Hello,

I've been looking at this for awhile, and, while I get the general idea behind the circuit, I can't quite break it down into what happens, step by step. If anyone could break this down for me, Barney-style (but still accurately), I'd really appreciate it. Thanks!

 

It has been covered here.

Here is a recap for this version.

On initial power on, capacitor C has no charge. C is pulled up to Vcc (+4.5V in your case). C starts charging, i.e. the left side starts to fall from Vcc towards GND.
When the base of the PNP transistor reaches Vcc - 0.7V, the PNP transistor is turned on. This turns on the NPN transistor also.

The collector on the NPN is suddenly pulled towards GND pulling C with it.
The base of the PNP goes negative. The base-emitter junction of the PNP is turned on hard and C rises back up towards Vcc.

When the base of PNP rises above Vcc - 0.6V, the PNP is turned off and so does the NPN. This pushes C back up to Vcc and the cycle repeats.

 

Or try read something here
http://www.talkingelectronics.com/p...er/TheTransistorAmplifier-P1.html#OSCILLATORS
http://electronics.stackexchange.co...npn-pnp-flasher-circuit-working/261407#261407

 

I think there is a problem with this circuit. When the transistors are switched on, there is nothing to limit the current flowing through the PNP and the base emitter of the NPN. It needs a resistor between the PNP collector and the NPN base, perhaps 1K.

 

Here is the same circuit using npn or pnp either way round.....
The frequency is set by the 330K and 10uF.


https://www.google.co.uk/search?q=p...AUIBygB&biw=360&bih=512#imgrc=qmJ1OjOOJNjJDM:

 

If anyone could break this down for me

Think of that 1k/1uf network as being alternately switched between the ground (when the NPN is conducting) and then out of the circuit (when the NPN is not conducting.

The circuit works this way:

1) when the NPN is conducting, the 1uf is being charged up through the PNP's base current -> the left side of the 1uf capacitor has a potential higher than its right side. The led is on.
1.5) as the 1uf capacitor continues to be charged up, the PNP's base potential goes up towards the rail, eventually shutting it off. When this happens, the capacitor's right side is connected to the 110R+LED, and begins to discharged through 1K + 1Meg+100K pot resistors -> the PNP's base potential goes beyond the 5v rail in this case, approaching 2*(5v - 0.7v).
2) as the 1uf capacitor is being discharged, the NPN's base potential goes down and once it is below 5v-0.7v, the PNP starts to conduct and you go back to 1).

Once that's clear, you get back to the conditions under which this circuit will work:

1) the NPN has very low leakage current;
2) the 1Meg + 100K should be sufficiently high;
3) low beta transistors work best here, because of 2).

 

Thank you all for your answers. Dannyf, that was exactly the kind of breakdown I was looking for. Thank you! I am a little unclear about the last three points you mentioned:
1) the NPN has very low leakage current;
2) the 1Meg + 100K should be sufficiently high;
3) low beta transistors work best here, because of 2).

2) is clear enough to me, because it is necessary that the capacitor take some time to discharge, but I'm not sure I get 1) and 3).

Thanks again.

 

The reason that the circuit can flip on and off is that the rc network is switched alternately to ground and rail. If the leakage current of that npn is high, it will never be switched to the rail. And the circuit will never flip states. No flashing of the led.

The 1meg resistors value is critical to the operation of this circuit. It has to be sufficient high so that the 1uf capacitor discharges slowly (led off time). You may find that you need to experiment it a little. For a high beta PNP, that 1meg resistor needs to be of very high value, potentially in excess of 10meg. A low beta transit or makes that value smaller.

 

Thank you!

 

Hello,

I've been looking at this for awhile, and, while I get the general idea behind the circuit, I can't quite break it down into what happens, step by step. If anyone could break this down for me, Barney-style (but still accurately), I'd really appreciate it. Thanks!
View attachment 115683

Same circuit simplified. 22 uF and 150K resistor gives about 1 bright, quick pulse per second. Timing is roughly R x C x 0.33. Yes, no current limiting through transistors or the LED.