I have troubles understanding the role of C1 capacitor in the oscillating circuit from the schematic below:

Schematic http://img823.imageshack.us/img823/7683/c1role.jpg

My question is: what is the role of C1 in this circuit?

Is it to ensure that the circuit will not latch? So that oscillation will continue?

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    \$\begingroup\$ When I drew this circuit in Falstad simulator falstad.com/circuit, nothing happened. No oscillation - output remained static. Why? \$\endgroup\$ – dext0rb Oct 30 '12 at 19:07
  • \$\begingroup\$ We expect oscillation at the resonant frequency at which the positive feedback is at 0 degrees (in phase) with the input. For oscillation to occur, the gain has to be at least 1 at that frequency. \$\endgroup\$ – Kaz Oct 30 '12 at 20:39
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    \$\begingroup\$ I still think the 10u cap has wrong polarization. The cathode of the cap can be 700mV at best, that means to polarize the cap correctly, the current in th 22R resistor at average should be 30mA. Didn't (virtually) build the circuit though to verify, but my feeling says it is wrong. \$\endgroup\$ – jippie Oct 30 '12 at 21:23

it's a speed up cap. It allows the edge through faster to allow the base drive to increase faster.

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    \$\begingroup\$ Could you give me a little more details please?Why is that speedup needed? I have tried this on a breadboard and it seems that with certain LEDs the circuit will not oscillate. It will latch to On. But if I change the value of C1 it will start oscillating again. \$\endgroup\$ – Beginner Oct 30 '12 at 18:12

Note: I somehow misread the question and wrote this answer for the bottom 10 µF capacitor. I'll leave it here as extra information about the circuit even though it doesn't pertain directly to the question asked:

The [bottom 10 µF] cap provides in-phase AC feedback, which is fundamental to a oscillator. Without the cap, you just have a amplifier from the base of the NPN to the collector of the PNP. Without the cap, it would find it's steady state level and stay there.

The cap provides positive feedback, but not at DC. When the PNP turns on a little more, its collector voltage goes up. Thru the cap, this put more current into the base of the NPN, which sinks more current thru the base of the PNP, which turns it on more, which makes its collector voltage go even higher. If the cap provided a connection all the way down to DC, this process would cause the circuit to latch. However, once the circuit starts to stabalize in what would be it's latched state, the capacitor starts to become a open circuit. The circuit will then relax, which now causes the collector of the PNP to drop, which drops the base of the NPN thru the capacitor, which then keeps feeding on itself to approach latching in the off state. Once again, the capacitor starts to act like a open as steady state is approached, the circuit goes a little the other way, the feedback via the capacitor amplifies this effect, and everything repeats.

AC feedback at a gain greater than 1 is fundamental to oscillation.

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  • \$\begingroup\$ Are you talking about C1 or the other capacitor? \$\endgroup\$ – Dave Tweed Oct 30 '12 at 20:01
  • \$\begingroup\$ @OlinLathrop I am a little confused about the "AC feedback". How does that work in this case? The circuit is powered from DC not AC. \$\endgroup\$ – Beginner Oct 30 '12 at 21:01
  • \$\begingroup\$ @Dave: The other capacitor. Oh, I just noticed that the OP asked about C1. Somehow I read that as the bottom capacitor in the diagram. Oh well, this answer explains the purpose of the bottom capacitor then. \$\endgroup\$ – Olin Lathrop Oct 30 '12 at 21:49
  • \$\begingroup\$ @OlinLathrop Could you please explain how can there be an AC feedback when the circuit is powered from DC? I am a beginner and I am confused about this. Thank you. \$\endgroup\$ – Beginner Oct 31 '12 at 8:40
  • \$\begingroup\$ @OlinLathrop Or could you please point me to the right direction from where I could learn more about this? Thank you. \$\endgroup\$ – Beginner Oct 31 '12 at 18:52

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