I was just designing this Common emitter transistor amplifier this morning.


simulate this circuit – Schematic created using CircuitLab

I have removed the input in the circuit intentionally. Here are a few words about the circuit:

  1. I constructed initially with only one transistor (BC107) and tested by giving audio signal as input. The output from the 8 ohm speaker was pretty much distorted, which I think was due to impedance mismatch.

  2. In order to overcome the distortion, I added a second stage using second transistor (2N3904). The second stage is a buffer circuit (Common collector mode) and it has been directly coupled to the first stage. I've tried the direct coupling because I studied it in my recent classes and want to test it practically.

Now, when I switched on the battery to give supply voltage, and connected the speaker at the output side, (without applying any input signal on the input side) I got a high frequency, audible static noise from the speaker like this:


When I changed the capacitor and tried out with values of 4.7uF, 10uF, 22uF and 100 uF, I found the frequency to be decreasing with increase in capacitance. I measured the sound frequency using an app in my cell phone (approximate frequency).

Now here are my questions: Why does this circuit produce such a noise? Can this circuit be further simplified in order to produce the same output. Also I've got the idea of developing this circuit as a musical instrument by adding several capacitor at the output so that, when I short the speaker to a capacitor manually it would produce a sound and then when I short it to another capacitor, it would produce a sound of another frequency. Given that I have the knowledge of the frequency of various musical notes, How do I calculate the capacitor values theoretically? Please provide a formula.

  • \$\begingroup\$ Generally "static" means "white noise", which that noise is not. \$\endgroup\$
    – user253751
    Feb 7, 2016 at 9:41

3 Answers 3


Some comments: I constructed initially with only one transistor (BC107) and tested by giving audio signal as input. The output from the 8 ohm speaker was pretty much distorted, which I think was due to impedance mismatch. This is NOT caused by impedance mismatch. An 8 ohm speaker is a very low impedance, one tiny BC107 will have BIG trouble to drive it. Not at low volumes but at even a moderate volume, a BC107 wil not cut it. What happens when the BC107 reaches its limits ? I starts to clip meaning it cannot provide more signal to the speaker. This clipping you will hear as distortion.

Concerning impedance mismatch: You do not need impedance matching to drive a speaker. 99.99% of audio amplifiers have a very low output impedance. When loaded with an 8 ohm load that means there is actually a large impedance mismatch ! But that is OK since what we want is to put a voltage across the speaker. You usually only need impedance matching for high-frequency signals like more than 10 MHz or so OR for transporting audio signals over long a distance (phonelines).

Now you have oscillations and I think that is because you increased the gain of your amplifier (since you added an extra stage). I hope you agree that the base of Q1 is the input of your amplifier. Note how the voltage at this input is just a divided down version of the supply voltage. If the supply voltage varies, the voltage at the input will also vary !

Now if there is a signal going to the speaker, the second stage draws current. I guess you're feeding this amplifier with a battery so the supply voltage will drop a little when there is current being drawn. This small drop in supply voltage will also appear at the input and it will be amplified and fed to the speaker. Then the speaker will draw some current. See the there is a loop here ? For an oscillator you need a loop. You build one.

I suggest to decrease the gain, you can do that not by changing the value of Cb but by placing a resistor in series with Cb. A value of 180 ohms is a good starting point.

  • \$\begingroup\$ I'd suggest testing with a bench-top power supply. Easy to see the difference and also the power required to run this circuit. Especially since you mention school. Schools have those things. \$\endgroup\$
    – Dave
    Feb 7, 2016 at 12:19

Looks like you forgot the 100uF (or so) decoupling capacitor across the power supply, so you built an oscillator.


Your 9 V battery has significant impedance. When the output stage draws current, the battery voltage goes down, which feeds back into the first stage, causing the whole thing to oscillate.

The way to fix this is to put a decent size capacitor across the battery. That lowers the impedance of the power supply, hopefully to the point where the remaining voltage drop due to output stage load doesn't feed enough into the input stage to oscillate.

It is also good practise to filter the supply to sensitive parts separately. Your first stage is such a sensitive part.

Another way this is dealt with is negative feedback. Make a bit more gain than you really need, then put negative feedback around the whole circuit to get a more controlled and flat gain. This will also make the device more immune to power supply or other noise that is injected after the input stage.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.