Investigating the Effects of a Common Emitter Amplifier

AI Thread Summary
The discussion focuses on the performance of a common emitter amplifier, specifically examining the effects of modifying gain and load impedance. The initial AC and DC gain was -10, which was intended to be increased to -50 by adding components C5 and R9. Participants are trying to understand discrepancies in voltage measurements, particularly why the output voltage drops significantly when a load impedance is connected. There is confusion regarding the calculations for voltage divider bias and the relationship between the emitter voltage (VE) and output voltage (Vout). The conversation also highlights the need for simulation files to clarify the circuit's behavior and resolve the issues discussed.
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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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