The time constant is T= 0.693*RC.
This is true in a special case when Vc(t) = 1/2 Vtotal (assuming Vc(0) = 0V)
Let us remember the general form of the capacitor (C) voltage while it is charged charged through a resistor R.
Vc(t) - Vc(0) = [ Vmax - Vc(0) ] * [ 1 - e^(-t/RC) ]
At time 0 we found that the voltage of C is -10.6 V
Vc(0) = -10.6
Vmax = Vcc - Vsat1 - Vd2
What is about Vc(t)?
Q2 will start to turn on again when:
Vb2 + Vd4 = Vc(t) + Vsat1 + Vd2
Vc(t) = Vb2 + Vd4 - Vsat1 - Vd2
Therefore the equation becomes:
Vb2 + Vd4 - Vsat1 - Vd2 - Vc(0) = [ Vcc - Vsat1 - Vd2 - Vc(0) ] * [ 1 - e^(-t/RC) ]
0.7 + 0.7 - 0.2 - 0.7 + 10.6 = [ 12 - 0.2 - 0.7 + 10.6 ] * [ 1 - e^(-t/RC) ]
11.1 = 21.7 * [ 1 - e^(-t/RC) ]
0.511 = [ 1 - e^(-t/RC) ]
This gives approximately t = 0.69 * RC as you said, since 0.511 is close to 0.5
Actually C2 could be chosen as 100n and Rb2 = 22K to produce 1.5 ms (approximately)
Now how Rc1 value can be determined? What is the function of Rc1 in this circuit?
Rc1 sets the recharging time of C2 to 10.6V when Q1 is off.
Could you find the time formula for which C2 is charged again to 95% of 10.6V?
Now t=0 is when Q1 is turned off (Q2 is on).
Vc(t) - Vc(0) = [ Vmax - Vc(0) ] * [ 1 - e^(-t/RC) ]
Vc(0) = - ( Vb2 + Vd4 - Vsat1 - Vd2 )
Vc(0) = - ( 0.7 + 0.7 - 0.2 - 0.7 )
Vc(0) = - 0.5 V
Note: Vc(0) = -Vc(t) of the previous calculation
Vmax = Vcc - Vbe2 - Vd4
Vmax = 12 - 0.7 - 0.7
Vmax = 10.6 V
Vc(t) = 0.95 * 10.6 = 10.07
10.07 + 0.5 = ( 10.6 + 0.5 ) * [ 1 - e^(-t/RC) ]
0.9522 = [ 1 - e^(-t/RC) ]
t = RC * ln [ 1 / ( 1 - 0.9522 ) ]
t = 3.04 * RC
Actually we can assume for a better recharging:
t = 10*RC
So if we know the minimum delay between the pulses:
Rc1 = Tmin/C2/10
Let us assume Tmin = 10ms
Rc1 = 0.01/100e-9/10
Rc1 = 10K
Note: This minimum delay time between two pulses should not be shorter than the time needed for the primary coil current to restore its expected maximum value (close to 6.25A in our example here).
Any comments or questions?