From: John Woodgate
Subject: Re: transformerless 5V power supply
Date: Sun, 15 Sep 2002 21:39:42 +0100
Organization: JMWA Electronics Consultancy
Reply-To: John Woodgate
NNTP-Posting-Date: Sun, 15 Sep 2002 20:49:52 +0000 (UTC)
X-Newsreader: Turnpike (32) Version 4.01 <5Z8C9wtxbnpWyFnyfFzqmVF739>
I read in sci.electronics.design that Nico Coesel
wrote (in <email@example.com>)
about 'transformerless 5V power supply', on Sun, 15 Sep 2002:
>Daniel Carignan wrote:
>>Anyone have any ideas how to get about 50 mA at 5V, from the mains,
>>without using a transformer, and with as little amount of components as
>>possible? I though there was an IC (switched capacitor or something like
>>it) that does just that but I can't remember the maker or part number.
>I've seen a circuit which simply uses a capacitor, a small resistor in
>series with the mains to reduce the line voltage. Next a rectifier
>diode, capacitor and zener diode are used to regulate the output
>voltage a bit. If you use choose an 8.2V or 10V zener diode you can
>use a 78L05 (100mA regulator) to stabilise the supply.
Even 50 mA is pushing the technology for this hobbyist-killer; it's very
inefficient except at very low currents. The capacitor converts the
mains into an approximate current source. The series resistor is there
to prevent a quasi-infinite inrush current on switch-on and MUST be of
adequate value and MUST be of high power rating. All the inrush has to
go through the zener, so it needs to be limited to well under 1 A. To
get 50 mA d.c. from a half-wave rectifier, you need about 130 mA a.c. So
the cap reactance needs to be about 850 ohms (120 V mains less 10 V
zener). That's 3.3 uF (allowing for some volts across the series
resistor), rated for continuous use at 120 V 60 Hz. Not small or cheap.
To limit the inrush even to 500 mA, the resistor needs to be 220 ohms,
and with 130 mA flowing it needs to be at least a 4 W part.
Regards, John Woodgate, OOO - Own Opinions Only. http://www.jmwa.demon.co.uk
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