CC - CE cascade

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polinski

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Hello everyone.
Im a newbie in electronics and i wanna study how a CC-CE cascade works.

I read some books for the formulas and the theory about it, but im confused a lil bit.
If i start with only the CE amplifier i have this circuit:
**broken link removed**
https://imgur.com/KTW7j

with random values


If i wanna write kirchoff formulas, there are:
Rc + Re = (Vcc - Vce)/Ic

where Vce = (Vcc)/2

and thevenin conditions are:
Rth = R1 // R2
Vth = (Rth/R1)*Vcc
and also:
Vth = Ib Rb + Vbe + Ic Rc with Ic ≈ β Ib

at least:
Av = - Rc/Re

However I have some problems with kirchoff formulas about CC-Ce cascade

There is my pspice model (gave by school):
https://adf.ly/3Z2SC

(with random values again)

-Vcc + (Vce)2 + I(re2)Re2 =0

where (Vce)2 is for the new resistor.

how must i continue?

i hope someone will help me, thanx.
 
Last edited by a moderator:
Why is your schematic as big as my neighbourhood?
I cropped it and used Ohm's Law and simple artithmatic to calculate the voltages.
 

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Ah sorry, i thought it was ok.
By the way, my problem is for cascade CE-CC in this picture:
https://imgur.com/4LqyD

Supposed R1,R2,Re and Rc with a value, i must find Re2, i have written these formulas:

- Vcc + Vce + Re2 *I(Re2) = 0

-Vcc + Rc *I(Rc) + Vbe + Re2 I(Re2)=0


are they correct?
 
Your equations looks good.

Ie1*Re + Vce2 = Vbe+ Ie1*Re2

Or

VRe2 = Vcc - (Ic2 + Ib1)*Rc - Vbe1

And don't forget that current that is flow through Rc is equal to:
IRc = Ic2 + Ib1
 
Last edited:
n p n junction
I want to understand how to 'put' the CE + CC configuration in cascade on a breadbord :/ im using a virtual one like that:
https://imgur.com/IS3Z9
I think i must use also the oscilloscope for misurating Vin, Vout and Av = Av1 *Av2
 
Yes I understand this, but what is the physical type of the BJT you are using in the breadboard (2N2222) ??
 
It really helped to me! Thanks a lot!
Another 2 questions:
1. in that figure i dont see the capacitor Cin, its normal or i must put it too on breadbord?
2. In this CC+CE configuration if i wanna find every resistor, i must write kirchoof formulas.
We write in the last topic some of them.

From: - Vcc + Vce + Re2 *I(Re2) = 0
i can find Re2, supposed that Vce = Vcc/2 ----> Re2 = Vcc/(2*I(re2))

From: -Vcc + Rc *I(Rc) + Vbe + Re2 I(Re2)=0
i cand find Rc so: Rc = (Vce/2I(Rc) ) - Vbe/I(Rc)
supposed Vbe = 0,7 Volts.

Is that correct?
[im trying also to write formulas for R1 and R2]
FOR THE CAPACITOR i use in every simulation 100 nF, i think its ok.

what do you think about all this?
 
polinski said:
It really helped to me! Thanks a lot!
Another 2 questions:
1. in that figure i dont see the capacitor Cin, its normal or i must put it too on breadbord?
Click to expand...
Yes you need to place Cin capacitor in the breadbord
See the attach file.


Well, you have done a small error in Rc equation

-Vcc + Ic*Rc + Vbe + Ie2*Re2 = 0

Rc = (Vcc - Vbe - Ie2*Re2)/Ic

But we can assume that your power supply voltage is equal 9V and start to choose the component value.

Re2 = 4.5V/4.5mA = 1KΩ

Vc1 = Ve2 + Vbe = 4.5V + 0.65V = 5.15V

Rc = (Vcc - Vc1)/ ( Ic1+Ib2 ) ----> assume Ic1 = 1mA we end up with

Ib2 = Ie2 / (hfe+1)

Hfe from the data sheet
https://www.electro-tech-online.com/custompdfs/2011/11/PN2222A.pdf

I chose Hfe_min = 50 for Ic>1mA

Rc = ( 9V - 5.15V) / (1mA + 90uA) ≈ 3.9KΩ

but if you want more standard resistor value, for example 10K, no problem at all.

Ic1 + Ib2 = (Vcc - Vc1)/Rc = 3.85V/10K = 385µA

I want the voltage gain to be equal 10 times.

Then Re = Rc/5 ≈ 2.2K

Vb = Ie*Re + Vbe = 350uA*2.2K + 0.65V = 1.42V

To find voltage divide resistor values we normal use this equations

R1 = ( Vcc - Vb) / ( 11*Ib)

R2 = Vb/(10*Ib)

But i chose R2 = 10K and

IR2 = 1.42V/10K = 142µA

R1 = ( 9V - 1.42V)/ ( 142uA + 10uA) = 49K ≈ 51K

Cin > 0.16/( F * Rin) = 1uF

F = 20Hz

Rin = R1||R2||(Hfe+1)*Re = 7.7K

I hope this helps
 

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Last edited:
Yes now i can watch your pic.
I noticed that you gave some random value, i think its ok
By the way, Cin that i use more, its 100 nF or 150 nF it depends.

IS Rin = R1||R2||(Hfe+1)*Re = >

1/Rin = 1/R1 + 1/R2 + 1/(hfe+1)*Re ?

For you Vb is Vbe?
 
polinski said:
By the way, Cin that i use more, its 100 nF or 150 nF it depends.
Click to expand...
The size of a capacitors is selected that Xc is much smaller then resistance of a cooperating resistor at the lowest frequency of operation Xc=0.1*R.

Xc = 1/(2*PI*F*C)=0.16/(F*C)

C = 0.16/(F*R)

So for Cin = 100nF the signal frequency must be greater then 160Hz


polinski said:
IS Rin = R1||R2||(Hfe+1)*Re = >

1/Rin = 1/R1 + 1/R2 + 1/(hfe+1)*Re ?

For you Vb is Vbe?
Click to expand...

Yes this symbol "||" mean connected in parallel resistors.

Vb is the voltage at base of a BJT

Vb = Ie*Re + Vbe
 
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