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  #1  
Old 03-12-2010, 07:08 PM
TexAvery TexAvery is offline
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Default Three 1.5 AAA vs. 4.5 wall wart

I am powering two LED fans with a 4.5 VDC 600 mA. wall wart instead of three AAA batteries.
The fans spin a bit faster compared to using three AAA batteries that the fans were originally designed for. After 3 hours the LED's burn out.
I think the current is too high compared to three AAA's.
If this is correct how can I reduce the current?

BTW... I added a zener diode but now the voltage is 3.7 VDC.. too low to make all the LED's light.

Last edited by TexAvery; 03-12-2010 at 07:23 PM.
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Old 03-12-2010, 07:31 PM
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Bill_Marsden Bill_Marsden is offline
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There is likely high ripple on the wallwart, you are only seeing the DC component. Add a large capacitance (for testing purposes) and measure the DC.

Remember, it is the load that dictates how much current is pulled. If there are pulses hidden in the pulses then they will be current spikes.

What is the value of the zener?

The zener is the best evidence that you aren't using filtered DC.
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Old 03-12-2010, 07:43 PM
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The wall wart outputs roughly 4.5v when there is a 300mA load applied. The problem is likely that your LED fans require far less than 300mA current, so the output voltage of the wall wart is too high, resulting in your LEDs getting fried.

What does the output of the wall wart measure with no load on it?

How much current does one of these fans use when running from three AAA batteries?
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Old 03-12-2010, 07:45 PM
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I like Wookie's answer better overall. Have you measure the voltages while running this?
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Old 03-12-2010, 07:54 PM
TexAvery TexAvery is offline
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4.5 Vdc 1600 ma. wall wart (I am worng about the 600 mA. original post)
No load = 4.66 v
Load = 4.48 v
Diode = 1N4148
When diode is in series voltage=3.7
Each fan current must be over 200mA. because my toy multimeter only goes to 200mA. and it reads OVERLOAD
with a rapid beep-beep
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Old 03-12-2010, 09:31 PM
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1N4148 diodes are rated for 100mA. At that current, they drop around 1.1v across themselves. At higher currents, they will quickly burn up.

If you want to try something cheap, try wiring a couple of 10 Ohm resistors in parallel between the supply and the fan's power lead. Radio Shack generally stocks these:
http://www.radioshack.com/product/in...watt+resistors
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Old 03-12-2010, 09:39 PM
someonesdad someonesdad is offline
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Quote:
Originally Posted by TexAvery View Post
Each fan current must be over 200mA. because my toy multimeter only goes to 200mA. and it reads OVERLOAD
You can get a suitable multimeter for less than 10 bucks from Harbor Freight (I've seen them on sale there for $4).
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Old 03-12-2010, 09:45 PM
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It is best to measure current by measuring the voltage drop across a 1 Ohm resistor.
If you measure current directly, you stand a very good chance of blowing your meter's fuse - or the meter itself.

If you place a 1 Ohm resistor in the current path, and measure the voltage drop across the resistor, you get a direct correlation from Volts to Amperes.
This is because Ohm's Law states:
I=E/R, or Current in Amperes = Voltage / Resistance in Ohms
If R=1, and I=E/1 ...
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Old 03-12-2010, 11:58 PM
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Well, the Harbor Freight meters DO have a 10A setting, but you have to plug the positive lead into the 10A jack instead of the V/Ohms/mA jack. That's pretty typical for any meter though.
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Old 03-13-2010, 06:25 PM
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The cheap fans were made in ***** which is a country where junk is made. You are lucky that they lasted 3 hours.
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