# Performing Mesh Analysis on h-parameter transistor models

So I'm fairly comfortable with Mesh and Nodal analysis with numerical values, but I am having problems solving this circuit involving an alternative h-parameter model of a transistor:

simulate this circuit – Schematic created using CircuitLab

First as a note, I could not incorporate subscripts in the editor. Everything including hi, hr, v2 etc is subscripted at the second digit.

The question is to find the voltage gain (v1/v2) and the resistance (Req =v1/i1)

So I have establashed two meshes I1 and I2. These meshes are in the clockwise direction of both loops:

I1

-V1 + hi(i1) + hr • v2 = 0

I2

(hf/ho)•i1 + (1/ho)i2 + v2 = 0

So I simply solve for V1 and V2 on each mesh and then get the answer

V1 = i1•hi + hr•v2

V2 = -(hf/ho)•i1 - (1/ho)i2

So V1/V2 would simply be

-(i1•hi + hr•v2)/((hf/ho)•i1 - (1/ho)i2)

This doesn't really help me on finding Req, and I think I need more simplification. Am I on the right track with this problem? What am I missing to solve this problem properly? Thanks

• Just for future reference, and it doesn't really matter if you do this in your question... just letting you know about MathJax. Subscripts like $h_o$ <--- can be written as $h_o$... or if there's more than one character in the subscript, like $h_{o1}$ can be written as $h_{o1}$. – KingDuken Sep 10 '17 at 22:55
• Also, is (hf/h0)i1 supposed to be a dependent current source, instead of a dependent voltage source? – KingDuken Sep 10 '17 at 22:59
• That's how it was written but I guess that does turn it into a dependent current source – skyler Sep 11 '17 at 0:47

You have a loading resistor That connects I2 to V2 by Ohms law. That's your missing equation. You can substitute I2 = -V2/RL. Now you have 2 equations still left which have variables V1, I1 and V2. Eliminate I1 and you get finally the wanted ratio of V2 and V1.

(BTW: We others like to see V2/V1 named as the voltage gain, not V1/V2)

As well you can eliminate V2 and get V1/I1 which was the other wanted ratio.