Date:         Mon, 31 Jul 2017 14:40:56 -0700
Reply-To:     Alistair Bell <albell@SHAW.CA>
Sender:       Vanagon Mailing List <vanagon@gerry.vanagon.com>
From:         Alistair Bell <albell@SHAW.CA>
Subject:      Re: Time Delay LED Lighting Circuit
Comments: To: David Beierl <dbeierl@ATTGLOBAL.NET>
In-Reply-To:  <CAMOH8LLU-FsKoi-VHpkhmbKu1Yh-K8foK_08ZDSMigkPFQf0og@mail.gmail.com>
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Haven't found your basic led to dim very well david. Back in the day I made up led and resistor combo and fitted into the bulb holders for the dash lights.  It dimmed to a certain point ( via headlight switch dimmer) then off. Not the same dimming as the incandescent.

Alistair 

> On Jul 31, 2017, at 1:27 PM, David Beierl <dbeierl@ATTGLOBAL.NET> wrote:
> 
> The resistor is to limit current through the LED which is a "constant
> voltage" device like any diode.
> 
> The LED will start dimming the instant power is removed.  I would expect it
> to dim rapidly at first and then more slowly.  It might retain a minimal
> glow for some time -- LEDs will visibly emit with microamps of current
> running through them.
> 
> LEDs dim perfectly well.  If you put 2-watt LEDs in your instrument panel
> they would dim similarly to the 2-watt incandescent lamps they replace --
> at high brightness.  At low brightness they would still dim much more
> slowly than the incandescents because the latter have to be red hot before

> they begin to emit, which takes considerable current.  But if you replace
> your two watt panel lamps with 0.2 watt LEDs, the dimming rheostat meant to
> work with a total of ~15 watts will have little effect on the 1.5 watt load.
> 
> The formula for energy in watt-seconds (aka joules) stored in a capacitor
> is Estored = Cfarads x Evolts^2 / 2.  So your 4,000 uF cap charged to
> twelve volts would in theory contain enough energy to run a two watt
> festoon bulb for about three tenths of a second.  In practice it would run

> longer than that because it would only be using the full two watts at the
> instant power was disconnected.
> 
> The way the commercial circuits work is by using a timing circuit (no doubt
> containing a timing capacitor) to operate a transistor that switches
> battery current on and off.
> 
> Yrs,
> d
> 
> 
> 
>> On Mon, Jul 31, 2017 at 3:45 PM, Neil N <musomuso@gmail.com> wrote:
>> 
>> Yes. I too think the resistor is for voltage drop. Parts list
>> specifies "an LED" so.....
>> 
>> My sense is that voltage stored in the cap would naturally taper off
>> but kind of moot if I used an LED(s).
>> 
>>