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Sci.Electronics.Basics -> 12v DC power supply read 18v
There are 4 messages in this thread.
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Author: EdDate: 10:19 28-04-08
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I bought a cheap 12v DC "wall wart" to operate a simple LED garage
door open alert. The circuit
has only a microswitch, a 1K ohm resistor, and 12v flashing LED.
However, the power supply
open-circuit voltage is actually almost 18v. Is this going to fry the
LED? Should I perhaps get
a 9v DC unit and hope it reads closer to 12v?
TIA
Ed
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Author: Bob EldDate: 11:04 28-04-08
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"Ed" <jag_manR__EM*-0_V_E653@hotmail.com> wrote in message
news:K1lRj.25$3O7.19@newssvr19.news.prodigy.net...
> I bought a cheap 12v DC "wall wart" to operate a simple LED garage
> door open alert. The circuit
> has only a microswitch, a 1K ohm resistor, and 12v flashing LED.
> However, the power supply
> open-circuit voltage is actually almost 18v. Is this going to fry the
> LED? Should I perhaps get
> a 9v DC unit and hope it reads closer to 12v?
>
> TIA
>
> Ed
No. This transformer will not hurt the LED.
The max current will be about 16mA which is way below the max rating of the
LED. Most LED's can take 50mA or more. Furthermore, any additional load on
the transformer will drop the voltage toward 12 Volts resulting in an even
lower LED current. The 12 Volt number is a loaded rating.
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Author: EdDate: 11:28 28-04-08
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"Bob Eld" <nsmontassoc@yahoo.com> wrote in message
news:LHlRj.876$1b7.453@newssvr13.news.prodigy.net...
>
> "Ed" <jag_manR__EM*-0_V_E653@hotmail.com> wrote in message
> news:K1lRj.25$3O7.19@newssvr19.news.prodigy.net...
>> I bought a cheap 12v DC "wall wart" to operate a simple LED garage
>> door open alert. The circuit
>> has only a microswitch, a 1K ohm resistor, and 12v flashing LED.
>> However, the power supply
>> open-circuit voltage is actually almost 18v. Is this going to fry
>> the
>> LED? Should I perhaps get
>> a 9v DC unit and hope it reads closer to 12v?
>>
>> TIA
>>
>> Ed
>
> No. This transformer will not hurt the LED.
>
> The max current will be about 16mA which is way below the max rating
> of the
> LED. Most LED's can take 50mA or more. Furthermore, any additional
> load on
> the transformer will drop the voltage toward 12 Volts resulting in
> an even
> lower LED current. The 12 Volt number is a loaded rating.
>
>
Thanks, Bob!
Ed
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Author: whit3rdDate: 19:53 04-05-08
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On Apr 28, 7:19=A0am, "Ed" <jag_manR__EM*-0_V_E...@hotmail.com> wrote:
> I bought a cheap 12v DC "wall wart" to operate a simple LED garage
> door open alert. The circuit
> has only a microswitch, a 1K ohm resistor, and 12v flashing LED.
> However, the power supply
> open-circuit voltage is actually almost 18v.
The 'wall wart' is probably filtered and unregulated. The
rating "12V" only indicates a minimum output at some specified load.
Thus, the wall wart is not fully specified by the posted info.
A 'flashing LED' isn't an LED at all, but a kind of oscillator
circuit with an LED lamp. What kind of circuit, I can't tell.
So, we can't be sure the wall wart and 'flashing LED' are
compatible.
Finally, if this is intended to indicate some kind of door-open fault,
the quiescent (idle) current in your wall-wart is a critical value,
because it could be wasting 3W just sitting there with no load.
The energy-smart place for your microswitch is on the INPUT side of
the
AC transformer, not the low-voltage output side. At $0.10 per
kilowatt-hour, a 3W drain costs you three dollars a year, mainly
for wasted power. A cheap wallwart is more expensive than
replacing flashlight batteries every few years, in that case.
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