n00b help - BJT circuits output resistance

Discussion in 'Homework Help' started by ankitmcgill, Oct 3, 2009.

  1. ankitmcgill

    Thread Starter New Member

    Oct 3, 2009
    Hello all,

    I am taking my first mircoelectronics course this semester and we are studying BJT's right now.

    I am having a hard time wrapping my head around the whole concept of 'ouput resistance'. in theory it seems simple enough that

    R_out = v_test / i_test

    but i cant seem to apply it to problems. so i thought i would start with the basics.

    would appreciate a lot if someone can help me with the following doubts :

    1. in the hybrid-pi model of a BJT when we short the base and collector terminals what is the output resistance when looking into the collector and how ?

    thank you for all the help
  2. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    I assume you have studied the nodal, mesh and loop methods for circuit analysis; is this the case?

    Show an attempt to solve the problem (set up nodal equations, for example), and someone will help you if you can't get a solution.
  3. kkazem

    Active Member

    Jul 23, 2009
    Hi ankitmcgill,
    Here's my stab at the problem you described. Using the Hybrid-Pi BJT model with the Base and Collector Shorted, we have h11=Rb, h12=V1/V2, therefore h12= 1, since V1=V2 (b shorted to c). h21 = beta (Ic/Ib), and h22=I2/V2 @ I1=0. Now, the output resistance would be 1/(h22), except that the base & collector are connected. So, to get Ro, the output impedance, we have to compute the parallel combination of Rb = h11 & Re=(1/h2). This is as follows: Ro=[(V1/I1)*(V2/I2)]/[(V1/I1)+(V2/I2)]. But since b & c are shorted, V1=V2, so we can simplify as follows: Ro=[V1^2/(I1*I2)]/[(V1/I1)+(V2/I2)]. We can further simplify byclearing the denominator by multiplying top & bottom by (I1*I2) as follows: Ro=V1^2/[(V1*I2)+(V1*I1)]. Factoring the denominator, we get: Ro=V1^2/[V1*(I1+I2)] Further simlifying we get Ro=V1/(I1+I2)= V1/(I1+beta*I1)= V1/(I1)*(Beta+1). However, i may make more sense to write it in terms of V2 & I2. So, well start with the above eqn: Ro=V1/(I1+I2) = V2/(I1+I2)=V2/[I2+(I2/beta)]. Finally, we get Ro=(V2*Beta)/[I2*(Beta+1)]. And since Beta is usually 100 or greater, we can ignore the additional "1" in the (Beta+1) factor and simplify to: Ro=(approx)V2/I2, which is basically what your R_out = v_test / i_test means. I hope that helps.