# Converting between model T and Hybrid Pi BJT models

#### blah2222

Joined May 3, 2010
582
Hi all, just looking through some old notes and I remember the prof going over the conversion between the two but how is the one made from the other? He moved around circuit components in a couple steps from T to Pi showing the transition, but I can't see it.

Any help would be great.
Thanks!

#### WBahn

Joined Mar 31, 2012
29,494
The key is in understanding where the two models come from in the first place and what each of the elements represent. From there it is a pretty simple matter (particularly if we are talking about the most basic versions consisting of a single resistor and a single dependent current source) of analyzing the two circuits to find the relationships for v_be, i_b, and i_c, setting them equal, and then solving for the parameters of one model in terms of the other.

#### blah2222

Joined May 3, 2010
582
Okay, so I understand in terms of the small-signal model that the incremental resistance re is modelled off of the BE diode and the dependent current source is just the forward active relationship between Ib and Ic.

From this I can see how the T-model is formed. Mathematically I can prove why both models are representing the same thing, but I am wondering how you are able to move the components from the T-model configuration into the Pi-model configuration.

#### WBahn

Joined Mar 31, 2012
29,494
I don't understand what you mean by "moving" components from one configuration into another. Let's say we each have a black box with three terminals and inside is a resistor and a dependent current source. I take my black box and choose the component parameters for the resistor and the current source and I tell you which model (pi or T) I have hooked my components up in. Your job is to hook your components up in the other configuration and choose your component parameters such that, with the boxes closed, no one can tell the difference (at least not based on current/voltage measurements taken at the terminals).

If that's not good enough, then I guess here's a way that you could imagin converting between the two:

Take the hybrid-pi model as the starting point. Now, we can insert a resistor between the current source and the emitter without changing anything since a current source hides the resistor from the collector terminal (i.e., it will generate whatever voltage across it is necessary to compensate for the presence of the resistor). So let's choose the value of this resistor such that the voltage across it is v_be. Since the current is gm*v_be, this resistor has to have a value of 1/gm. Now, since the resistors are tied together at the emitter and since they always have the same voltage drop across them, we can install a shorting jumper between the base terminal (top of r_pi) and the bottom of the current source and no current will flow in it (zero voltage across is), meaning that we haven't changed anything. These two resistors are now in parallel and we can combine them into a single resistor, r_e. We have just converted from the hybrid-pi to one of the common versions of the T model. Starting from the T-model, we would just follow this process in reverse.

#### blah2222

Joined May 3, 2010
582
Thanks for the visual interpretation!

I think another way of looking at it and what I'm realizing now is that my prof displayed each model as a two-port model.

Where the Hybrid-Pi is a common-emitter two-port and the T-model is a common-base.

I'll look through my notes again and try to do the proofs myself, but it is becoming a lot more clear.

Thanks again!

#### blah2222

Joined May 3, 2010
582
Hybrid-Pi Two-Port Model (small-signal):

$$i_{b} = g_{\pi}v_{be} + g_{r}v_{ce}$$

$$i_{c} = g_{m}v_{be} + g_{o}v_{ce}$$

$$g_{\pi} = \frac{i_{b}}{v_{be}}|_{v_{ce=0} = \frac{\partial i_{B}}{\partial v_{BE}}|_{Q-point}$$

$$g_{r} = \frac{i_{b}}{v_{ce}}|_{v_{be=0} = \frac{\partial i_{B}}{\partial v_{CE}}|_{Q-point}$$

$$g_{m} = \frac{i_{c}}{v_{be}}|_{v_{ce=0} = \frac{\partial i_{C}}{\partial v_{BE}}|_{Q-point}$$

$$g_{o} = \frac{i_{c}}{v_{ce}}|_{v_{be=0} = \frac{\partial i_{C}}{\partial v_{CE}}|_{Q-point}$$

Where:

$$i_{C} = I_{S}[exp{\frac{v_{BE}}{V_{T}}}][1 + \frac{v_{CE}}{V_{A}}]$$

$$i_{B} = \frac{I_{S}}{\beta_{F}}[exp{\frac{v_{BE}}{V_{T}}}][1 + \frac{v_{CE}}{V_{A}}]$$

<tedious calculations>

$$g_{m} = \frac{I_{C}}{V_{T}} ~= 40I_{C}$$

$$g_{r} = 0$$

$$g_{\pi} = \frac{g_{m}}{\beta_{0}}$$

$$g_{o} = \frac{I_{C}}{V_{A} + V_{CE}} ~= \frac{I_{C}}{V_{A}}$$

Do the same thing for the common-base configuration, you get gm=gm and re=alpha/gm, which is correct!

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