commented Jan 30, 2021 at 19:46
3
\$\begingroup\$

Use an INA4181. Quad bidirectional current sense amplifier with a reference level input. 4 of them can handle all 16 transformers.

The output will be offset by the reference level. So no other IC's will be required and your output voltage will be centered on the reference level you supply.

Generate the reference with a voltage divider from your 5V ADC reference to keep it center rail.

The lowest gain available is 20. You have miscalculated the amplitude of your signal by a factor of 10. With a burden resistor of 10 Ohms you will get an amplitude 0.5V at the amplifier input. With a gain of 20, that would put the input to the ADC off-scale.

So you can use a 2 Ohm burden resistor. Then you will have +/- 100mV at the amplifier input, and 2.5 +/- 2V at the amplifier output (assuming you use 2.5V as the reference). If you wanted to you could use a burden resistor exactly tuned to get a 0-5v swing, but I think it is not a bad idea to leave a little dead space at the top and bottom.

The input to the current sense amplifier is differential. You can bias it to any convenient point. You can even ground one side of the burden resistor because inputs as much as 200mV below ground are acceptable.

\$\endgroup\$
2
  • \$\begingroup\$ This does look ideal, but the INA4181 is unfortunately not available in the assembly parts catalogue I'm using (JLCPCB). \$\endgroup\$
    – Polynomial
    Commented Feb 1, 2021 at 15:13
  • \$\begingroup\$ If you able to get the INA4181 delivered to your premises, maybe you could leave it off of the BOM and install it yourself after the PCBA's are delivered to you with all other parts installed. \$\endgroup\$
    – user57037
    Commented Feb 1, 2021 at 17:26
2
\$\begingroup\$

Andy has already replied straight to the point, i.e. how to solve your requirements in terms of basic components (op-amps and the surrounding networks), i.e. at the PCB level - excellent as usual, and I have hardly anything to add at that level :-)

If this is your league and allows you to save costs, I'd hardly object. So... let me mention the following as a broader context. For the record, for people maybe coming later, who are not in the league of "design your own PCB".

In the broader business of "industrial process control", the area that you're asking about is called roughly "discrete signals I/O". There are add-on cards for the PC's doing multi-channel ADC (typically by a multiplexer prepended to the ADC), multi-channel DAC, parallel binary input/output at various electrical levels. I come from the area of industrial PC's, but actually the more typical device here is a PLC. And there are "distributed I/O modules" = small boxes, accessible via some digital "fieldbus" (Modbus appears to be the cheap and open option, if that suits your application). And as an alternative to the PCI etc. cards, there are also USB-based modules (although I'd generally recommend against USB in an industrial environment). So much for ready-made off-the-shelf hardware, that "contains an ADC" = the part you did not ask about :-)

I believe the very point of your question is "signal conditioning". I.e. your "sensors" don't give the right signal level or format, or you need to add isolation etc. There are off-the-shelf gadgets for this role too, sometimes called barriers. Decades ago, there used to be a popular product range for high-density signal conditioning, the 5B series product range by Analog Devices (google query | particular download link) - phased out by AD years ago, without an in-house replacement. There are 3rd-party replacements or work-alikes: the OM5 series by Omega or the SCM5B series by Dataforth. Those 5B series modules used to have a standardized pinout and came with an 8-way passive backplane / carrier board, called the 5B08 I believe. And there are/were work-alike backplanes from other vendors, I am told not necessarily pin-compatible at the outer interfaces (5V and GND swapped in the ribbon cable connector).

Another brand that rings a bell in the context of signal conditioning is Opto22, apparently with their proprietary module form factors.

Nowadays, it seems that the business at large has moved towards individual slim DIN-rail-mounted signal conditioners, where each analog channel occupies a dedicated 6mm "slice" with a dedicated set of wiring terminals. Or some of them are in a wider box... I've seen these from Weidmueller, but the other vendors are probably down the same path.

More to the point, you'd like to measure RMS voltage (or current) and maybe you'd prefer to convert it to DC first, so that you don't need to sample the AC at high speed? There are/were signal conditioning barriers with that function. Namely, I've noticed the SCM5B33 by Dataforth. Interestingly, there probably never was a 5B33 by Analog Devices :-) nor do I see an equivalent module by Omega. Dataforth also make a stand-alone DIN-rail-mounted version called the DSCA33.

Now if I should address the possible sub-topic, of how to convert RMS to DC in terms of detailed circuitry, down to op-amps and stuff... that's an interesting question in itself :-) Apparently, Analog Devices used to make or still do make dedicated chips for that. They also have an app-note on the topic, numbered the MT-081. Apparently you need to calculate, in your analog circuit, the square (2nd power) of your instantaneous input signal, average that over time, and take a square root of the filtered product, to get the desired RMS output. The AD appnote suggests that you can do the square using a multiplying amplifier (multiply the input by itself), and do the square root by a multiplying amplifier tied in the negative feedback of an op-amp... My impression is, that if you can come up with an ADC+DSP system fast enough for your application, and it's within your capability to program the very basic code for that task, it is a cheaper solution, less demanding on manufacturing tolerances. If OTOH you were a digital/MCU design engineer at heart, what you may find intriguing is the 16-way multiplex of your input signal, coupled to S&H and ADC. You need to do it properly to avoid cross-talk between your analog input channels, stemming from the mux settling after each switch-over (parasitic capacitance of the node connecting mux output to S&H and ADC input). Of course if you can get 16 dedicated ADC channels, you're fine... how about 16 dedicated tiny MCU's, each with a dedicated on-chip ADC, good enough for 50-60 Hz AC, and opto-isolate their digital output stream? :-)

\$\endgroup\$
2
\$\begingroup\$

I would be inclined to use something that doesn't use a common rail for the mid point voltage. This is mainly for ease of building, debug and maintenance.

With a common mid-point rail, its failure, or a short in any one of the amplifiers, could stop the whole system working. Having completely independent channels allows you to compare voltages around a working and a non-working one. It avoids the possibility of inter-channel crosstalk.

The CT secondary is ground referenced. You can probe it with a scope without upsetting the amplifier biassing, or connect it with grounded coax.

This first circuit uses AC coupling to handle the voltage shift. It's simple, but introduces a small phase shift, which may or may not be permissible.

I've redrawn the amplifier so that it looks like the non-inverting gain stage that it is. It also avoids the unnecessary cross-over, for tidiness.

schematic

simulate this circuit – Schematic created using CircuitLab

The second schematic is DC coupled.

I don't like working out resistors for differential gain stages with algebra, as I'm too likely to make a mistake. That's why I've drawn the circuit twice. The first version shows the differential gain stage in all its symmetrical glory. R1 and R2 connect to the differential input, R3 and R4 define the gain to the differential output, referenced to 2.5 V. Even with -0.5 V input, the opamp inputs will stay above GND, so in the amplifier's common mode range.

The second diagram shows how we remove the 2.5 V output reference. Using the Power Of GreyskullThevenin, the 2.5 V with a series impedance of 390k can be replaced by a voltage divider from 5 V using 2x value resistors.

The third refinement would be to replace R102 and R104b with a single 88.6k resistor to minimise parts count.

Presenting it in these three steps allows you to see where the values come from, so should you want to alter the gain by changing R3 and R4, you can work out new values.

Of course, if these values aren't quite right, you'll still get gain and voltage shift, but the output may not be centred around exactly 2.5 V.

schematic

simulate this circuit

A note to anyone else who is thinking of asking a circuit question. Using a schematic produced by the site's built-in schematic editor like this makes it much easier for responders. We can copy and edit the original schematic, preserving reference designators and layout where possible.

\$\endgroup\$
6
  • \$\begingroup\$ Some good points here. Am I right in thinking that the divider actually might provide more accurate results than a fixed 2.5V voltage reference, assuming they're 1% parts or better, since it'll follow any variation in the +5V rail that might affect the ADC readings? Also, is C1 doing anything here, since the transformer is already AC coupled anyway? \$\endgroup\$
    – Polynomial
    Commented Feb 1, 2021 at 15:19
  • \$\begingroup\$ Sorry, just realised exactly why C1 is there. \$\endgroup\$
    – Polynomial
    Commented Feb 1, 2021 at 15:30
  • \$\begingroup\$ One curiosity - when I simulate this with a real model opamp instead of ideal, the initial signal drifts up and causes clipping for a short while, before settling back down to the correct midpoint after about a second. With higher gain (for a 32A range) this takes significantly longer - I stopped simulating after about 3 seconds and the output was only just dipping back in after being stuck at 5V DC. Is this a limitation of the simulation, or something that would happen in practice? \$\endgroup\$
    – Polynomial
    Commented Feb 1, 2021 at 15:59
  • 1
    \$\begingroup\$ @Polynomial The simulation is showing you exactly what would happen in practice from switch-on with uncharged capacitors. If you want the simulation to start faster, you could precharge them all to 2.5 V. I'm assuming that you'd leave this powered up the whole time, and the switch-on transient would not be an issue. If you want to sleep it between readings, say for battery operation, and need a fast start-up, you'd need to DC couple. Capacitors are convenient and self-biasing when you're not interested in DC. A current transformer (any transformer) has exactly zero DC output! \$\endgroup\$
    – Neil_UK
    Commented Feb 2, 2021 at 9:27
  • \$\begingroup\$ Thanks, that makes sense. I think what I'll end up doing is designing the board so that it's initially DC coupled like Andy showed, but with DNP and 0R parts included so that I can rework it to AC coupling if I need it. The main benefit I see of the DC coupled version is that there's negligible phase shifting, which is convenient for other features of the design (although I appreciate that I didn't mention this before). Your note about scope probing is very helpful - I definitely would've fallen for that issue! Luckily my scope supports pseudo-differential probing. \$\endgroup\$
    – Polynomial
    Commented Feb 3, 2021 at 14:17
1
\$\begingroup\$

One less common, but in my opinion superior, option is to use a transimpedance amplifier. This presents a virtually 0 ohm load to the transformer secondary winding, while providing an output voltage according to the current.

schematic

simulate this circuit – Schematic created using CircuitLab

To work with a single-sided supply, I've added a virtual ground circuit in the bottom section with OA2. It just provides a stable +2.5V voltage as a reference. You should divide this from the same reference voltage as your ADC uses, to avoid any offset drift.

The upper section with OA1 is the transimpedance amplifier. The gain of the circuit is determined by R3. For example, if the transformer secondary current is 10 mA, a 100 ohm resistor causes OUT voltage to be VGND + 100 ohm * 10 mA = 2.5 V + 1 V = 3.5 V.

The real cool part of using transimpedance amplifiers with transformers is that the magnetic flux of the transformer will remain close to 0 all the time. This improves linearity and low-frequency gain flatness a lot.

A downside in the simple version is that the transformer is directly DC-coupled to the opamp, which requires more careful analysis of impulse tolerance and behavior when the circuit is switched off. Also, as shown, the opamp should be of low offset voltage type. This Maxim IC appnote gives an AC-coupled version that tolerates larger opamp input offset voltages also.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ The problem with a transimpedance amp for this project, especially with 16 channels, is the current consumption in that R3 load resistor. With the output swing the OP wants, it means a lot of opamps simply wouldn't be able to drive it, and you can forget about battery powered. \$\endgroup\$
    – Neil_UK
    Commented Feb 4, 2021 at 6:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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