15C LE spike fix

[jebem]

(04-30-2014 04:59 PM)Jeff O. Wrote:  (...) Adding the capacitor will have no impact on current draw from the batteries during program execution (...)

(...) As soon as the program starts, the capacitor will discharge in a very short time, then the load will be carried solely by the batteries (...)

No offense, but I beg do differ
I would agree with Your first sentence if we were talking about DC current on passive (ie, resistive) loads. But this is not the case (explanation below).

Actually capacitors have been used since more than 100 years ago in every single power supply, and everywhere as bypassing circuits to allow digital circuits to operate properly. In fact they are present in every single calculator, HP or any other brand.
A capacitor stores energy that can be delivered at a very fast rate when compared to any battery, of cell or any other type.
And because the capacitor is connected in parallel with the voltage source (the batteries) it never "discharges": if it did, then the batteries would be totally "discharged" as well. What happens here is that the capacitor is pulled up by the battery all the time.

One characteristic of the capacitor is to present a extremely low impedance to the load when compared with the very high impedance of those button cells. That is the "secret" behind this kind of capacitor application.

We are dealing here with AC (sinusoidal series) function signals of extremely high frequencies resulting from the gate's switching operations. When a flip-flop circuit transitions from 0 to 1 or 1 to 0, the transition time is not Zero in the real world (if it was, that would be an ideal square wave), so during that amount of very short transition time the instantaneous AC current is extremely high and would require an ideal Zero Ohm impedance to maintain the required voltage source.
Because the cell batteries have high internal impedance, the voltage drop will happen instantaneously as well to a low value, and that is the cause to crash the digital circuit/calculator.
So the solution is to use a capacitor in parallel with the battery, because a good capacitor will present a extremely low impedance to these high frequency signals, allowing the required transient current to be delivered without appreciable voltage drop.

This is not about DC theory, it is about AC transient signals of high current demand, that can only be supplied by a capacitor, or in a much more efficient way, using a regulated power supply (now, that should be the proper design for a good calculator, as HP and others did in the good old days. But we all know why they don't do it anymore for such types of calculators, don't we?).

I other words, the battery efficiency will be superior when using a capacitor.
That's what several studies have demonstrated (please have a look on my initial post for some references).

Again, I do not own a HP-15C LE, so I can not:
- Check for the issues myself;
- Take experimental readings to find out the root cause of such alleged issues;
- Make any experiments and readings.
So I'm not telling that a capacitor will fix anything in your calculators.
Keep your mind open, accepts nothing for granted, make your own experiments. That is my life orientation.

One final remark: If I had such calculator, I would use it as a pocket calculator, meaning normal operations that can be resolved in less than a couple of seconds, and I would leave the batch processing consuming minutes and hours to finish to a more powerful and efficient machine for that job