hi how are you i have a question and i tryed to solve it but i couldnt i hope you can help me
i have a design common emitter amplifier with emitter resister and emitter bypass capacitor for the follwoing specifications :
i have got Voltage gain is : 70
and Lower cut off frequency: 135 Hz
and WE take RL as 10k and quiescent operating points ,Vce=6v And Ic = 2mA

i need to get the Input and Output impedances of the Amplifier
Those are my questions i need some one to help me
i tryed to take this rule :
Vcc-RcIc-Vce=0
Vce=Vcc-IcRc
but we have got already the Vce i couldnt know how to figuer out
PLEAS HELP ME

 

Can you post a schematic of your transistor circuit so that we can better answer your question?

This sounds like a homework question. If so, it should be in the homework section.

hgmjr

 

i just got those theres no schematic of the
transistor .... yes its ahomework i put it here by mistake :S

 

This looks like a homework problem, so I don't want to give a full answer. However, just to point you in the right direction, consider the following.

Input and output impedances are AC specifications, and they will not depend directly on DC variables like Vcc or Vbe (but DC variables show up indirectly as constant parameters for the Q-point). What you need to do is make an equivalent AC linearized circuit model that describes how small AC signals behave around a DC operating point (Q-point). Then you solve for the ratio of input AC voltage to input AC current to get input impedance. And, solve for the ratio of output AC voltage to output AC current to get output impedance.

Note, this is not trivial to do, and AC impedance calculations always confuse people more than AC gain (i.e. output/input ratio) calculations, but the principles are the same.

 

Thank you so much but can u help me in getting the ruls like what i should make step1??
coz what u wirtten is a information but i need how i can findthe ruls like are they VBB-IBRB?

 

Can you tell us what level class you are taking. The procedure I'm describing is typically done in a college junior level electronics design course in electrical engineering. If you are at this level, your book should describe how to determine Q-point (DC conditions) and set up an AC equivalent circuit. If you are not at this level, then perhaps you are expected just to look up some formula.

 

am in level two i have book called (electronic device) and i tryed to search but i couldnt ok can u give me example ? for this question?

 

any help??

 

hii tryed to take this rule :
Vcc-RcIc-Vce=0
Vce=Vcc-IcRc
but we have got already the Vce i couldnt know how to figuer out
PLEAS HELP ME

In the equations above You keep forgetting to add the IcRe

Vcc-IcRc-Vce-IcRe=0
Vce=Vcc-IcRc-IcRe

Have you learned yet about the inherent emitter resistance "re" where re=26/Ic (ma)
then Zin=(Beta+1) x (Re+re) ?

And with the emitter bypass cap. depending on the Xc can sometimes eleiminate Re out of the equation?

These are things you need to look at in solving this problem.

If you didn't study these yet, then I'm steering you in the wrong direction. Whoops..

 

Ops nonon we didnt study those still
OMG its too hard what u written re=26/Ic (ma)
then Zin=(Beta+1) x (Re+re) i didnt study notyet...

 

Ops nonon we didnt study those still
OMG its too hard what u written re=26/Ic (ma)
then Zin=(Beta+1) x (Re+re) i didnt study notyet...

That my be a bit more advanced, But to give you an idea of what that means, is Ib and Ic flow through the same resistor Re. That is the physical emittor resistor used to bias the emitter of the transistor.

Because this resistor is in both the input and output currents, then because (Ic is Beta x Ib) then any signal current flowing into the base of the trans. will see two resistances, Re and re.

First re is a inherent resistance in a diode, it is not a stable value, but it changes with respect to the current flowing through it. So the equ. to solve for this value is approx. 26 divided by Ic where Ic is given in (ma)
Known as "shockleys relation"

Now the (Beta+1) x all this,

Youll learn more about this in depth, but for now juist know that a signal current into the base of a transistor will see Re as a resistance of Beta x Re. so if Beta =100 and Re=68 ohms then the signal current (ac) will not see 68 ohms but rather around 6.8K ohms for Re and even more than this when you include re as well, but at least 6.8k ohms. BUT>>>
this may be shunted by your bias base rersistors as well as any capacitive reactance.

When you learn this it will become more clear to you.

 

Why doesn't your teacher teach you anything??

 

Thank you so much

 

i have got those answers but i dont know what els ihave to get
vcc=vce*2=6*2=12
vre=10% of vcc =vre=1
VRC=VCC-VRE-VCE
12-1-6=5
Rc=5\2=2.5
IE=Ib=1/2.5=0.5
Rc=26mv/2=13k
Ro=RL//Ro
10*2.5/10+2.5=2k
26/2=13
re1+RE1=RO/AV=
2*10^3/70=28.57
RE1=2*10^3/70-13=15.57

THIS WHAT I GOT FROM THIS EQUATION I HOPE ANY BODY CAN HELP ME TO KNOW ELS WHAT I SHOULD DO..THANK YOU SO MUCH

 

Quoting "and WE take RL as 10k and quiescent operating points ,Vce=6v And Ic = 2mA"


When you say Q points I assume being an amplifier it is the half supply voltage at Vc which is respect to ground.. When you have a emitter resistor Re then Vc is taken as the Q point, because it couples the output signal to the load, with respect to ground. ... without Re then Vce could then be the Q point, because the emitter is grounded.

If by chance your looking at the emitter capacitor as coupling the emitter to ground that is for calculating Ac gain and impedances,

But all Q point bias voltages are DC.

This make sense????

Just trying to help clarify your question in the first post.

 

Thank you all for helping me

 

sorry to point this out... though it has nothing to do with the question, but you should seriously type better, you don't seem to be able to speak or write proper English, so at least when you type something, make it clear....