HP-97 Charging circuit not working

[swperk]

I just picked up a pristine HP-97 that will not operate from nor charge the battery from the AC adapter. Digging into it I found an open 8.2 ohm, 2 watt fusible resistor in the charging circuit.

On the schematic, there are two power resistors in the charging circuit, R3 and R4. R3 is a charging current limiter that is shorted out by the power switch to allow full current from the battery when the calculator is running. R4 appears to be a current limiter for both charging, and operation of the calculator from the AC adapter.

Here's the issue: the parts list says that R3 is an 8.2 ohm 2 watt resistor, and that R4 is a 4.7 ohm, 2 watt resistor. On my calculator, the 4.7 ohm resistor is in the R3 position and the 8.2 ohm resistor is in the R4 position.

Is this just an error in the parts list, or was my calculator not assembled correctly? It makes sense to me that the fusible resistor should be protecting the entire calculator, not just the battery, and that my calculator has been assembled correctly. I believe it's just an error in the parts list. Also, there is absolutely no evidence of any post-production work having been done on this calculator.

It's too much for me to ask anyone to open up their -97 and have a look, but if anyone can shed some light on this I'd appreciate it!

I've attached the relevant pages from the HP-97 service manual.

Thanks!

[rprosperi]

There are at least 2 board designs for the 97 (maybe more) and both are not covered in the original service manual. I don't know if the later design was ever documented in a later revision of the service manual, but take at look at this one (June 1982), it may be different from yours:

https://literature.hpcalc.org/items/1127

[swperk]

Hi Bob,

Thanks for the info! The manual you pointed me to is the same one that I already have, so any potential documentation errors are still there.

I was not aware of the other revisions of the -97 boards. My A2 keyboard PCB has p/n 00097-60019, which is the same as is listed in the replaceable parts section of the manual, so I figure this manual must be the one that is applicable to my calculator.

One thing I noticed is that the R2 location on my A2 PCB has no resistor installed. The service manual says that this is either a 200K or a 330K resistor that is "selected." I guess HP "selected" not to install one in my calculator! :-)

Even without that R2 resistor, the calculator appears to work just fine, as long as I keep the battery pack charged up with an external charger.

Barring any replies that advise me to the contrary, I will simply replace the open 8.2 ohm resistor once my Digi-Key shipment arrives. The data sheet I have for the fusible resistor does not say what amperage will cause it to go open. Assuming that it will open if more than 2 watts are dissipated, that implies it will fail open at a current of 350 mA (P/R=I^2). Does that seem reasonable, or am I making some bad assumptions?

Regards,
Stan

[swperk]

Sorry, I must have fat fingered the calculator when deriving the assumed maximum current draw for the 2 watt, 8.2 ohm fusible resistor. It's actually SQRT(2/8.2), or .494 A.

[teenix]

Attached is an image of the resistor positions. It seems, as you mention, R3 and R4 reversed in the parts list as the 8R2 resistor is across Q1 and is shown as R4, but R3 in the parts list.

The resistor you mention may not have been required. It would have been for the cathode driver and low battery detection.

The lower image half also shows a change to a mistake in the original schematic. This monitors the current through the displays and when required shunts R4 by turning on Q1 to allow more battery current.

If the 8R2 resistor is open circuit, that will stop the charger current reaching the battery. A 500mA fusible 8R2 resistor seems feasible.

cheers

Tony

[AndiGer]

R2 for battery low detection was obviously replaced by a tantalum capacitor. I once wrote down: red with a green dot. Maybe 2.2uF.