# non linear bjt amplification problem

#### acelectr

Joined Aug 28, 2010
73
Hi I am working on a design project. Actually it is a multistage amplifier design but I've stucked at the first stage. I've designed this common emitter bjt amplifier. The output graph and the circuit schematic is attached. I think the dc operating point is fine but still my output is not fine. firstly it is not symmetric, like it has an offset but it cant since there are capacitors. When I am changing Rc to 7.5k ohm Vc becomes 7.5V which i think would've been better but when doing it then the output is a clipped amplifier signal. What is it that I am missing here?

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#### Jony130

Joined Feb 17, 2009
5,316
First of all you should design the circuit that Rc << RL.
Secondly, your emitter voltages is to high.
Usually we set Ve = 1V and chose voltage divider current ten times greater then base current.

#### acelectr

Joined Aug 28, 2010
73
Can you explain the reasons of these please? Shouldn't this depend specifically on the dc operating point?

Joined Dec 26, 2010
2,148
There is a fundamental limitation of linearity with transistor amplifiers, but some aspects of your circuit probably do make this worse.

Having a high emitter voltage restricts the available collector signal swing: as the previous poster suggested, 1V is probably sufficient.

Reducing the collector load relative to the external load will allow a bigger voltage swing to be obtained for a given percentage swing of the collector current. The percentage swing (fractional swing, if you prefer) of the collector current is a key issue here. The voltage gain will vary more or less directly with the collector current, so that for an NPN amplifier the negative - going half-cycles appear sharper, as in your waveform. This effect becomes particularly noticeable if the transistor is driven from a pure voltage source, as here.

Some mitigation of the distortion may be obtained with a higher source impedance, and of course by restricting the fractional current swing.

To obtain a more linear response, consider adding some negative feedback. Leaving some portion of the emitter resistance un-bypassed is one of the simplest ways of doing this.

Finally, I note that your input potential divider resistors are very low. The chain consumption is 50mA, which seems extremely wasteful, 50 times more than the collector current. Typically, this chain is made to consume only about 10 times the BASE current.

#### Jony130

Joined Feb 17, 2009
5,316
CE amplifier has a very high output resistance (and this is not good).
Connecting the load resistor reduces the voltage gain but also decrease positive voltage swing.
For example if you have CE amplifier Vc = 7.5V; Ic = 1mA and Rc = 7.5K.
The voltage gain is equal Av = 40*Ic*Rc = 300V/V
And the max positive swing is equal to 7.5V (RL = ∞ )
But if we connect RL = 7.5V
then the gain Will be equal to Av = 40*Ic* RC||RL = 150V/V
And max positive swing 3.75V.
So it is always good to have Rc<<RL in practical circuit Rc = 0.1*RL.
The max negative voltage swing occurs when BJt is start to saturation.
So for your circuit Vc = 10V and Ve = 5V.
The voltage on collector cannot be less then 5.2V so max negative voltage swing is less then 5V.
And you should also remember that this type of circuit with large voltage gain and lack of negative feedback always have very high distortion.

#### acelectr

Joined Aug 28, 2010
73
Firstly why would the emitter voltage restrict the avaliable collector signal swing? I think here we are talking about the dc operating point. As far as I know this is related with the collector voltage not the emitter. Or is it actually related with the collector-emitter voltage?
Secondly what would cause a difference in making Rload beind larger then Rc. Please do not misunderstand me I am just trying to understand the concept as much as possible. Since Rload is parallel to Rc no matter what will happen they will share the same voltage values. So why would I want to make Rc lower then Rload and kill all my bjt biasing settings.
I did not understand the part where you explained about the collector current effect on the gain. Well since the collector current will change the transconductance will also change and of course the gain will be affected but still I did not get why would it affect my negative cycles particularly and sharpen them.
Well actually there are alot to ask about your reply but I don't really want to cause much effort on your side But what I've been told is that when designing an amplifier one should take in to consideration these points: A good dc operating point (for the output signal to swing freely) and the dependence on the Beta constant. I've lowered the equivalent resistance of the base as much as possible so that the circuit would not vary dramatically with small changes of Beta. But I think there are a lot more to take in to account. I think those what I've just wrote are only related with biasing, not amplification.

#### acelectr

Joined Aug 28, 2010
73
CE amplifier has a very high output resistance (and this is not good).
Connecting the load resistor reduces the voltage gain but also decrease positive voltage swing.
For example if you have CE amplifier Vc = 7.5V; Ic = 1mA and Rc = 7.5K.
The voltage gain is equal Av = 40*Ic*Rc = 300V/V
And the max positive swing is equal to 7.5V (RL = ∞ )
But if we connect RL = 7.5V
then the gain Will be equal to Av = 40*Ic* RC||RL = 150V/V
And max positive swing 3.75V.
So it is always good to have Rc<<RL in practical circuit Rc = 0.1*RL.
The max negative voltage swing occurs when BJt is start to saturation.
So for your circuit Vc = 10V and Ve = 5V.
The voltage on collector cannot be less then 5.2V so max negative voltage swing is less then 5V.
And you should also remember that this type of circuit with large voltage gain and lack of negative feedback always have very high distortion.
How did you write the gain equation? Where did this 40 came from. Wouldn't be Av= -(transconductance)*Rc//RL*rpi. I am obtainin all my equation from the low frequency equivalent circuit model, particularly pi model circuit model.
I still did not understand how did you get to the max positive swing voltage value. I actually do believe in what is written but I do not understand logic behind it. Still I am missin somethings.

#### Jony130

Joined Feb 17, 2009
5,316
Try to analysis this diagram more carefully.

Vcc = 12V; Vc = 6V (red plot ); Ve = 2V (green plot) ; Vin = 1V (blue plot).
An I hope that it is clear to see that voltage on collector can swing from Ve to Vcc = 12V.
The positive half from Vc to Vcc and negative form Vc to Ve+Vin

As for voltage gain (with Ce capacitor)

$$Av = gm*Rc = \frac{Ic}{26mV} * Rc = \frac {1}{26mV} * Ic * Rc = 38.46 * Ic * Rc \approx 40 * Ic* Rc$$

Or simply

$$Av = \frac{Rc}{re} = \frac{Rc}{\frac{26mV}{Ic}}$$

$$re = \frac{1}{gm}$$

An

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#### Audioguru

Joined Dec 20, 2007
11,248
A single transistor with a very high voltage gain and no negative feedback is a distortion machine. That is why opamps have been used for low distortion audio for the past 30 years. Transistors are still used to produce guitar FUZZ which is very high distortion.

Here is a transistor with no load (a load would simply reduce its output level). When its voltage gain is 160 then its high level distortion is an awful 40%.
When it has plenty of negative feedback to reduce its gain to 9.2 then its distortion is still bad at about 3% but is much better.

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#### acelectr

Joined Aug 28, 2010
73
Well I've set a negative feedback and the resulting output had quite less distortion than before without the negative feedback. Bud my gain have decrease dramatically, like from 250 to 20 (( . I gues I'll need more transistor. What I don't understand is that I've told that Re is needed for biasing and was shorted (for ac) with a cap so to avoid the negative feedback. Because negative feedback is the devil. But since it kills the distortion I guess it is also quite needed. What is the logic behind this. Why would negative feedback decrease the distortion affect?

#### Audioguru

Joined Dec 20, 2007
11,248
Negative feedback reduces gain and reduces distortion.

An OPA134 opamp has an open-loop gain of 10000 at 1kHz.
When it has negative feedback so that its gain is 1 then its distortion is 0.00008%.
When it has less negative feedback so that its gain is 100 then its distortion is 0.008%.

#### acelectr

Joined Aug 28, 2010
73
Well I've nearly done with this multistage amplifier design but I'm having some troubles at the last parts. The output will be taken from a 200ohm load and the output impedance should be maximum 100 ohm. I think I've managed the output resistance issue but when I am taking the output over a 200 ohm resistance I'm getting crazy signals. But when I am exchanging the load resistance with a 20k ohm then the result is more satisfying both in its peak values and its shape. Why do I get this distortion over a small resistance load and not over a high resistance load. How one can solve this issue?????

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