Blink a LED for 2 Seconds
Your job is to analyze my design of a Darlington Astable Oscillator and compute why each value is chosen. The ramps are ~ 2 s each. The load current target was 10mA.



Due to component tolerances and simple design, 10% accuracy is about as good to can expect from calculations.  Temperature , hFE, Vcc and RC tolerances will contribute to this.

This "astable" oscillates due to positive feedback loop formed by the "figure 8" of Q1,C1,Q2,C2.  The necessary condition include that the base current Ib=Vcc/Rb  must be amplified by hFE and projected to exceed the emitter-collector current in order for it to saturate as a switch. When it does, Vc drives the base negative and is pulled by the T=C1*Rb2 which is the turn off delay time of 2 seconds. Since dV/dt=Ic/C for 64% of full swing and 4M*1uF=4s, you expect that the Darlington base voltage switches at 0.65V*2=1.3V.  This is shown on both Vbe's on right side scope traces. 

One positive ramp goes from -2.5 to +1.3 as the traces display the min and mix values.  This difference **dV = 3.8 in 2 second off time while the 2M base resistor is tied to 5V is clamped by Vbe would otherwise have a ramp of 4 s going from -2.5 to 5V {or a target of dV=7.5V} which is twice the actual dV=3.8 hence the Off-time is 1/2 of 4 s.  

hFE=100*100=10k for a Darlington is a reasonable assumption.  The Ic*hFE/Ib overdrive ratio must be >=2 to saturate Q1. 

I had a target current of 10mA min. in each Q, 

Ic=(Vcc-Vce)/Rc  Q1: (5-0.7V)/390= 11 mA  ... Q2: (Vcc-Vf-Vce)/Rc= (5-1.8V(red)-0.6V)=2.6/220ohm=12mA    this balance is required to balance Vce voltages and thus balance ramp sizes and the duty cycle.

When the base and collector R pullups go to the same supply we can simply look at the R ratio to see if there is enough current to saturate the transistor.  MOT used call this base "overdrive ratio" in the '70's.

Bias ratio= (Rc*hFE)/Rb = (390*10k)/2M= 3.9/20 ~ 2 which is the minimum to overdrive the base and saturate Q1. As hFE drops to 10% of rated hFE at the rated Vce(sat). We don't mind if Vce is a bit (0.1V) higher as this reduces the size of the Cap needed to make along time constant of 4 seconds with a marginally low "overdrive" factor.  

If you use KVL accurately to account for Vbe, Vce drops this will improve accuracy a bit as Vce can drop to 0.7V with 10mA while Vbe is 1.3V makes a difference of 0.6V on a target swing of dv=7.5V or &lt; 10% error.   

I apologize for the brevity.

Falstad Sim