I'm not too familiar with "current transfomers" I bought one a few years ago to experiment with, but haven't been able to find the time to do that yet.
What I understand is that you need to provide a current path on the secondary side or otherwise the voltage on the current transformer will become very high.
As the primary function of the current transformer is to transform the current, it will adapt the voltage to make that happen. If the diode is conducting: no problem. 0.6V extra on the secondary side. maybe about 0.6mV extra on the primary side.
But when the diode blocks. no current can flow on the secondary side. So the voltage goes up until the diode starts conducting regardless of its efforts to try to block.... Not good. Provide an extra diode for that path. It would be pointing upwards and would be connected directly across the pins of the current transformer. Moving the zener to the other side of the diode would probably also work. You'd restrict the range by 0.6V, so if you can find a slightly larger-value zener, that'd be nice.
Actually though I don't think the 0V6 reflects back into the primary, the current is what matters and that is not affected by the diode voltage
In reverse bias? As you say not good. Does the diode go open or short? If short the Zener is forward biased If open then what? This is apparently a commercial product. Did you look at the frequency range of the CT?
The circuit is a bit small to see clearly, but as I understand, the need is to detect the presence of mains, and a measurement is not required?
If so, there are a few ways, such as using resistor and opto-coupler, electric field or transformer. There are also some ICs for this with built-in
isolation, but possibly not in a DIL-friendly package.
Some air gap is good. There used to be lots of high impedance volt detector circuits using JFETs (discrete JFET circuits are not as popular these days)
such as 2N3819/4416 I can try to dig some up if required.
Recently I needed to detect mains (well, actually to detect zero-crossing in my case) and I just used a small transformer, Farnell have some extremely
compact ones, I used one with 6V output which is 22x22x15mm and a comparator circuit. But then people need to lay out a PCB for mains, and
this starts getting difficult - see this discussion on mains control. But really if the OP just wants to detect mains, he could use any pre-built mains adapter
as gordon77 mentioned in that thread.
EDIT: Looks like Avago HCPL-3700 is available in DIL package. It needs series resistance for mains detection, but it has built-in isolation for the
remote side. The user still needs to lay out a PCB for mains though.
I hope I did not mark my own comment as "the correct answer" I did not mean to do so. And I don't seem to be able to mark any more a helpful - that's not helpful.
Actually though I don't think the 0V6 reflects back into the primary, the current is what matters and that is not affected by the diode voltage
It is a transformer. It transforms both current and voltage, but in opposite directions. So while a normal transformer is used in "voltage mode", you transform say 240V to 12V (say 20:1). When you load it with a current of say 1A, the primary current will be 1/20th of that.
With a current transformer you do the opposite: you transform the current down. So if you have 1A primary, you get (with a 1000:1 current transformer) you'll get 1mA on the secondary side. In theory (with a perfect transformer): If you open circuit the secondary side, the transformer will work in voltage mode: zero current on the secondary side, so zero current on the primary side. So with zero current through the load, no voltage drop over the load, so, 220V over the transformer. i.e. an output voltage of 220V * 1000 = 220kV! This situation is to be avoided. :-)
So for example you load the transformer with 1kOhm, so 1mA becomes 1V which is comfortably measurable. So now the output voltage is 1V, and therefore the input voltage must be 1:1000th, or 1mV. So this is the so-called burden that the current-measurement causes. Adding a 0.6V diode in series with the 1k resistor on the secondary side will add 0.6mV to the burden. Completely irrelevant compared to the 220V for the load, but still it's there.
When the transformer is NOT perfect the resistance of the wire in the primary will probably cause a bigger voltage drop than the calculated burden from the theoretical transformer.....
shabaz If you click on that picture you get a version that's plenty large. :-)
If you click on that picture you get a version that's plenty large. :-)
Ah, maybe that's only if you're a member, not sure..
I registered on the day I saw a blog post that showed dangerous mains wiring, to suggest that perhaps the link should be removed or at least put up a warning.
My comment was not published, and membership was not granted : (
What I was thinking, and what lead me astray, was that the diode drop does not affect the current, the voltage simply rises to overcome it. But yes a CT is a transformer and must obey the rules. If there is more voltage in the secondary, more heat is being generated, the VI product has risen, and that energy must come from somewhere - there must be more energy extracted from the primary - more VI product - more burden in the primary.
Thank you for correcting my false thinking.
I like the idea of moving the Zener, a diode pointing down as you say. I was thinking of a bridge which might, or might not, be better but certainly costs more. I will LTspice it. I guess I will use a 1:500 transformer and ignore the frequency response of the 56100C for now - www.murata-ps.com/data/magnetics/kmp_5600.pdf
