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Question about charger circuit

I've got an old ICOM Marine radio that has a bunch of small NiCads in it; they're in series-- seven of them, for an 8.4V battery.

The battery pack can be fed 13.8VDC through a small plug, this is supposed to charge the batteries (and/or power the radio if the batteries are dead). Well, the batteries are very dead, and I'm planning to replace them. On the way, I traced out the charging circuit:

||100pf ------------------------||----------->|----- to B+ | | | || | | V -> LED | _180 | | | |-|___|-| | .-. | | | | |1200 | _180 | | | | |-|___|-| | '-' | | | | | V | - | | o)----------------------------------------- to B-

13.8V in "jack"

There are a couple of things I don't get. First, it seems to me that even if the diode just before the positive side of the NiCads is silicon, I get something like 13.1V into the nicads -- that seems like way too much.

The left half of the circuit looks like it's just a current limiter for the LED, with a diode to protect the LED from reverse-voltage issues; the diode in the upper right prevents the charge-light from running when the charger's not connected.

But what about the capacitor and two resistors? The capacitor is rated at 10V, and I just don't understand what that part of the circuit is doing at all.

The resistors make SOME sense -- the combined resistance is 90 ohms; with the Nicads full charged, we get 13.1 - 8.4 = 4.7 V; at 90 ohms I get a current of 50mA, which seems...well, maybe OK.

If anyone can help make sense of this circuit for me, I'd be delighted. In some sense, it's all moot -- I'm just gonna replace the nicads and use the thing -- but I'd *like* to understand it if possible.

--John

Reply to
John F. Hughes
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Or look at it the other way? When they're empty, there will be full voltage, and a big current to charge. As the batteries charge, more of the voltage lies over the battery, and current decrease. This reduces current to the battery, which might in fact give it the desired maintenance charge current/voltage.

This is just guessing, from a quick peek at the schema. I'm not a wizard at analoge electronics...

--
MVH,
Vidar

www.bitsex.net
Reply to
Vidar Løkken

Somewhat. Empty batteries still ahve voltage, and the 50 mA is still a lot for a permanent charge. But this is the general way the circuit works. Now I wonder: what is the 100 pF capacitor across the current limiting resistors doing there? The capacitor is more likely 100 uF (given its rating at 10 volts) but also more probably connected across teh battery, not across the resistors.

Thomas

Reply to
Zak

No the battery "drags" the output voltage down. A battery acts like a voltage source remember. The voltage of a NiCad battery on charge will typically reach 1.3V per cell when it's full (More for old cells). This means that a 7 cell pack will be at around 9.1V not 8.4V when full AND still on charge.

See the graph..

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Yes thats correct.

Forget the capacitor. It's too small to do anything important!

You are right about the resistors. They limit the charging current exactly as you describe but I'd assume slightly different numbers..

The voltage drop across the resistors when the battery is full is...

13.5 - 9.1 - Vdiode = about 3.7V. So the current through the resistors is 3.7/90 = 41mA

At the start of charge the battery voltage is lower. Perhaps 1.1V per cell or 7.7V for the pack. This means the voltage across the resistors is

13.5-7.7 - Vdiode = 5.1V and the current is 5.1/90 = 56mA

So if you connect an empty battery the current may start high. Fall quickly to about 50mA then near the end of the charge when the pack is full it falls to about 40mA. These numbers may vary quite a bit depending on the condition of the battery and the exact diode used.

Seems to me you have it right.

Not sure about that capacitor though. Sure you traced it right?

Reply to
CWatters

My mistake on that -- it's 100 microfarad at 10V (and electrlytic to boot). When I saw that an d the two resistors, I instantly though "some sort of pi-filter in case the 13.8V feed (coming from some wall-wart like source) has ripple in it." But when I traced the wires, the capacitor really IS parallel to the two resistors. Go figure!

Right -- I was just winging it :-)

Thanks for the confirmation of my suspicious.

-John

Reply to
John F. Hughes

Humm Perhaps they figure that "hitting" the battery with a short duration current spike somehow helped it live longer. Sometimes cells fail short circuit (caused by some form of crystal growth between the plates which I visualise to be like stalagmites!). Hitting them with a current spike blows this "fuse"... at least thats a theory..

Reply to
CWatters

Can anyone help me figure out how to read the schematic in th original post of this thread

As with many diagrams in posts, all I see is 15 mostly very shor

lines consisting mostly of vertical bars (|), and starting out wit "||100pf", and ending with "o)-----------------to B+.

Judging from the comments, the rest of you can read a schematic that

can't see

Appreciate any advice

awrigh

Reply to
awright

I can't find the original post but try this...

Copy the text into notepad (or possibly MSWord), Select all and change the fint to a fixed pitch font like "System" or similar.

Reply to
CWatters

--- I can't find that post, but it you just want to see whether you can display ASCII schematics, I've enclosed a simple comparator circuit that you ought to be able to read if you set your newsreader to display messages in a non-proportional font like Courier or Courier New.

Schematic follows:

+5V>-------+---------+--------+ | | | | | [10k] | | | | +---|--[1M]--+ | | | | VIN>--[1K]-------+--|+\ | | | >------+-->VOUT [10k]-------+-----+---+----------->GND

Howzat?

-- John Fields Professional Circuit Designer

Reply to
John Fields

I see you posted at Google this time. If you are continuing to read there, this is the answer:

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(redbelly thinks it's your newsserver modifying the contents of posts, which would definitely make the Google Archive the answer for you.)

Reply to
JeffM

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