At 11:59 PM 4/11/01 +0100, you wrote:
>> >Some machines use the battery as a shunt regulator -- that is to say they
>> >rely on the fact that the voltage across the battery pack, even when
>> >charged from a supply _capable_ of giving out a higher voltage, will be
>> >limited. This is one reason why many HP and TI calculators may be damaged
>> >if you connect the charger without a battery pack in place.
>>
>> Tony's right. You should have the battery in place when trying this.
>> You'd probably be safe leaving the battery out if you used a power supply
>> with the output set at a safe voltage (~ 1.2 to 1.5 Volts/cell) but don't
>
>I would be careful using anything above 1.2V/cell. I've seen machines
>(the HX20 is one such) were there are 4 NiCd cells (nominally 4.8V, of
>course) connected to the +5V line. And that's the _only_ regulation in
>the machine.
Hmmm. How do they charge the battereis? 5 vdc isn't enough to charge 4
NiCads. Do they take them out and charge them outside the circuit.
>
>Now admittedly, the machines that do this generally use CMOS chips which
>are reasonably forgiving about mild overvoltage. But still I'd not want
>to risk it.
>
>
>> try this with a standard charger. They're seldom regulated and you could
>> fry something. Leaving the dead battery out and trying to run a HP 2xC
>> calcualtor on the charger alone is THE biggest cause of dead HP 2xC
>> caculators.
>
>Is this the case with HP 3x machines (Spice/Spike) as well?
No, I don't think so. At least not a series problem. However the Spice
and Woodstock chargers will not supply enough current to operate the
calculators without the batteries.
The problem with the HP 2xC models is that the RAM is connected directly
to the battery in order to keep the Continous memory alive. When you remove
the battery and try to run the calculator off the standard charger you're
putting up to 18 volts directly into the RAM. Poof! There goes all the
magic smoke! AFIK this is only a problem in the 2xC models since they are
they only only models that connect the circuits directly to the battery
(and the chargng circuits) and not through an internal power supply.
The charger
>circuit and battery pack seems to be electrically identical, and the
>machine's PSU is similar. In any case, I am not going to risk it.
>
>Of ocurse in the -C (continuous memory) models, there is a backup feed
>from the battery pack to (one of) the power pins on the RAM chip.
>Normally, the battery voltage (+3V) is lower than the running voltage
>(+6V-ish, from the switching coverter circuit), so it's OK. But plug the
>charger in without the battery and that line will rise to about 10V (and
>there's almost no load on it to pull the output down). That is going to cook
>that RAM chip.
Exactly. That's why there are so many 2xC models around that have a
display but no memory, not even the X, Y and Z registers.
>
>I have a nice bench PSU -- 0-30V, 0-10A. I use that for testing machines
>with the battery pack removed -- set to 1.2V/cell of course. This seems
>to be as safe as using the original battery.
It's even safer IMO since you can set the current limit low enough to
prevent doing any damage even in the event of a short. I have a small 0-20
volt 400 Ma HP supply that I use for testing all my calculators.
>
>
>> >I am told that some HP calculators (the Woodstock HP21 series, for
>> >example) can be damaged if you put primary batteries into the battery
>> >pack. Now, I know how the Woodstock PSU should work (it's a little
>> >transformer-based transistor oscillator with regulation taken from one of
>> >the outputs), and I don't see why it wouldn't regulate properly at 3V
>> >input rather than 2.4V input. But I certainly am not going to risk it :-)
>>
>> FWIW I don't recommend it but I've run Woodstocks at 3 VDC and even a
>
>From what I remember, the Woodstock PSU outputs a +ve Vss (about
>+6.something voltes), a -ve bias supply and a lower +ve supply for the
>display driver. The first output can't be less than the battery voltage
>(It's a switching converter -- if the transistor is turned off all the
>time, the Vss output is effectively connected to the battery by one
>winding on the transformer). The same might apply to the display output
>(which is lower, but still around 4V IIRC). I doubt very much if the
>circuit will have problems running on 3V input.
>
>However, I have heard that Woodstocks don't like 3V batteries. And to be
>honest, even though I can't see why there'd be a problem, I am not going
>to try it. All the chips in the Woodstocks are HP-custom and
>unobtainable, so I have to preserve all I can.
I've never tried to run a Woodstock off of primary batteries so I don't
know if there's a problem with that or not. I have run them off of
regulated power supplies set at 3 V with no problem although I usually set
the supply to about 2.4 volts. I'm guessing that perhaps primary batteries
won't supply enough current for the Woodstock. They have an LED display so
current draw is a bit high (~150Ma). The other odd thing that I've noticed
is that is you have to have more current available or it won't start. I
have to set my PS up to about 250 to 300 Ma if I remember correctly before
the Woodstock will come on. After that the curerent drops to about 100 to
150 ma.
Joe
>
>-tony
>
>
Received on Wed Apr 11 2001 - 20:59:15 BST
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