Electronic – How to solve BJT amplifier clipping a signal

amplifieraudiosaturationspeakerstransistors

I'm trying to make an audio amplifier for an 8 ohm speaker. I'm new in electronics area and having a problem with it.
I already had other issues before arriving where I am, like the need of a high power stage because the low impedance of the load.

To the high power stage my tentative after some reading was to use a transformer at the transistor collector to match the low impedance of the speaker with the high impedance of the amplifier output.

The first problem were, if I use an 9V DC source for biasing the transistor, the voltage at transformer primary will float at maximum in 0 to 9V (In an extreme case). So, if the speaker can work at a maximum peak to peak voltage of for example 6V, the ratio n1/n2 should be at the maximum 1.5, because the peak voltage at secondary will be 9/1.5 = 6V. Is that a right reasoning?

Going ahead, if I use a transformer with 1.5:1, the impedance seen in the primary will be 8*(1.5^2) = 18 ohm.
So the configuration of this stage of the amplifier would be like this:

Schematic showing a BJT and a transformer on the collector

The transistor is biased with 4.5V at emitter resistor, 4.5V at collector-emitter (Vce), and the voltage at the primary is almost 0V for DC.
The capacitor at emitter has low impedance and it's almost a short circuit for AC.
So, in my mind, the voltages will vary from 0V in the primary and 9V at Vce, to 9V in primary and 0V at Vce (in an extreme case).

But when I run it, these waveforms appears. Yellow curve represents the input voltage (250 mV peak), and purple is the voltage at the primary of the transformer (not the output). It have a positive peak of 130mV, but at the bottom seems not to be what I expected. The curve is being cut out at -20mV. It doesn't should saturate only in a relative high voltage, next to the 9V of CC source?

What is wrong and how can I solve this?

EDIT:

I have changed the bias of the transistor for a collector current of 50mA, with a base resistor of 27k instead of 1.35M. Also removed the emitter capacitor (which really reduced the distortion) and changed the emitter resistor to 10 ohm. The clipping has went away for these low input voltage (250 Vpk), and the voltage at primary is apparently a perfect senoid, with amplitude a little higher than the input voltage. But if I increase the input voltage to 3V peak, the clipping reappears at the bottom. I still can't figure out why this happens.

EDIT 2:

I'm adding the image of the waves for the last update:

the clipping voltage has increased, but it is still happening. It starts to appear at about 1V peak at the input.

And I don't know if I have made it clear but this is the end stage of my amplifier, with a low gain, only for drive the speaker. I'm making the stages separately for easy understanding of each part. As this stage has a high input impedance I will use a common emitter before for high gain of voltage.

And sorry for my bad english 🙁

EDIT 3 (solved):

The problem was the transistor must be biased with a collector DC current of at least the same value that the peak to peak current which load requires.
Other thing I noticed is that the BC548 I was using fails with a relative big current (>200mA) even in the simulation, the base-emitter voltage increases significantly, also making the wave to clip.

Best Answer

Second answer: getting back to basics

  1. You cannot analyze anything completely, unless all impedances are known.
    • Learn what impedance is Z(f,dc), how to measure and compute it for every component and looking in & out of any branch
  2. This is just a simulation, BUT what is real and what is ideal and what is time scale steady state or ? e.g. is LdI/dt steady state?
    • e.g what is transformer Inductance, hFE of Q1, What is Z(f)?
  3. What is the basic configuration "common emitter" What does ideal common emitter do?

    • amplify base current with infinite collector impedance at source
    • voltage gain is Impedance Ratio *I = Collector/Emitter * Current.
    • What is load Z(f) on collector ?
      • Np/Ns = n, turns ratio, n² * R_Load or Ic= Iout/n
  4. What is gain from base to emitter current?

    • Is AC current less than DC current so Ie does not drop to zero?

      1. Does Vce drop to zero?
    • then AC current exceeds DC current bias.

    • then output appears to "clip". (your problem)
    • except wrong name. Clipping is when Ic goes to 0 with max Vce.
    • Here Vce goes to 0 thus we call it "saturated"
  5. Why saturated?

    • because 8 Ohm AC Load draws more current than Collector DC current
    • Ip=Vp/Rload Idc > Ip is mandatory to keep Ic>0.
    • thus for 8Vp out Idc=8V/8Ω = 1A then Ic must be >2Adc to supply AC 2App This is crucial to understand for Class A
    • Remember this is a Class A amplifier
    • What is overall gain of Class A
    • Vin to Ie AC gain Ie= Ve/Re , for AC Vb=Ve (if Vbe>0.6)
    • Vc = Ic * Rc for high hFE Ic=Ie thus
    • Av = (Rc/Re)
    • collector load = Rc = ωL // 8Ω , if ωL>> 8Ω Rc=8Ω
    • Thus overall gain = Rc/Re = 8/10 assuming 1:1 XFMR
  6. Why is a single Rb to V+ poor?

    • DC collector current depends on hFE which can be >+/-50%
  7. How would Rbase be improved if it connects to collector for negative feedback?

    • it lowers output impedance and input impedance
    • it regulates gain better instead of being a current source
    • now it becomes a voltage gain device with R ratio Av= Rf/Rin
      • where Rf is the collector to base feedback
      • and Rin is source series impedance.

Here is what it takes to drive 8 Ω with 8Vp or 16Vpp in Class A, using negative feedback with Av<10. Note the difference from your values as the impedance ratio is essential to proper Dc bias, gain and avoidance of saturation, clipping and significant distortion. Due to efficiency Class AB is far better and Class E even better.

Note the Java Sim link above shows scope +/- peak readings (snap below) enter image description here