HP Forums

Full Version: My HP25 really dead?
You're currently viewing a stripped down version of our content. View the full version with proper formatting.

Hi there,
when i tried to use my HP25 days ago, nothing happened.
Thinking of low nicads i hooked it on a 5V DC-supply that i
used to charge the HP25. (with 1400mAh NiMh cells)
Switching on the HP, only erratic Stuff ist displayed.

Did I kill my HP by bad battery contacts? The display only
comes up when the supply voltage is near 5V. And then it is
flickering or displaying an 8 in the second rightmost digit.

Is there some help or do I have an eyecatcher for the bookshelf? :-(

73
Peter

Hi;

chances are your HP25 is alive and well, O.K.? Now, let us know how long does it take since the last time you used it. If it takes too long (maybe years or even months) chances are you have a battery leakage problem.

Please, be carefull with chargers for these calculators. The avarage output for the Woodstocks AC adapters is 10Vac, but you should only use a charger when good batteries, even if discharged, are correctly placed AND in perfect connection with the calculator's terminals. Batteries are used as voltage reference and using an AC adapter when they are not installed OR not correctly mounted OR if you are using old, dried and high-impedance dammaged batteries, chances are you'll have a burnt circuitry. Please, be carefull.

Success and best regards.

Luiz C. Vieira - Brazil

Don't forget this important HP-25 problem.......
the battery charger puts out a high voltage,
and it relies on the presence of the battery
to clamp the voltage to something safe.

If you pull the battery pack while the charger
is plugged in, the voltage shoots up and you
can blow the whole unit. Or like the prior
post, if the batteries get dried out
(or just dirt on the contacts) again its like
there is no battery and the voltage can shoot up.

Atrocious design principle ?? Sure! But there's
not many other design mistakes like this.

You could modify such a calculator with parts
that clamp the voltage to something safe. It is
really quite shameful that the original design
didn't sport protective clamp parts. Some
zener diodes would suffice. They cost mere dimes
even at the earlier time. If i was an HP-25C
fan I'd be inside of there adding those parts.

Hi, Norm;

I did not measure voltage across batteries when they are charging, but I guess its not far beyond 3.2Vcc; what about 3.3Vcc zener diode, 1Watt dissipation power? I am not sure about test current, but I remember some types need about 5ma to 8mA to reach zener voltage. As NiCad's deliver about 2.5Vcc when fully charged, this 3.3Vcc zenner is invisible while charging and in normal operation with batteries in. As it does not have the zener as standard component, would adding a zener diode cause troubles to the charging circuit? I have an HP25, not a 25C unit.

I've been thinking the same: why not to protect the unit against the possibility of having a simple bad contact? If we think over the design itself, the number of components to add is minor.

And Woodstocks were desined when HP used to be HP.

Cheers.

Luiz C. Vieira - Brazil

About protecting an HP-25C against missing battery pack:

There's a reason I threw out the idea but
did not pick your parts. It has to be done
carefully !! Now that you're asking me
"ok what parts" well, lets take a look at that.
I'm an analog designer who deals with the esoteric
qualities of discretes, so "don't worry, trust me"
(said the Enron executive).

For starters, NO, you can't just go grab a 3.3V
1 Watt Motorola Zener. That old family of
discrete zeners (1N4733, etc) performs very nicely
at 6.2V (abrupt conduction transition) but down
at 3.3V the zener family performs badly. Such a
model will shunt too much current at 3.3V, and not
only that, it will draw current at 3V or even 2.5V
The Current/voltage characterization graph does NOT have a sharp knee.

Anyway, you don't want it conducting at 3.3V.
Figure the battery voltage may go higher than that
while charging. NEXT, go ahead and remove the
battery pack, what happens? All of the current
excess goes thru the zener. How much is that??
I don't know you tell me. Say its 700 milliamps.
How much wattage in the 3.3V zener?
3.3 x 0.7 = 2.3 watts. NOPE, you've cooked the
zener, melted the solder where you reworked it,
and probably put a dimple into the plastic case
on your fine nostalgic calculator.

3.3V Zener WILL NOT WORK.

We've even got other issues to worry about....
if the zener is leaking (drawing current) while
you are using the battery only, you're going to
shorten the run-time. And when the calculator is
sitting in the fur-pouch, a leaky zener across
the battery is going to drain the battery down.

Here's what I would do..........

I would grab about 3 zener diodes of
value 4.7V Part #1N4732A.
Put all those parts in parallel, not series,
across the switched voltage on the circuit board,
NOT directly across the battery.

They won't conduct noticeably at the battery voltage
of 3.0 - 3.6V . However, they WILL conduct heavily
if the voltage gets to 4.7V.
NEXT, I am deliberately choosing a lower voltage zener.
WHY? Because they are soft in transition.
The zener family performs with a very sharp knee if the parts are doped to 6.2V.
But if doped to 4.7V, the knee is still soft.

What's that spell? CURRENT SHARING! The excess
current will divide between these zeners and none
of them overheats.

Even this solution is a little reckless for sure, but this
is not a moon-launch. I'm throwing it out on 5 minutes
of consideration. One could keep designing better but
then the circuit requires a schematic, further parts,
further foolin' around, and I don't even have an HP-25C.
This idea will protect your calculator
when the battery pack is bad or missing. BTW
the circuits in the HP-25 appear to be from the 5V
logic era and won't be blown unless you get to 7V,
8V, or beyond and the zeners will prevent that.

AGAIN, Dr. Analog recommends (3)
Zener diodes, P/N 1N4732A, in parallel.

One other reason for this recommend..... if you
somehow reverse the voltage on the main supply,
the zeners will protect nicely by conducting
in reverse, and since they are in parallel they
will all conduct.

www.onsemi.com/productSummary/0,4317,1N4732A,00.html

The only thing that bugs me is that the arrangement
I suggest you add may draw an extra 30-40mA just
when you are running the battery pack, BUT, these
older units are power-hogs with way more than
that going into the calculator, I doubt you'd notice
an obvious shortening of battery life. Meanwhile
I presume all you enthusiasts for the "REAL" HP
calculators figured out how to put NiMH or
rechargeable Lithium into a red-LED unit? It would
run all day anyway on that.

Hi;

Thank you, Norm. This is what I call the best of two words: analog as power source for a digital device. Not that analog is only for power sources, far from me thinking like this. I learnt analog electronics first, digital electronics (not digital design) came after.

Just to figure out the charging current. If batteries are 250mAh and you need 12 to 14 hours to charge them, power supply should not deliver more than 25mA at the batteries contacts, should it? If the calculator is ON, that's another story.

AC adapters specs (Voyagers) read .8Watt, 10 Vac. output. This means they cannot deliver more than 80mA. So, when calcualtor is switched to ON and charging batteries, this is all they can offer.

Look at this image.

This is the power supply for the HP21. The HP25 is about the same, but there is an interesting circuit: an addition if you want to use the calculator without batteries. What is used as voltage reference? Voltage drop over four NP junctions (five diodes plus the b-e junction), what gives us about 2.5 to 2.8 Vcc. This is exactly what we need! So, I believe instead of zeners we could try four switching diodes connected in a series arrangment, and this set connected parallel to the batteries. They are fast and will remain quiet until conductive voltage is reached.

I think this is a better, easier solution. What do you think, Norm? Something to be tested?

Please, I'd like this proposal to be questioned and, if applicable, replaced by another, better one.

Best regards.

Luiz C. Vieira - Brazil

Hi;
---------------------------
Luiz Viera wrote:
> I learnt analog electronics first, digital electronics
> (not digital design) came after.

Well I'm pretty much all analog, which is a
neglected specialty these days. As you saw by
my prior writeup, even a simple thing like a diode
has a considerable amount of complexity in its
application (i.e inadvertently drain the battery
while trying to add some protection). Obviously
the digital guys did not do analog, at HP, when
designing the HP-25C or we wouldn't discuss a convenient
protective circuit. It gets real humorous these days
when a company needs a bit of classical control theory
involving feedback. They let the digital software guys
run wild with a stepper motor, they can't solve the
problem to save their life, they totally refuse the
trusted and true classical linear man (me) and are
blowing a $10Million dollar opportunity as we speak
and can't even be bothered to sit down for lunch to
talk about it. So, analog has its place.
Even today in the microsoft culture. For specialty
topics only, of course, the rest goes digital I admit.

-----------------
Let me firstly answer about the 5 or 6 diodes in
a row. Uh-uh, don't do that. Again, the diodes
are too smooooshy in their turn on characteristic.
They might measure at the casually phrased "0.6V"
if they have a quarter amp going thru them.
But take them down to 0.4V and they may still have
10-20 mA going thru. That is the problem. They
don't turn on & off at an abrupt voltage threshold.
The strong temperature variation is a further concern
which points you back towards a zener.

Remember this is a clamp application. You want the
protector to be as inert as possible when its voltage
is not exceeded. Above its threshold, you want it
to conduct heavily. Regular diodes are a relatively
poor performer for abrupt conductivity on the I/V curve.

BTW you could substitute a transistor wired
with Base to Collector, in lieu of a diode. Such
a transistor is a 2-terminal device (BC and E)
and the beta kicks in and it has a much sharper
threshold where it starts to conduct current.
But even then it varies with temp and operating current.

Lastly, although you might not include it as a criteria,
I really like that the thing should conduct in reverse
should somebody ever stick a battery in backwards.
I think that important. You don't get that benefit
if your choice of protector is 5 diodes in a row.

-------------------

Let me chug along presuming a 4.7V zener for
a moment.

Luiz Vieira wrote:

> If batteries are 250mAh and you need 12 to 14 hours to
> charge them, power supply should not deliver more than
> 25mA at the batteries contacts, should it?
> AC adapters specs (Voyagers) read .8Watt, 10 Vac. output. > This means they cannot deliver more than 80mA.


That's a very miniscule wattage rating for an AC
wall transformer. U sure the POINT 8 watt is not
just a little ding in the plastic ? I mean, 8 to 10 VA
(i.e. 8 to 10 watts) is a very common wallbug rating.

Well let's consider a small number like 80 mA @ 10VAC
I'll temporarily revise my proposal to a single 4.7V zener
as a rework offering protection.

If you had a 4.7V zener in there (one only)
then the dissipation is P = E x I = 4.7 x .08
.08 ENTER
4.7 TIMES
gives 376 mW zener power for an 80mA charger current.
This is an acceptably low value for a 1 Watt zener.

I would suggest up to 0.6 W is OK in a 1 Watt zener,
which would suggest a current of
I = P / E
0.6 ENTER
4.7 DIVIDE
127 milliAmps max.
(try that on your 49G, blecch)

SO, if you have more than 127 milliAmps in that
charger, you've got a problem if using a single
1N4732A zener.

My concern is, how could the adapter current output
possibly be so low.

The charger would be so weak that it wouldn't
simultaneously run the calculator and charge as well.
I just naturally presumed the output capability
was around 1/2 ampere or so, which is why the
talk of several parallel zeners.

HMMMMMMM some of this is coming back to me. I knew
a guy with HP-25C. And I think he pulled the pack
a few times with the charger on,
and the display got DAMN BRIGHT (but nothing blew up).
AND, I'm remembering that 8.2 ohm resistor. 0
But can you run the calculator while the battery
simultaneously recharges? It's fair game ? Or no?
At 0.8W I don't see how it could do both. And BTW
10-14 hrs is a rotten recharge time, I'd want to
speed that up.

---------------------------

Your schematic is helpful. Nice job.
Is it an authentic HP diagram? Unclear who
is providing the 5-diodes on a transistor base added
regulator circuit (that's 5 diodes MINUS the Vbe
of the transistor, not Vbe added). Like is that
a non-HP circuit, or, did they sell you a little module, etc. So you'd presume 4 x 0.6V = 2.4 V output.

There appears to be a small oscillating
switcher regulator, but without some explanatory
circuit notes I didn't follow that immediately.
It is sufficient to see that B+ can be overvoltaged
if the battery disappears, and the schematic is
clear in that regard.

------------------------

Luiz Vieira wrote:

> I think this is a better, easier solution. What do you
> think, Norm? Something to be tested?


By the schematic, our focus point would
be to add a protective clamp item to the
calculator itself, electrically across the 60uF 6V
B+ capacitor. With that in mind consider the
following circuit choices:


A. At this juncture I still would not consider
a 3.3V zener to add there. It will shunt away too
much current when the battery voltage is normal
at about 3.3V , as well as concern about its
power handling.

B. I'd recommend against the 5 diodes in a row.
Again, it may shunt away too much current when
the battery voltage is normal at about 3.3V , and,
nothing I was proposing brought the discrete count to 5.
Lastly, it does not offer reverse protection if
the B+ line somehow gets hooked up backwards.

C. If the charger current is less than 125mA,
a single 1N4732A diode will suffice (link provided earlier).

D. If the charger current is 125 mA - 250mA
then (2) 1N4732A diodes in parallel will suffice.

E. And you can triple or quadruple the diodes for
higher power handling. The slightly reckless argument
for current sharing is valid for a 4.7V zener
(soft turn-on knee) though not true for higher voltage units. The units should at least be from the
same box from the same mfr., to encourage sharing
if paired together in parallel (not a mil-spec technique
but OK for this situation).


--------------

I hope this exchange is useful, knowing that HP-25C
has a design error that plagued it from day #1.
It would be good if we could settle in on a
'best rework' thru this forum so that an 'official
recommended' rework could be established. I'd sure
want one if I had an HP-25C.

I am trying to avoid the most serious analog error,
which is very common and undermines the whole
specialty of analog, which is to say 'use the circuit
I said, because I thought of it not you'. This is
very common engineering practice in the analog area
but does not stack up. Instead, any circuit proposal
should be approved or rejected based on its merits.

It is hoped that we are considering different
types of clamping circuit rework, only by how well the rework will perform, in terms of cost, rework-ability,
simplicity, and reliability all as issues.

The fruit of this, is that maybe enthusiasts of this
calculator can make the rework and avoid irreversible
damage that probably destroys specimens to this day.


I have no HP classic equipment here whatsoever.
Actually thru this forum I made progress in buying
an HP-34C but it is not delivered yet. There is no
way I can check the charger or its current if I don't
have any.

If I provide some theory and practice (and I have)
you could provide the test-out on the bench, Luiz,
see if the rework does the job, and then
we would get it all squared away thru use of
this forum.

Regards,

- Norm
analogee@gte.net

For the protection to work in the 25C and 29C machines it MUST be on the battery side of the switch. Basically these machines connect the battery directly (or mayby through a low drop Schottkey diode) to the RAM chip so that it will still be powered up while the machine is off. I suspect that a 6V diode would do a perfectly good job of protecting the machine. The killer seems to be the peak voltages generated by the half wave rectfied AC transformer.

Interestingly, almost all dead Woodstock machines have a bad ACT chip. For the 25C, the RAM chip is usually toast also. For the plain HP25, mayby half the machines have a bad RAM chip also. The ROM chips and display driver are almost always good.

Hey, Norm; good words!

Spice AC adapters run around .8Watt. Topcats and portable HPIL devices (82162A printer, tape driver) have the same 3Watt AC adapter. I use a Spice AC adapter to charge the HP25 batteries (I use an original Woodstock pack rebuilt with original Spice batteries, and they are the same NiCad's for both), and I know the original Woodstock charger delivers 10Vac instead of the 9Vac available at the Spice AC adapter. As seen in the circuit, it seems this .5Vac (upper half wave) is neglectible. I know that the slow passive (dummy, overnight) charger and the fast (inteligent) charger are two charging options, but I do not know which one is the best for battery life extend. The HP21 (and many Woodstocks, if not all of them) charger circuit is limited to the 8.2 ohm/1W resistor and the half-wave rectifier. A small "pumping" action over the batteries will be observed as we see there is no filter capacitor right after the diode, only in paralell with the battery. And this is a critical value for a capacitor: about 20% over average voltage. Without the batteries this capacitor will be in this voltage limits... As you mention, no clamping device to protect anything.

Other fact is that we should reall increase the charging current to reduce charging time, and consequently use another transformer. I think the max current is limited also having in mind the low current for the charging period. And I am almost sure these AC adapters deliver more than the 80mA in the beginning of the charge period, for an almost completely discharged battery, because they get somewhat hot (not to burn). As the battery is being charged, charging current reduces and may go below specified 25mA, unless the calculator is switched to ON.

And you are right: many moer may be written about diodes and their use. Im an Electrical Engineer and I deal with electronics (analog and digital) because of pasion. I love this...

Oh, the schematics. I'm not the author, I simply took part of an "uncredited" GIF image dated Jully 17th, 1997. Strange fact I did not notice: date stamp reads 17 Jul 1997; it's not a north-american standard for date stamp. Also, replacement power supply is noted as inserted, say, not availble in original diagram. And this insertion will use a few miliamps to run, and these miliamps are probably less than the 25mA supposed to charge batteries.

About David's comments: the +B from the power supply in the HP21, that comes straight from the battery power line, goes right to the LED drivers. As expected, these IC's will deal with more current than the rest of the circuit and some are design to "survive" even to external short circuits. I imagine why these guys survive under a "no batteries" operation...

Wow! Good words, Norm. Thank you for the enlightment. This is good, very good stuff.

Best regards.

Luiz C. Vieira - Brazil

Norm wrote:

It gets real humorous these days when a company needs a bit of classical control theory involving feedback. They let the digital software guys run wild with a stepper motor, they can't solve the problem to save their life, they totally refuse the trusted and true classical linear man (me) and are blowing a $10Million dollar opportunity as we speak and can't even be bothered to sit down for lunch to talk about it. So, analog has its place. Even today in the microsoft culture. For specialty topics only, of course, the rest goes digital I admit.

I cannot agree more. You're right. I got what I could from both worlds, and I must admit both analog and digital theory may become hard and heavy if limits for both are not well defined. I myself feel as if I'm lazy for analog concerns and I tend to use digital as much as I can, but I know this is because of lack of my own resources.

What you mention is completely true. Let's face it: there are moments when analog should allow some digital action and also the opposite is true. All digital and all analog system will always have their crytical points and in some cases a mixed technology will be somewhat mor reliable. Let's take one classical, current example: digitally controlled UPS. There is no way to have them in all analog or all digital design.

Best regards.

Luiz C. Vieira

Greetings Luiz, David:

I like that we are continuing to come up, thru teamwork,
discussion, and a consensus, for a "best fix"
for the HP-25C overvoltage problem.

--------------------------

David objects to placing a clamp at the 60uF 6V capacitor.
He said that the RAM chip gets fried on overvoltage even
if the unit is turned off, as it's connected separate from the B+ node I focused on. OK fair 'nuff
I'm accepting of new data, HOWEVER, that was not
shown on the schematic in the earlier post "LOOK AT THIS". I take it that the RAM power comes the switch where
the wire is labelled "A2" .

With the new data in mind, where are we putting the
clamp then? At the "A2" node, to ground, then.
Works for me.

-----------------

Now, we are still getting hung up on part # selection.
David said 'I feel that a 6V diode will do a perfectly
good job of protecting the machine'.

Well OK, the 4.7V is out, because I would still be
concerned about leakage thru that part when the
battery is sitting at 3.3V which it can. You might
lose a couple mA and find the battery doesn't hold
charge as long as w/o the protector. That's unacceptable.

OK, what say we select a 5.6V, 1Watt Zener.
Part #1N4734A .

http://www.onsemi.com/productSummary/0,4317,1N4734A,00.html
A substitute by "Diodes Inc" out of Digi-Key is
inferior by the I/V curve tracer but would still suffice.

By jumping the voltage higher in our choice, I am no
longer concerned that it will leak at 3.3V. It will not.

OK, but still that final concern of cooking the zener!!!!!!

I live near Boeing, you have to consider worst case
(the plane is really high up, the cabin pressure is
set really high, the air outside is really cold,
the turbulence is really bad, etc etc)

So our worst case is if the battery pack is pulled out,
the charger is connected (and strongly powered)
and the calculator is "OFF". In that case, the
zener diode sinks heap plenty current. You will
have exceeded 0.75W if you exceed 130mA.

SO, it's now on you Luiz. Drop in the 5.6V part,
1N4734A, where David said. Connect your strongest
charger, pull the battery pack, and turn off
the calculator. Wait 5 minutes, see if the zener
is unacceptably hot. (We are winging this rather
than calculating because we're still non-committal
on the true current delivery of the charger with
a 5.6V clamp on its output).


---------------------------

IF the zener is too hot, you can parallel TWO
5.6V zeners, but only as follows:

Add a 1/4W 3.3 ohm in series with EACH ZENER
(an R, then a D) THEN, parallel such sub-assemble
resistors and zeners. A 5.6V zener is sharp enough
that the extra resistance must be added to each
zener to encourage current sharing, otherwise
one gets hot and cooks itself.

--------------------

Another good 'bench test' would be to take
TWO 5.6V zeners and connect them "back-to-back"
(stripes facing outwards) to create a 2-terminal
device. Then, simply attach this load to your
charger, but with an AC ammeter in series as well
(a true-RMS is best, but I'm not fussy, any
multimeter will do). If you note the current output
into a set of diodes arrayed thus, then that
is the current rating we are looking for,
and then we can get very scientific rather than
just guess about zener overheating.

---------------------

You've now got specific things to try, I hope
that Luiz can check this out, so that it becomes
a 'standardized' recommendation for the Woodstock
calculator enthusiasts.

---------------------

I wonder is there ANY merit to adding TWO diodes
one on either side of the switch. That way if the
power switch is set to "off" and
somebody attaches power backwards to the main board
(an errant open-back test?) you've got a diode-drop
for reverse protection. Probably too fussy to
talk about, probably just the A2 addition is sufficient
but wanted to bring it up anyway.


- Norm

You can always use a 5W or 10W zener...

When I rebuild the three pin version of the 82120A rechargeable HP41 pack, I place a 6V zener directly across the output of the 4-cell battery string. The circuit of this particular pack can dump a lot of voltage into the system if the battery string opens up. Hopefully if the battery string opens up, the system will act flakey and the user will stop using the pack. If the diode fries, so be it... it is certainly better than no protection at all. There is not enough room for a more elaborate fix.

I have checked battery life compared with unmodified packs, and can't tell much difference. The self-discharge rate of the cells seems to swamp the diode leakage current farly well.

Hi, Analog folks (as I am, too)

I read some stuff about batteries a few years ago and I try to keep track with inovative products, but I watn to know your thoughts and concrete knowledge about two subjets:

- charging methods - charging current

By charging methods I mean slow against fast, constant current against "pumping" constant current. I saw an inteligent charger project that use charging periods and monitors battery voltage and current between periods. It charges batteries in the shortest requested time, but battery life may be reduced if it is charged with current that's bigger than 20% the recomended load current. Say, if you have a 1,000 mAh battery and the recommended load current average is 250mA, than you should never charge it with more than 50mA.

I know these aspect depend mostly on battery specs and recomended charging procedures. Is there a "rule of thumb" (I always remember Steve Oedekerk) that will lead to the least pain for the batteries? I mean that because not everytime we have access to the battery model and type (old ones), neither the chance to test battery's current and voltage before charging it in deep.

Just to add more knowledge in my "knowledge" bank...

Cheers and best regards.

Luiz C. Vieira - Brazil

Topic:You can always use a 5W or 10W zener...

Reply: Those are quite large, with a #10-32 bolt-mount
stud on them. When you get into even bigger power
diodes, there is a big pigtail on the top that
looks like a rope. Those get stuck into arc-welding
equipment and stuff. Therefore was thinking that
the 1W zener diode would be a good form factor
for the HP-25C. Did you click on the link to the
data sheet ??

---------------------

Topic: When I rebuild the three pin version of the 82120A rechargeable HP41 pack, I place a 6V zener directly across the output of the 4-cell battery string. If the diode fries, so be it... it is certainly better than
no protection at all. There is not enough room for a more elaborate fix.


Reply: Again, again, again, a "6V" zener, often
found as "6.2V" (i.e. 1N4735A) will have a VERY sharp knee
and if the battery stays below about 6.1V, fine, it
won't conduct current, at 6.3V it will conduct a ton.
Any of these is a +-5% tolerance just for threshold,
so 4x1.5 =6V seems like you are cutting it too close.
Meanwhile, plz consider that you said 'if the diode fries'
however, what I'd like to do is come up with a recommended
rework that has just as much integrity as if the factory
designers had done it back in '76 . They wouldn't sell
you things that would overheat & melt back then, we should
be able to avoid that now. That's why I'm trying to choose
a zener voltage-value and wattage-value that actually
are compatible with the application. If word gets around
it could become a reasonably standardized rework.

Topic: The self-discharge rate of the cells seems to
swamp the diode leakage current fairly well.

Reply: Well, sure, there's some milli-amperes going
thru the battery internally when it self-discharges.
We want to arrange that the diode clamp leaks significantly
less than the battery, when it is just sitting there.
I didn't run numbers but I think you want LESS THAN 1 milliampere going thru this added protective clamp
when it is in parallel across the battery.

Something like 200 microamps would be nice. See, I think
we might get that with the recommendation from yesterday.
What was it, a 1N4734A ?? Anybody check a data sheet
or order some parts ?


Regards,

- Norm

Hi Luiz Vieira:

Well your question was fairly general in how it was asked;
let me be fairly general in the reply.

Firstly, I'd not heard that rule about charging at
20% the rate of discharge. And yet, there's a lot
of appeal to that. I'm not disagreeable. Meantime
there are plenty of systems these days that charge
batteries fast. Go to Home Depot and look at the
cordless power tools, bet you'll be surprised.

Here is an answer you can count on: Do whatever
the mfr. allows you to do. If a company releases
battery cells and state a certain allowable charge
rate then go ahead and do it.

If you don't have data handy, yeah, that 20% figure
sounds like a darn good rule of thumb. Think I learned
something myself there, thanks for bring it up.


The other topic I'd bring up with you, is just that
effective charging for batteries is very, very,
VERY fickle. It's like a recipe for getting a high
quality paint finish, or a good batch of cookies.
Do one thing wrong and you blow it.

One thing that amazed me once was I tried to charge
batteries with a pure DC current. The battery voltage
went up real fast and it refused to take any more charge.

Upon discharge, the battery had no real oomph in it
and discharged almost immediately.

TURNS OUT THAT the ripple-charging from 60 cycle
is NECESSARY. (You know, you get 120 bumps per second
if there is no filter capacitor). It turns out that
that nasty AC ripple sort of 'shakes up' the charge in
the electrolyte and allows the charge to soak in properly.

So sometimes the best things in life are free. Its very
easy to charge w/ the ripple present; remove it by using
an expensive regulated supply, and use
pure DC and it works WORSE.

Once a recipe is working then don't mess with it. BTW
for HP Classics I find the battery topic VERY intriguing
because we now have more powerful battery cells which
could be substituted and make the unit run longer/brighter/better. NiMh and rechargeable Lithium
come to mind. BUT I don't take it lightly, because you
don't know if it will work properly until its been tested
out with the presumed HP charger, tested over many
charge-discharge cycles, etc etc.

I sincerely feel this clamp-zener thing for HP-25C is
very important, and will be disapointed if nobody
tests the actual parts which might solve the problem.
It is part of having a better charge/discharge system
put into a classic. It is a total waste to just use
the same lousy Ni-Cd cells that were available in 1960.

- Norm

Norm wrote:

. BTW for HP Classics I find the battery topic VERY intriguing because we now have more powerful battery cells which could be substituted and make the unit run longer/brighter/better. NiMh and rechargeable Lithium come to mind. BUT I don't take it lightly, because you don't know if it will work properly until its been tested out with the presumed HP charger, tested over many charge-discharge cycles, etc etc.

A few weeks ago I wrote about this same concerns with almost the same approach. One of the sollutions pointed out was using AAA-size NiCad's that have the same 250mAh capacity shown in the earlier 60's NiCads used in the Woodstocks and later in the Spices. I bought some of them and I am reasoning about one particular matter: physical size. I think I'll build an AA-size "suit" that can hold an AAA-rechargeable in side so I'll use them in an Woodstock. Charging and "discharging" time should obbey existing predictions: 12 to 14 hours charging (maybe a little less) and 2 to 4 hours using time average. Anyway, the idead (suggested by Renato, Brazil) is a good one and it seems it will work fine. Based on this same principle, I'll rebuild an HP82143A battery pack with four AA-size NiCads with 850mAh capacity for testing. I rebuilt one original of thse packs with two not-working ones, and it seems original bateries are, in fact, exhausted.

I did not read this post yet, but it seems... tastefull!

Best regards, Norm.

Luiz C. Vieira - Brazil

Hi;

I remember completely rebuilding an HP82120A, including the hand-drawing of a new PCB and replacing the rectifiers, and I am sure my unit has a zener diode accros the bateries. As the bateries never go beyond 6.2Vcc, it is probably a 6.2V zener. And I see it as completely necessary, because the zener diode is not only at the batteries poles, it is indeed at the calculator's batteries connectors, a lot-to-worried-about point. The 82120A circuit is merely a full-bridge rectifier that delivers 18VCC to a 12Vcc regulator (SOT package). Regulator output goes to a 1W resistor (don't have the value) and a zener to reduce its output to 6.2Vcc. If you want (I think it's not necessary) I can add the 82120A diagram in a post.

Just to add this info.

Luiz C. Vieira - Brazil

Hi Luiz Vieira ,

I lied about something but not on purpose.
I actually DO have an HP-25C around here.
When I realized that, I made efforts to find a buyer
and I think that somebody has now agree'd.
Point is, its a fairly authentic old 25C package
in the black/green box.

So I pull out the charger, and it says "10VAC 1.8VA".
Why did you say "0.8" Sure you didn't mean "1.8" ??

In terms of the analog dialect, that "1.8VA"
might as well say "1.8Watts" . The reason they
say "VA" (watts = volts x amps) is because it is
the transformer factory talking. Transformer guys
always say "VA". It doesnt mean "Volts AC" it
means "VOLT AMPS". The reason is that they don't care
about the phase relationship between the volts and
amps, they only care about how much voltage, and how
much amps, either one, because that is what sets
the maximum capabilities of the transformer (ie how
much before it gets fried). It's really just the
current that cooks the transformer. The voltage
is spec'd at full load. Multiply together to get "VA".

NOW THEN, that means that this thing can deliver
180mA max (10VAC x .18 mA = 1.8 VA)

OK, 180mA into a 6.2V zener is 1.11 W .
NOW, I freely admit that we should be less than
that, and should not melt the zener.
Why? Well there seems to be half-wave
rectification, and a zener is not even a full
load, its a clamping load, conducting momentarily.

SO, we could presume the dissipation is
SIGNIFICANTLY LESS than 1.11W on a 6.2V zener,
with the charger in front of me. That's
getting to be OK.

Personally, would I use the 4.7V zener for
the HP-25C ?? Absolutely yes, I would. The 6.2V
is still pretty high for the clamp point. The
calculator is supposed to run on 2.4 - 3.4 volts
not 6.2V .

The big problem is, what if somebody uses a completely
different charger ? The 1.8VA rating is really, really
low. Is it even enough to run the calculator and
charge the battery at the same time ? If somebody
puts on a bigger charger, we're going to have
cook-the-zener problems again.


I'd feel safer with the 4.7V zener for HP-25C
but use a 6.2V if you must.

You do have these parts kicking around? If not
they can be ordered from Digi-Key. Or i have some
in my basement.

Don't know why you would rebuild a battery pack
with AAA. I thought all this stuff ran on 2 AA
batteries.

If this thread-link disappears off the 2-day
forum, I suggest we post anew back at the top !

Regards,

- Norm

Luiz:

The capacity of a battery is usually denoted by "C". If a battery has a certain "capacity" (i.e.: charge) of, say 1500 mAh (milliampere hours), this value divided by ten (C/10 -> 150 mA) has been suggested many times as an appropriate average charging current fo a slow (or trickle) charger. Batteries are supposed to adequately tolerate C/10 charging currents without overheating. This seems adequate for NiCd and perhaps NiMH, I would not apply this rule of thumb to Lithium cells without more reassurance from the cell manufacturer.

In a perfect world, C/10 charging will produce 10-hour recharging cycles. In the universe we live in, some 20% to 40% losses are usual, so C/10 recharging cycles last about 12 to 14 hours (assuming a fully depleted cell). Fast chargers need some kind of feedback (i.e.: cell temperature and voltage) to modulate larger currents while avoiding overheating (or venting, explosions or fire)

The cells in the 82120A pack are 70maH. A 1mA leakage rate would deplete the cells in less than three days. I get a couple of MONTHS before recharging. The zener is drawing VERY little standby current. Most 41CV and CX machines have a standby current of around 10uA. The 41C can go up to 100uA. Most nicads self discharge around 1-2% per day...

Yes, you could do a better job of protecting the early three pin 82120A packs. Just layout and build a new PC board for them. Not really practical considering the number that are floating around out there. That leaves the simple fix of adding the emergency backup protection zener.

Hi,
If any of us post on this again, lets go
back to the top of the forum.

The 70mAH specification was very interesting.
It suggests that 100-200uA flowing thru a
protective clamp is not acceptable, which was
not what I originally expected.

I agree that maybe we aren't done yet, maybe
that does point towards your choice of a 6.2V
zener over anything else.

I think we should post again,
but at the top of the board.

- Norm