Re: testing for 24vac

On Fri, 13 Jan 2006 10:15:55 -0800, Walter Harley wrote:

> "George Economos" <nospam@xxxxxxxxxx> wrote in message
> news:2acfs1p99ntog1cm869mtdd7untbdgjbjm@xxxxxxxxxx
>> [...]
>>
>> Okay I am almost ready! One more question:
>>
>> Does it matter if I use a 1/4W or 1/2 W resisitor?
>
> The resistor will be dropping 22V. Current flow will be whatever you
> choose, e.g., 10mA. Since the diode is blocking half of the AC cycle, power
> dissipation is halved.
>
> So, power dissipation will be P = EI/2 = 22V * 10mA / 2 = 0.11W. You'll be
> fine with either a 1/4W or 1/2W resistor, assuming you choose 10mA of
> current (that is, a 2.2k resistor).
>
> Note that if you hook it up the way that Jasen suggests, with the diode in
> parallel with the LED but in the opposite direction, then current is flowing
> in both half-waves, so the power dissipation in the resistor doubles
> compared to the above calculation. The LED is still only turned on half the
> time, though, so its brightness does not change. For that reason, I think
> that Noway2's suggested circuit is preferable to Jasen's.


No, because when the diode and LED in series are reverse-biased, there's
no telling if the reverse leakage of the diode will allow the LED's
reverse breakdown to be exceeded. There's no facility for them to "share"
the reverse voltage, and if they did, the LED would pop.

Use the diode in series if you need to, but put another diode in
antiparallel with the LED, just to protect it.

Good Luck!
Rich


.

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