Which stepdown transformer give equivalent of 18V DC - diyAudio
 Which stepdown transformer give equivalent of 18V DC
 User Name Stay logged in? Password
 Home Forums Rules Articles diyAudio Store Gallery Wiki Blogs Register Donations FAQ Calendar Search Today's Posts Mark Forums Read Search

 Please consider donating to help us continue to serve you. Ads on/off / Custom Title / More PMs / More album space / Advanced printing & mass image saving
 4th March 2010, 02:24 PM #1 diyAudio Member   Join Date: May 2008 Which stepdown transformer give equivalent of 18V DC 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 Last edited by HankMcSpank; 4th March 2010 at 02:32 PM.
 4th March 2010, 03:04 PM #2 diyAudio Member   Join Date: Jan 2002 Location: Tyrone Ga. U.S.A. 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.
 4th March 2010, 03:13 PM #3 diyAudio Member   Join Date: May 2008 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?
 4th March 2010, 03:36 PM #4 diyAudio Member   Join Date: Jan 2002 Location: Tyrone Ga. U.S.A. 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.
diyAudio Member

Join Date: May 2008
Quote:
 Originally Posted by woody 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)

 4th March 2010, 04:17 PM #6 diyAudio Member   Join Date: Feb 2010 Location: Bucharest 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. Last edited by Th3 uN1Qu3; 4th March 2010 at 04:19 PM.
 4th March 2010, 04:19 PM #7 diyAudio Member   Join Date: Feb 2010 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!
diyAudio Member

Join Date: May 2002
Location: The great city of Turnhout, BE
Blog Entries: 8
Quote:
 Originally Posted by 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...[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
__________________
Music is dither to the brain; lets me think below the usual chaos - me
Linear Audio Vol 13 is out! Check out my Autoranger and SilentSwitcher

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is OffTrackbacks are Off Pingbacks are Off Refbacks are Off Forum Rules

 Similar Threads Thread Thread Starter Forum Replies Last Post argonrepublic Power Supplies 1 7th January 2009 06:47 PM Learnincurve Power Supplies 10 3rd December 2007 07:46 PM finestocean Parts 9 29th July 2005 08:46 PM cowanrg Parts 2 3rd October 2003 03:41 AM

 New To Site? Need Help?

All times are GMT. The time now is 10:18 AM.

vBulletin Optimisation provided by vB Optimise (Pro) - vBulletin Mods & Addons Copyright © 2017 DragonByte Technologies Ltd.