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Author Topic: Charging the Zen Micro with a home-made charger - isolating the problem...  (Read 4002 times)

viridius

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I suppose you could do that; dropout voltage should be around two volts, so don't go any lower than seven volts.  They don't need to be LEDs, either.  Regular silicon or germanium diodes will drop about 0.7 and 0.3 volts each, respectively.
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trials_modder

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So you don't think there'd be much problem having up to 500mA of current flowing through an LED? In series, the voltage is "split" between the components and the current is the same.
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viridius

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I'd go with a rectifier diode for this application, yes.
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trials_modder

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Any specification on what kind of rectifier diode I'd need? Would it just be in series with the voltage regulator, and would it be able to handle the current?
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viridius

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It would be in series with the regulator.  Any diode capable of handling half an amp or more will work, like the 1n4001.
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trials_modder

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I tested the charger again today, and I can confirm that the Zen IS in fact receiving enough current to charge as it both displays the charging symbol and glows blue (a post on the web noted it wouldn't glow blue if it wasn't being provided enough power). However, the voltage regulator becomes quite hot quickly. I didn't want to risk damaging the regulator but I suppose a prolonged usage would trip the temperature guard on the regulator and greatly limit current flow, thus disabling the charger. I think a heatsink would only buy minutes of charge time; I doubt it would distribute and radiate the excess heat that fast.
I noticed the 1n4001 's application is just to prevent reverse current, and does not (as the datasheet indicates) drop much voltage. Would a simple resistor work? I know it also converts to heat energy, but perhaps it will help take the burden off. Zener diode? I am looking at my only source for components near me http://www.rpelectronics.com/, aside from Radio Shack.

EDIT: If I get one of those fat 5W resistors, maybe overheating won't be a problem.
« Last Edit: July 07, 2006, 11:52:22 PM by trials_modder »
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trials_modder

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Another thought: why not have several of the voltage regulators (>=2) in parallel with each other? How many would I need so that the thing stays cool: 7volt drop * 500mA max draw (USB regulation) = 3.5W max of heat needing to be distributed. Then again, having a huge heatsink might do the same thing.
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trials_modder

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Tested charging with the heatsink off a Pentium 100 chip. I think it prolonged the time the chip could function, but soon was also quite hot. Idea: Mount a small case fan on it, connected to 12v input source. That or use the diodes. I'm thinking of using a combination of both, diodes to lower the voltage to ~7, then a heatsink fixed to the outside of the case w/o a fan. Will the diodes lower the voltage independantly of the input (always by x amount of volts?)
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viridius

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Yeah, diodes will drop a pretty constant 0.7 volts.  While they are normally used for rectifying a signal (eliminating negative voltage swings), this characteristic makes them useful if a quick and dirty voltage adjustment is needed.  Question: why don't you just use a 5 volt wall wart?  They're probably a dollar at a thrift store or garage sale.
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trials_modder

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It seems like the Zen Micro also has two charging modes, trickle and full. Using my circuit on trickle keeps the regulator cool, but takes forever to complete the charge (I haven't tried to fully charge it using that method yet). Full charge can demand up to an amp, or whatever it can get. This is the method that causes the overheating - the regulator is dumping too many watts too fast. What value of a resistor could I use that could knock down the voltage to 7?
The reason I'm not using a 5v wall plug is so that I can also use the circuit in a car.
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