How does this circuit work? (two transistor amplifier)

Thread Starter

newguy193

Joined Jun 26, 2021
14
Hello again

I want to trouble you guys for another assignment. I need to provide what it does and calculate its lower and upper cut frequency.
I know that a common collector(emitter follower) is used for impedance matching but I can't say what happens if two of them are joined in series.

A very brief explanation of how it works is all I need. thank you

Edit: hopefully this will be the last trouble from me haha
 

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Audioguru again

Joined Oct 21, 2019
3,866
Your transistors have no part numbers then you cannot determine the upper cutoff frequency.
Your capacitors have no values then you cannot calculate the lower cutoff frequency.
If you match impedances then you throw away half the signal voltage level.
 

Thread Starter

newguy193

Joined Jun 26, 2021
14
Your transistors have no part numbers then you cannot determine the upper cutoff frequency.
Your capacitors have no values then you cannot calculate the lower cutoff frequency.
If you match impedances then you throw away half the signal voltage level.
I have those values in its description. I'll type it below it may help.

C(pi)=24 pF , C(u)=4 pF , beta=150 , V(A)= infinite , V(BE)=0.7

by the way I don't get your final statement!
 

Audioguru again

Joined Oct 21, 2019
3,866
The output impedance of a 10W audio amplifier is normally very low at 0.02 ohms so it damps speaker resonances and produces a low voltage loss of 2V (10W into 8 ohms= 8.95V RMS).

If the amplifier has an output impedance of 8 ohms then if its unloaded output level is 8.95V RMS then its output level into an 8 ohm speaker is half at 4.48V RMS producing an output power of only 2.5W and the speaker will resonate like a bongo drum.

Almost all signal source and load impedances in audio are never matched. Usually a load is at least 10 times the impedance of a source to avoid a signal voltage loss.

The capacitance of the transistor will change with added stray capacitance causing the upper cutoff frequency lower than your calculations.
I do not know what your V(A) is. If it is the "voltage amplification" then it definitely is not infinite. It might be 250 times with distortion that is so high that the output level cannot be measured.
 

Thread Starter

newguy193

Joined Jun 26, 2021
14
The output impedance of a 10W audio amplifier is normally very low at 0.02 ohms so it damps speaker resonances and produces a low voltage loss of 2V (10W into 8 ohms= 8.95V RMS).

If the amplifier has an output impedance of 8 ohms then if its unloaded output level is 8.95V RMS then its output level into an 8 ohm speaker is half at 4.48V RMS producing an output power of only 2.5W and the speaker will resonate like a bongo drum.

Almost all signal source and load impedances in audio are never matched. Usually a load is at least 10 times the impedance of a source to avoid a signal voltage loss.

The capacitance of the transistor will change with added stray capacitance causing the upper cutoff frequency lower than your calculations.
I do not know what your V(A) is. If it is the "voltage amplification" then it definitely is not infinite. It might be 250 times with distortion that is so high that the output level cannot be measured.
its "early voltage"
 

ericgibbs

Joined Jan 29, 2010
14,207
hi 193,
Can you say what electronics course you are studying and at what level, so that we can adjust our replies to suit.?

E
 

ericgibbs

Joined Jan 29, 2010
14,207
"A very brief explanation of how it works is all I need. thank you "
hi 123,
Look at this marked up image.
Would you say the amplifier is designed for AC or DC amplification.?
At the points indicated, what do you calculate the DC impedances to be.?
Also what is the approx DC gain of 1st Stage of the amplifier and also the DC gain of the 2nd Stage.?

beta=150 , V(BE)=0.7

We can consider the other parameters of the circuit later.

E
proj.png
 

Thread Starter

newguy193

Joined Jun 26, 2021
14
hi 123,
Look at this marked up image.
Would you say the amplifier is designed for AC or DC amplification.?
At the points indicated, what do you calculate the DC impedances to be.?
Also what is the approx DC gain of 1st Stage of the amplifier and also the DC gain of the 2nd Stage.?

beta=150 , V(BE)=0.7

We can consider the other parameters of the circuit later.

E
View attachment 242537
Honestly? I can't answer to any of your questions. I haven't studied electronics for a few semesters so I've forgotten many things.

Here's the simulation I've done. result was unexpected. output is cut off, I wonder why. It became like a rectangular as if there are diodes in the circuit!
 

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ericgibbs

Joined Jan 29, 2010
14,207
hi 123.
Surely you can answer this question.:)

Would you say the amplifier is designed for AC or DC amplification.?


The square wave is due to over driving the input.
 

Thread Starter

newguy193

Joined Jun 26, 2021
14
hi 123.
Surely you can answer this question.:)

Would you say the amplifier is designed for AC or DC amplification.?


The square wave is due to over driving the input.
the way transistors amplify is by giving them small signals(usually to base) and in return they give larger signals at output. So id say its an AC amplification. btw I hadn't have heard about dc amplification before.

Also thanks for hint on input. the square like output problem now solved!
 

ericgibbs

Joined Jan 29, 2010
14,207
yes I'm studying bachelors in electronics engineering. college student
hi 123,
Sorry to be blunt, but if you don't understand these simple circuits now, you are going have future problems in your Bachelors study work.

Perhaps some revision of your early work would be a good starting point

E
 

dcbingaman

Joined Jun 30, 2021
498
Because of Ce, this circuit is going to have an unpredictable and insanely high AC gain and every transistor you use will give you a different gain the gain will even vary with temperature! This is not a well designed amplifier in any way.
 
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