AC Clamp Meter measuring DC

[drewh]

Hi all,

Great forum! Have read posts from time to time but only registering now.

I have a quick question about my AC Clamp Meter. I wanted to measure the current my pool Chlorinator cell was drawing from the power unit - its rated at 7V DC / 30 A. Seeing as I only had the AC clamp meter I gave it a try to see, and it read the current fine (about 14.5A, it's several years old now).

I'm slightly confused by this, the current generated from the power unit would be DC would it not (probably full wave rectification?), so how is it reading it? Just taking the top peak each time?

I have an intermediate knowledge of electronics but haven't really looked into this side of things at all.

 

[Lucien Nunes]

I did measure the current draw on the AC feed side, it was 1.06A

Unfortunately you can't calculate it effectively from that reading either as the waveform will be badly distorted by the half-wave rectification.

Does your multimeter count to 2999? I won't recommend this but the 10A range on many meters is rated to 20A for some short duration like 30 seconds, and if so, might withstand 30A for long enough to take a reading. Not recommended because at 30A the shunt will get nine times as hot as at 10A and could be rapidly damaged, and the reading might not be accurate.

You could make a shunt, take a reading of something stable under 10A like a lamp load and make a shunt that halves the reading, then use it to measure the cell current without changing the configuration. Of course if it has a different tempco to the meter's own shunt the two will diverge as they warm up under load. And you'll still be at the mercy of the meter's response to the waveform.

It's just unfortunate that you have a particularly unfriendly waveform to measure, at a current in excess of a normal DMM's capabilities, at a voltage that won't tolerate much voltage burden from the ammeter.

Actually, I wonder about the half-wave rectification. If I were making the power unit I'd use full-wave, because you can use a smaller transformer that would make more of a saving than the extra cost of the rectifier. Why half-wave, or is that a red herring?

 

[PEG]

^^^^ Love it when you talk dirty,Lucien :bow:

 

[drewh]

Unfortunately you can't calculate it effectively from that reading either as the waveform will be badly distorted by the half-wave rectification.

Does your multimeter count to 2999? I won't recommend this but the 10A range on many meters is rated to 20A for some short duration like 30 seconds, and if so, might withstand 30A for long enough to take a reading. Not recommended because at 30A the shunt will get nine times as hot as at 10A and could be rapidly damaged, and the reading might not be accurate.

You could make a shunt, take a reading of something stable under 10A like a lamp load and make a shunt that halves the reading, then use it to measure the cell current without changing the configuration. Of course if it has a different tempco to the meter's own shunt the two will diverge as they warm up under load. And you'll still be at the mercy of the meter's response to the waveform.

It's just unfortunate that you have a particularly unfriendly waveform to measure, at a current in excess of a normal DMM's capabilities, at a voltage that won't tolerate much voltage burden from the ammeter.

Actually, I wonder about the half-wave rectification. If I were making the power unit I'd use full-wave, because you can use a smaller transformer that would make more of a saving than the extra cost of the rectifier. Why half-wave, or is that a red herring?

Great to read all that and very interesting.

My meter reads to 3999 but I'm not that confident on making a shunt and wouldn't want to risk it in this case especially if it's not entirely correctly done.

I know so annoying that it's above what my meter can handle. Just a question, given the measurement would literally take 5 seconds (I'd switch it on, check measurement and switch it off), could I connect the meter in series with the cell lead? Ie chlorinator power unit from positive, wire to one meter lead (alligator clips on end to meter), then from other meter lead to respective connection on the cell wiring. Then literally turn it on for 5 seconds, read, off. The cell being 4 years old will more than likely draw under 20A, given its 30A rated for new.

Just a question also given the half wave non smoothed source. Could this 14.5A actually be in any way a close approximation - is the AC clamp meter ignoring the missing lower half of the waveform and just measuring the upper peaks? If it was close, it really wouldn't matter.

Good point re half wave but no reason for the guy to lie - if anything he could have said full wave rectification with smoothing to make it sound more impressive. So I'd give the benefit of the doubt there I think.