# Help with oscillator circuit

Discussion in 'General Electronics Chat' started by Ack, Oct 4, 2008.

1. ### Ack Thread Starter New Member

Sep 30, 2008
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I have attached a circuit that I am having trouble understanding. Here is what I think about the circuit so far. C1 charges through R1 and R2. Q1 is active for a short time until C1 is charged. Q1 takes Q2's base to ground activating Q2 since it is a PNP. The speaker is active during this time. I do not understand what / how discharges C1. Does Q2 discharge it when it activates, thus discharging back through R1 and R2? This circuit is out of a kit and does not give any real explanation to its operation. Any explanation or corrections to my thoughts would be greatly appreciated.

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2. ### scubasteve_911 Senior Member

Dec 27, 2007
1,202
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Your thoughts seem a lot like mine when looking at this circuit. I believe the 1.1M resistance and the 10nF capacitor play a balancing act with the transistor. As soon as Vbe is exceeded, the transistor turns on, then a lot of current goes into the base to support the load. I think this would increase the demand for base current beyond what the 1.1M resistance is supplying, effectively shutting of Q1 until it is finally charged again.

I think the only way to really understand is through actual mathematical analysis, since the currents are being balanced in a way that isn't intuitive.

Steve

3. ### hgmjr Moderator

Jan 28, 2005
9,030
214
The circuit you are using is similar to the one in this document from Forrest Mims.

Figure 1.6 in this document is basically what I think you were trying to build. There is one major difference you will notice. In your circuit, the feedback capacitor is fed from the collector of the PNP whereas in the circuit in Figure 1.6, the feedback capacitor is connected to the junction of the 100 ohm resistor and the speaker.

I would recommend you try making your circuit agree with the one in Figure 1.6. That way the inductance of the speaker will play the roll of a frequency dependent element in the circuit.

hgmjr

4. ### happyfpga New Member

Oct 4, 2008
7
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I think it works in this way

1. C1 is charged through R1+R2
2. Q1 base reach to 0.7V, Q1 is turned on, then Q2 is turned on
3. Once Q2 is on, the collector of Q2 will be around 4V, assume Vce(sat) = 0.3V
4. C1 will be charged pump, Q1 base will be 4+0.7 = 4.7
5. C1 begin to discharged through b-e of Q1. Since C1's another end is kept around 4, there is no charging from R1+R2.
6. The discharging is very fast, Q1 is shut down, Q2 is off
7. Then changing from R1+R2 again.
8. Speaker has some kind of AC, playing a tone.

5. ### Audioguru New Member

Dec 20, 2007
9,411
896
How can the base voltage of Q1 exceed +0.7V when its emitter is at 0V??????

The charge in C1 discharges at a very high current at first into the base of Q1.

6. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
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Notice that when Q1 turns on, there is a path from the power supply, through the base emitter junction of Q2, and through the on resistance of Q1 to ground, without anything to limit the current. If powered by a dry cell battery, there might not be sufficient current to hurt the tranistors, but a stiff supply might just damage them.

I tried simulating the circuit from the Forrest Mims notebook. I couldn't get it to oscillate at first, but I discovered a few things that make it work. For the speaker, I used 500 uH inductance in series with 5 ohms. I increased the resistor R2 from 100 ohms to 200 ohms. And, I added 250 ohms in series with the collector of Q2; this really helped.

Notice that when the oscillator is running, a reverse bias is applied to the base emitter junction of Q1, sufficient to break it down in zener mode. Some simulators don't have a defined reverse breakdown voltage in their transistor models; mine didn't, and the oscillator behaves very differently without it. I added a pair of 5.6 volt zeners in series from the base of Q1 to ground. You need two zeners because if you only have one its forward characteristic will shunt the base current; the second zener can also be an ordinary rectifier diode such as 1N4004.

This reverse bias is due to the inductance of the speaker. If I leave out the inductance in the load, the oscillator won't work.