# Calculate Capacitor for dimming 25 LEDs after cutting power source

Discussion in 'Electronic Basics' started by BeachDude, May 5, 2004.

1. ### BeachDudeGuest

I'm 100% amature all the way, but I figured out how to hook up a bunch
of blue/green LED's under the seats of my honda.

Everything is on a 12volt toggle switch so I can turn the LEDs on or
off as I choose (while the car is running). The power is spliced off
the cig.ligher wires.

The effect I want is the LEDs to slowly dim out when the power supply
has been cut (ie. toggle the on/off switch).

The power supply is regulated from 14volts to 7.95volts using a 7808
voltage regulator.There are 25 LEDS, running at 23mAmps each (in
parallel). ANd I have a 220 ohm resistor ahead of each LED so they
run comfortably at 3.6 volts.

What size of capacitor and diodes will I need to make them dim when
the power is cut. I understand from reading other posts, that they
will die as they dim when the forward voltage drops below the arch
point in the LEDs.

I'm not looking for anything fancy... just something to dim them out
automatically (lasting about 5 seconds) as the power diminishes.

Any suggestions?

I was told by the local electronics guy to put 3x 4700 uf Capacitors
.... that should be enought to do 25 LEDS??? HOW do people come up
with these calculations??? HELP.

2. ### SørenGuest

(BeachDude) wrote in

25 x 23 mA = 0.575 A
You want it to last 5 seconds

C = As/V =>

C = 0.575A * 5s / 4.35V = 0.66 F (Farad) or 660,000 µF !

You should go slap your local electronics guy silly (if he weren't

You could use a couple of transistors (or a power darlington) to amplify
the effect of a small capacitor. With eg. a (total) gain of 3,000 you
would only need 220 µF to accomplish the same thing.
And then you don't have to worry about a huge inrush current

3. ### Robert C MonsenGuest

You need to get the electrons from something. If he 'cuts power', the
only source is that capacitor... so he is out of luck with the
amplification idea. If he wants to power his LEDs from the battery,
and use the switch as a signal, then its not too hard to design a
circuit that will decrease the current in a linear fashion.
<http://www.ElektronikTeknolog.dk/cgi-bin/LM317/>

4. ### SoerenGuest

Hi Robert,

I assume his battery will still be available for supply of electrons,
like in any other car, even if he flicks a switch

Exactly

This will do it:

+Batt O---+-----+------+-----------+
| | | |
[100] [470] [470] [470]
| _|_ _|_ etc. _|_
| _\_/_ _\_/_ _\_/_
O | | |
\ +------*-----------+
O | |
| b|/c |
+---| NPN |
| |\e |
220µF _|_ | b|/c
[___] +----| Medium Power NPN
--- |\e
| |
Gnd O-----+------------+

The 100 Ohms on th switch gives a soft on (as well as protect the switch
against inrush current of the capacitor). It could be changed for other

The capacitor is just an example value, since I cannot know the total
gain in the transistor (darlington) stage, as I have no control over
which transistors is selected for the job.

It is easy to build up, so play around with the value of the capacitor
when it is build.

5. ### BeachDudeGuest

Whoa... This just got complicated... a power darlington??? I
kinda sorta get what your saying, as the largest CAP I can get is
4700µF and if I needed over 66,000 µF to get the 5second dim/bleed of
power effect - thats ALOT of Caps!!!

At \$5 a piece, and trying to hid them under my dash board... this
power darlington thing sounds better.

Now please bear with me, I only learned about LEDs, Voltage Regulators
and Resistors a month ago. I know exactly how each on fits into the
"big picture" and the physics behind their operation.

I think the diode and capactor concepts are comming to me... so how do
I use this power darlington to reduce the amount of Caps - keep costs
low... and my brain from getting confused

Thanks.

6. ### John PopelishGuest

You can replace your 8 volt regulator with a 1.2 volt regulator, the
LM317, and use it as the power darlington. It produces an output that
is 1.2 volts more positive than its reference pin, so you pull the
reference pin to the battery with your switch, through a resistor and
limit how positive the reference pin can go with a 6.8 volt zener
diode to ground. To get it to turn off, you have to parallel the
zener with enough resistor to absorb the small current the reference
pin puts out while keeping the voltage low enough that the LEDS are
off. This will work just like what you have now, except that the
switch does not power the regulator, only its reference pin.

If you parallel the zener with a capacitor, the reference pin will sag
slowly when the switch is open, instead of snapping low, but the cap
supplies current only to the resistor around the zener, instead of to
the LEDS, so a much smaller one will do the job. But all the current
is supplied by the regulator so no additional darlington is needed.

7. ### cornytheclownGuest

Look for computer grade caps at surplus electronic component sellers
like all electronics or meci.....They make 10 volt computer grade caps
into the 100,000 uf range....you could series parallel to get your
voltage and capacitance requirements.

5 seconds seems like an awlful long time though....looks like even 2
seconds would look "neat" fading out the lights.

8. ### SørenGuest

Hi,

(BeachDude) wrote in
Actually, it was 10 times that capacity - 660,000 µF.

Yes I can imagine

Message-ID: <[email protected]>
for a complete solution (without voltage regulator).

9. ### BeachDudeGuest

I'm starting to understand, but I need a visual.
Not sure if the attachment will work - but here is the exsiting setup I
have. Its in MS.Excel format - and they say a picture is worth a thousand
words...
This darlington thing has my attention (because 660,000 µF caps is not
possible!!!)

http://users.efni.com/~twm/BasicLEDs.xls
Take this attachment and email it back to me personally with the
modifications so I can clearly see how it works. I'm a point a-to-b kinda
circut beginner.

Thanks.
BeachDude.

PS. If anyone else wants a stab at the whole diagram thing - that would be
awesome too!!!
PPS. Where does one aquire good cirucit drawing software (cheap)...
ms.excel is ok...but slow!!!

You can replace your 8 volt regulator with a 1.2 volt regulator, the
LM317, and use it as the power darlington. It produces an output that
is 1.2 volts more positive than its reference pin, so you pull the
reference pin to the battery with your switch, through a resistor and
limit how positive the reference pin can go with a 6.8 volt zener
diode to ground. To get it to turn off, you have to parallel the
zener with enough resistor to absorb the small current the reference
pin puts out while keeping the voltage low enough that the LEDS are
off. This will work just like what you have now, except that the
switch does not power the regulator, only its reference pin.

If you parallel the zener with a capacitor, the reference pin will sag
slowly when the switch is open, instead of snapping low, but the cap
supplies current only to the resistor around the zener, instead of to
the LEDS, so a much smaller one will do the job. But all the current
is supplied by the regulator so no additional darlington is needed.

10. ### BeachDudeGuest

How does the power darlington calculate the rate at which the LEDs dim?

Can you easily modifiy it so they take longer or shorter to blink out?

Thanks.

You can replace your 8 volt regulator with a 1.2 volt regulator, the
LM317, and use it as the power darlington. It produces an output that
is 1.2 volts more positive than its reference pin, so you pull the
reference pin to the battery with your switch, through a resistor and
limit how positive the reference pin can go with a 6.8 volt zener
diode to ground. To get it to turn off, you have to parallel the
zener with enough resistor to absorb the small current the reference
pin puts out while keeping the voltage low enough that the LEDS are
off. This will work just like what you have now, except that the
switch does not power the regulator, only its reference pin.

If you parallel the zener with a capacitor, the reference pin will sag
slowly when the switch is open, instead of snapping low, but the cap
supplies current only to the resistor around the zener, instead of to
the LEDS, so a much smaller one will do the job. But all the current
is supplied by the regulator so no additional darlington is needed.

11. ### Robert C MonsenGuest

The power darlington can be seen as a 'current multiplier'. Assuming a
current gain of 1,000 (which isn't too unreasonable) you have a cap
discharging into your base, allowing 1,000 times the current to be
passed between the collector and emitter.

With this design, you choose the capacitor that works best with the
load you have, since you can't predict exactly what the current gain
of the transistor will be.

Regards
Bob Monsen

12. ### John PopelishGuest

The regulator as I describe it, above, is just a current amplifier
that makes a copy of what ever voltage is applied to its reference pin
with the addition of 1.2 volts. A mechanical analogy would be power
steering that is off center. The regulator supplies all the LED
current but does this while copying the reference pin voltage. You
could easily program the off time by switching between capacitors
connected to the reference pin, or disconnecting them, altogether or
instant on, off operation.

Here is the data sheet for the LM317:
http://cache.national.com/ds/LM/LM117.pdf

Note on page 4 that the adjustment pin (what I have been calling the
reference pin) current is not more than 100 ua (that's micro amperes)
and the adjustment pin current change as the output current swings
from 10 ma to Imax is not more than 5 ua.

So the resistor in parallel with the zener diode must pass at least
100 ua at a voltage drop that allows almost no current to pass through
your LEDS (lets guess that they are off at 2 volts, so 1.2 volts less
than this would be .8 volts) so .8 volts / 100 ua = 8000 ohms. Let's
say we pick 4700 (a common value) just to be on the safe side.

The on switch will have to provide enough current to this resistor to
pull it up well above 6.8 volts to have the zener diode turn on and
hold the on voltage steady. With 12 volts in and something like 10
volts across the 4700 ground resistor (to provide the extra current
for the zener) the switch resistor would have to be about 2/10 * 4700,
since it drops 2 volts when the 4700 drops 10.
This is about 940 ohms. Lets say we go with 1k, since the regulator
will also dump its less than 100 ua into the zener.

Now, to calculate the fade time constant in seconds, you just divide
the time by 4700 ohms to come up with the capacitance in farads.
5 seconds / 4700 ohms = .001 farad = 1000 uf, a very cheap and
available part.

There will also be a slowing of the on voltage but this is much faster
because the 1000 ohm resister is effectively in parallel with the 4700
ohm resistor during the charge up, so they have an equivalent parallel
resistance of about 825 ohms so the turn on time will be about 6 times
faster than the turn off.

13. ### John PopelishGuest

Did you receive my email response?