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Help needed with Logitech charger circuitry

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Old 04-03-06, 06:24 PM Thread Starter   #1
voodoomelon
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Help needed with Logitech charger circuitry


Hey all.
Thought i might get better help in this section than the general hardware section.

Pic below is of a Logitech LX700 mouse base charger, the small circuit is simply the part that contains the gold contacts, which connects directly to the 6v DC source. My mouse refuses to charge in the station.



Heres the problem. Right up to the end of the red and black wire, there is a reading of 6v, no problem there, but there is nothing measuring at the gold contacts.
The circuit is really simple as you can see, with just 3 components connected on a daisy chain.

The order goes like this:
6v wire (1000mA) -> black capacitor -> orange inductor(I've been told, & its 500mA) -> blue component (resistor? only has 225 marked on it) -> gold contacts.

There is no current measured after the inductor, so that must be the problem.
I showed it to my friends dad, who is the head of the electronics dept in my college, and he says the inductor isn't needed, and that they merely include them to make sure that consumers dont plug in a transformer other than the 6v one that comes with it, and cause damage.

He said it should be fine to just bypass the inductor with a piece of copper.

I'm just slightly concerned about charging my mouse without that component, does anyone agree with my friends dad?

Ps, what is the small blue component?

Any help appreciated, thanks!
Steve.

PS Logitech recommend using rechargeable batteries rated between 1500-2300 mAh.

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Old 04-03-06, 07:25 PM   #2
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The inductor is not for making sure people dont plug in the wrong supply. Even with the inductor it would still cause damage. The inductor is for filtering purposes to get the cleanest DC voltage under load. The small blue component is a capacitor. 225 is a mesurement for its capacitance. 22 is the significant factor, and 5 indicates the number of 0s after the 22 (think of it like Scientific notation) It is mesured it pf (Peco Farads) 2200000pf
1pf = 1*10^-12
1uf (microfarad) = 1*10^-6
so the capacitor you are looking at is a 2.2uf capacitor wich is common in regulation and filtering circuits.
Try and replace the inductor with an identical one. (your sure its an inductor right?)
If you can give me the rating in uH (micro henrys) and I can probably guide you to finding a similar one.

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Old 04-04-06, 05:52 AM Thread Starter   #3
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I was told in another section of the forums that the orange component is an inductor. The only marking on it is "500mA".

So are you saying that the circuit can't operate without it?
Is it a bad idea to bypass it?

Thanks for the reply.

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Old 04-06-06, 07:07 PM   #4
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if it is an inductor i wouldnt be too concerned about bypassing it, sure there might be a little more noise but is it that critical for chrginf a couple batteries?

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Old 04-07-06, 07:03 PM   #5
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It depends on the battery. Some batteries will suffer shorter life span from noise on the line.

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Old 04-25-06, 03:20 AM   #6
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I would say order a new component for every one that's on the board and replace them. Digikey or Mouser would have the components you're looking for, and for dirt cheap.

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Old 04-25-06, 07:51 PM Thread Starter   #7
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Thanks Celeron.

I actually tried by-passing the damaged component, and there is now 6v coming out of the two contacts, but the mouse still wont charge with any batteries.

Think i might go on a component search.


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Old 05-10-06, 11:13 AM   #8
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How did you manage to measure amps (you mentionned "current" if your first post)?

As far as I know, you need to disconnect something to measure amps, if you did not do that, you've probably blown the fuse of your amp-meter, or made a false reading by short-circuit'ing a component.

Right, wrong ?
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Old 05-10-06, 01:51 PM Thread Starter   #9
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Hi.
By current, I meant voltage. There was no voltage measured at the particular point.

Well, I bypassed the orange component and there is now a voltage coming through to the gold contacts.

But here's the weird thing, the transformer is outputting 10.1v.
It's only rated at 6v. Is this normal?

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Old 05-10-06, 01:54 PM   #10
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Hi,

A transformer could provide more voltage than rated when it is not under any load. It's normal I guess if it's not under load.
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