Hi everybody,
As promised, here are some measurements on the circuit :
The first image is the original output at the collector of T2 (Point E) :
This doesn’t quite resemble a sine wave, isn’t it ?
So, we need to bias T2 properly, so these are some calculations :
To act like a “perfect” amplifier, T2 should divide the battery voltage. Let’s say about 5V. This voltage should be across the E and the C. To be safe, we assume that the DC current (without signal) flowing through E and C is about 20mA max.
So, 5V / 20mA = 250 Ohm. The resistance of the transformer is 30Ohm, thus 250Ohm – 30Ohm = 220Ohm
Therefore we take a 220 Ohm resistor and place it into the Emitter circuit (if we would put it in series with the transformer, we could get a loss of amplified signal across the resistor)
The Base current Ib = Ic / Hfe. So Ib = 20mA / 180 = 0,1mA.
To achieve this, we use a resistor divider. Now comes the cool part. We use the volume potentiometer of 10K for volume but also for bias adjustment.
So, when turned at max. the volume is the loudest and the current through the E and C of T2 = 20mA.
If we turn it down, less signal is fed to the base, but also less bias. This means that when the volume is at zero, our transistor T2 is completely cut off, which means that no current flows through the transformer, saving the battery.
The resistor divider gives this 0,1mA. Now it’s up to you to calculate Ib !
Please note down your calculations in a comment.
Now, for some oscilloscope pictures :
The picture at the left shows a 60° phase shift in the oscillator between A and B.
The picture at the right shows a 180° phase shift in the oscillator between A and C. This phase shift causes the circuit to oscillate. We call this type of oscillator a phase shift oscillator.
Here we see the input (D) of amplifier T2 and it’s output (E).
Note how the sine wave is now correct.
Greetings