CT measuring circuit

[DominicB]

I am trying to calculate and measure the electrical load for a property that consist of several buildings, to find out if our generator can adequately handle the load. In the main entrance there are CT's on each phase going to a box with a circuit board that contains the burden resistors, small transformers, rectifiers and a DC output.

The only info on this board says that has a 0-10vdc output. The DC output disappears down a conduit that nobody seems to know the end of. So I'd like to know if I can use this DC voltage to conveniently measure the real time current draw.

Here's the numbers:

CT ratio 1600/5, or 320/1
DC voltage output about 3.0v
Burden resistor 0.2 Ohms
I measured the voltage drop across the burden resistors and it was about 0.05v on each phase, which translates to 0.25A, or about 80A per phase on the mains supply.

I'm not sure where to go to next. My questions:

Can I calculate a factor from these numbers that can convert the DC voltage to an AC current reading? and is the relationship linear?
How does the circuit combine the 3 phase currents into one DC voltage, is it added, or averaged, or something else?
I was thinking that the 10vDC output would represent 5A output from the CT's, but the numbers don't seem to fit

Thanks for your expertise

 

[ptonsparky]

Why not just ask the owner or POCO for billing history?

 

[mull982]

I am trying to calculate and measure the electrical load for a property that consist of several buildings, to find out if our generator can adequately handle the load. In the main entrance there are CT's on each phase going to a box with a circuit board that contains the burden resistors, small transformers, rectifiers and a DC output.

The only info on this board says that has a 0-10vdc output. The DC output disappears down a conduit that nobody seems to know the end of. So I'd like to know if I can use this DC voltage to conveniently measure the real time current draw.

Here's the numbers:

CT ratio 1600/5, or 320/1
DC voltage output about 3.0v
Burden resistor 0.2 Ohms
I measured the voltage drop across the burden resistors and it was about 0.05v on each phase, which translates to 0.25A, or about 80A per phase on the mains supply.

I'm not sure where to go to next. My questions:

Can I calculate a factor from these numbers that can convert the DC voltage to an AC current reading? and is the relationship linear?
How does the circuit combine the 3 phase currents into one DC voltage, is it added, or averaged, or something else?
I was thinking that the 10vDC output would represent 5A output from the CT's, but the numbers don't seem to fit

Thanks for your expertise

One thing to keep in mind is that when measuring the DC voltage, it will have a multiplier to convert the AC RMS value into the equvelent DC voltage value. This multiplier will depend on what type of rectifier is used half wave, full wave, six pulse, etc... Maybe if you can provide this we can make more sense of equating the DC voltage to is equivelent AC RMS value.

 

[SG-1]

Is this low voltage (<600V ) so you can place a clamp-on ammeter around the conductors going through the CTs ? You can then see if your DC output conversion factor is correct.

 

[DominicB]

Thanks for the replies.

The object of the metering is so that I can measure the load for the entire property, and shed certain loads before powering on the generator, and maybe turn on more things when I see that I have enough power available. The generator that was installed was not big enough to power everything.

As for the circuit board and the rectifiers, I was hoping that this was a standard piece of equipment that some of you may have seen before. It has small transformer for each phase, and what looks like a full wave rectifier (for each phase), and somehow it is combined together to produce a single DC voltage output. I guess it's not that common after all.

And as for measuring the current to the property, it's definitely do-able, just a bit inconvenient to get a newer meter with the extra flex probe, crawl around among the crickets, and wrap the meter around bundles of 4 500MCM.

 

[mivey]

Why not use the CT current?

Also, why not trace the circuit board output? It must run to some kind of energy monitoring equipment and you might just get the info you are looking for in some kind of report, graph, etc.

 

[ATSman]

If the output is 0-10VDC then it sounds like there is a transducer in the circuit that converts the 5 amp CT secondary current into primary current at the meter connected downstream.
If it is a 3 phase transducer it may average the phase current values.
So calculating:
3v/10v x 5 amps = 1.5 amps

1.5 amps x 320A = 480A per phase.

A 1600A service would require (4) 500MCM cables per phase. Use a flexible
split core CT instrument (like FLUKE) around a phase to double check that the 480A reading is correct.

Tony

 

[SAC]

CT ratio 1600/5, or 320/1
DC voltage output about 3.0v
Burden resistor 0.2 Ohms
I measured the voltage drop across the burden resistors and it was about 0.05v on each phase, which translates to 0.25A, or about 80A per phase on the mains supply.

If I were going to take a stab at it, I'd assume the DC output is intended to represent KVA.

For a balanced 3ph load, (I think) VA = V * I * sqrt(3) = 208 (?) * 80 * 1.73 = 28.8KVA

Let's assume the output voltage is linear with KVA, so 28.8KVA/3.0v = 9.6KVA/volt. So take the DC voltage, multiply by 28.8KVA to get the total load. It is close enough that it sounds like it might be 10KVA/volt (was it 2.88v instead of 3.0v?).

If you can vary the load in a significant way, you can calculate this at a couple of points and check if that is the correct.

 

[DominicB]

If I were going to take a stab at it, I'd assume the DC output is intended to represent KVA.

For a balanced 3ph load, (I think) VA = V * I * sqrt(3) = 208 (?) * 80 * 1.73 = 28.8KVA

Let's assume the output voltage is linear with KVA, so 28.8KVA/3.0v = 9.6KVA/volt. So take the DC voltage, multiply by 28.8KVA to get the total load. It is close enough that it sounds like it might be 10KVA/volt (was it 2.88v instead of 3.0v?).

If you can vary the load in a significant way, you can calculate this at a couple of points and check if that is the correct.

This idea sounds promising. When I was measuring the DC voltage it was varying between 2.69v and 3.15v, and then I'd go back and measure the AC across the burden resistors, which also varied around 50mV. I'll have to do some more measurements and, as you say, vary the load.

Those wires disappear up a conduit, and I think they were intended to be part of an energy management system that was never completed. Strange that I can't seem to find them at least terminated in a box somewhere.

Thanks for all the responses, they've got me going in a better direction.

Dominic