Common-Emitter Behavior

Question

I built circuit shown bellow (NPN BJT):

^{simulate this circuit – Schematic created using CircuitLab}

Everything worked normal as it should (except of unwanted distortion). There is that one strange thing I'm seeing lately when I connect oscilloscope's probe between emitter and ground. It is the amplitude of the signal measured - when the amplitude of F.G. is some mV, the emitter amplitude more or less follows the input amplitude measured between base and ground; things change when the input voltage is increased for few 10mV, where the emitter's amplitude looks like this:

Eventually proceeding to more sharp shape of peak.

Even stranger thing is that, after those triangular waves are shaped, the voltage simply stops at this point as the input voltage is increased.

• Does this behavior has to do anything with almost constant Vbe which is approx. 0.7V?

• Or is this another unwanted distortion produced by transistor?

Answer 1

It's a bit hard to say, given that we can't see any of the scope settings (vertical or horizontal).

But the waveform you're showing is a characteristic of how the emitter bypass capacitor gets charged and discharged. The charging impedance (through the transistor) is much lower than the discharging impedance (through the resistor). As you drive the base harder, the charging current goes up (faster rise time), but the discharging current cannot change (same fall time).

Bit what this really indicates is that the emitter bypass capacitor is too small for the signal frequency — the voltage should not be changing significantly at all. Either pick a higher signal frequency or use a larger bypass capacitor.

Answer 2

Impedance of 10uF at 1kHz is about 16 Ohms , and emitter can only pull up, while emitter resistor pulls down. You do the math and scrap this circuit.

1st design by specs with gain, input & output impedance and include your load. 2nd choose a topology that supports these requirements. Then after your analysis , ask yourself a better question.

Answer 3

@Bimpelrekkie Why would it? because at one point voltage of emitter just stops increasing as the input voltage on the base is increased - don't you find this a bit strange? – Keno

Wrong DC operating point. Look:

R1 to R2 sets (16.9/84.6)*12=2.4V on the base, or approx 1.8V on emiter. So emiter current would be approx 1.8V / 300 Ohm = 6 mA 6 mA thru 1.5k collector resistor would set 9V over it.

That's only 12V-1.8V-9V=1,2V constant voltage between collector and emiter. Select the correct ratio between the collector and emiter resistors to set the DC operating point in the middle between 12V and 1.8V. (That is approx 5v over collector-emiter)