Investigating the Effects of a Common Emitter Amplifier

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SUMMARY

The discussion centers on the analysis of a common emitter amplifier circuit, specifically focusing on the effects of load impedance and voltage divider biasing. The circuit initially exhibited an AC and DC gain of -10, which was modified to achieve an AC gain of -50 by adding capacitor C5 and resistor R9. The user encountered discrepancies between calculated and simulated voltages, particularly regarding the output voltage after C6 and the emitter voltage VE. The conversation highlights the importance of understanding load impedance effects and voltage divider calculations in amplifier circuits.

PREREQUISITES
  • Common Emitter Amplifier configuration
  • Voltage Divider Biasing principles
  • AC and DC Gain calculations
  • LTspice simulation software
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  • Study the impact of load impedance on common emitter amplifier performance
  • Learn how to calculate voltage divider biasing in amplifier circuits
  • Explore AC analysis techniques for transistor amplifiers
  • Familiarize with LTspice for circuit simulation and analysis
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Electronics engineers, students studying amplifier design, and anyone involved in circuit simulation and analysis will benefit from this discussion.

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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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.
 
I think you mentioned that you were going to upload your simulation results? (at least PDF screenshots)...
 
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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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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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.
 
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
 
Did changing that fix the problem?
 

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