Auto shut-off circuit for light

[nscott]

I have a small battery operated lamp that I would like to build an auto shut
off circuit for. I want to switch the lamp on and then have it shut off (or
dim out) after about 30 seconds.

Perhaps a circuit that goes on long enough to charge a cap, shuts off, and
then lets the cap power the lamp for a few seconds until it bleeds off?

The lamp is powered by 6 VDC (4 * AA) and the only components in it are the
batteries, switch, and bulb.

Thanks in advance for any help.

nscott

 

[Robert Monsen]

I have a small battery operated lamp that I would like to build an auto shut
off circuit for. I want to switch the lamp on and then have it shut off (or
dim out) after about 30 seconds.

Perhaps a circuit that goes on long enough to charge a cap, shuts off, and
then lets the cap power the lamp for a few seconds until it bleeds off?

The lamp is powered by 6 VDC (4 * AA) and the only components in it are the
batteries, switch, and bulb.

Thanks in advance for any help.

nscott

This one will keep the light running while the button is pressed, and for 30
seconds after the button is released.

http://home.comcast.net/~rcmonsen/misc/timedlight.jpg

The mosfet should be a fairly low on resistance one, and its Vgs(th), which
is the point at which it just starts to conduct, will determine the timeout.
The one in the circuit has a Vgs(th) of about 3.2V, so with a 6V start, the
drop will take -ln(3.2/6)*220k*220uF = about 30 seconds. The timeout formula
is

t = -ln(Vgs(th)/6)*R*C

Thats assuming perfect components.

Regards,
Bob Monsen

 

[Walter Harley]

This one will keep the light running while the button is pressed, and for 30
seconds after the button is released.

http://home.comcast.net/~rcmonsen/misc/timedlight.jpg

The mosfet should be a fairly low on resistance one, and its Vgs(th), which
is the point at which it just starts to conduct, will determine the timeout.
The one in the circuit has a Vgs(th) of about 3.2V, so with a 6V start, the
drop will take -ln(3.2/6)*220k*220uF = about 30 seconds.

If you used a 1M resistor, you could use a 47uF cap, which would be
considerably smaller. (Wouldn't want to go higher than 1M or it would start
turning on when you waved your hand near it.)

When you press the button, the cap charges through the diode. The only
thing limiting current is the ESR of the cap and of the batteries - probably
a total of 1 or 2 ohms. So, peak current (for a brief moment) of let's say
5A, fading with an exponential time constant of t = RC = 220usec. The
Philips datasheet (first one I looked at) says the 1N914 can take a one-time
peak of 4A for 1usec, or 1A for 1msec. Maybe something like a 1N4001 would
be better? Also, using a smaller cap would shorten the current pulse, plus
it would probably have higher ESR.

 

[Robert Monsen]

for

If you used a 1M resistor, you could use a 47uF cap, which would be
considerably smaller. (Wouldn't want to go higher than 1M or it would start
turning on when you waved your hand near it.)

When you press the button, the cap charges through the diode. The only
thing limiting current is the ESR of the cap and of the batteries - probably
a total of 1 or 2 ohms. So, peak current (for a brief moment) of let's say
5A, fading with an exponential time constant of t = RC = 220usec. The
Philips datasheet (first one I looked at) says the 1N914 can take a one-time
peak of 4A for 1usec, or 1A for 1msec. Maybe something like a 1N4001 would
be better? Also, using a smaller cap would shorten the current pulse, plus
it would probably have higher ESR.

Actually, the diode is not necessary at all. It was left over from when I
was using a signal generator to generate the pulse while simulating the
circuit.

A little resistor would probably be a better choice, now that I think about
it. Maybe 10 ohms 1/4W?

And, I think the mosfet the OP buys will probably have a different Vgs(th)
than the simulated one, so he'll have to use the formula anyway, and pick
whatever components are convienient. Thats why I included the formula.

Regards,
Bob Monsen