Which is better to get a symmetric swing of the output signal?

[samy555]

For the following voltage divider config.


Is it better to make VC= 0.5 VCC or VCE= 0.5VCC?
thanks

 

[MikeMl]

It is not a "voltage divider"; it is a "stabilized common emitter class A amplifier". For largest undistorted output swing, I would bias it so that V(c) =~V(cc)/2, but it is not very critical, since the drop across the emitter resistor will be a small fraction of V(cc) anyway.

 

[samy555]

Thank you Mike
Why all the electronic books say that for symmetric VCE must = VCC/2?
is that approximation?

 

[MikeMl]

Here is a sim of a gain of 20 amp. The sim is run with different values of R3; where the lt. blue trace seems to have the lowest distortion (R3=90K).

Note the plot of V(c)-V(e). The lt blue (R3=90K) trace starts out at 5V (half V(cc)), so it looks like the textbook is right.

Did you doubt the text book? What did the text book say about the peak amplitude at V(c) before serious distortion appears there? Only someone who didn't care about distortion would operate this amplifier with an amplitude approaching these levels. At lower amplitudes, the bias point is less critical...

 

[crutschow]

....................
Why all the electronic books say that for symmetric VCE must = VCC/2?
is that approximation?

That's so the maximum symmetrical output swing can go from near Vcc to near ground without clipping.

 

[Winterstone]

Thank you Mike
Why all the electronic books say that for symmetric VCE must = VCC/2?

Because it`s correct (assuming Vce,sat=0).
Watch the load line in the Ic=f(Vce) characteristics.

Additional explanation:

For Ic=0 the voltage VcCE across the transistor is VCE=VCC.
This explains why VCE=0.5*VCC gives symmetrical conditions - independent on RE.

 

[samy555]

Here is a sim of a gain of 20 amp. The sim is run with different values of R3; where the lt. blue trace seems to have the lowest distortion (R3=90K).

Note the plot of V(c)-V(e). The lt blue (R3=90K) trace starts out at 5V (half V(cc)), so it looks like the textbook is right.

Did you doubt the text book? What did the text book say about the peak amplitude at V(c) before serious distortion appears there? Only someone who didn't care about distortion would operate this amplifier with an amplitude approaching these levels. At lower amplitudes, the bias point is less critical...

There is no sim

 

[MikeMl]

There is no sim

There is now...

 

[samy555]

OK
another question: If I have two designs (like the above one) which

are the same except that one designed around IC=0.1 mA and the other has IC =10mA
I know that the first maintains the battery and has larger input impedance and larger output impedance than the second.
are there other differences between?
thanks

 

[MrAl]

Hi,


To show linearity a bit better, try using a triangle. If you dont mind doing a little more work, use a stepped wave with say 10 steps up and 10 down or more. Sines make it harder to judge.

 

[Winterstone]

OK
another question: If I have two designs (like the above one) which

are the same except that one designed around IC=0.1 mA and the other has IC =10mA
I know that the first maintains the battery and has larger input impedance and larger output impedance than the second.
are there other differences between?
thanks

Yes, of course. The gain of the stage is prop. to Ic (because the transconductance gm is related to Ic).

 

[MikeMl]

Yes, of course. The gain of the stage is prop. to Ic (because the transconductance gm is related to Ic).

However, if the gain is set by having an unbypassed emitter resistor such as in the sim, this matters not.

 

[samy555]

Can I understand from your words that the design with a small IC is better for a small signal common emitter bjt?

 

[Winterstone]

However, if the gain is set by having an unbypassed emitter resistor such as in the sim, this matters not.

I think it matters because samy555 has mentioned a current Ic=0.1mA.
For such small currents the condition RE>>1/g is not always fulfilled.
(1/g=Vt/Ic=26/0.1=260 Ohms).

 

[MrAl]

Can I understand from your words that the design with a small IC is better for a small signal common emitter bjt?

Hello there,


If you have an amp that has good linearity and swing and scale all the resistances up, you get a little less gain but still good linearity and swing, but a little more swing if the transistor was operating in the higher saturation area. That's because the saturation voltage is less for less collector current.

The reason the amplifier has good linearity is because the forced gain is less than the circuit gain, and the emitter resistor provides for negative feedback. A smaller emitter resistor would result in larger swing, but not as good linearity.

Here's a shot of a transistor amplifier showing how good the linearity is. The upper trace is actually two traces:
1. The actual transistor output, and
2. A computer generated trace of a perfectly linear transistor amplifier with zero non linearity.

You cant tell the difference so well that's how good it is. Transistor is 2N4401.

 

[MikeMl]

Here is the circuit rebiased for a 2N4401, and an Icq of 100uA. The red trace (R3=260K) linearity is not too bad.

 

[MrAl]

Hi Mike,


Did you try setting the 'gain' to 10 instead of 20? Also, comparison using different values of collector resistor (and thus emitter resistor also and possibly bias) would be nice to see too.

The best view for the linearity is to see a 'perfect' triangle of the same amplitude very near to the actual output. That immediately shows any curve (non linearity) very clearly. I would imagine a careful hand could even draw a straight line right next to one or both sides of the output triangle, but an internal generated wave is probably better such as the scaled and offset input waveform itself.

 

[atferrari]

The best view for the linearity is to see a 'perfect' triangle of the same amplitude very near to the actual output. That immediately shows any curve (non linearity) very clearly.

So evident and I never considered that! Always used sinusoidal input...!