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From: "Da Man"
References: <%AEP9.18271$Jb.firstname.lastname@example.org> <email@example.com> <22MP9.firstname.lastname@example.org> <email@example.com> <firstname.lastname@example.org>
Subject: Re: LED Ballast
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Date: Tue, 31 Dec 2002 15:17:17 GMT
NNTP-Posting-Date: Tue, 31 Dec 2002 11:17:17 AST
Use a fuseable resistor. Two parts in one! I wonder if they are avalible in
UL / CSA / CE approved versions for 120 / 220 V AC?
"Bill Sloman" wrote in message
> Spehro Pefhany wrote in message
> > On Mon, 30 Dec 2002 00:02:06 GMT, the renowned "Dave M"
> > wrote:
> > >"cpemma" wrote in message
> > >news:email@example.com...
> > >> Mark Mark wrote:
> > >> > Yup I know a resistor and a transformer would work just fine but I
> > >> > actually saw a ballast that powered LEDs and the owner didn't feel
> > >> > like telling me anything about it's manufacturer.
> > >>
> > >> A 12k (240v) or 5k6 (110v) resistor will do, giving 20mA led current,
> > >> but you need a diode (1N4004, 400v PIV, just in case the led fails;)
> > >> inverse-parallel with the led so it doesn't see any significant
> > >> voltage.
> > >>
> > >>
> > >I would put the 1N4004 in series with the LED and resistor. That way,
> > >power is being wasted in the half cycle that the LED doesn't conduct.
> > >LED is protected from reverse breakdown by the 1N4004.
> > You could also use a bridge of 4x 1N4148s at the LED or 1x WOx bridge
> > and generally be better off. If using the series 1N4004, I'd still put
> > a parallel 1N4148 etc. across the LED.
> > Note that the LED is probably not UL approved insulation for the mains
> > voltage and you may have real or compliance-related safety issues on
> > that regard.
> Chip Shults'capacitor and resistor would let you tie one end of the
> LED to ground, which should allow you to guarantee extra-low safety
> voltage (less than 24V) between the LEd and any damn-fool humnan with
> a very thin finger, which should deal with any safety issues. You
> might have to add a zener to protect against the LED failing open
> The capacitor would be chosen such that it dropped most of the AC
> voltage from the mains, and the resistor - ideally a part guaranteed
> to fail to open circuit, thus metal film or metal oxide - would limit
> the current in the event of a mains spike. I've always liked the
> Philips - now BC Components - MRS25 parts, which can dissipate 600mW.
> With 20mA through the LED you can afford to drop 30V (rms) across a
> 1k5 MRS25. By the time you have toleranced everything - power film
> capacitors tend to be be +/-10% or worse - that 1k5 probably falls
> back to 1k.
> The capacitance for 120V/60Hz would be about 0.47uF - Farnell doesn't
> seem to stock any class Y supression capacitors as big as that, and a
> 0.47uF X1 costs about $1.50 in small quantities. You might need to add
> a fuse in series with the resistor and the X1 capacitor to make
> everybody perfectly happy that the circuit would fail safe if the
> capacitor failed to a short circuit, even if the resistor was more or
> less guaranteed to blow first.
> John Woodgate may well find flaws in this superficially attractive
> scheme, but it looks reasonably neat to me. Using a reactive impedance
> for your current limiting reduces the heat dissipated in the ballast
> from about 4.8W (20mA at 240V) to the 500mW you can dissipate in a
> moderately cheap resistor.This would make construction a bit easier,
> in that you don't have to worry about providing a low thermal
> resistance heat path to ambient.
> Bill Sloman, Nijmegen
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