I am a starting hobbyist and just built this circuit for testing:

http://circuit-diagram.hqew.net/Simple-Tone-Oscillator-Generator-by-2N2222_10535.html

but I don't get how the cap is discharging each cycle. If someone could point out & explain my thinking error I'd appreciate it:

1-9V pushes holes to top of the PNP while at the same time pushing holes through both resistors to push holes to the NPN gate while also pushing holes to the + side of the capacitor

2- ground pushes electrons to - side of capacitor, while pushing electrons up through NPN emitter, which triggers(amplified flow) through the NPN then the PNP, which triggers holes to flow down from PNP emitter to collector to (speaker)

So once the cap is charged, the flow goes through the transistors, eventually flowing straight through the PNP to speaker to ground. I don't get at what point/how the cap empties and then recharges again (creating the oscillation)

Would appreciate any help on this -



ps. I used very loose wording ('holes' 'pushes' etc) rather than exact terms i.e. just looking for a conceptual answer if possible

 

Here is the schematic image;

I understand your point, it looks like once the two transistors turn fully on the speaker will have almost 9v volts across it and everything will just sit there.

(If the circuit works) then I think the key to oscillation will be the speaker mechanical and magnetic properties, and the very effective coupling of the speaker to the base of Q1 through the 22uF cap.

So it turns on, speaker cone moves against its spring, the voltage rises, but due to mechanical issues and speaker coil inductance the current rises and that can cause a drop in the voltage on the speaker. It only takes a tiny drop there to start to cause Q1 to turn off (via the cap) and the positive feedback through that cap is huge, so any mild turning-off effect in Q1 will become a massive turning-off effect.

 

Thanks for the reply. So from what you said, here is how I understand the flow now:

When Q2 opens up, all the holes now see an (easier/less-resistive) path to ground through Q2, so they 'stop' (or slow?) going through the (resistors/Q1) paths. The holes sitting on the cap then see an available pathway to ground via Q1 (no longer supplied by holes from the source), so they discharge from the cap, go through Q1, to ground, thus emptying the cap.

Once the cap is empty, Q1 now has no holes coming from either (the source or the cap), so it shuts off, thereby shutting off Q2. Holes from the source now see no easy path to ground, so they start going through the resistors / charging the cap again etc.

So increasing cap size would change the tone pitch/frequency because there would be more holes flowing from it through Q1 during discharge, thus keeping Q2 open a bit longer, thus slowing the speaker cone up/down rate.

If the above logic is correct, I have two questions:

-Why not then just have one 60k resistor?
-Why do we need 2 transistors, i.e. why doesn't one NPN accomplish the same thing?

(I have tried both the above and neither adjustment generates any tones but I don't understand why not)




p.s. yes the circuit works/I have it built here with elenco items

 

Thru the base of the 2n2222 transistor. When voltage across cap reaches transistor B-E forward bias of ~ 0.6 volts there is a discharge path to ground.

 

Some points to note:

1) 22mF is not the same as 22μF.

22mF is 0.022F
22μF is 0.000022F

2) Don't try to analyze circuits using both holes and electrons. It will only add to confusion. You can use electrons alone or conventional current alone.

3) Do not refer to transistors as a valve that "opens".
"open" in electrical terms suggests that no current flows.

Why not then just have one 60k resistor?

You can use a single fixed resistor. The variable resistor allows you to adjust the frequency of oscillation.

Why do we need 2 transistors, i.e. why doesn't one NPN accomplish the same thing?

The dual NPN/PNP transistor circuit is a classic oscillator.
As the capacitor charges through the resistor, the voltage at the base of the NPN transistor gradually increases until the NPN transistor conducts which also causes the PNP transistor to conduct. The capacitor dumps its charge through the NPN transistor and the cycle repeats.

 

Noted on the language advice, thanks. I think I get it now. Its the voltage buildup on Q1 base that triggers the cap dump. Once dumped, Q1/Q2 go off, triggering flow back through the resistors/to the cap to charge again etc. This also explains why too-small caps were not working: they couldn't build up enough voltage against Q1 base to trigger it.

But I still don't quite see the need for the PNP. The cap dumps through the NPN; the NPN can conduct (while triggered) directly from the source on down through the speaker. Why is the PNP needed? just for double-amplification?

 

... I think I get it now. Its the voltage buildup on Q1 base that triggers the cap dump. Once dumped, Q1/Q2 go off, triggering flow back through the resistors/to the cap to charge again etc.
...

Not exactly. It's the voltage rise on the speaker that pushes current through the cap into the NPN base, causing it to turn on harder (and stay on) while the speaker voltage is rising. That's called positive feedback.

And the turn off mechanism is not the cap "dumping" but occurs when the speaker voltage stops rising, at that point there is no current pushed through the cap and any noise or mechanical/magnetic issue of the speaker causes that voltage to drop even the tiniest amount causes some current to be pulled back through the cap, causing some turning off of the NPN. And once the NPN starts to turn off (even the tiniest amount) the speaker voltage drops and then the NPN is forced to turn off hard. Positive feedback again (but in the other direction).

...
But I still don't quite see the need for the PNP. The cap dumps through the NPN; the NPN can conduct (while triggered) directly from the source on down through the speaker. Why is the PNP needed? just for double-amplification?

No. The NPN emitter is to ground. If you connect the emitter to the speaker then it cannot work because the speaker will not be at a higher voltage than the base (which is needed to force current through the cap into the base as speaker voltage rises).

So it needs two transistors.

 

so then if I swapped the speaker for say an equivalent-load light-bulb, it wouldn't flash on/off right? (i.e. the circuit probably wouldn't work as it is dependent on the physics of the speaker to create the on/off triggers)

 

I can't say for sure but I doubt it would work.

The turn off phase seems to require something that lowers the voltage on the - pin of the cap to initiate the turnoff. Speaker inductance/mechanical issues etc can do that.

Try a light bulb and see.

 

I guess I don't understand as much as I thought. From everything I read online this dual-transistor concept was a general-oscillation concept, i.e. people use it to create pulse-counters, alarm triggers, all sorts of things. I was very interested interested in finding the simplest oscillator-creator I could build, this example just happened to have a speaker as the load. But now it seems the speaker itself is the key.

Could this circuit be modified (with speaker removed) to just create timed output pulses of say a few volts every second? i.e. I'd like to built the world's simplest oscillator so I can nail the concept down.

 

This is not correct.The circuit will oscillate with the speaker replaced with a resistive load or an LED and series resistor. Try 100Ω for starters.

You can also build a simple oscillator circuit using a UJT (unijunction transistor).

 

I wonder if the turn off mechanism is related to something more fundamental than random excursions of the load voltage. Perhaps the reverse recovery current in the NPN base-emitter junction capacitance...?

 

No, it's more likely the beta of the NPN.

Turn off is the weakest part of this circuit which explains why it has issues with small cap size and load type.

If the OP wants a proper reliable two transistor oscillator then a 2-tran multivibrator is a much better choice. Or a hysteretic oscillator with two transistors as a schmidt trigger and some posiitve feedback through a cap. Both of those oscillator types should have reliable turnon and turnoff characteristics.

 

I simulated this in Ltspice and it seems to be a pulse oscillator.
On/off ratio of the output pulse is about 0.6%
The pulsed base current of the PNP is about 700mA and should be limited by a resistor.
As RB said there are better oscillator circuits.

 

A search for simple found this.

Some may not work.

http://members.shaw.ca/roma/twenty-three.html

Interesting none the less.

 

My component selection is limited, so I swapped the 50kVar & 10k for one 100k resistor, changed input V to 6v, changed the cap to 10uf, and replaced the speaker with a 2.5V/.3A lightbulb.

The circuit successfully flashes the bulb about once per second. Changing the resistor and/or capacitor changes the bulb flash speed, though already I notice there is a very small range of cap/res combinations that trigger flashing; all others either keep the light on or off. Also I note the flash speed is not fully steady, sometimes it randomly sort of speeds up for a second then slows down again, then resumes steady. Strange.

Thanks for everybody's posts. I will be going through all the examples/links to try them if possible. (I only have 1 NPN and 1 PNP right available right now, hoping to later add some of those (other parts/transistors) mentioned)

 

I wonder if the turn off mechanism is related to something more fundamental than random excursions of the load voltage. Perhaps the reverse recovery current in the NPN base-emitter junction capacitance...?

The positive feedback and charged capacitor will turn-off the transistors.
To make this circuit to work as a multivibrator we need to select 10K+Pot so that the base current is so small that immediately after C1 is fully charged the transistors comes out from saturation into the linear region.
And this mean that voltage at pnp collector start to drop. And this drop is feedback into NPN transistor base. So the NPN base voltage also start to drop the magic of a positive feedback comes to play. And eventually capacitor that was previously charged will "pulls-down" the voltage on the NPN base below ground.


Short circuit description
The speaker inductance has no influence on how this classics multivibrator work.
When we first connect this circuit into supply. The C1 capacitor is initial discharge.
So the voltage across capacitor is 0V (capacitor act just like a short circuit).
Immediately after we power up the circuit. The current start to flow via R1 and R2 resistor. But because C1 capacitor is discharge (0V across capacitor) all this current will flow through C1 and speaker to ground. No current will flow into Q1 base and this is why Q1 and Q2 or OFF.
But as you know voltage at charging capacitor start to rising. And when this voltage at pin X reach Q1 threshold voltage (about 0.6V) Q1 start to conduct a current. If Q1 start to conduct the Q2 all so immediately will start conduct current. This means the voltage on Q2 collector starts to rise (at point Y). And this rise in Y voltage is feedback to point X via capacitor C1.
But as you know the voltage at Q1 base (point X) can not be higher then 0.7V because of the diode in the base-emitter junction Q1. This means that there will be a huge Q1 base current. Which will ensure that Q2 is in saturation region, and the voltage at point Y is close to 9V.
Also at the same time C1 will quickly discharge C1 and then C1 will start recharged phase in the opposite polarity. All this will happens very quickly because of a large Q1 base current (charging current). Also at this phase the spearer "gives" a shot pulse of sound.

The C1 was rapidly charged to V_C1 = Vcc - Vbe - Vce(sat) ≈ 8.6V So cap no longer conducts any current. So when C1 capacitor is full charged. The Q1 base current is provided only by R1 + R2 resistor. R1 + R2 don't provide enough base current on Q1 to keep Q2 in saturation region.
So Q2 immediately after C1 is full charged comes out from saturation to the linear region.
So voltage at Q2 collector is start to fall. And this drop in Y voltage again is feedback via C1.
For example if voltage at point Y drops to 8.9V the voltage at Q1 base will also drop to 8.8-8.4V = 0.5V thanks to the positive feedback provided by C1 capacitor.
So eventually C1 that was previously charged to 8.4V "pulls-down" the voltage on the Q1 below ground (-8.4V). So Q1 and Q2 will immediately goes into cut-off.
And now C1 cap will act very similarity to a voltage source and start the discharge phase in the circuit:
C1 right plate ---> speaker---> power supply ----> R1 + R1---> C1 "left" plate.
Eventually cap will be full discharge (0V across the cap) but the current will still flow in the same direction and capacitor will start the charging phase, but I described this at the beginning.
So the cap will charge to 0.6V ( left plate more positive than the right plate) and Turn ON Q1 and Q2 . Also notice that in this phase the speaker current is to small to produce any "loud sound" on the speaker.
That is one complete cycle.
So as you can see Q1 and Q2 are ON for the "very short" time ( C1 charging phase).
And they are OFF for "long time" (C1 discharging phase). And the speaker play the sound only when Q1 and Q2 are ON.

http://forum.allaboutcircuits.com/showthread.php?p=501235#post501235
http://www.talkingelectronics.com/p...er/TheTransistorAmplifier-P1.html#OSCILLATORS

 

fantastic explanation Jony130 ty

 

...
Short circuit description
The speaker inductance has no influence on how this classics multivibrator work.
...

It is NOT a multivibrator, they use two transistors in a configuration where Q1 is on and turns Q2 off. Then Q2 turns on and it turns Q1 off.

A multivibrator might even have three or more transistors, each vibrating in turn like a ring counter. That's where the "multi" comes from. There is multiple vibrating.

This circuit is not a multivibrator because both its transistors turn on together. It's just a vibrator. Or an oscillator.

...
The C1 was rapidly charged to V_C1 = Vcc - Vbe - Vce(sat) ˜ 8.6V So cap no longer conducts any current. So when C1 capacitor is full charged. The Q1 base current is provided only by R1 + R2 resistor. R1 + R2 don't provide enough base current on Q1 to keep Q2 in saturation region.
...

That's close but it's not the whole story. What you've said is that the beta of the transistor is the cause of turnoff (as I said in post #13).

But the speaker inductance is large, so the speaker voltage will rise very quickly but it's current slowly builds up over time. That current rise (when high enough) has a dramatic effect on the transistor saturation, as does the speaker magnetic/mechanical effects like resonance so those two things play a large part in the oscillation. That also means the circuit works better with a speaker than it does with a resistive load.

 

It is NOT a multivibrator, they use two transistors in a configuration where Q1 is on and turns Q2 off. Then Q2 turns on and it turns Q1 off.

A multivibrator might even have three or more transistors, each vibrating in turn like a ring counter. That's where the "multi" comes from. There is multiple vibrating.

This circuit is not a multivibrator because both its transistors turn on together. It's just a vibrator. Or an oscillator.

Yes maybe you are right, This circuit is astable oscillator or half-multivibrator .

That's close but it's not the whole story. What you've said is that the beta of the transistor is the cause of turnoff (as I said in post #13).
But the speaker inductance is large, so the speaker voltage will rise very quickly but it's current slowly builds up over time. That current rise (when high enough) has a dramatic effect on the transistor saturation, as does the speaker magnetic/mechanical effects like resonance so those two things play a large part in the oscillation. That also means the circuit works better with a speaker than it does with a resistive load.

But this dos not change the fact that this circuit will work even if you replace the speaker with a resistor. If you choose resistor wisely this circuit will work without the speaker.
And the similar principle as I described in post 17 is use in this oscillator and his improve version.

And the the effect the coil inductance has been discussed here in this thread
http://www.electro-tech-online.com/...current-or-voltage.109630/page-19#post-903396