differential amplifier

Not easy. I assume you want to measure the voltage drop across something like a one ohm resistor. The best way is to have like a V/F converter running floating, then send the pulses down to ground level with an optocoupler or capacitor, then back to voltage with a F/V converter. You can get better than 1% accuracy and basically infinite CMMR up to a few kilovolts.

There are also special chips for high-side current sensing and also isolated op-amps, but they're a bit on the spendy side.

Not easy. I assume you want to measure the voltage drop across something like a one ohm resistor. The best way is to have like a V/F converter running floating, then send the pulses down to ground level with an optocoupler or capacitor, then back to voltage with a F/V converter. You can get better than 1% accuracy and basically infinite CMMR up to a few kilovolts.

There are also special chips for high-side current sensing and also isolated op-amps, but they're a bit on the spendy side.

I want to read the voltage across the boost converter's DC bus voltage and use the sample as a feedback. it is not a simple boost converter and because of this I must read the DC BUS+ DC BUS- differentially.

I know those chips and they are quite expensive unfortunately. I have found out a circuit but the circuit response is not good for voltages between 0-40V.

Very thnx bythe way.

Ancient_Hacker wrote:

You can probably do it with a regular op-amp and some 1% resistors, and maybe a trimpot. If you trim the input resistor balance for maximum CMMR, you can probably get 1% accuracy and good enough CMMR. The alternative is to use V/F, optocoupler, and F/V converter chips, which will cost a bit more but give much more accuracy.

CMRR plus my dyslexia = CMMR. Sorry

Maybe an out of the box solution is best. Can you devise a circuit where the op amp positive rail is at this high voltage, while the negative rail is set with a voltage regulator in a manner where the voltage across the op amp is within its specifications. Then V to I the differential voltage down to a resistor with one side at ground.

Note this is not isolation, but just a scheme to get the differential voltage back down to a potential near ground.

hmm, maybe i am lost here but i thought is was CMRR and mot CMMR ? CMRR = "Common Mode Rejection Ratio". or are we talking something else ?

well don't feel sorry, i looked it up on the net thinking that i was falling behind the tech scene.. as it is, your not the only one using it! ,... its seem to be an epidemic

Why not use a simple linear optocouple with the diode side actoss the output voltage and the tranny side feeding the control circuit.

this is pretty standard stuff for true "isolated" power supplies. Should work in this application as well.

great CMRR!

Could you advise a linear optocoupler. I may try to see if it is linear enoguh for me.

As far as the out of the box solution:

The "high" voltage can change between 370 to 450V. so I can not use that approach.

Mook Johns> Why not use a simple linear optocouple with the diode side actoss the output

You could try a dual optocoupler to improve linearity (making the assumption that they are reasonably well matched, and behave similarly).

Regards Ian

Without specifying the cmrr and the frequency responce required it's inpossible to give to an answer however the 3 opamp intrumentation amp layout should do what you need.

You can hit an accuracy point somewhere in the middle if you float an amplifier up at the 400V to boost the signal from the 1 ohm to some larger swing and then use the simple diff amp with 1% resistors to receive the signal from that.