Common Emitter with emitter resistance small signal model

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SUMMARY

The discussion focuses on calculating the input resistance (Rin) and voltage gain (Av) of a common emitter amplifier with emitter resistance using the pi model. The user successfully derived the voltage gain as Av = -B(R_C||R_L)/(r_pi) but struggled with the input resistance calculation. The correct formula for Rin is confirmed to be Rin = RB || (r_pi + R_E(β + 1)), clarifying that Vbe is across the base-emitter junction only, not including external resistors.

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Electrical_kid
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[Mentor's Note: Thread Moved from EE Forum]

This link has the picture, for some reason it wasnt showing in this post. http://i.imgur.com/TYlb32t.jpg
Here is my problem, I am supposed to use this circuit and solve for Av and Rin. My problem is that I have solved for Av but I can't solve for Rin no matter what I use. Its very frustrating. I need to use the pi model.

Here is my work for the Av gain.

i2yiMT6.jpg


Please let me now if this is correct and then a tip on how I can solve for Rin which should be Rin = RB|| r_pi (R_E(1+B)). Is Vbe only across r_pi or r_pi and R_E?

Av = -B(R_C||R_L)/(r_pi) which is what I have in my work in the upper right middle
 
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Welcome to PF.
Is this either coursework or homework ?

small_circuit.png
 
Last edited:
Baluncore said:
Welcome to PF.
Is this either coursework or homework ?
Its homework, do you think there is a typo in the Rin given? Or is the emitter resistance in the wrong place?
 
I shrank and posted the circuit.
Rin is shown as an indeterminate arrow, Is it measured at Vs of at the base terminal ?
Rin is frequency dependent because of the series C in the base circuit.
Re will appear higher resistance to Rin because it is seen through the base current gain, beta.
 
Electrical_kid said:
Please let me know if this is correct and then a tip on how I can solve for Rin which should be Rin = RB|| r_pi (R_E(1+B)). Is Vbe only across r_pi or r_pi and R_E?
Vbe is always the voltage across the active b-e junction; it doesn't embrace voltage across externally placed resistors.

Your formula for Rin looks awkward. The expressions on each side of the "||" symbol should have units of Ohms.
 
Should Rin be Rin = RB || (rpi + RE*(beta+1)) ?
 
Electrical_kid said:
Should Rin be Rin = RB || (rpi + RE*(beta+1)) ?
That looks better.
 
NascentOxygen said:
Your formula for Rin looks awkward. The expressions on each side of the "||" symbol should have units of Ohms.
If R_pi has dimension ohms then your Rin = RB || r_pi (R_E(1+B)) has dimensions of ohm || ohm2
Where has your R_pi come from ?

See schematic in post #2
Do you think Rbase = R1 || R2 ?

How is beta defined ?
Maybe; Rin = Rbase || (Re * beta)
or; Rin = Rbase || (Re * (1+beta) )
 
Baluncore, I think the reason I was having trouble is because the answer they provided was missing the plus sign, a previous professor of mine provided me a pdf file explaining how they get Rin and it definitely shows that it should be Rin = RB || ( rpi + RE(beta +1)).
 

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