HP 65 with capacitor wired to battery terminals - why?

[dmh]

I recently acquired a 65 and a 35 that both have capacitors wired to the battery terminals as per the attached (65 shown).

Why would this have been done?

Something to do with running on the AC adaptor without a battery (even though the 35 runs fine on AC without a battery)?

[Ren]

For whatever reason, I was unable to view the picture.
My SWAG is the previous owner had a lot of noise on their power grid, or maybe the power supply they had was weak and so the capacitor provided enough power for intermediate demands of the calcs.

[Dave Britten]

(07-19-2021 03:18 PM)Ren Wrote:  For whatever reason, I was unable to view the picture.
My SWAG is the previous owner had a lot of noise on their power grid, or maybe the power supply they had was weak and so the capacitor provided enough power for intermediate demands of the calcs.

I bet it's something like that. These calculators aren't supposed to be run without a good battery, as it acts as a regulator. This looks like someone stuck a cap in there to run it off wall power with no battery. Maybe the previous owner wasn't impressed with battery runtime and wanted to run it purely from external power? No idea if it's safe to do that with a big cap instead of a battery!

[EdS2]

Any idea how long an HP-65 would run from a 4700uF capacitor? Would it, for example, allow you to unplug your mains adaptor and take the calculator to another room, without losing state?

[Hlib]

Quote:Any idea how long an HP-65 would run from a 4700uF capacitor?

According to my approximate calculations, this time will be no more than 0.2_sec.
If this calculator consumes a current of 0.05_A at 4_V (I don`t remember exactly) and turns off at a voltage of 2.4_V, then it will be:
t=c×r×Ln(Us/Uf)=0.192_sec.
c=4700_μ, r=80_Ω, Us=4_V, Uf=2.4_V.

[jonese]

For the HP-65, I can only imagine it was used to carry the current spikes from the motor while it ran on the AC power pack. Similar to the HP-97 needing a battery for proper printer operation.

Dunno about the HP-35.

[teenix]

(07-20-2021 08:17 PM)jonese Wrote:  For the HP-65, I can only imagine it was used to carry the current spikes from the motor while it ran on the AC power pack. Similar to the HP-97 needing a battery for proper printer operation.

Dunno about the HP-35.

The original AC charger would not run the card motor by itself and needed a charged battery. Same deal for the 97 and as mentioned its printer.

The motor is not driven by PWM for speed control and has filtering, so is not noisy enough (in normal operation) to cause problems.

I cannot see a valid reason to have the cap like that unless the owner had a simple rectified AC supply powering the calculator. It would appear batteries weren't used so this setup could be damaging, although I believe the calculator is actually working.

Hopefully the displaced battery contacts have not broken away from the mounts under the case.

cheers

Tony

[EdS2]

(07-20-2021 04:26 PM)Hlib Wrote:  

Quote:Any idea how long an HP-65 would run from a 4700uF capacitor?

According to my approximate calculations, this time will be no more than 0.2_sec.
If this calculator consumes a current of 0.05_A at 4_V (I don`t remember exactly) and turns off at a voltage of 2.4_V, then it will be:
t=c×r×Ln(Us/Uf)=0.192_sec.
c=4700_μ, r=80_Ω, Us=4_V, Uf=2.4_V.

Yikes! Those big capacitors are really not so big. Thanks for the numbers.

[Ren]

(07-21-2021 09:00 AM)EdS2 Wrote:  Yikes! Those big capacitors are really not so big. Thanks for the numbers.

I'm not sure how long it could run on today's "supercapacitors" >= 1 Farad

[cruff]

(07-21-2021 05:56 PM)Ren Wrote:  

(07-21-2021 09:00 AM)EdS2 Wrote:  Yikes! Those big capacitors are really not so big. Thanks for the numbers.

I'm not sure how long it could run on today's "supercapacitors" >= 1 Farad

Just plug 1 into the formula given above for c and you get 40.8 seconds.

[Ren]

(07-21-2021 07:01 PM)cruff Wrote:  

(07-21-2021 05:56 PM)Ren Wrote:  I'm not sure how long it could run on today's "supercapacitors" >= 1 Farad

Just plug 1 into the formula given above for c and you get 40.8 seconds.

That would be the "theoretical" answer, but supercaps may not allow the calc to draw the necessary power to run.

So, I guess I was being rhetorical, not theoretical!
B^)

[cruff]

(07-22-2021 04:56 PM)Ren Wrote:  That would be the "theoretical" answer, but supercaps may not allow the calc to draw the necessary power to run.

Very true, but with a sufficient number in parallel, you could probably do it. Don't know about the physical volume they would use though, might not be very portable.