this is a part of a circuit shown in the following link:hi, just wondering, what does this circuit do?
is it just to change the output voltage of the 555?
if so what does the transistor for?
5.6/6.8 = 82%The supply voltage is 6.8V if the battery is fully charged.
Then the output high of the 555 is about only +5.6V.
No. It is about 1.3V, not a percentage of the supply voltage.walid said:hi guru
5.6/6.8 = 82%
is the output high of the 555 is always about 82% of Vcc?
part of the 555 schematic is shown below.It is two base-emitter junctions in series. 0.6V + 0.7V= 1.3V.