# Why does the voltage across the speaker oscillate in this circuit?

For my final exam, I decided to work on this circuit design available online. Sorry, I forgot the website. No credit is due to me.

When I simulated this over at MultiSim 11, the voltage across the speaker oscillates when the switch is in its OPEN state. I don't know why. I haven't dealt with speakers before, just RLC, diodes, and BJT components.

As a side question, is it possible to use Laplace transforms on BJTs? If yes, can someone point me to a good reference on how to do it?

Speaker resistance is at 8 Ohms. This is the simulation result. Red means Vin. Vout is the orange. • it's web-site is this: extremecircuit.blogspot.com/2010/01/…
– Roh
Oct 9, 2013 at 6:53
• Isn't it meant to do that? Oct 9, 2013 at 7:14
• It oscillates because the output is fed back to the input through C1, and the phase delay of the feedback loop is 360 degrees/ 0 degrees /1 cycle long. What is the question? Oct 9, 2013 at 7:16
• What method of circuit analysis can I use to show that the output is indeed oscillating? That would be the key concern after knowing why it oscillates. Oct 9, 2013 at 7:27

Whatever AC voltage you get on the base of Q1 will cause the dc bias current of Q1 (set by the 330kohm resistor) to rise up and down in sync with that ac voltage. This modulation of dc bias current will be amplified by Q1 and results in a larger undulating current in and out of Q1's collector.

As this current has a path directly through Q2's base the same will happen with Q2. Q2's collector will amplify its base current undulations and force that current through the loudspeaker. This results in a voltage across the speaker but doesn't necessarily mean sustained oscillations. You have to examine the circuit to see if the voltage on the speaker reinforces the original ac voltage on Q1's base.

You do this by looking at the way the transistor current in the collector is affected by the base current. If Q1's base signal rises, collector current also rises and this forces more current through Q2's collector which increases the voltage across the speaker.

This is called positive feedback and results in oscillation.

• Is there a way to show the oscillating output voltage computationally? I mean you don't have to solve it. I just need to know what methods I can use to solve it. My professor wants me to show the oscillation computationally, so I'm having a harder time than anyone else in my class. Oct 9, 2013 at 7:38
• @solitude - the problem is the speaker - it might have serious effects on how Q2's collector current gets turned into a voltage. I'd try simulating it first to see how similar real-life is with sim results. If similar then the frequency of operation is mainly determined by the 330k and the feedback capacitor - if not similar then your model of the speaker is not accurate enough and sorry I can't help you with uncovering speaker models. Oct 9, 2013 at 7:44
• It's fine. I have never faced this kind of problem. There are no computational models for speakers as far as I have researched. I don't think considering the transistors in to small signal models can help? Oct 9, 2013 at 7:59
• Does your school have a simulator - you can learn a lot from that. There are generic models for speakers around on the net but probably won't help you. Oct 9, 2013 at 8:03
• If you're referring to a school-made simulator, then no. But we do have simulators in general. The generic models didn't help. If all else fails, I have to submit the report stating that there were no suitable speaker models available, thus making it unsolvable. Oct 9, 2013 at 8:13

is it possible to use Laplace transforms on BJTs?

Yes, if you linearize the BJT about an operating point, i.e., do the s-domain analysis of the circuit with the BJT replaced by its small-signal model.

However, I don't think that's what you want in this case. This is non-linear, large-signal oscillator and so, you must do a large-signal analysis in the time-domain.

What you essentially have with this circuit is a relaxation oscillator.

Consider the initial condition that C1 is uncharged and Q1 / Q2 are off.

The voltage at the Q2's collector is zero volts and, since C1 is uncharged, the voltage at Q1's base is zero volts which is consistent with Q1 / Q2 off.

The capacitor will charge via current through R1 and, after some time, the voltage on the base of Q1 will increase to the point that Q1 begins to conduct which will, in turn, cause Q2 to conduct.

Now, we have positive feedback since, as Q2 starts to conduct, the voltage at Q2's collector will begin to rise which will add to the voltage at the base of Q1. This will act to further turn on Q1 and Q2 and, in a very short time, Q2 will saturate.

At this point, the Q2's collector voltage is fixed at about 2.8V and the capacitor current reverses; the path being "down" through Q2, through the capacitor, into the base of Q1. This acts to build a voltage across the capacitor that subtracts from the 2.8V at Q2's collector.

After some time, the capacitor voltage increases to the point that Q1 begins to turn off which acts to begin turning off Q2 and thus decreasing the voltage on Q2's collector.

Once again, we have positive feedback since, as voltage on Q2's collector begins to decrease, the voltage on Q1's base is decreased further which accelerates the turning off of Q2 and, in short order, both transistors are off.

Now, C1 can once again begin charging through R1 and the cycle repeats.

Below are plots of the output voltage and the voltage across C1 (positive referenced to Q1's base). Note that when the slope of the voltage is positive, the current is "left to right" through C1 and, when the slope of the voltage is negative, the current is "right to left" through C1. Do you see a way to show this computationally? Hint: There are essentially two different regions of operation: (1) both transistors off and (2) both transistors on (Q2 saturated).

• That looks like an integrator. Do correct me if I'm wrong. We're trained with the most basic components. There's a lack of attention in applying them. Also, using the large-signal model would be a problem. As much as I'd love to use it, I don't have time anymore to study and use it. Oct 10, 2013 at 9:56