I need to design a circuit which will make a LED ON for two seconds and then OFF for two seconds and so on. You can use capacitors, resistors ,transistors and relays for the circuit only (no 555 timer or controller can be used). I also need formula and calculated values of resistors, capacitors for the above mentioned blinking scenario as I have to defend the components. Can someone please provide the link or share the circuit with values calculation. There are a lot of circuits available online but I cannot find calculations for them!

  • \$\begingroup\$ What ideas have you considered using? How precise must the timing be? Can the LED be any LED? Or some particular LED? What's the power source(s) you are allowed to use? And are you not only asking for a circuit, but also a complete defense of the circuit, in writing, to be posted here for you? \$\endgroup\$
    – jonk
    Dec 1, 2019 at 20:12
  • \$\begingroup\$ If you're just told that you can use "relays", you might be interested in what's called a "time-delay relay". I suspect that might be cheating, though. \$\endgroup\$
    – Hearth
    Dec 1, 2019 at 20:23
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    \$\begingroup\$ is this a school assignment? \$\endgroup\$
    – jsotola
    Dec 1, 2019 at 20:25
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    \$\begingroup\$ There are a lot of circuits available online but I cannot find calculations for them! I find that incredibly hard to believe but perhaps you are not searching for the right thing. Could you explain what resources you've referred to thus far? \$\endgroup\$
    – user103380
    Dec 1, 2019 at 20:26
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    \$\begingroup\$ Welcome to the site. I'm sure you can see that it's not a free design house, homework-answering service or on-line technical encyclopedia, copied out to you on demand. People will help you take the next step if your question shows you've done as much as you possibly could on your own - which yours doesn't, I'm afraid. Please edit your question and greatly improve it. Show your work and findings so far in considerable detail with a schematic. The better the quality of question, the better the quality of the answers you will attract. Again, a warm welcome. \$\endgroup\$
    – TonyM
    Dec 1, 2019 at 22:14

1 Answer 1


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.

enter image description here

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 < 10% error.

I apologize for the brevity.

Falstad Sim

  • \$\begingroup\$ was -1 a troll or a just a nitpicking silent objector or a person too frightened to comment and say something intelligent? 250mHz Flashing LED square wave within 5% was done as required. Homework not done. \$\endgroup\$ Dec 2, 2019 at 18:22

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