Using current sensor at high voltages

[hkBattousai]

I want to sense the current drawn from the source in the picture. I need an interface to connect this IC to the supply. I don't want to use halleffect sensors since their are too expensive.

Supply voltage range of the IC is [5V, 20V], how can I reduce 5000V to this range in a cheap, efficient and non-space-wasting way?

 

[alphacat]

Is the ground of the source connected to the ground of the ZXCT1081?

 

[dougy83]

What is the 5000V referenced to? If it's ground- (-ve supply-) referenced then you can sense the current on the low side and not worry about translating the 5kV (as you will have a small ground-referenced voltage proportional to the current drawn).

 

[alphacat]

By saying "ground-referenced", you mean that the '-' (minus) of the source is connected to the GND pin of the IC (ZXCT1081), right?

 

[hkBattousai]

By saying "ground-referenced", you mean that the '-' (minus) of the source is connected to the GND pin of the IC (ZXCT1081), right?

Yes.

What is the 5000V referenced to? If it's ground- (-ve supply-) referenced then you can sense the current on the low side and not worry about translating the 5kV (as you will have a small ground-referenced voltage proportional to the current drawn).

Please refer to the internal circuit of the IC package:

The current sensing section of the circuit is not isolated. The voltage drop between two current sensing pins will be lower than 2.5V (maximum rated value mensioned in the datasheet), but the voltage drop between pins "Input" and "GND" will be about 5kV. That is a problem for sure.

 

[alphacat]

Do you know what is low-side configuration?

In low-side configuration, you connect the Load to the '+' of the 5KV source, and the current sense resistor to ground.
That way, the Input is not connected pin SENSE_4, and SENSE_5 is connected to ground.

You got there a high-side configuration, which is just the opposite.

 

[hkBattousai]

Do you know what is low-side configuration?

In low-side configuration, you connect the Load to the '+' of the 5KV source, and the current sense resistor to ground.
That way, the Input is not connected pin SENSE_4, and SENSE_5 is connected to ground.

You got there a high-side configuration, which is just the opposite.

Unfortunately, I can not do "low-side configuration", because the device I'm designing will be conected directly to the power outlet, and the load will be connected to my device. Therefore, the load will be at the low-side in my desing.

 

[Reloadron]

Taken from the specification sheet:

Description
Ordering information
The ZXCT1081 is a high side current sense
monitor with a gain of 10 and a voltage
output. Using this device eliminates the need
to disrupt the ground plane when sensing a
load current.
The wide input voltage range of 40V down to
as low as 3V make it suitable for a range of
applications; including systems operating
from industrial 24-28V rails and power
supplies.

The device you are wanting to use is designed as a high side current sensor to work with voltages in the 3 to 30 Volt range. My take from reading the data sheet is that it cannot be used in low side measuring applications. They seem pretty clear about its use and configuration. My take is that it won't work for your application.

Ron

 

[crutschow]

The best way would be to measure the current in the ground lead, low side of the circuit as others have suggested. Measuring the current in a circuit 5000V above ground is very difficult. You would need a circuit with very high isolation, such as optical couplers to provide power to the circuit and output the signal.

Is it not possible to put a resistor in series with the return lead of the 5000V circuit to measure the current? Show us a schematic of the circuit.

 

[hkBattousai]

The best way would be to measure the current in the ground lead, low side of the circuit as others have suggested. Measuring the current in a circuit 5000V above ground is very difficult. You would need a circuit with very high isolation, such as optical couplers to provide power to the circuit and output the signal.

Is it not possible to put a resistor in series with the return lead of the 5000V circuit to measure the current? Show us a schematic of the circuit.

It is not possible to place a series resistor in the return lead, because the input voltage my be AC as well. 5000V voltage level is the worst case, The input may be in any amplitude and signal form, of course there will be a limit in the frequency.

 

[tcmtech]

There has to be some point in the circuit where the 5 KV comes back from or is referenced to.

 

[crutschow]

Since you have excluded the simple solution, then only the more complex ones are left. You may have to reconsider a hall device or some similar isolated sensor in spite of it's cost. What is the current level and what type of accuracy do you need?

 

[hkBattousai]

Since you have excluded the simple solution, then only the more complex ones are left. You may have to reconsider a hall device or some similar isolated sensor in spite of it's cost. What is the current level and what type of accuracy do you need?

50A minimum and 1% tolerance at most.

 

[alphacat]

It is not possible to place a series resistor in the return lead, because the input voltage my be AC as well. 5000V voltage level is the worst case, The input may be in any amplitude and signal form, of course there will be a limit in the frequency.

I dont see what is the problem with low side configuration - connecting the sense resistor in the return lead.

If you connect the '-' (minus) of the source to the GND pin of the IC, then you'd never have 5KV on any pin, besides GND which might receive -5KV (if its AC), but the GND is just the IC's reference voltage, so that should be fine.

 

[crutschow]

50A minimum and 1% tolerance at most.

250,000 Watts minimum!(?)

 

[hkBattousai]

250,000 Watts minimum!(?)

Hmm, ok, I have exaggerated a bit, say it is 50A maximum.
250kW may sound ridiculous, but it is an extreme condition and my circuit isn't supposed to drain any power in the first place.

 

[Reloadron]

Hmm, ok, I have exaggerated a bit, say it is 50A maximum.
250kW may sound ridiculous, but it is an extreme condition and my circuit isn't supposed to drain any power in the first place.

My initial thought based on your first post was you had a device like a HiPot test set where 5 KV isn't unusual and you wanted to measure mA. Then you come up with 5 KV @ 50 Amp minimum which as was pointed out is 250 KW. Now you claim you "exaggerated" a little and the current is not to exceed 50 Amps? Exaggerated a little is a heck of an understatement I would say. You just redefined the terms greater than and less than considerably.

Your circuit is going to draw some power, there is no magic with this. I suggest you begin with a Google of High Voltage Current Transformers and work from there. Next it isn't going to be cheap or inexpensive to say the least. The power loss is "Burden".

I have no clue what your "extreme condition" is and if you choose not to share exactly what your goal and application is cool with me. I would say 5 KV at a quarter million watts is most definitely extreme though.

Best Of Luck On This One
Ron