1
\$\begingroup\$

In a previous question, I asked about the detailed construction of in-amps. Now, I'm curious about a feature that some of them have that I'd like to know how they achieve:

I used an INA286 in a recent project for current monitoring. It worked perfectly according to its datasheet specifications, but I'm left wondering how they managed to make it not only survive, but function with inputs well above or below the power rails.

To pull out the important parts from the datasheet, this in-amp is powered by a single-ended supply as low as 2.7V, but can function with input common-mode voltages of as much as 80 volts, fully 77V above the positive rail, or as low as -14V, far below the negative rail.

I can see in the block diagram provided in the datasheet some sort of switching magic, but I don't follow how it works, or how the input switches are driven considering the high voltage required.

The block diagram, along with the abysmal bandwidth characteristic of isolation amplifiers, makes me think that this is essentially an "isolation amplifier lite edition", using capacitive isolation, but any further detail than that I can't seem to work out.

How does this device (and similar ones, for that matter) manage to handle such voltages? How exactly does that array of switches work?

(As an aside, this question was originally going to be about how the output offset feature works, but it turns out that's actually quite simple to understand just by looking at the block diagram.)

\$\endgroup\$
  • \$\begingroup\$ It can't function at Vcm=80...it just won't break. Recommended operating conditions is Vcm at nominal 12V \$\endgroup\$ – Scott Seidman Mar 12 '19 at 23:22
  • \$\begingroup\$ It can, however, function at much higher common mode voltages, as evidenced by my using it to measure current on a 50V rail. 12V is a typical value, no maximum is listed. \$\endgroup\$ – Hearth Mar 12 '19 at 23:24
  • \$\begingroup\$ The INA28x series is not really a series of instrumentation amplifiers, despite the "INA" prefix. TI calls them "current shunt monitors" in the datasheet and lists them under current sense amplifiers but not instrumentation amplifiers. The architecture is quite different -- in amps are usually rail to rail at the input at best whereas current sense amplifiers are often designed to handle a common mode that exceeds the rail(s). \$\endgroup\$ – Null Mar 13 '19 at 13:55
  • \$\begingroup\$ @ScottSeidman It functions just fine at Vcm up to 80V. I've used them myself in applications with such a high Vcm. \$\endgroup\$ – Null Mar 13 '19 at 13:57
  • \$\begingroup\$ @Null From a black-box perspective, they are pretty similar to in-amps, though; they have all the usual in-amp characteristics. But I see what you mean architecture-wise; I'm sure the die for an INA28x looks very different from that of any conventional in-amp. \$\endgroup\$ – Hearth Mar 13 '19 at 14:21
1
\$\begingroup\$

They're drawing a rather large current- 25uA from each input typically, that's enough current to power the high-side switches provided the chopping isn't too high frequency. It's probably something like 100kHz.

So the components exposed to the 80V would be the capacitors and some kind of high voltage current sinks, possibly lateral JFET type structures, maybe someone with high voltage monolithic IC design experience can comment.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ That makes sense--power the switches from the voltage they're switching. I have to wonder how it all works with either positive or negative voltages, though, given that this particular example works with negative CM voltage too, and you'd need different polarity devices for switching that. I think. \$\endgroup\$ – Hearth Mar 13 '19 at 2:16
  • \$\begingroup\$ it switches at 200kHz, from the zeroes in the transfer function. 'Current sinks'? It appears to have switches to switch those capacitors in the standard 'emulating a resistor' way, which would presumably be easy to drive through level shifters from the low voltage part of the device. \$\endgroup\$ – Neil_UK Mar 13 '19 at 7:07
  • \$\begingroup\$ @Neil_UK Where do you reckon the 25uA input bias current goes? \$\endgroup\$ – Spehro Pefhany Mar 13 '19 at 14:14
  • \$\begingroup\$ @SpehroPefhany I guess I wouldn't call the circuitry 'current sinks'. I presume there's a level shifting biassing chain taking some of the current, and dynamic charging of the gates of the switches. I also think I'm wrong suggesting an 'emulating a resistor' configuration, as that would require good matching for good CMRR (unless the same cap is used on both nodes alternately). I would guess it's more likely to be a flying capacitor, sampling the differential high voltage, and transferring it to a low voltage. AT least, that's how I'd try to design it first. \$\endgroup\$ – Neil_UK Mar 13 '19 at 15:15

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.