# What determines the real world transformer ratio of a clamp on AC power meter?

The ideal step-up transformer, when the primary coil has 1 winding and the secondary coil has 60 windings would step up the voltage 60 times.

With this improvised clamp on AC power meter, which is essentially a transformer, they used 60 windings for the secondary coil, and the live wire is essentially wound once (or half?). I would expect that to step up the voltage 60 times, but instead they only get 0.2 mV - 2 mV AC of output per 1 amp of current of mains flowing through, depending on the core (various cores used were an aluminum carabiner, a zinc/steel key clamp, a key ring). Why is this so low, does the core make such a big difference that what should be a step up of 60 is instead a step down of 100,000? And why is it proportional to the current flowing?

Edit: A couple clarifications: in this setup the primary is getting residential mains voltage, so 120v. A space heater (1kw, so several amps) is turned on, and the transformer is clamped on to one wire (a split extension cord).

• There may be worse ways of making a current transformer, but none come to mind right now. Think of the alli crab as a shorting support for a Rogowski coil , and read carefully how that's supposed to work, as opposed to a how a current transformer is supposed to. Feb 4, 2019 at 19:56
• Why do you think 0.2~2mV is a step down? What's the voltage across the primary? Feb 4, 2019 at 20:04
• Unless the core cross section area is huge or the mains side voltage is minuscule, you saturated the core hard. Did you have any source impedance or blew a fuse? You need to calculate the number of primary turns according to Urms=4.44fNAB. Feb 4, 2019 at 20:41
• @Hearth updated the question, the voltage across the primary is 120v. Feb 4, 2019 at 20:59
• @JohnSmith If there's 120V across the primary of the current transformer, that leaves 0V for the heater. Remember KVL. Feb 4, 2019 at 21:01

I would expect that to step up the voltage 60 times, but instead they only get 0.2 mV - 2 mV AC of output per 1 amp of current of mains flowing through, ... Why is this so low, does the core make such a big difference that what should be a step up of 60 is instead a step down of 100,000?

simulate this circuit – Schematic created using CircuitLab

Figure 1. The setup.

If you're measuring voltage on the output then you need to measure voltage on the input. If your transformer is of decent quality then 1 mV on the output suggests a 1/602 mV drop on the primary. This will be measured by VM1 and will exclude the voltage drop across the load.

And why is it proportional to the current flowing?

Because that's how transformers work. The primary current induces an alternating magnetic flux in the core proportional to the current in the primary and that induces a proportional current in the secondary.

Response to the update:

in this setup the primary is getting residential mains voltage, so 120v.

No it's not. The lamp is getting 120 V. There is a piece of wire in the primary of the CT will extremely low resistance. The point of my answer is that VM2 will have a 120 V reading but VM1 will have a fraction of a millivolt.

The primary sees the resistance of the secondary divided by the turns ratio squared. Since this is very low the voltage drop in that part of the primary circuit is very low.

• I've updated the question with a bit more info, using your example the voltage of the input is 120v, and the lamp (really a space heater) is using several amps. For the 2nd part, I don't understand why the voltage VM1 is proportional to the current through LAMP1, which is what happens when using a clamp-on ammeter. Feb 4, 2019 at 21:06
• @transistor voltages are induced and not currents. Current flows due to the induced voltage and a path of conductivity. Feb 5, 2019 at 12:01
• Thanks, Andy. I was afraid I'd get pulled up on that. You have a far superior grasp of magnetics than me so I'm quite happy for you to correct mine or post your own answer. Feb 5, 2019 at 13:52