Rpi model Of Transistor

Until now we have seen the h-model of BJT and based on that we had designed an amplifier. Now we are going to see the rπ model of BJT in depth. But first of all, you need to understand why we are using the rπ model if there is an h-model of the transistor. This is because the h-model fails if you are working at higher frequencies at around 1Mhz, also the parameters which are used in the rπ-model define the actual operating conditions while the parameters present in the h-model don't reflect the actual operating conditions.

Here is the diagram of rpi-model:-

were rπ:-internal impedance of BJT
gm:-transconductance
Vpi:-voltage generated at rπ
ro:-resistance of the current source (gm*Vπ).....(Which is ideally infinite therefore we neglect it.)

Now we will see some derivation of some parameters which are present in the rπ model.
1) rπ:- Since it is an internal impedance of transistor, therefore, it is given as rπ=Vπ*Ib
rπ = β/gm
2) gm (Transconductance):-
For understanding this first you have to understand its definition. We will break it into 2 words (Trans)  and (Conductance). First, we will understand What is the definition of conductance and then the meaning of trans.
So as you know conductance [g] is inverse of resistance [r] i.e.
if r = v/i
then g = 1/r = i/v
and trans means relation between input and output parameters.Therefore we can say that
gm = i(output) / v(input).........(1)
Definition of transconductance(gm):- Change in Output current(Ic) with respect to the Input voltage(Vbe)
i.e. (gm = dIc/dVbe)..........(2)  {From 1}
But let's see its equation when we are dealing with transistors.
In diode, there was 1 equation Id=Io*[e^Vd/Vt - 1]
In the case of BJT transistors, the equation would be Ic=Icbo*[e^Vbe/Vt - 1]
Where Ic is an output current
ICBO is current due to minority charge carriers.
Vt is a thermal voltage which is around 26mv.
If you want to know why is minority carriers and how does it affect the working of transistor than watch this video is given below.

Now,Ic = Icbo*e^Vbe/Vt - Icbo....(3)
Icbo*e^Vbe/Vt = Ic + Icbo....(4)
Now we will differentiate equation (3) with respect to Vbe
dIc/dVbe = d/dVbe * (Icbo*e^Vbe/Vt)
dIc/dVbe = (Icbo*e^Vbe/Vt) / Vt
dIc/dVbe = (Ic + Icbo) / Vt .......From equation(3)
Now, As I discussed in my earlier post where I had explained the concept of current due to minority charge carriers there we saw that Icbo will be only in a few nA(nano ampere) or uA(microampere). So in the above equation, we can neglect it.
dIc/dVbe = Ic/Vt......(5)
therefore gm = Ic/Vt.....(From 2 and 5)
Now you would be thinking that the h-model and rπ model are totally different but they are the same up to some extent.

As you can see in the h-model the current source is given by (hfe*Ib) and in the rπ model, it is given as (gm*Vpi)
so gm*Vpi = gm*(rπ*Ib) = gm*(β/gm)*(Ib) = b*Ib = hfe*Ib....Sine(hfe = β)
So you saw that the rπ model and h-model are almost the same.
I didn't cover the T-model of BJT so I'm attaching one video which will explain the significance of the

T-model over the r-pi model and the h-pi model.

I hope you understood this concept in the next post we will move for the next type of transistor i.e. JFET.