Univ. Charger K7300: How to set alternative charging times?

Hello,

The programmable timer CD4536 in the K7300 kit is set to provide either 52’ or 14h charging time. This is achieved by setting either H or L value in pin #11.

In order to set alternative charging times, different coding is required for the other coding pins #9, #10 and #12.

Does anybody know what the timer values would be for each of the 16 combinations of H/L settings for the 4 pins? (e.g. What would be a way to calculate them?)

CD4536 datasheet provides a “decode out selection table”, but the output they provide is the “number of stages in divider chain”, and I don’t know how to translate it into a time scale.

I would appreciate any clues on solving this issue.
Regards,

Hello,

I found the answer myself. Let me share it with you.

On page 16 of the assembly manual for the K6200:

velleman.be/downloads/0/illu … _k6200.pdf

you can find the requested table that translates pin assignments into time periods for the CD4536 timer.

With this information we can go back to the K7300 charger. By playing around with the 16 combinations of coding pins #9, #10, #11 and #12 you can set the most convenient charging time for you application.

In my case, I found the default 52min or 14h either too short or too long charging times. What I did was setting pin #9 to LOW while keeping #11 to HIGH, which enables setting a charging time of 7h. Obviously, in order to achieve full charge, charging current has to be set accordingly, as follows:

I = Q / (t / f) = Q x f / t

I: Charging current, in mA.
Q: Battery capacity, in mAh.
t: Charging time, in hours.
f: Inefficiency factor, in the 1.1-1.4 range.

Examples:

  1. Charging 2 AA NiMH batteries of 2100mAh, 14h setting.

Voltage set to 2.4V
Pin assigments #9=H, #10=L, #11=H & #12=H.
Inefficiency factor: 1.4.
Charging current: I=2100x1.4/14=210mA, we select 200mA.

  1. Charging 7.2V NiMH battery pack of 3300mAh, 7h setting.

Voltage set to 7.2V
Pin assignments: #9=L, #10=L, #11=H & #12=H.
Inefficiency factor: 1.2 (it decreases as charging time is lower).
Charging current: I=3300x1.2/7=565mA, we select 550mA.

I hope I made it clear enough. Feel free to contact me for further advise.

Regards,