ikalogic said:
hello,
Now, I KNOW what components decide of all the timings, this is not my problem, but i cannot figure out exaclt how it works.. and i did not find good sites to explain this, i think it is an important ussue.
2nd: while i know R2 and R3 will slow down the discharging process of the 2 capacitors, i dont know actually WHY ???, i would have understood if they were in parallel with the capacitors...am going crazy! :twisted: :evil:
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The others have explained everything except your question quoted. The following explains how the timing works. Refer to the circuit below.
When T1 is on, T2 is off and vice versa. While T1 is on, C2 charges rapidly via R4, T1 B/E and D1 to the difference in potential between the 12 Volt supply and the voltage at the base of T1 (about 1.4 Volt so the difference is about 10.6 Volt). I have not shown the charging in the diagram as it is assumed to be rapid.
At the moment T2 turns on, the Collector voltage of T2 falls to about 0.9 Volt (allowing 0.7 Volt across D2 and 0.2 Volt across T2 C/E). Thus the base of T1 goes to about 0.9 - 10.6 = -9.7 Volt and so T1 is turned off.
C2 charges towards the supply voltage (+12 Volt as indicated by the dashed red line in the diagram) but when the base of T1 reaches about +1.4 Volt, T1 turns on and T2 off.
During the charge of C2, the Base voltage of T1 can be represented by the equation v = 12 +(-9.7 - 12) e^-t/tr where tr is the time constant
tr = R2 * C2. See the timing diagram.
When v reaches +1.4 Volt, t = T and so
1.4 = 12 + (-9.7 - 12) e^-T/tr
Thus 21.7 e^-T/tr = 12 - 1.4
Hence e^-T/tr = 10.8/21.7 Invert both sides gives
e^T/tr = 21.7/10.8 = 2.009
Thus T = tr ln 2.009 = 0.7 tr
But this is one half cycle. The second half cycle is approximately the same length so the period of oscillation is
2 T = 2 x 0.7 tr = 1.4 tr = 1.4 R2 C2.
Note that I have used the word charge to include both charge and discharge.