Why does a strange waveform appear in the RC coupled common emitter amplifier circuit?

Celefor

Why does a strange waveform appear in the RC coupled common emitter amplifier circuit? [Copy link]

 

When I was simulating the saturation distortion of the RC coupled common emitter amplifier circuit, I found a strange waveform as shown in the figure below:

In the above figure, when R1 is adjusted to 10kΩ, the voltage suddenly drops from 387mV to 280mV at 26ms. After that, the saturation distortion gradually weakens and is no longer noticeable after 260ms.

In the figure above, when R1 is adjusted to 5kΩ, the voltage suddenly drops from 670mV to 495mV at 17ms. Then the saturation distortion gradually weakens and is no longer noticeable after 130ms.

In the figure above, when R1 is adjusted to 1kΩ, the circuit works normally, but at 26 milliseconds the voltage suddenly jumps upward from 740mV to 795mV.

I initially thought that the phenomenon was caused by the coupling capacitor, and tried to change the size of the coupling capacitor, but found that the above data did not change significantly, so I ruled out this idea. At present, it should be the cause of the transistor itself, but I can't figure out the specific cause of this phenomenon, so I posted it to the forum and asked the big guys to help me with advice.

jpwu10

The static working point setting is unreasonable.

maychang

This is a completely normal waveform.

输出波形在正半周期尖锐负半周期圆钝，那是因为你使用电压源作为输入信号，但三极管输入特性有显著的非线性(电压和电流不是线性关系)。但三极管输出电流与输入电流之间的关系却接近线性。于是输出电压和输入电压之间也就成了非线性关系。

maychang

If a current source is used as the input signal, this nonlinear distortion will be reduced a lot. Of course, you can also connect a relatively large resistor in series with the input voltage source, such as several thousand ohms or even larger, to make the signal closer to the current source, and the relationship between the output waveform and the input voltage will be more linear. It is easy to simulate adding a resistor, so why not give it a try. Of course, the premise is that the bias current is set correctly, that is, the top and bottom of the output waveform appear at the same time when the input signal amplitude increases.

Celefor

maychang posted on 2023-2-4 09:48 If a current source is used as the input signal, this nonlinear distortion will be greatly reduced. Of course, a relatively large resistor can also be connected in series with the input voltage source...

Um, I have another question. Please look at the simulation waveform. There is a jump in the waveform at 26 milliseconds. What's going on?

maychang

As for the upward or downward jumps in the waveform, that is a problem with the simulation software. Newer simulation software has solved this problem, so there is no need to worry about it.

maychang

Celefor posted on 2023-2-4 09:51 Well, I have another question. Please look at the simulation waveform. There is a sudden jump in the waveform at 26 milliseconds. What is going on?

I have just replied on the 6th floor. Maybe it is a little later than your question on the 5th floor, not timely enough. But it takes a little time to type dozens of words on the keyboard, please understand.

Celefor

maychang posted on 2023-2-4 09:53 I have just replied on the 6th floor. It may be a little later than your question on the 5th floor, not timely enough. But it takes a little time to type so many dozens of words on the keyboard, ...

Thank you! Can you recommend the latest simulation software you mentioned?

maychang

Celefor posted on 2023-2-4 09:55 Thank you! Can you recommend the latest simulation software you mentioned?

[Can you recommend the latest simulation software you mentioned?]

I have no recommendation.

Celefor

jpwu10 posted on 2023-2-4 07:06 The static working point setting is unreasonable.

Well, I did this on purpose to create saturation distortion. But I don't know how to explain the sudden jump in the waveform at 26ms.

maychang

Celefor posted on 2023-2-4 10:01 Well, I did this on purpose to create saturation distortion. But I don't know how to explain the sudden jump in the waveform at 26ms

You did not notice what state the circuit was in when the simulation started.

Before the simulation starts, the 12V power supply voltage is not applied by default (zero), and the voltage across C2 and C1 is also zero by default. Then, when the simulation starts, the voltage across C2 is still zero, the base voltage is zero, and the base current is also zero. As the 12V power supply charges C2 through R2, the voltage across C2 gradually increases. Finally, it stabilizes. The time required from the start of the simulation to reaching stability (usually it is considered stable if the difference from the final value does not exceed 5%) is quite long in this circuit because you use a large coupling capacitor, reaching 800uF.

maychang

Celefor posted on 2023-2-4 10:01 Well, I did this on purpose to create saturation distortion. But I don't know how to explain the sudden jump in the waveform at 26ms

As the voltage across C2 gradually increases, the simulation software will adjust the operating point after a period of time, and then adjust the operating point again after a period of time. Adjusting the operating point is the reason why the waveform changes suddenly during simulation.

maychang

Celefor posted on 2023-2-4 10:01 Well, I did this on purpose to create saturation distortion. But I don't know how to explain the sudden jump in the waveform at 26ms

Note: The collector waveform is very different before and after the sudden change at 26ms in the first post. Before, the area of the waveform above and below the horizontal axis was not equal, the area above was large and the area below was small. After the sudden change, the areas above and below were almost the same. In fact, such a sudden change will not occur in the circuit. You can know this by taking an oscilloscope and looking at the waveform after the actual circuit is powered on. Therefore, this sudden change is a problem with the simulation software.

Celefor

maychang posted on 2023-2-4 10:55 As the voltage across C2 gradually increases, the simulation software will have to adjust the operating point after a period of time, and then adjust the operating point again after a period of time. Adjust...

Thank you for your detailed answer!! Now I completely understand! Thank you!!

tomas_li

If there is no resistor at the emitter, although the amplification factor is relatively large, it is easy to produce waveform distortion because the emitter transconductance gm is greatly affected by the output current. It would be better to add a resistor to the emitter.