If you are not aware, the current transformer goes around one wire (either one).
It won't work going around both.

The output from one with adequate turns should be sufficient to light an LED if that is all the indication you need.
You would need a diode (could be a second LED) across the LED in reverse direction to avoid a high voltage for the opposite half-cycle.

 

If you are not aware, the current transformer goes around one wire (either one).
It won't work going around both.

The output from one with adequate turns should be sufficient to light an LED if that is all the indication you need.
You would need a diode (could be a second LED) across the LED in reverse direction to avoid a high voltage for the opposite half-cycle.

I was going to get to that part eventually, after describing how to make a CT. And certainly a means to pprevent reverse voltage failure in the LED will be needed.

 

Get one of these and it just clips around ONE wire.
Then the output signal is used to drive whatever sensor you want. I added a bridge rectifier feeding an ultra bright LED with a suitable series resistor for a pump running sensing device for a customer some years ago.

 

Oh, another thing, if you need more output, just wind the wire around through the core a couple of times.

 

At my place there are two heating cables, one placed along the exterior of water pipes, the other is inside the line (pipe) from the well to the pressure tank. The heating cable exterior to pipes I control remotely as there isn't any provision of a manual switch for it at this house that I bought second-hand. So especially with the one that I turn off when the air temperature is high enough that it isn't needed, for peace of mind, I like an indicator showing that it is actually heating. It is crucial that the one inside the pipe is working when it's cold, so I want to keep an eye on that one also.
-Pete.

How much current does it use?

 

How much current does it use?

If you mean the current draw of the heating cables, I know that the one inside the pipe (immersed in water!) is 1 A. Amperage of the exterior heating cable, I don't know, but it is also probably about 1 A.

 

If you exceed the maximum current of a rectifier diode, does it always become an open, or is there the possibility that it might rather become a short?

 

If you exceed the maximum current of a rectifier diode, does it always become an open, or is there the possibility that it might rather become a short?

I have seen diodes fail both open and shorted. So the short answer is "Yes". Most often, overcurrent leads to an open, but not every time.

 

added a bridge rectifier feeding an ultra bright LED with a suitable series resistor

Since it's a current transformer the resistor will not have much effect on the LED current.

 

If you exceed the maximum current of a rectifier diode, does it always become an open, or is there the possibility that it might rather become a short?

If it's a moderate overload, then the junction may melt and cause a short.
If the current is high enough then it can open the connection from the lead to the chip.

 

One warning required when using the basic style current transformer, it should never be left open circuit due the ability to produce a dangerously high voltage.

 

If the current is about 1A, then you could wire a 6V 6W lamp in series with the live supply (provided you could find a lampholder which meets the insulation requirements for mains). That‘s about the simplest indicator you could make.

 

If you exceed the maximum current of a rectifier diode, does it always become an open, or is there the possibility that it might rather become a short?

The dead status of a diode that has been over currente4d is not part of the normal specification.
In my limited experience, they blow open, but also make a mess of the PCB, and the carbon they deposit is conductive .

 

One warning required when using the basic style current transformer, it should never be left open circuit due the ability to produce a dangerously high voltage.

Is that because its a current transformer, and with an infinite resistance across its terminals ( open ) the current can make an infinite ish voltage ?

I'd never thought of that but makes sense to me . good old V=IR, strikes again.

 

See this old post of mine #2 also an excerpt from the link.
https://forum.allaboutcircuits.com/...er-danger-when-secondary-is-left-open.101695/

"However, if the ammeter was removed, the secondary winding effectively becomes open-circuited, and thus the transformer acts as a step-up transformer.
This due in part to the very large increase in magnetising flux in the secondary core as the the secondary leakage reactance influences the secondary induced voltage
because there is no opposing current in the secondary winding to prevent this.

The results is a very high voltage induced in the secondary winding equal to the ratio of: Vp(Ns/Np) being developed across the secondary winding."

 

Is that because its a current transformer, and with an infinite resistance across its terminals ( open ) the current can make an infinite ish voltage ?

I'd never thought of that but makes sense to me . good old V=IR, strikes again.

It does have some parallel inductance, which is why you don’t tend to get that infinite voltage in practice!

 

It does have some parallel inductance, which is why you don’t tend to get that infinite voltage in practice!

ta
It was just I've used current transformers for decades, just never considered what happens when left open circuit !
Good job I got into habit of fitting ESD diodes on connectors.

 

Good job I got into habit of fitting ESD diodes on connectors.

Another surprising thing is just how much power they can deliver! Intuition tends to suggest that currents way above the rated current might cause them to saturate and limit the output. Not so! Saturation is limited by volts x time, so overloads <5ms come sailing through and blow your protection diodes apart. They need series impedance between the burden resistor and the a/d.

 

My suggested home-made CT would be a quarter inch section of half-inch steel water pipe with perhaps 20 turns of #24 wire. I suppose that even that few turns could produce a lot of volts. We did use a bunch of commercially produced CTs to verify that all of the die heaters were working in one project. Those may have had an integral resistor, they were made just for that sort of application. They connected directly to an input module of a PLC, as I recall, so they must have delivered at least 5 volts at whatever current.

 

My suggested home-made CT would be a quarter inch section of half-inch steel water pipe with perhaps 20 turns of #24 wire. I suppose that even that few turns could produce a lot of volts. We did use a bunch of commercially produced CTs to verify that all of the die heaters were working in one project. Those may have had an integral resistor, they were made just for that sort of application. They connected directly to an input module of a PLC, as I recall, so they must have delivered at least 5 volts at whatever current.

Certainly I could be wrong as so far I have never attempted to construct a current transformer, but my impression is that only 20 turns of #24 wire would produce very little voltage. An "AC Line Current Detector" that I found on the web by Bill (?) Bowden calls for 800 turns or more of #30- #40 wire wound on a1 inch diameter U- bolt.

He writes, "The magnetic pick-up [his coil on a U-bolt] produces about 4 milli-Volts peak for a AC line current of 250 mA."

Regards,
Pete