# Hi!

Discussion in 'Electronic Basics' started by Nils Magnus Englund, Sep 7, 2004.

1. ### Nils Magnus EnglundGuest

I have a 10W 12V light bulb which I want to install a "fader" on - so when
the switch is turned on, the bulb uses around 1-2 seconds to fade up, and
then fades down again in the same amount of time when you turn the switch
off.

How could I do this? Shouldn't I be able to use a capacitor? (How much
capacitance would it need?) Any other things I could do?

Regards,
Nils Magnus Englund

2. ### Robert C MonsenGuest

You won't be able to fade a 10W bulb with a capacitor; there is simply
too much current.

Now, bulbs have very low resistance when they are starting up, while
they are heating. However, assume that the bulb is heated already.

Power = V^2/R

so

10 = 144/R

so

R = 14.4

You want a 2 second fade, and it generally takes 5 time constants
(R*C) for it to come up to full voltage. Thus

C * 14.4 = 10

so

C = 694mF

Now, this is assuming a constant resistance; unfortunately, the
resistance will be orders of magnitude when you are starting the bulb,
and will go down as the bulb cools. Thus, you'll need a much larger
capacitor, which will be quite expensive if you can even get it.

A better way would be to control the current through the bulb using a
pass transistor and an opamp or two. This assumes you can get 12V to
the bulb and opamp all the time, since you aren't really storing the
energy in the circuit, and you'll need to power the circuit even after
the switch is off.

Here is a circuit that will work for you (view with courier font):

12V
o---------------------o---------------o---------,
| | | |
| | | |
+--+-o | | |
__--o-| | | |
+--+-o | | | |
| | | | |
| .-. ,----|---. | .-.
| 100k | | | | | | ( X ) 10W
| | | | |\| | | '-'
| '-' '--|-\ | | |
| | | >--o | |
| Vc o--------|+/ | | |
| | |/| .-. | |
| --- | | | 82k | |
| 10uF --- | | | |\| ||-+
| | | '-' ,--|-\ ||<- IRL510
| | | | | | >-----||-+
| | | o------|--|+/ |
| | | | | |/| |
| | | .-. '----|---------o
| | | | | | |
| | | | | 1k | .-.
| | | '-' | | | 0.1R
| | | | | | |
| | | | | '-'
| | | | | |
-----------o----------o---o-----------o---------'
GND

OPAMP = LM324 or other single rail opamp.

created by Andy´s ASCII-Circuit v1.25.250804 www.tech-chat.de

The timing is created by the resistor and capacitor on the left side.
The single pole double throw switch alternately grounds or charges the
node into the first opamp's non-inverting input. The opamp is a
follower, so the voltage at the output is the same as at the
non-inverting input. The divider 82k/1k scales this output, and the
right opamp ensures that the voltage at the + side of the 0.1 ohm
resistor is equal to the scaled output. That means the current through
the 10W bulb will follow the same shape curve as the charge/discharge
curve of the control resistor/cap on the left side. The current will
be Vc/8.3 amps. Thus, it'll go from 0 to about 10V, allowing the
transistor to nearly saturate, which means it'll stay relatively cool
when the switch is on.

You can probably get everything you need at radioshack or maplins. You
may need a heatsink for the mosfet, because it dissipates something
like 4W when the circuit is turning on or turning off. The 0.1R sense
resistor should be at least 1/4 W.

Regards,
Bob Monsen

3. ### Graham KnottGuest

Wont you have to discharge the capacitor thro a resistor at switch off?

4. ### JamieGuest

use a common NPN tranny found at radio shaft that can handle
at least 2 amps or more.
most likely found in a T-220 style case.
use a heatShink on it.
you will need a resistor and CAP on the Base .
12 Volts into the resistor.
12 Volts to one side of the Bulb.
the other side of the resistor is connected to the
Base of the tranny.
the Cap (+) side is also connected to the base of the
tranny..
the emitter to ground, the collector to the other side of the
bulb.

first the size of the resistor..
since its obvious the bulb is going to draw aprox.
.8 amps and the average Hfe on a Npn transistor is around
45..70.
so i am will assume 50.
after doing some cals i can see that using a common
560 Ohm resistor should do it around a 1/4 watt or maybe
a 1/2 watt to be safe.
and using a 1000 Uf should be enough to give you aprox
2..3 second rate.

P.s.
you may want to consider running the Transistor in/near saturation
level if you plan on keeping the bulb on.
this will help reduce heating in the transistor.
using a 470 ohm resistor may be better.
as it is basically near that point already.
you should get 0.6 volts less than your source at the bulb if all goes
well.

this is off the top of my head but should do you.