Re: 12V regulator


"Steve" <s@xxx> wrote in message
news:oktnf4h6dce4cc3dhafjpvtovermnbv6qu@xxxxxxxxxx
On Sun, 19 Oct 2008 21:35:39 -0400, legg <legg@xxxxxxxxxxxxxxx> wrote:

On Sun, 19 Oct 2008 13:19:11 -0500, Steve <s@xxx> wrote:

I have a spst lighted switch that uses a 12vdc lamp (led w/ resistor).
It's just a basic three terminal switch, one is n.o., one is common,
the other is +12vdc for the lamp. I'm trying to design a circuit
that's a basic as possible to drop the 115vac down to 12v.

A .047uF 400Vdc or 250Vac (X2) will limit led current to ~2mA,
approximately half of which goes through the led. Placing a 100R 1/2W
resistor in series will limit capacitive surge current through the
diode required to route reverse polarity current around the indicator
terminals.

Due to switch construction, the whole shebang will be live when the
switch is on, so suitable precautions are required to isolate this
from surrounding circuitry and hardware and to fix the parts securely
in position.

Brightness may be an issue - larger capacitors, at this voltage, tend
to get physically large.

This is probably only cheaper than obtaining the appropriate switch if
all of the parts required are already on hand.

RL
I think you're right, probably best to just go get a neon switch. I
was trying to use parts I had on hand, this was a nice looking blue
led switch, but it's proving a bit more difficult than I wanted for
this particular application. Thanks for the replies,
Steve

You should be able to use about a 10k resistor (about 1-2 watt), and a
1N4004 or higher rated diode, in series with the LED. About 12 mA will flow
at a 50% duty cycle, for an average of 6 mA, which might not be bright
enough, but should be OK. A 5 k 2W resistor will give twice the brightness
with more heat. It should be OK if it is away from any plastic parts, as it
will get hot. Capacitors are more efficient, but can produce high surge
currents through the LED if there are spikes on the line.

Paul


.

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