amplifier quality waveform for different transistors

Question

I have tested two transistors in multisim, the first is the default(more of generic) and the second is the 2N3904

here is the first picture with default transistor in placed

and here is the corresponding waveform

as you can see, its smooth, but if we use the 2N3904 (available in my local electronic store)

the waveform looks like below

the waveform looks heavily distorted. What should we change in the circuit so that the waveform is the same as the original transistor placed. Thanks

Also, as a side note, what modifications are needed, so that the waveform is not inverted. Thanks

Answer 1

Transistors are inherently nonlinear. The local feedback from the emitter resistor in a single stage is not enough to fix this, particularly if its value is low.

If you substitute one transistor for another, the load line will change. (And by the way, you have to consider the AC load line separately from the DC one, because AC sees different impedances, both through the base of the transistor and through the surrounding components, such as obviously the capacitor bypass on the emitter).

Although the first trace might look like it is undistorted, looks can be deceiving. A spectrum analysis will show you the distortion products.

You will not get low distortion out of a single stage, unless you keep the signal swing very small, far away from the the voltage rails. Perhaps the ultimate example of "cleanliness" in a single BJT stage would be an emitter follower, with a reasonably small input signal.

If you want a virtually distortion-free amplifier, the usual approach is to combine several stages which have modest gain to produce a high overall gain, and then tame that gain with global negative feedback. Global feedback will correct for all kinds of distortion, such as nonlinearities in the individual stages. It can even nearly eliminate crossover distortion from a class B output stage, so that only a small kink or discontinuity is seen on the waveform.

Answer 2

To answer the second question first:

The basic single-transistor voltage amplifier you have simulated is inherently an inverting configuration. This is not actually much of a concern for audio purposes, because our ears can not identify that a waveform is inverted, after all!

Further, if you follow this circuit block with a power amplification block, you have the option of inverting the waveform again. Problem solved.

Now the waveform distortion: This is most probably due to the input signal not being suitably biased to fall within the linear portion of the particular transistor's transfer graph (Please check the datasheet for details, I can't do it right now). By changing the bias resistors R1 and R2 suitably, the input signal can be squarely placed midway in the linear part of the curve, that should eliminate the distortion.

Answer 3

What, for me, jumps out immediately is the unusually small value for the emitter resistor \$R_e\$.

In order to stabilize the operating point against variations in transistor parameters, \$R_e\$ needs to be "large enough".

The DC bias equation for this circuit is:

\$I_C = \dfrac{V_{BB} - V_{BE}}{\frac{R_{BB}}{\beta}+\frac{R_{e}}{\alpha}}\$

\$V_{BB} = 40V \dfrac{R_1}{R_1 + R_2}\$

\$R_{BB} = R_1 || R_2 \$

For the values you have,

\$I_C = \dfrac{9.35V}{\frac{4.05k \Omega}{\beta}+\frac{8 \Omega}{\alpha}} \$

To have good \$I_C \$ stability against variations in \$\beta\$, the right hand term in the denominator should be much larger than the left hand term.

For example, in your case, look at the variation in \$I_C\$ when \$\beta \$ is doubled:

\$I_C = 192mA \ , \beta = 100 \$

\$I_C = 330mA \ , \beta = 200 \$

This is unacceptable and most likely the reason for the vast difference in behaviour when you changed to a different transistor.

You should increase the value of \$R_e \$ so that variations in \$\beta\$ make insignificant changes to the bias point.

This will reduce \$I_C\$, the DC collector current, which is desirable as Richman points out in his comment.

With a reduced \$I_C\$, you'll want to adjust upwards the value of \$R_C\$ to get the desired DC collector voltage.

Answer 4

I selected a lower base voltage and removed the bypass capacitor on the emitter then ensured the collector does not saturate for input level. I chose a gain of 20x (Rc/Re) but matched the collector to the load, so the gain is ~ 10. Note this SIM shows the positive and negative peaks are matched due to optimal biasing.

See what happens when you saturate the collector by raising Vb by lowering 60K (right mouse>edit> adjust slider>apply)