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I'm just starting out learning about electronics. I have a project in mind which is really just a variation on this circuit:

schematic

I've built this on a breadboard, used different values for R1 and C1 and observed change in frequency of the LED flash.

My question is about how the circuit works.

Is it the case that when the power is turned on, C1 starts charging and 'prevents' current from flowing to the base of Q1 until it is fully charged, at which point Q1 is 'turned on', causing Q2 to also be turned on, which then allows C1 to discharge?

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    \$\begingroup\$ Yup, that's about it. The LED produces a short pulse of light and the process restarts. \$\endgroup\$ – Andy aka Nov 30 '20 at 12:38
  • \$\begingroup\$ Thanks - so it's just a fundamental 'thing' about C1 charging that makes the current not flow to Q1? \$\endgroup\$ – bins Nov 30 '20 at 12:53
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    \$\begingroup\$ The voltage between the base and emitter \$V_{BE}\$ needs to reach about 0.6V before significant current will flow into the base. R1, C1, and R2 divide the supply voltage. When C1, is uncharged, the voltage across R2 is 3V*47/(100k+47) = 1.4mV. Not enough to cause base current. \$\endgroup\$ – Math Keeps Me Busy Nov 30 '20 at 13:22
  • \$\begingroup\$ Try read this electronics.stackexchange.com/questions/338128/… or this electronics.stackexchange.com/questions/261288/… \$\endgroup\$ – G36 Nov 30 '20 at 14:26
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    \$\begingroup\$ Not "until it's fully charged" but "until it's charged enough (0.5V or so) to start turning Q1 on". Otherwise, that's about right. \$\endgroup\$ – Brian Drummond Nov 30 '20 at 14:44

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