Voltage Imbalance Detection Using Difference Amplifier

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kami7jun

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Hi,
I'm trying to detect a voltage imbalance between two points in the circuit. The common voltage would be around 285V and the differential voltage goes up to 95V.
I tried to used a three stage current sense configuration using LT1990 same as the one mentioned in this tutorial:
https://www.analog.com/static/imported-files/tutorials/MT-041.pdf
I know this circuit is originally used for current sense and not for voltage imbalance but that was the best option that came to my mind.

Can anyone help me here?

Thanks,
 

The LT1990 only goes to 250V common-mode range so won't work for your requirements. Don't know of any general purpose op amps that go to 300V.

The tutorial you referenced does not mention current measurement.

What accuracy and resolution do you need for this differential voltage measurement?
 

I assume the common mode is the average of the inputs. The highest average I do have in my circuit is -237.5 so it should work OK. Right?
I'm trying to measure currents as low as 6mA based on that voltage imbalance.
 

You can make your own differential amplifier from a long-tail pair.

I'm not sure if it must be a current mirror in all cases.

There are transistors which will endure the volt levels you're talking about.
 

kami7jun said:
I assume the common mode is the average of the inputs. The highest average I do have in my circuit is -237.5 so it should work OK. Right?
I'm trying to measure currents as low as 6mA based on that voltage imbalance.
Click to expand...
No. You assume incorrectly. It's the maximum voltage at the inputs, not the average.

I don't understand measuring current with a large voltage imbalance. That makes no sense. How exactly are you planning to do this measurement?
 

Hi,

It's not clear what you want to do. It sounds a bit confused.

Please show us a schematic or a draft and some more detailed description.

********
If this is for a three phase system, then just connect a resistor from each phase in star connection.
This gives the voltage inbalance. With an extra resistor to neutral you get a voltage divider.

Klaus
 

crutschow said:
No. You assume incorrectly. It's the maximum voltage at the inputs, not the average.
I don't understand measuring current with a large voltage imbalance. That makes no sense. How exactly are you planning to do this measurement?
Click to expand...

I think, both explanations are not correct.
Each voltage difference can be split into two parts: One part that is common to both voltages (Vc) and another part that has different signs (differential part Vd).
Example: V1=12V and V2=4V with Vc=(12+4)/2=8V and Vd=(12-4)/2=4

That means: V1=Vc+Vd and V2=Vc-Vd.
 

In operation amplifier datasheet, common mode voltage range is often used as a synonym for the absolute input voltage range, considering the only small differential voltage amount. But this assumption isn't correct for instrumentation or difference amplifiers with low finite gain.

In LT1990 datasheet, the term common mode voltage is used literally incorrectly.

The comment, that the instantaneous voltage at amplifier inputs must keep the specified absolute voltage range apllies for all kinds of amplifiers.
 

The obvious way is to use a 100:1 resistive attenuator to bring the voltages down to a reasonable value. As the extreme value of "unbalance" is 95 V, this can easily be limited by a catching or zener diode to reduce the effective dynamic range.
Another way would be to establish an independent stabilised supply of , say 280V with a string of zeners, then feed each supply via , say, 10K into a opto coupler, connected to the common supply. the transistor outputs can then drive the inputs of a differential op amp. The dynamic range would have to be checked and limited (perhaps) with a zener diode.
Frank
 

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