# Discovering what kind of CT clamp I have and what the readings are

Hello. Newbie here. I am trying to use these CT clamps from a broken Home Energy Meter – Aeotec system. (The clamps are not broken, but the base zwave unit is.)

According to their docs, I have the 200A clamps (based on dimensions). I have no idea how many turns there are inside the clamp and their documentation doesn’t say either. Nor do I know the step-down ratio of the clamp (docs don’t say this either). Taking apart one of the clamps, there are 2, 1N4761A Zener Diodes soldered together from each wire lead, before exiting the clamp.

I’ve exposed the 3 wires from a power strip, and clamped around one of them. Using a hair dryer, I multimeter’d the two wires coming from the clamp. I got 2.5V at max dryer usage (high/hot air). The dryer says 1825W 125V 60Hz on the handle. My multimeter says I have 120V (USA).

I have several questions. 1) Do the diodes function as burden resistors, or do I still need to put a resistor between the clamp leads? 2) The reading of 2.5V with the hair dryer on means what exactly? Can I infer anything or am I still missing some critical piece of information (like A->A ratio / turns ratio)? 3) If I can measure the amps the dryer is using, does that correlate to the 2.5V reading and help determine anything?

Welcome, @jackt, to the OEM forum.

I think what you have there is a genuine current transformer, and its output will be a current, not a voltage. What makes me think this is the presence of the zener diodes. There are there for safety - without them, a c.t. will try its best to drive the current it wants to output into an infinitely high resistance load, which implies an infinitely high voltage. The danger is obvious, and the zener diodes clamp the voltage to a safe level - if 75 V can be considered ‘safe’. They should do nothing in normal use.

It’s always safe to short-circuit a current transformer, even a voltage-output type - just as it’s always safe to leave a (familiar) voltage transformer open circuit - so I think you’ll get a much more meaningful result if you repeat your test but with your multimeter reading a.c. amps (or more likely, milliamps). What you ideally need is the secondary current at 200 A primary current - you can put multiple turns of wire through the c.t. to multiply the 15 A or so that you drier takes, e.g. 6 passes through equals ×6, equivalent to about 90 A . If you do this, I suggest you start with 1 pass and record the current you see, adding a pass each time. What you should see is a straight-line graph.

If it’s possible, you need to measure the primary current (the drier) and the secondary current (with the meter on amps and short-circuiting the c.t. secondary winding) at the same time.

And remember, c.t’s are specified by the current ratio, not by the number of turns, so a 2 kA : 5A c.t. is a totally different beast to a 20 A : 50 mA, even though they have the same nominal number of secondary turns.

Given the current ratio, and a knowledge of the voltage whatever you’re using the c.t. with needs (assuming of course it does need a voltage and not a current), you can easily calculate the value of the burden.

My bet is the output current will be 50 mA or 100 mA at 200 A (if this is the rated current).