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Re: current transformer winding wire help please

>

>>>>I am thinking of using 0.125mm wire for the secondary of a current >>>>transformer but don't know its current rating. DAGS and wire tables >> >> ... >>>> >>> >>>A guess for iron transformers is 4A/mm^2 or 4E6A/m^2 and that's for a >> ... >>> >> >> Thank you Genome and John Fields for your replies, I guess I should have >> given more information. Hope this is not too long and I don't show my >> ignorance too much. >> ... >> Rectified/filtered output came out at 71V and 11mA with a 6K resistive

load

> and 47V and 30mA with a 1.5K load. >> I can easily reduce the turns to lower the voltage but what current

should I

> aim for so that this 'thing' will live happily ever after ? The

transformer

> will be in a ventilated plastic box under the house. Ambient sometimes

gets

> to 35 °C. >> ... > >C'mon, guys, the detailed theory may be a bit involved, but the practice >of CT's is pretty simple! > >If you have the secondary evenly wound with N turns of low-resistance >wire, and short it, the current through the secondary opposes the >current through the primary to exclude all flux from the core. So the >secondary current is 1/N the primary current, to a very good >approximation. > >The farther you get from a perfect sheet of secondary current, >zero-resistance wire, and a perfect short circuit, the more flux gets >into the core, and generally the less accurate the current ratio. But >the approximation is still very good up to core saturation, insulation >breakdown, and secondary power dissipation limitation. > >The only thing about the OP's CT that really concerns me is the very >thin, 0.125mm wire. Otherwise, he could just short a couple of LED's >back-to-back in series with another pair of back-to-back LED's and put >the resulting voltage into his input. > >If he doesn't have enough turns, he gets too much current. 47V at 30ma >into 1.5K could make for too much current into a LED. > >He's obviously using his core evenly enough to get 71V at 11ma. The 71V >is probaly limited by core saturation, causing the unequal >voltage-current ratio. The 47V at 30ma may be close enough to the >linear range of his CT that we can expect 300ma at 5v, but then he may >have winding resistance issues bringing on extra secondary voltage drop, >and maybe too much secondary power dissipation. > >I'd suggest running the CT into a 10-ohm power resistor and measuring >the resulting voltage. Or better yet, directly into an ammeter. This >would tell you very nearly the actual turns ratio, and whether the >ultimate current is low enough for a LED circuit (or a Zener, or even a >specially selected low-value resistor for your needs). Run it for a >while to see if the current transformer heats up. If it stays >reasonably cool for half an hour or so, and gives a reasonable current >into the LED circuit, you may be in good shape already. > >Of course, you get a free power indicator with the LED's, too :-). > >John Perry

Thank you to everyone, you've been very helpful and I appreciate your advice, much of which I'm sorry to say, I didn't understand.

After reading (many times) what you all have said I did as John Perry suggested. I used a fan heater as a load on the primary (9.5A) and a 100 ohm

5W resistor on the secondary and ran it for an hour. I used the 100 ohm because it gave a good sine wave output without distortion. The drop across the primary was 320mV, the voltage on the secondary 11V with a current of 108mA.

After an hour I could hold my finger on the transformer core without a worry but of course I couldn't do that with the resistor. The windings remained cool to the touch.

I'm not sure what all this means but it seems like it will do my job. I will run it soon for a 4 hour test to see if I've 'tuned for smoke' :o) If anyone thinks I'm an accident waiting to happen please let me know.

I'm thinking of rectifying and filtering to feed DC to an optocoupler for the feedback to the computer so I imagine the operation of the transformer will change again.

Thankyou again for your help and your time and if anyone has a better way of doing all this I'd love to hear about it.

Gordon

PS. I know what a WAG is but Barry and Chair ?

Reply to
Gordon W
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OK, Gordon,

I assume you've seen my later correction to some of the silliness of my original post, so I'll go from there.

100 ohms is more than I'd like to see, but it could have gotten you close enough to the true picture. The nice sine wave is a good sign, but eyeballing a scope trace is not ideal. Assuming you're not saturating a little (that is, 100 ohms is not too much resistance), you have 9500/108=88 turns on your secondary. That 108ma is going to be far too much for most LED's, so you'll need a much lower parallel resistance to get within the voltage rating of your optocoupler.

Think of the secondary as an AC current source. No matter what impedance you put across its output, it will put out 1/88 the primary current, up to the linear limit of the core. A 10-ohm resistor will dissipate 1/10th the power of your 100-ohm resistor, and your primary voltage will be 1/10 its present value. Likewise, your secondary voltage will be 1.1V, 1/10 what it was with the 100-ohm resistor.

Now, consider that you'll likely need 20ma to get reliable signal from your optocoupler. You need to shunt 90'ish ma from the 1.8'ish volts the optocoupler's LED will drop: 1.8V/.09A=20 ohm in parallel with your optocoupler. Add your rectifier bridge (or don't, unless you really want a DC on-off signal), and you're done.

Of course, I'd really like to see your measurements with a 10-ohm resistor on the secondary, so we can be really sure you have the turns ratio nailed down. The closer you are to saturating, the less confidence we can have in your measurements. And you've already shown that the CT can handle the expected current, so things only get better from here on.

Oh, remember that all the calculations above (except the turns ratio) depend on the 9.5A primary current. If that changes, the secondary current changes accordingly, requiring recalculation of the shunt resistor. If you expect varying primary current, you'll have to regulate the signal to the optocoupler.

jp

Reply to
John Perry

ohm

with

worry

remained

I did it again with the 100 and a 10 ohm resistor and there is a very slight flattening on the peaks of the sine wave more on the 100 so I guess I'm just on saturation. With the 10 ohm I measured 1.1V on the secondary and 110mA as you predicted. The couple of points difference in the turns ratio I guess is due either to nearing saturation or the accuracy of my DMM. What I've done now is connect a bridge to the secondary with a capacitor and a 22 ohm resistor and a red LED across the output of the bridge. This gives

1.63V and 20mA through the LED and 230mV on the primary.

Yes I will have to do it again as the real deal pulls 7.5A and I'm hoping the current through the element won't change much as the water heats up.

Thank you John you've spent a lot of time on this and I appreciate it.

Gordon

Reply to
Gordon W

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