Which stepdown transformer give equivalent of 18V DC

[HankMcSpank]

sorry for the somewhat strange title.

I'm running a DC current through an coil to heat it up (don't ask!) ...which for my purporses is about right at circa 18V DC.

Now I'd rather run an AC voltage through the coil, but here's my dilemna.

to get the equivalent 18V DC voltage I'd need to find what AC signal RMS value would yield that, a quick look on here...

RMS Calculator

reveals that 25V pk (or 50V peak to peak), would be about 17.9V RMS.

I live in the UK where the mains is 240V AC. Therefore is it as simple as trying to source a 5:1 step down transformer?

or have I missed an important point?!



Also...I have a fair few transformers inside 'Wall Warts' lying around - to identify a suitable one to rip apart to source a transformer from, I'm figuring to get a DC voltage of 20V, it'd need a raw AC signal of about 65Vpk-pk (ie full wave rectified = 25V peak, then regulated down to give 20V) ...if my basic calculations are right, I need to look about for a DC Wall wart in the region of 20V DC

 

[woody]

If you were simply heating a resistor the ac voltage neaded to heat it and the dc voltage would be the same in your case 18V. BUT because you are heating a coil with ac things get a LOT more complicated. Is this an air core coil or an iron core
coil? What is the inductance, what is the dc resistance? In any case the value of ac
needed to heat the coil will be higher than the dc value. Supose the inductance is .1MH then the ac value will only need to be a fraction of a volt higher than the
dc voltage. But supose the inductance is 100H you could need thousands of volts
to get the same heating as using dc.

 

[HankMcSpank]

Hi Woody,

Thanks for the quick reply.

Yes, I should have said about the coil's inductance - it's under 1mH (probably about 800uH), so at 50 Hz, the reactance is going to be pretty negligible...and I've all but disregarded it (like I say, my target is the AC equivalent of 'circa' 18V DC not exactly 18V)

With this new info in mind, are my original calculations in the ball park?

 

[woody]

Yes that's my take on it. 800uh would be about .25 ohm of reactance. So 18v plus
what ever voltage would be lost across .25 ohms in seriese with the coil. But back
to your figure of 17.9v ac. The reason you got 17.9V is a rounding error. 18v ac provides the same energy into a resistive load as 18v dc.

 

[HankMcSpank]

18v ac provides the same energy into a resistive load as 18v dc.

i'm no expert but is that right? I thought the point of converting AC into RMS is to allow DC -esque calculations?

So to get the same energy as 18V DC into a resistive load, would it not take 25pk AC signal? (50V peak to peak)

 

[Th3 uN1Qu3]

The first assumption is wrong. A diode bridge converts the one way peak AC voltage to DC, minus the voltage dropped on the diodes of course. But when it comes to heating you are right in a way, the AC RMS value required to heat up something to a temperature is equal to the DC value required to heat that something to the same temperature. This is because in DC peak voltage is equal to RMS voltage.

 

[xdpereira]

Here is an easy way to compare energies: Connect a 24v automobile turn indicator lamp to your dc source and observe the brightness. Now connect the same lamp to each of your available ac sources. Select the one that comes closest in brightness.
Guaranteed to work!

 

[jan.didden]

sorry for the somewhat strange title.

I'm running a DC current through an coil to heat it up (don't ask!) ...which for my purporses is about right at circa 18V DC.

Now I'd rather run an AC voltage through the coil, but here's my dilemna.

to get the equivalent 18V DC voltage I'd need to find what AC signal RMS value would yield that, a quick look on here...[SNIP]

It's even simpler. 18V RMS gives the same heating as 18VDC. By definition.

Mains transformes are always spec'ed in terms of V RMS. You need a 18V secondary transformer.

jd