PDA

View Full Version : Barrel battery replacement



generic486
June 22nd, 2012, 12:27 AM
Yesterday, i removed an infamous ni-cd from my XT clone. It was not leaking but it leaking eventually is inevitable. Anyway, the XT does not have the battery on the main board. It is on a seperate card` and as such has no 4 pins that will allow me to put a non-rechargable in. What should I replace it with or should I just leave it off? I don't think it served any perpose as DOS 3.3 didn't save the time anyway. Did many XT boards not have a battery on the mainboard or is this just a rare occurence?

SpidersWeb
June 22nd, 2012, 12:52 AM
Chances are it's to remember the time - but it'll need a battery that isn't dead and a DOS driver to set / get the time.
So if you leave it disconnected, it's not a great loss (up to you). Hopefully someone else on here can recommend a good replacement.

Most XT's don't have a built in battery-backup because configuration was usually done with switches, on AT and later machines the battery usually keeps the time and remembers the configuration (which was set by software).

Maverick1978
June 22nd, 2012, 10:12 AM
Take pictures of where the old battery was at. It's always possible to just solder bare wires to the positive and negative terminals or solder points, letting these run out to a battery pack with a few 1.5v AA batteries loaded up. At best, you might be able to affix a coin cell enclosure.

MikeS
June 22nd, 2012, 10:41 AM
Yesterday, i removed an infamous ni-cd from my XT clone. ... What should I replace it with or should I just leave it off? I don't think it served any perpose as DOS 3.3 didn't save the time anyway. Did many XT boards not have a battery on the mainboard or is this just a rare occurence?As mentioned, PCs and XTs had no need for a battery on the main board; they were only needed on third-party boards with a real time clock.

DOS 3.3 (or any version of DOS) doesn't "save" the time; the clock is running all the time (that's what the battery is for) and DOS just looks up the time when it needs it, but it does require enabling it in your startup files.

If you're OK with manually setting the time every time you boot, just leave the battery out; otherwise, assuming it's a standard 3.6V NiCd battery then a cordless phone battery is a good replacement, usually available for a buck or two at your local Dollar store.

It's NOT a good idea to replace a NiCd with a coin cell or regular AA or AAA batteries unless you disable charging!

Chuck(G)
June 22nd, 2012, 10:51 AM
You can still get replacements. For example this one (http://www.cables4computer.com/Power_Products/3.6V-Nicad_Batteries_Barrel_2Pin_3-51FT-A_3.6V_Nicad_Battery.html). I suspect that a local store, such as Batteries Plus would also be able to get one for you.

NiMH (http://www.batteriesinaflash.com/plc-computer-backup/plc-2-pin-computer-backup-battery-comp-16-2nmh-replaces-v-2-v60r) rechargeables will work as well.

MikeS
June 22nd, 2012, 10:58 AM
You can still get replacements. For example this one (http://www.cables4computer.com/Power_Products/3.6V-Nicad_Batteries_Barrel_2Pin_3-51FT-A_3.6V_Nicad_Battery.html). Like you, I prefer to keep batteries off the boards, which is why I replace all mine with cordless phone batteries that have leads; they're also much cheaper.

pearce_jj
June 22nd, 2012, 11:01 AM
I've replaced a few NiCds in old machines, but the problem is that they then don't get enough run time to keep them charged so are dead anyway and then leak again just for good measure! So personally I either just cut them out and be done with it, or if it's critical to the operation of the machine (non-PC I'm thinking here, for example a fruit machine or something like that) then replace with AA batteries with charging disabled - either by cutting charging supply if it's that easy, or adding a simple blocking diode.

Chuck(G)
June 22nd, 2012, 11:05 AM
Yeah, but you have to give a nod to the purists who want things to look original--and I guess I can't blame them.

If I were replacing a barrel battery with another barrel battery, I suspect it might be a good idea to slip the thing inside a plastic bag before soldering it in--or to use socket pins to enable easy replacement of the battery. The original battery on my Quadram board lasted only about 3 years before springing a leak.

Most early clock chips will work down to about 2.2V (check datasheets), so I suppose that even a 3V lithium coil cell in series with a Schottky diode will do the job and last for years.

bozimmerman
September 30th, 2015, 11:49 PM
Yeah, but you have to give a nod to the purists who want things to look original--and I guess I can't blame them.

If I were replacing a barrel battery with another barrel battery, I suspect it might be a good idea to slip the thing inside a plastic bag before soldering it in--or to use socket pins to enable easy replacement of the battery. The original battery on my Quadram board lasted only about 3 years before springing a leak.

Most early clock chips will work down to about 2.2V (check datasheets), so I suppose that even a 3V lithium coil cell in series with a Schottky diode will do the job and last for years.

Hi Chuck,

How do you use a diode with the battery cell holder? I assume you put the diode between the Positive + side and the motherboard, and then test you still get the 3V out of the diode. Is this correct?

- Bo

modem7
October 1st, 2015, 03:45 AM
How do you use a diode with the battery cell holder? I assume you put the diode between the Positive + side and the motherboard,

http://www.minuszerodegrees.net/images2/blocking_diode.jpg


and then test you still get the 3V out of the diode. Is this correct
With the above in place on the motherboard, and the motherboard powered off, measuring the voltage between the - and + solder pads on the motherboard will get something less that the battery voltage. That is because the diode will have some voltage over it.

Chuck used "Schottky diode" because that type of diode will 'drop' less voltage than a standard (silicon) diode. For example, if the battery itself measures 3V, and the Schottky diode drops 0.3V, then the motherboard will receive 2.7V

Chuck(G)
October 1st, 2015, 08:57 AM
To add to this, after more than 3 years, I should add that you can also use a MOSFET as a blocking diode with essentially no voltage drop. I covered it in a a couple of other posts. Perhaps a search on "MOSFET" will turn it up.

modem7
October 1st, 2015, 01:34 PM
To add to this, after more than 3 years, I should add that you can also use a MOSFET as a blocking diode with essentially no voltage drop. I covered it in a a couple of other posts. Perhaps a search on "MOSFET" will turn it up.
A copy of the document that you pointed to in that post is at [here (http://www.minuszerodegrees.net/misc/Reverse%20Current%20Battery%20Protection%20Circuit s.pdf)].

bozimmerman
October 1st, 2015, 01:40 PM
http://www.minuszerodegrees.net/images2/blocking_diode.jpg


With the above in place on the motherboard, and the motherboard powered off, measuring the voltage between the - and + solder pads on the motherboard will get something less that the battery voltage. That is because the diode will have some voltage over it.

Chuck used "Schottky diode" because that type of diode will 'drop' less voltage than a standard (silicon) diode. For example, if the battery itself measures 3V, and the Schottky diode drops 0.3V, then the motherboard will receive 2.7V

Thanks!

I tried this out and, indeed, the voltage drop was very small. For fun, I then turned the diode around backwards and checked again. The voltage drop was much greater (3.6V -> 2.0), but I was surprised that it was still as high as it was. I guess I expected 0. Am I misunderstand what a diode does?

- Bo

modem7
October 1st, 2015, 02:21 PM
For fun, I then turned the diode around backwards and checked again.
Careful. Doing those kinds of things with electronics can result in smoke then tears.


The voltage drop was much greater (3.6V -> 2.0), but I was surprised that it was still as high as it was. I guess I expected 0. Am I misunderstand what a diode does?
Sometimes, to explain an observation, you need to take into consideration, the remainder of the circuitry.

Chuck(G)
October 1st, 2015, 02:40 PM
I tried this out and, indeed, the voltage drop was very small. For fun, I then turned the diode around backwards and checked again. The voltage drop was much greater (3.6V -> 2.0), but I was surprised that it was still as high as it was. I guess I expected 0. Am I misunderstand what a diode does?

It depends on the type of diode.

bozimmerman
October 1st, 2015, 02:58 PM
It depends on the type of diode.

Which is precisely why I keep asking questions. :) I have several of these Schotsky diodes: 1N5819 DC.

What I did was exactly what the circuit showed, but I intentionally turned the diode around. I did know that I putting a battery and one of many identical diodes at risk, but better to understand exactly how this all worked before putting my PC at risk, right? My understanding is that the computer would, when powered on, jack up the voltage to over 5V in order to recharge the old NiCd. Add to this my mistaken impression that diodes block voltage in one direction, and you can understand my dismay at seeing nothing more than a larger voltage drop.

Your advice and responses are greatly appreciated. I will literally be sitting here staring at my screen, as replacing these batteries is my current project, and I appear to have the parts, without the confidence that comes from understanding.

- Bo

SpidersWeb
October 1st, 2015, 03:57 PM
I'm not completely sure where you were measuring to get the 2V reading, but with a normal diode it will prevent current from flowing in the wrong direction. If the battery is connected, and computer on, you'd always expect to see a voltage reading on both sides of the diode though.

And since they share the same ground - there will be a voltage difference across the diode as well.

5V minus 3V = 2V.

The multimeter really shows the difference (probably the wrong word to use, but it'll do) between two voltages. A resistor is always on, so that difference can be also called the drop, but a diode can switch off - so the drop can only be accurately measured when the thing is actually on and flowing - if it's in backwards, it's not going to be flowing.

e.g. you could remove it and you'd still see 2V across those two points (assumption being you were measuring across the diode to determine the "drop" and the computer was on at the time).

That's a neat trick for noisy +12V fans - power them using +5 and +12 (instead of +12 and GND)- you get a difference of 7V DC which is enough to spin but not suck up a cat walking past.

(Sorry for waffly, not technical post)

bozimmerman
October 1st, 2015, 04:06 PM
I'm not completely sure where you were measuring to get the 2V reading, but with a normal diode it will prevent current from flowing in the wrong direction. If the battery is connected, and computer on, you'd always expect to see a voltage reading on both sides of the diode though.

And since they share the same ground - there will be a voltage difference across the diode as well.

5V minus 3V = 2V.

That's a neat trick for noisy +12V fans - power them using +5 and +12 (instead of +12 and GND)- you get a difference of 7V DC which is enough to spin but not suck up a cat walking past.

(Sorry for waffly, not technical post)

Thanks for your response. I should have been more clear about where that number came from.

I have a lithium battery 3.6V.

When I put the 1N5819 in the Correct direction on the positive poll of the battery, and measure across the diode to the negative side of the battery, I read like 3.3 or 3.4 or so.. very small drop. However, when I put the 1N5819 in the Wrong direction on the positive poll of the battery, and measure the diode to the negative side of the battery, I read 2 volts.

This concerns me, because I thought the point of using the diode was to prevent voltage from crossing the wrong way.

For example, if I *were* to put the battery and diode into the PC circuit the Correct way, I would expect to read 3.3V going into the PC from the battery+diode circuit. However, when I turn the PC on, I expect the circuit to suddenly be fed 5V from the PC, which would cross the diode the "wrong way", and possibly still cause the lithium battery to explode or whatever.

My ignorance is leading to fears, you see. :)

- Bo

SpidersWeb
October 1st, 2015, 04:11 PM
When you measured the 2V, was the computer on at the time?
Or is this just on your workbench?

bozimmerman
October 1st, 2015, 04:17 PM
When you measured the 2V, was the computer on at the time?
Or is this just on your workbench?

This is before the circuit is even touching the computer. I was trying to gain some confidence in the diode before hooking it up to the PC.

I measured like this to get 2V:

--- negative multimeter probe ---> [negative 3.6V battery terminal]:!:!:!:!:![positive 3.6V battery terminal]----[ !! DIODE the wrong way] <----positive multimeter probe----

SpidersWeb
October 1st, 2015, 04:26 PM
Yeah I don't have an explanation for that. I would've expected 0 also.

Edit: although datasheet scanning, does indicate that some variants of this part may have a reverse DC voltage rating as low as 2V - e.g. if you put 3.6V on them, it'll start leaking through. I suspect that's what Chuck may have meant. May just need to get a higher rated part.

3pcedev
October 1st, 2015, 04:41 PM
As Chuck mentioned before you can use a MOSFET to protect the battery with ~0 voltage drop.

I did it on my 486 - Some details are in the thread here: http://www.vintage-computer.com/vcforum/showthread.php?45721-My-first-(hand-assembled)-computer-486DX2-80&highlight=3pcedev

H-A-L-9000
October 2nd, 2015, 11:42 AM
This cirquit helps to protect against reverse polarity, but not against reverse current. There is no way the MOSFET can turn off when the direction of current flow changes (when you power on the computer).

You can however use a P-MOS and switch that off with the 5V from the power supply:

BATT- -> GND
BATT+ -> S MOS
G MOS -> R1k -> GND
G MOS -> -1N4148+ -> R100ohms -> +5V
D MOS -> +RTC
Maybe not needed: GND -> -C100F+ -> +RTC