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 22nd December 2003, 09:31 AM #1 wrl   diyAudio Member   Join Date: Jan 2003 Location: USA Power Supply Questions Ok, I'm trying to design a power supply to power a standby LED, temp sensing IC, and momentary switch. I was planning on using the circuit presented by kristijan-k here, http://www.diyaudio.com/forums/showt...&threadid=6628 I was going to use a regulated power supply like this one... http://www.hut.fi/Misc/Electronics/circuits/psu_5v.html but with a 9V IC. What I was wondering is what value transformer I need to get in order to get 15V after the bridge rectifier. Is there a formula for calculating the DC voltage after rectification compared to the secondary AC volatages? Thanks, Wes __________________ Anything worth doing is worth doing right... and redoing to make it better... and again to fix it back the way it was.
 22nd December 2003, 09:45 AM #2 Richard C   diyAudio Member   Join Date: Jul 2003 Location: Nottingham, England WIth no load connected to the power supply the peak secondary voltage will appear at the capacitors immediately after the bridge. V peak = Vrms x 1.414 The final value with load connected depends on the transformer used and the current drawn by the load.
 23rd December 2003, 04:18 AM #3 wrl   diyAudio Member   Join Date: Jan 2003 Location: USA OK, so I'm a little confused. Are the values listed under the secondary caption at digikey the peak voltages or the rms voltages? -Wes __________________ Anything worth doing is worth doing right... and redoing to make it better... and again to fix it back the way it was.
sreten
R.I.P.

Join Date: Nov 2003
Location: Brighton UK
Quote:
 Originally posted by Richard C WIth no load connected to the power supply the peak secondary voltage will appear at the capacitors immediately after the bridge. V peak = Vrms x 1.414 The final value with load connected depends on the transformer used and the current drawn by the load.
V peak = Vrms x 1.414 - diodes drop (1.6 to 2V for a bridge) - losses.

The diode drop is especially important at low voltages.

sreten.

 23rd December 2003, 04:30 AM #5 wrl   diyAudio Member   Join Date: Jan 2003 Location: USA Also, if I just need a +15 volts to regulate down to + 9V with respect to ground, would I get a transformer to put out +-15 volts after the bridge rectifier and then just not use half of it? This seems a little wasteful, is there a better way to do this? Thanks -Wes __________________ Anything worth doing is worth doing right... and redoing to make it better... and again to fix it back the way it was.
 23rd December 2003, 06:54 AM #6 peranders   Electrons are yellow and more is better! diyAudio Member     Join Date: Apr 2002 Location: Göteborg, Sweden You should have 12 volt AC at least and 15 V AC seems not either to be a waste, quite normal. __________________ /Per-Anders (my first name) or P-A as my friends call me. SSR03 group buy. Sign up for interest HERE. 21 pcb's in order. 0 paid.
 23rd December 2003, 07:47 AM #7 wrl   diyAudio Member   Join Date: Jan 2003 Location: USA Right, I know at least 3 V more than the regulated voltage. But what I was wondering is that would give me +- 15V DC, as in a 30 V differential. Would I just then tie the -15 to ground and use only the + 15, or is there a way I can utilize both the positive and negative terminals off the rectifier to give a + 9V (after a regulator IC) with respect to ground? Or would it just be easier to use only the positive terminal. I'm used to the PSU on my leach amp which utilizes both the positive and negative voltages. But I think that I only need a +9V to power the momentary switch. Thanks, -Wes __________________ Anything worth doing is worth doing right... and redoing to make it better... and again to fix it back the way it was.
 29th December 2003, 06:06 AM #8 wrl   diyAudio Member   Join Date: Jan 2003 Location: USA bump sorry, just wanting to get started... My main questions are 1) Do I just use the output from half of the bridge rectifier and tie the negative DC voltage tab to ground? 2) or is their a way I can invert the negative voltage and then add it in series to the positive voltage? Either option is good, although I already ordered a tranformer to give me +-15 V and was planning on pursuing option (1). Any input on these questions is appreciated! Thanks, Wes __________________ Anything worth doing is worth doing right... and redoing to make it better... and again to fix it back the way it was.

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