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Old Aug 21, 2003, 11:57 PM
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13.5V Power Supply

I took an old power supply from a 386 laptop and with some help found the +/- wires that supply 13.5V. I neatened it up with some shrink wrap and now have a power supply for my FMA lipo202 charger. this charger runs at 11.5V - 13.5V, but I checked voltage on the power supply and it is always 13.59 - 13.62 and I am wondering if there is a simple circuit I could hook the output of the power supply to get the voltage down to like 12.5V? I realize v=ir, but something tells me it is not that simple.
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Old Aug 22, 2003, 04:21 AM
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The forward characteristics of silicon diodes are nonlinear. They have voltage drops of about 0.6 volts under load over a range of currents. Two silicon power diodes in series, polarized for forward conduction, will give you close to the voltage drop you are looking for except under no load conditions. The diodes should be sized to carry more than the maximum rated current of the power supply. You can add a load resistor to the circuit that is sized to use only a small fraction of the power supply current available. This small load resistor will bias the diodes so that their voltage drop doesn't go to zero when there is no other load on the supply.
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Old Aug 22, 2003, 07:35 PM
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Thanks Ollie,
I'm looking through the radioshack catalog and went to the diode section and boy there are all kinds of diodes for all kinds of purposes. Looking at the general purpose diodes, I see alot of different specs like Vrr, VF. It seems Vf refers to that voltage drop you mentioned, like 1.2V@3.0A. My supply is 1.3A, so would the drop be more like .4V@1.3A? So, using 2 diodes in series would give a .8V drop? I'm not quite sure how to select a resistor.

Whoops, you said nonlinear. Anyway, here's what I decided to get: 2 1N5400 diodes and a 1k resistor. Does that sound about right?

I welcome any advice
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Last edited by ElectroStorch; Aug 22, 2003 at 11:25 PM.
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Old Aug 25, 2003, 06:41 PM
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ElectroStorch


What Ollie is saying is that pretty-much what ever current you put through a silicon diode the voltage drop is always about 0.6V, They don't follow Ohm's law. You do, however, have to put a small amount of current through a diode to ensure the voltage drop.

1N5400 diodes are rated at 3A so no problem at 1.3A.

Select the resistor to take about 1mA this is all you will need to bleed through the diodes, at 13.5V you require 13.5kOhms this is not a preferred value but 12k is so use this value it's not critical so 10k would work just fine.

Andy.
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Old Aug 25, 2003, 11:19 PM
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Thanks Andy, Ollie,

I got 2 diodes and 1k resistor and tried them on an old dusty breadboard - didnt work. On that same breadboard from long ago I found some parts from who knows what - gave up on that stuff until now when need arose. I found a potentiometer that said real small letters 1Meg. I hooked that up with the diodes and didnt work. I twisted the screw CW and the power supply light came on steady, stopped blinking. This was all before your reply, Andy, so apparently the first 1k resistor and the last setting on the potentiometer did not meet the minimum threshold. With only a Voltmeter it is kind of hard to work with circuits, but I was able to see the 0.6 Volt drop for each diode. It seems all I have to do is connect charger to input side and output side of diodes to get 12.~ volts I wanted.

What kind of change in current should I expect with this circuit, unchanged since diodes rated at 3.0A?

Also, I picked up a small piece of protoboard from RadioShack and the holes are just a bit too small for those diodes. Are these holes Linked in some systematic way? Just drill em out, right? The board is marked with a letter and number grid system.
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Old Aug 26, 2003, 12:24 PM
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ElectroStorch

If as you suggest, the charger takes some current when it is switched on but not charging, you don't need the resistor as the current draw at idle will be enough to make sure you get the diodes to drop their characteristic voltage of 0.6V and therefore you won't over-voltage the input of your charger circuit.

There are different paterns of proto-board so before starting to build your circuit get a rough idea how it is going to fit onto the board and place the components then remove any unwanted connections that short any parts of the circuit together. If you plan it just right sometimes you don't need to remove any connections.
The easiest way to remove connections is to "drill" the copper away. Don't drill all the way through the board, only go far enough to remove the copper, check with a magnifying glass when complete and before powering up to ensure there are no whisps of copper left where you drilled.

Andy.
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