Common Emitter Amplifier Analysis

  • Thread starter SaruMihai
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  • #1
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This is my analysis of common emitter amplifier.

Ic = βIb
Ic ≈ Ie (because Ib is very small) => Ie = βIb
Vcc = IcRc + Vce + IeRe
Vcc = IeRc + Vce + IeRe
Vce = Vcc - βIb(Rc + Re)

Vb = IbRb + Vbe + IeRe
Vb = Ib(Rb + βRe) + Vbe

Gain = Vce/Vb
Gain = (Vcc - βIb(Rc + Re))/(Ib(Rb + βRe) + Vbe)
Gain = - βIb(Rc + Re)/Ib(Rb + βRe)
Gain = - β(Rc + Re)/(Rb + βRe)

but everywhere I found only Gain = -Rc/Re

Please if someone can tell me what's wrong,I've tryed to understand this amplifier
for some time,and this is the analysis that i've came up with,and sorry for bad english
 

Answers and Replies

  • #2
5,573
203
This is my analysis of common emitter amplifier.

Ic = βIb
Ic ≈ Ie (because Ib is very small) => Ie = βIb
Vcc = IcRc + Vce + IeRe
Vcc = IeRc + Vce + IeRe assuming Ic=Ie
Vce = Vcc - βIb(Rc + Re)

Vb = IbRb + Vbe + IeRe
Vb = Ib(Rb + βRe) + Vbe

Gain = Vce/Vb only in AC analysis and you miss the -ve sign
Gain = (Vcc - βIb(Rc + Re))/(Ib(Rb + βRe) + Vbe)
Gain = - βIb(Rc + Re)/Ib(Rb + βRe)
Gain = - β(Rc + Re)/(Rb + βRe)

but everywhere I found only Gain = -Rc/Re

Please if someone can tell me what's wrong,I've tryed to understand this amplifier
for some time,and this is the analysis that i've came up with,and sorry for bad english

You are doing DC analysis in all the equations. The Gain =-Rc/Re is AC analysis. You cannot use

Gain = Vce/Vb

You cannot use DC parameter for this. Vce as you wrote depend on Vcc, Rc etc. You have to do the AC analysis for this.
 
  • #3
vk6kro
Science Advisor
4,081
40
Just to cut down on the typing, assume
beta= 300 (small signal current gain or hfe)
intrinsic base resistance =200 ohms,
emitter resistor = 350 ohms,
collector resistor = 2000 ohms
ib = small signal base current

Input voltage = ib * 200 + ib * 300 * 350 ie base voltage plus emitter voltage
Output voltage = ib * 300 * 2000 ie base current times hfe times collector resistor.

Gain = output voltage / input voltage = ib * 300 * 2000 / ib * 200 + ib * 300 * 350

Gain = (ib * 600000 / ib *200 + ib * 105000) or (ib * 600000 / + ib * 105200)

ib cancels, so the gain = 600000 / 105200 = 5.7034

Notice that the ratio of collector resistance to emitter resistance = 2000 / 350 = 5.7143, so the simple formula is an excellent shortcut if the value of the emitter resistor is fairly large and hfe is large.
 
Last edited:
  • #4
6
0
Thanks for the replies,I understand now why I was wrong,and sorry for posting in the wrong category,I'm new to this forum.
 
  • #5
psparky
Gold Member
884
32
Thanks for the replies,I understand now why I was wrong,and sorry for posting in the wrong category,I'm new to this forum.

I found your thread to be most educational.
 
  • #6
6
0
But beside the wrong calculated Gain,the rest of the analysis is right for q active point calculation?
And I found on the internet that Rb < βRe so that Re feedback to work wich means that
VRb < VRe
assuming that Ic ≈ Ie,could anyone please tell me why?
 
  • #7
6
0
I found your thread to be most educational.

I'm happy to know that this thread is useful for you :)
 
  • #8
5,573
203
From years of design transistor circuits including IC designs, I never use this way of analysis. As long as β>100 or so, you can in the first past assume Ib≈0, Ic≈Ie, Rb≈0 and really simplify the thinking and calculations. After all, all your parameters are only approximations anyway. There are temperature dependence, process dependence that likely blow out all your effort to predict the result. I suggest you to get a used book by Malvino. It is a very easy book to read, it is cheap. It show you how to look at transistor in DC and AC and super impose them together.

It is my opinion that you should not get stuck on this way to work with BJT.

1) Ib is important if you DC input impedance is very high that the little current drawn is causing voltage drop at the input circuit.

2)Rb is important mainly in noise calculation where the thermal noise of the Rb become important. Also when in some power application where Ib can get high and need a pre driver to even drive the transistor. Other than this, mostly, we don't even think of the Ib.

How close is Ic≈Ie very seldom even come to mind in circuit design.

Transistor circuit by default is not very accurate, if you need more accurate result, you need negative feedback to stabilize the circuit. With negative feedback, this will make a lot of the concern irrelevant.

One thing that is important you never mention. The approximated input impedance of the common emitter stage. Input impedance equal β(re+Re). Where re is 1/Vt which is about 25Ω for Ie=1mA. 250Ω if Ie=0.1mA etc. You have not even include this in and this turn out to be important for low current design like what I am doing in music electronics at the moment.

Get the Malvino book for cheap, then learn what is important in design. I am still using the materials for designing very advanced transistor circuits.
 
  • #9
6
0
From years of design transistor circuits including IC designs, I never use this way of analysis. As long as β>100 or so, you can in the first past assume Ib≈0, Ic≈Ie, Rb≈0 and really simplify the thinking and calculations. After all, all your parameters are only approximations anyway. There are temperature dependence, process dependence that likely blow out all your effort to predict the result. I suggest you to get a used book by Malvino. It is a very easy book to read, it is cheap. It show you how to look at transistor in DC and AC and super impose them together.

It is my opinion that you should not get stuck on this way to work with BJT.

1) Ib is important if you DC input impedance is very high that the little current drawn is causing voltage drop at the input circuit.

2)Rb is important mainly in noise calculation where the thermal noise of the Rb become important. Also when in some power application where Ib can get high and need a pre driver to even drive the transistor. Other than this, mostly, we don't even think of the Ib.

How close is Ic≈Ie very seldom even come to mind in circuit design.

Transistor circuit by default is not very accurate, if you need more accurate result, you need negative feedback to stabilize the circuit. With negative feedback, this will make a lot of the concern irrelevant.

One thing that is important you never mention. The approximated input impedance of the common emitter stage. Input impedance equal β(re+Re). Where re is 1/Vt which is about 25Ω for Ie=1mA. 250Ω if Ie=0.1mA etc. You have not even include this in and this turn out to be important for low current design like what I am doing in music electronics at the moment.

Get the Malvino book for cheap, then learn what is important in design. I am still using the materials for designing very advanced transistor circuits.

Thank you for this ,i will follow your advice,and I'm very glad that I joined this forum,people are very nice here :)
 
  • #10
5,573
203
Thank you for this ,i will follow your advice,and I'm very glad that I joined this forum,people are very nice here :)

I forgot to mention:

Gain ≈ Rc/(re+Re).

re=1/gm=Vt/Ic

Vt≈25mV at 25 deg C. It is only approximation. In common design, this is plenty good.

And this is more important than the other parameters. I am designing circuit this very moment that run 50uA collector current. re≈500Ω, this turn out to be a big part of the gain calculations.
 
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