# Common Emitter amplifier -frequency analysis

Discussion in 'Mathematics and Physics' started by aani, Jan 27, 2016.

1. ### aaniNew Member

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Hi anyone ,

Q: Frequency analysis of transistor amplifiers

A. For circuit (a) below, calculate: Re and Rc to make Ic ≈ 0.2 mA and VCE ≈ 6V.
Ce to give a pole frequency at about 400 Hz, and Cc to give a pole at about 50 Hz.
The oscilloscope input resistance is 1 MΩ. For initial calculations assume Cµ = 5 pF and CCRO + cable = 110 pF. (These will be measured in the lab.) The generator resistance is 50 Ω, which may be taken as a good approximation to zero for this experiment. Moreover, the input will be monitored across the generator and its voltage kept constant, so it may be treated as an ideal voltage source.
Estimate the voltage gain, the frequency of the zero introduced by Ce, and the upper 3-dB frequency fh. What is the voltage gain if Ce is omitted?

B. Circuit (b) is to be inserted in series at point B to buffer the load capacitance. Calculate R to give a collector current of about 1.2 mA. Assuming a 'typical' value for hfe, recalculate fh. Is the pole frequency introduced by Cc significantly affected?

C. Leaving circuit (b) in place, a 10 kΩ resistor is inserted at point A (simulating a higher source resistance than that of the generator). Recalculate fh. Is the pole frequency introduced by Ce affected?

(Circuit is attached in the pdf file.)

Regards,
Anila

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• ###### Question_Linear-electronics.pdf
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2. ### aaniNew Member

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Hi anyone,
Pls help on this .Any clue?

3. ### MrAlWell-Known MemberMost Helpful Member

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Hello,

I can help you get started using common assumptions for a circuit like this.

First, since iC should be 0.2ma then you can assume that iE is also 0.2ma because the base current is usually small.
Since the AC generator is zero for DC, the voltage at the emitter will be about -0.7v which puts 4.3v across the emitter resistor.
To get 0.2ma through that resistor and thus iE=0.2ma, you need a resistor of value:
Re=4.3/0.0002=21.5k ohms.

To get 6v for vCE you know that the emitter is at -0.7v so you need vC=6-0.7=5.3v.
The supply is 12v so to get 5.3v at the collector you need to drop 12-5.3=6.7v.
To drop 6.7v with a current of 0.0002 amps you need a resistance of:
Rc=6.7/0.0002=33.5k ohms.

So it's mostly Ohm's Law. See if you can take it from here.

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5. ### aaniNew Member

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Hi ,

Yea, I have too managed to work out the Rc and Re by DC analysis.
But I have difficulties in the later stages below.Could you help pls,...

Ce to give a pole frequency at about 400 Hz, and Cc to give a pole at about 50 Hz.
The oscilloscope input resistance is 1 MΩ. For initial calculations assume Cµ = 5 pF and CCRO + cable = 110 pF. (These will be measured in the lab.) The generator resistance is 50 Ω, which may be taken as a good approximation to zero for this experiment. Moreover, the input will be monitored across the generator and its voltage kept constant, so it may be treated as an ideal voltage source.
Estimate the voltage gain, the frequency of the zero introduced by Ce, and the upper 3-dB frequency fh. What is the voltage gain if Ce is omitted?

B. Circuit (b) is to be inserted in series at point B to buffer the load capacitance. Calculate R to give a collector current of about 1.2 mA. Assuming a 'typical' value for hfe, recalculate fh. Is the pole frequency introduced by Cc significantly affected?

C. Leaving circuit (b) in place, a 10 kΩ resistor is inserted at point A (simulating a higher source resistance than that of the generator). Recalculate fh. Is the pole frequency introduced by Ce affected?

(Circuit is attached in the pdf file.)

6. ### MrAlWell-Known MemberMost Helpful Member

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Hello,

Well again we will look at this in the simplest possible manner for now. If this is part of a course however you should read your textbook and refer to your notes in case the professor wants you to do it a certain way. For a good example, sometimes iE=iC*Beta is acceptable and sometimes only iE=iC*(Beta+1) is acceptable. In real life either way is usually acceptable but that's not how courses in electronics usually work because often they want you to follow a specific procedure or technique. They are giving approximations however so that's a good sign they are not too particular.

For Ce, the frequency would be:
w=1/(Re*Ce)

and in Hertz:
f=1/(2*pi*Re*Ce)

The gain changes with frequency because of that Ce.

Do you know what the typical Beta is for this transistor or how to find it?

Have you done anything like this before?
If you have another worked out example from a previous assignment we can go by that to be sure we are doing it the way that the instructor wants to see it done.

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7. ### aaniNew Member

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Could you check my work outs and advise for any corrections?

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8. ### aaniNew Member

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Hi MrAL,

Last edited: Feb 1, 2016

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Hi MrAL,

10. ### MrAlWell-Known MemberMost Helpful Member

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Hello,

I downloaded all your pages but i'll need time to go over them all as i came down with the Flu about a week ago and it seems to be quite persistent.
i'll try to look at them again later today or tonight.
There are other members here too that might be able to help also.

LATER:
On page 2 why did you use re in the calculation for Ce?
Why not just the external RE?

Last edited: Feb 1, 2016
11. ### aaniNew Member

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Hi Al ,

Thanks for correcting me.I should use Re straight away.Do u agree with my currency calculations for Re,Rc,Cc,Ce,and voltage gain?

12. ### MrAlWell-Known MemberMost Helpful Member

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Hello again,

Well ok lets start with Rc then, get that out of the way, then proceed to calculate Ce.

We agree that Re (external) is 21.5k.

Keeping in mind that the supply voltages are 12v and -5v also, and that we have the emitter biased to -0.7v constant with 200ua flow.

The first way we can use the -0.7v emitter voltage:
12-i*Rc-6-(-0.7)=0

We know that i=0.0002 amps, and solving this for Rc we get:
Rc=33.5k

The second way is to recognize that the base current is zero because of our approximation of Ic=Ie, but it still keeps the emitter at -0.7v and the collector and emitter current at 0.0002 amps:
12-i*Rc-6-i*Re-(-5)=0

and again knowing that i=0.0002 amps and also that Re=21.5k we solve this for Rc and get:
Rc=33.5k

So either way we do this we get 33.5k.

Do you see how this works?

Unfortunately i may take a little longer than usual to reply as i am sleeping quite a bit these past few days. Hoping to get over this stupid cold or flu or whatever the heck it is soon. I believe it is better that we take this one step at a time though.