Originally posted by sucahyo
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me a lot, otherwise i could still hit the wall with this relays stuff...
Originally posted by sucahyo
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most important, other values were completely wrong, just 220 uf works
well...second : pull out that diode that goes from pin 7 to electrolytic plus,
10 nf capacitor you do not need too....third : resistor that goes from batt.
plus to pin 7 could be 10-100 kohms, but the most important is resistor
that goes from pin 7 to electrolytic plus : there you can put resistor from
1 Mohm or resistor with higher values(2,3 M or so), and then your circuit
is going to work well !
Originally posted by sucahyo
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Originally posted by sucahyo
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Cause i do not think it is good to use battery as primary (source)
battery when voltage is below 12 V !
Originally posted by sucahyo
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and before charging was 11, 85 i think, and primary was about 12,75
and now is 12,74 so according that formula it would be like this :
0,50*5 Ah(i think it is 5 Ah,not 7 Ah) = 2,5
0,01 * 41 Ah = 0,41
2,5/0,41 = 6,09
If it is so, than i must congratulate both of us !
But i must ask you : do you think that this method could
assure us that this calculating result is real reflexion of
reality ? And is 5 hours after charging enough for secondary
battery to stop droping, because this old one still droping
a little bit ?
Cheers !
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