I've used following schematics to build blinking LED circuit and it worked.

Blinking LED circuit

I tried to understand how it works qualitatively without solving differential equations. By this I mean understanding voltage dynamics across each element (when and why it goes up or down and whether faster or slower in comparison to previous time interval) and same about current.

Using KVL, KCL and basic properties of circuit elements I come up to the following (time interval from the start and until the LED turns off for the first time):

  • The process starts with maximum current that is split between NPN E-K and PNP E-K branches and therefore source voltage is less then 3V.
  • The source voltage is increasing and approaching 3V value.
  • The current through NPN B-E path is decreasing.
  • Voltage across LED is increasing.

By initial voltage and current values I take those when I replace capacitor with short circuit. Removing capacitor removes dynamics and for me it sounds logical that these values are initial when we put capacitor back.

My question is the following: Is there an approach to understand how each element voltage and current change qualitatively in time so I could draw approximate plots?

I think this is important because solving complex differential equations just give you magic result - it just doesn't let you "feel" the circuit. Also I doubt that the one who created this circuit for the very first time draw dozens of circuits, solved dozens of diff equations and finally found the one that makes LED blinking. I think this person could "feel" the circuit the way I want to.

  • \$\begingroup\$ You should start with the voltage over C1, not the supply voltage. \$\endgroup\$
    – Janka
    Commented Feb 18, 2018 at 18:53
  • 1
    \$\begingroup\$ 1/ Normally schematics are drawn with the positive rail at the top. 2/ You can calculate it but much simpler is to put the circuit in a simulator. \$\endgroup\$
    – Oldfart
    Commented Feb 18, 2018 at 19:08
  • \$\begingroup\$ C1 is backwards too. \$\endgroup\$
    – Trevor_G
    Commented Feb 18, 2018 at 19:18
  • \$\begingroup\$ @Trevor_G measurements show opposite. Capacitor is being charged by current from PNP collector to NPN base. \$\endgroup\$ Commented Feb 18, 2018 at 19:50
  • 1
    \$\begingroup\$ Your schematic is not drawn following standard convention. That makes it tricky to see what's going on, once you are used to looking at schematics. Check this Link \$\endgroup\$
    – Tyler
    Commented Feb 18, 2018 at 20:05

1 Answer 1


this is not an answer .... here is a re-draw of your schematic ... the C1 is in the same orientation as yours


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ The problem with this circuit is that it will never blink. Well, it blinks on and then it stays on. This happens because the \$ U_{AK} \$ of a LED is mostly larger than the threshold voltage for conducting of the base-emitter junction of a BJT. The result will be that upon connecting the battery, \$ C_1 \$ will start charging through \$ L_1 \$, thereby raising it's voltage \$ U_{AK} \$ to above \$ U_{BE_{th}} \$ of \$ Q_1 \$ so that \$ Q_1 \$ immediately starts conducting, pulling \$ Q_2 \$ into conduction. \$\endgroup\$ Commented Nov 27, 2018 at 12:49
  • \$\begingroup\$ Because the base-emitter junction of \$ Q_1 \$ loads \$ R_1 \$ to a level of about \$ 0.7 \$ V, which is probably less than \$ L_1 \$'s \$ U_{AK} \$, the at first sight apparently wrong polarity of \$ C_1 \$ is hereby justified. However, I don't see the circuit blinking. \$\endgroup\$ Commented Nov 27, 2018 at 12:50

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