the HP-25C protector circuit
Message #7 Posted by Norm on 13 Mar 2003, 3:16 p.m.,
in response to message #6 by Vieira, Luiz C. (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
| 
Go back to the main exhibit hall