Finding a suitable CT sensor with emonTx

For my project (in US), I need to measure three wires with loads of around 250-300A. Since there’s three wires that need measuring, I’d assume using the emonTx would be the way to go about it. From the “Use in North America” page, it sounds like this will work? I’ll need to put on my own 3.5mm plug, correct? How does one go about replacing the burden resistor on an emonTx?

Thanks

Hi Chris,

That particular CT has an output of only 250 mV. Ideally, you want 1.1 Volts AC at the ADC input of an emonTx.

This CT has an output of 1 Volt. That falls a bit short of the 1.1 Volt ideal, but it will yield much better performance compared to the 250 mV CT.

Yes. The Wattcore CTs are not supplied with a connector.

Since your proposed CT has an internal burden, you won’t need the on-board burden.
Simply flick the SMT resistor off the PC board with a hot soldering iron.

Thank you very much! Out of curiosity, since I’m still learning all this, where does the CT you linked talk about the burden, so that I know what to look for in the future? Also, pardon what is probably a stupid question, but where on the board is the SMT resistor? And I just remove it and I’m good to go with the new CT?

Thanks

Not a problem. Ya gotta start somewhere. :wink:

The key is the description. An unburdened CT puts out current.
To develop a voltage, a burden resistor is used. e.g. if the description says the CT puts out 333 mV
(a very common CT) it has a burden.

An unburdened CT can put out voltage high enough to be a saftey hazard. To mitigate that hazard, some manufacturers use back-to-back Zener diodes (mounted inside the CT case) to clamp the output to a safe voltage. The output from a CT “clamped” in such a manner is current

Here’s a picture emonTx3 User Guide — OpenEnergyMonitor 0.0.1 documentation
The resistors are just above each of the jacks. There is a pair of holes on either side of each resistor.

Almost. You’ll need perform the calibration procedure.

Alright, thank you for you help! Hopefully that should be all I need to know for now at least.