Community
OpenEnergyMonitor

Community

400A service, CTs in parallel with a common burden using a single input

Getting Started
,
Tags: #<Tag:0x00007f6e03c1f9a8> #<Tag:0x00007f6e03c1f7a0>
#1

I am trying to setup whole house monitoring on a friend’s house. They have a huge house with a 400A service (400A disconnect feeding to two 200A panels). Optical Sensor is not an option as the meter is on a pole about 200m away from the house. I have ordered 4 Wattcore WC3 200A:100mA CTs. Because I have so many lines to watch I am hoping to set them up as described in 5b of EmonTx-in-North-America (wired in parallel with a common burden using a single input).

If anyone was interested in checking my calculations I would be grateful. I know just enough to get myself into trouble.

So the CTs I have ordered are 200A:100mA and the specs say nothing about turns, but I am going on the assumption the turn ratio is going to be 2000:1

Because I intend to wire two CTs in parallel to measure both legs and save on inputs I know that my burden resister needs to expect double the current. I calculate a SINGLE of these CTs as needing 11 ohms and that two hooked up in parallel should require a 5.5 ohm value for the burden.

I want to have the burden resister hooked up before the jack so that when unplugged it is safe. The EmonPi and EmonTx V3 have a 22 ohm burden resistor. If I were to keep the existing burden resistors in place and just add something to lower total resistance I calculate that I would need a 7 ohm resistor.

Am I on the right track? With this I can use one EmonPi input for the first 200A panel and the second EmonPi input for the second panel for a combined house power. I would then use an EmonTx V3 for measuring consumption on a few other circuits.

With a burden of 7 in this particular configuration it works out to MAX current of 220 each leg. With a burden of 8 in this particular configuration it works out to a MAX primary current of 198A each leg with both legs maxed. I’m inclined to go with an 8 ohm 1W 1% metal film resistor.

Thank you for your time! I’ve been reading and reading for about a month on various topics. I’m just looking for some confirmation that I am on the right track from someone else who has done this.

#2

First, forget the turns ratio. The manufacturer will actually tweak it to get the current ratio closest to specification over the widest range. All you need are the currents, that is how a CT is specified. (A 2000:1 CT would have a 1 A secondary current at the maximum primary current of 2 kA.)

Your sums appear to be correct (although I don’t actually do them the way you did), but you won’t buy a 7 Ω or an 8 Ω resistor. You actually need 7.33 Ω. If you go up to the next E24 value, that’s 7.5 Ω, you’ll reduce your maximum total current to a little over 391 A. But you might find 7.5 Ω hard to get. The next value above that, which should be readily available as it’s an E12 value, will be 8.2 Ω. If you go down to a 6.8 Ω your maximum current will be 431 A, up to 8.2 Ω and it’ll be 358 A. Or you could use another parallel resistor - say 68 Ω in parallel with the 8.2 Ω. That will give you almost exactly your 400 A.
A 1 W resistor is overkill, a ¼ W resistor is plenty big enough, the actual maximum power is just over 0.16 - 0.18 W.

You’ll need to change the "scales = " value in emonhub.conf to give the correct calibration, depending on the burden you end up with.

#3

Thank you for all the information and help Robert. I appreciate it.

I found a 1% tolerance wirewound resister in 7.5 Ω value from a local supplier and it doesn’t cost much.

Thanks again!