Is this the way this circuit works ?


Vcc = 15v

With vref = 5v on the pin #5, then Vthreshold = 5v, and Vtrigger = 2.5v, meaning that

Vo = 0    when the capacitor voltage on pin #6 is Vc > 5v

Vo = VCC  when the voltage on pin #2 is DCC < 2.5v

The capacitor doesnt start to charge until T/2, when the pin #2 triggers, and the timer puts Vo = Vcc.

When the voltage on the capacitor Vc = 5v for pin #6, the timer puts Vo = 0, and the capacitor discharges through pin #7.


If there´s a constant current of:

 VbaseQ1 = (15*5.6k)/(5.6k + 3.6k)
 VbaseQ1 = 9.1304v

 V330 = 15 - 0.7 - VbaseQ1
 V330 = 5.1696v

 Ie = V330 / 330
 Ic = Ie = 15.6654 mA

Finally the tcharge, with Vc = I*t/c, for Vc = 5v

tcharge = 3.1918ms

If the charging starts at 8.3333ms, the capacitor charges until t = 11.5251ms

  • 1
    \$\begingroup\$ Do you have a question? \$\endgroup\$ – Dave Tweed Oct 26 at 11:56
  • 1
    \$\begingroup\$ If you have additional information related to your original question, use the "edit" button to add the information there. Do not create a duplicate question. \$\endgroup\$ – Dave Tweed Oct 26 at 12:12
  • \$\begingroup\$ Hi, the question is how the circuit works?, im posting how i think i works so i can get some feedback, but there´s no answer. Should i delete both posts and ask a simpler question ? \$\endgroup\$ – Samu R Oct 26 at 12:46
  • \$\begingroup\$ Q1 and those 3 Rs form a constant current circuit. This gives a linear rising voltage on C1 until THR triggers a falling edge on DIS. \$\endgroup\$ – Brian Drummond Oct 26 at 15:07

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