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'Signal Diodes'
2000\03\09@074903 by Werner Soekoe

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Hi

I wish to know if anyone knows exactly what the characteristics are of the
1N4148 signal diode (a.k.a. 1N914). I have seen a circuit where the signal
diode was connected in series with a LED to limit the LED's current. Is this
possible? How well does it work?

Thanks

Werner Soekoe
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2000\03\10@015209 by Nigel Goodwin

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In message <000f01bf89c5$b7592aa0$0a01a8c0@werner>, Werner Soekoe
<.....wsoekoeKILLspamspam@spam@GLOBAL.CO.ZA> writes
>Hi
>
>I wish to know if anyone knows exactly what the characteristics are of the
>1N4148 signal diode (a.k.a. 1N914). I have seen a circuit where the signal
>diode was connected in series with a LED to limit the LED's current. Is this
>possible? How well does it work?

I don't see as it's possible, the only reasons I can think of for having
a diode in series is to feed it from an AC supply, protect it against
reverse DC connection, or to drop the voltage fed to it by 0.7 volts or
so (but why is a different matter!).
--

Nigel.

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2000\03\10@061033 by Joe Hamilton

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By dropping V you limit current (ohms law).  Although I don't think this is
a practical application of a diode, I have used them to drop V for various
reasons, but not exactly for current limiting, that is just a bi-product.
Anyway the 1n914 is obviously not the only diode that can do this, just a
very common diode in any project.

My 2 cents,

Joe

{Original Message removed}

2000\03\11@015849 by Nigel Goodwin

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In message <004c01bf8a81$6fb1c9c0$9341fc9e@oemcomputer>, Joe Hamilton
<.....joeh100KILLspamspam.....CROSSWINDS.NET> writes
>By dropping V you limit current (ohms law).  Although I don't think this is
>a practical application of a diode, I have used them to drop V for various
>reasons, but not exactly for current limiting, that is just a bi-product.
>Anyway the 1n914 is obviously not the only diode that can do this, just a
>very common diode in any project.

Sorry, I disagree!. Applying Ohms law shows that by dropping V you
reduce current, you don't limit it, a 5 volt supply feeding through a
silicon diode to an LED is going to pass a serious amount of current and
destroy the LED.
--

Nigel.

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2000\03\11@021028 by Spehro Pefhany

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At 06:50 PM 3/10/00 +0000, you wrote:

>Sorry, I disagree!. Applying Ohms law shows that by dropping V you
>reduce current, you don't limit it, a 5 volt supply feeding through a
>silicon diode to an LED is going to pass a serious amount of current and
>destroy the LED.
>

Yes, in general. It may be possible to operate an LED from a supply
of the right voltage with no dropping resistor (they do have
significant internal resistance, meaning the I/V curve does not
go straight up when you get to the usual Vf, it has a significant
slope).

The original LED calculator from Clive Sinclair used this method to
save power (and money and space).

Personally, I don't use this method and I don't recommend it.
It's too sensitive to the LED characteristics (which are not
guaranteed closely enough) and the exact supply voltage and
the temperature.


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2000\03\11@070049 by Joe Hamilton

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> In message <004c01bf8a81$6fb1c9c0$9341fc9e@oemcomputer>, Joe Hamilton
> <@spam@joeh100KILLspamspamCROSSWINDS.NET> writes
> >By dropping V you limit current (ohms law).  Although I don't think this
is
> >a practical application of a diode, I have used them to drop V for
various
> >reasons, but not exactly for current limiting, that is just a bi-product.
> >Anyway the 1n914 is obviously not the only diode that can do this, just a
> >very common diode in any project.
>
> Sorry, I disagree!. Applying Ohms law shows that by dropping V you
> reduce current, you don't limit it, a 5 volt supply feeding through a
> silicon diode to an LED is going to pass a serious amount of current and
> destroy the LED.

"REDUCE, LIMIT, Geee, I'm Sorry"

Hey I didn't say to go and burn up your led.  I meant that by dropping, say
.7 volts you "limit" current by decreasing the V to the load.  5 - .7 =
4.3. --- Now 4.3 - 1.9 (led V) = 2.4 ---
2.4 / .008 (led I) = 300.  So for the led I just described you would use a
300 ohm resistor in series with the LED and diode.  I never said to throw
out ohm's law.
{Quote hidden}

2000\03\12@054946 by Nigel Goodwin

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In message <003601bf8b51$bce50240$9a41fc9e@oemcomputer>, Joe Hamilton
<RemoveMEjoeh100TakeThisOuTspamCROSSWINDS.NET> writes
>"REDUCE, LIMIT, Geee, I'm Sorry"

Two very different words, with entirely different meanings :-). If a
diode could limit the current, that would be ideal, but they don't -
obviously one of those FET constant current devices would work even
better than a series resistor.

>Hey I didn't say to go and burn up your led.  I meant that by dropping, say
>.7 volts you "limit" current by decreasing the V to the load.  5 - .7 =
>4.3. --- Now 4.3 - 1.9 (led V) = 2.4 ---
>2.4 / .008 (led I) = 300.  So for the led I just described you would use a
>300 ohm resistor in series with the LED and diode.  I never said to throw
>out ohm's law.

The thread was about using JUST! a diode to feed an LED, no resistor at
all, which is why I said it wouldn't be a workable solution.
--

Nigel.

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