Investigating the Effects of a Common Emitter Amplifier

In summary, the conversation discusses using a common emitter amplifier to increase the input voltage and studying its effects. The AC and DC gain were adjusted by adding C5 and R9, and there is some observed non-linearity in the AC gain. The main focus is on studying the effects of connecting a load impedance and measuring the output voltage. There is confusion about the difference between VE and Vout when the load impedance is not connected, and how to calculate the voltage divider bias in this scenario. The conversation also mentions uploading simulation results and fixing an issue with the load resistor value.
  • #1
PhysicsTruth
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Thread moved from the technical forums to the schoolwork forums
Summary:: Using a common emitter amplifier to pump up the input voltage, and studying the effects henceforth.

Here is a circuit with 0.6V pk-pk input voltage (v_in) for the CE amplifier, 1.17 kHz frequency-
end9.png


For this, without C5 and R9, the AC and DC gain was -10. I was supposed to increase the AC gain to -50 while keeping DC gain intact, so I added C5 and R9 in order to get that. I was supposed to play around with only the emitter leg. It would be nice if someone would confirm if I did it correctly, keeping in mind that frequency for the high pass filter at 3dB is 100 Hz. Also, a lot of non-linearity is observed in the AC gain, which I don' t seem to understand.

The main thing is that, I'm supposed to study the effect of connecting a load impedance from C6 to ground and measuring the output voltage there. Without R10, ##V_{E}## had a peak of 8.8V approx, which is justified by the quiescent collector current, but why did the voltage after C6 drop to around 4.9V? I can't seem to find out. Also, according to me, R10 creates a voltage divider bias, thus ##V_{E}## should be thus 8.8 * 4.7/9.4, which is roughly 4.4 V. Surprisingly, the simulation gives me a peak of about 6.5V for ##V_{E}##, and the voltage measured between C6 and R10 gives me around 2.4 V. This is supposed to be a voltage divider, since I observed that when R10 is very high, normalcy is retained. But, how to calculate these voltages using appropriate equations, as my calculations don't comply with the simulation's?
 
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  • #2
Someone please atleast give me some insight as to why does ##V_{E}## differ from ##V_{out}##, when the load impedance isn't connected, and how to calculate the voltage divider bias when the load impedance is connected. My calculations don't seem to match with the simulation results.
 
  • #3
I think you mentioned that you were going to upload your simulation results? (at least PDF screenshots)...
 
  • #4
berkeman said:
I think you mentioned that you were going to upload your simulation results? (at least PDF screenshots)...
Yeah sure. By the way, I have realized it now, after drawing the AC model, where ##R_{L}## appears in parallel with the collector resistance. I would upload the results once I'm free :)
 
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  • #5
PhysicsTruth said:
Someone please atleast give me some insight as to why does VE differ from Vout, when the load impedance isn't connected, and how to calculate the voltage divider bias when the load impedance is connected. My calculations don't seem to match with the simulation results.
It looks like LTspice. Can you please upload your file.asc as a text file.
Rename it as file.asc.txt so it can be attached to your next post.
We can then run the simulation and do the numbers.
 
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  • #6
It is impossible to read the diagram so we need the file.asc
The load resistor looks like it has a value of R10 = 1m = 1 milliohm.
Maybe you intended R10 = 1MEG.
 
  • #7
Baluncore said:
It is impossible to read the diagram so we need the file.asc
The load resistor looks like it has a value of R10 = 1m = 1 milliohm.
Maybe you intended R10 = 1MEG.
Yeah, 1 mega ohm
 
  • #8
Did changing that fix the problem?
 

FAQ: Investigating the Effects of a Common Emitter Amplifier

What is a common emitter amplifier?

A common emitter amplifier is a type of electronic circuit that is used to amplify a small input signal to a larger output signal. It is commonly used in audio amplifiers, radio frequency amplifiers, and other electronic devices.

How does a common emitter amplifier work?

A common emitter amplifier works by using a transistor to amplify the input signal. The transistor acts as a switch, allowing a larger current to flow through the circuit when a small input signal is applied. This amplified current is then used to produce a larger output signal.

What are the advantages of using a common emitter amplifier?

Some advantages of using a common emitter amplifier include its high gain, low output impedance, and its ability to amplify both AC and DC signals. It also has a simple and inexpensive design, making it a popular choice for many electronic applications.

What are the potential applications of a common emitter amplifier?

Common emitter amplifiers are commonly used in audio and radio frequency applications, as well as in electronic circuits that require amplification of small signals. They can also be used in sensor circuits, signal processing, and in other electronic devices that require amplification.

How can the effects of a common emitter amplifier be investigated?

The effects of a common emitter amplifier can be investigated by analyzing its input and output characteristics, such as its gain, frequency response, and distortion. This can be done through simulation software, or by building and testing a physical circuit. The effects can also be studied by varying the components used in the amplifier and observing how they affect its performance.

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