Date: Mon, 31 Jul 2017 17:28:10 -0400
Reply-To: David Beierl <dbeierl@ATTGLOBAL.NET>
Sender: Vanagon Mailing List <vanagon@gerry.vanagon.com>
From: David Beierl <dbeierl@ATTGLOBAL.NET>
Subject: Re: Time Delay LED Lighting Circuit
In-Reply-To: <CAMOH8LLU-FsKoi-VHpkhmbKu1Yh-K8foK_08ZDSMigkPFQf0og@mail.gmail.com>
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Addendum: The way you size a current-limiting resistor for a diode circuit
is to subtract the diode forward voltage from the supply voltage, and then
calculate the resistance that would allow the desired/rated current to flow
in a circuit with that voltage, using some safety margin for increasing
temperature or supply voltage.
Suppose you have a one watt LED with rated forward voltage of 2.6 volts and
rated current of 385 mA, to be driven from an automotive supply with
specified maximum voltage (like our vans) of sixteen volts.
Vresistor = Vmax - Vfwd = 16V - 2.6V = 13.4V.
Rresistor = Vresistor / Ifwd = 13.4V / 0.385A = 34.8 ohms. Next larger
standard 5% value is 36 ohms.
To find how much power the resistor will dissipate,
Presistor = Vresistor x Ifwd = 13.4V x 0.385A = 5.16 watts, so use a ten
watt resistor.
Clearly this application would be better served by three or four LEDs in
series if that would fit in with the optical requirements. Four in series
would give Vfwd of 10.4V and Vresistor of only 3.6V. A ten ohm two watt
resistor would serve.
Yrs,
d
On Mon, Jul 31, 2017 at 4:27 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).
>>
>>
