How do I get a KCL equation involving Vin in this circuit?

Question

^{simulate this circuit – Schematic created using CircuitLab}

I've been scratching my head with this one. The circuit below is a small signal equivalent circuit of a PNP CB amplifier. I'm trying to get the transfer function Vout/Vin. I was already able to find Vout/Vbe using KCL across the node Vout. Now i'm stuck with the equation for Vin. Could anybody help me get started?

Here is what I got for Vout/Vbe where Rout = R3//R4:

$$\frac{v_{out}}{v_{be}}= \frac{g_{m}R_{out}*1-\frac{sC_{2}}{g_{m}} }{1+sR_{out}C_{2}} $$

Edit: I made quite the mistake in the schematic.

Edit(2): Here is the old schematic:

Answer 1

How do I get a KCL equation involving Vin in this circuit?

KCL equations involve currents, not voltages, that said this might be the answer you are looking for..

In the case of your circuit there is only one KCL equation which can be made to involve \$ Vin \$, namely the sum of the currents into the node between \$ R1 \$, \$ Rin \$ and the current source: \$ I_{R1} + I_{Rin} + gm V_{be} = 0\$ (1)

now from ohm's-law; \$ I_{R1} = \frac{V_{R1}}{R1} \$ (2) and \$ I_{Rin} = \frac{V_{Rin}}{R_{in}} \$ (3)

substituting (2) and (3) into (1); \$ \frac{V_{R1}}{R_{1}} + \frac{V_{Rin}}{R_{in}} + gm V_{be} = 0 \$ (4)

to get \$ V_{in} \$ involved we can apply KVL to the mesh involving the voltages across the two resisors; \$ V_{in} - V_{Rin} + V_{R1} - V_{be} = 0 \Leftrightarrow V_{in} + V_{R1} - V_{be} = V_{Rin} \$ (5)

substituting (5) into (4); \$ \frac{V_{R1}}{R_{1}} + \frac{V_{in} + V_{R1} - V_{be}}{R_{in}} + gm V_{be} = 0 \$ (6)

And now you have a KCL equation (6) involving \$ V_{in} \$

Here the reference directions are taken such that all currents into the node are possitive, and the voltage reference directions follow from the standard conventions.

EDIT; The OP made a mistake in the schematic,

after the correction equation (1) has to be; \$ I_{C1} + I_{R1} + I_{Rin} + gm V_{be} = 0\$ (1b)

and equation (6) becomes; \$ I_{C1} + \frac{V_{R1}}{R_{1}} + \frac{V_{in} + V_{R1} - V_{be}}{R_{in}} + gm V_{be} = 0 \$ (6b),

to find \$ I_{C1} \$ we need to use that; \$ I_{C1} = C_1 \frac{d}{dt} V_{C1} \$ (7),

since \$ V_{C1} = V_{R1} \$ (8) substituting (8) into (7) we can write this as; \$ I_{C1} = C_1 \frac{d}{dt} V_{R1} \$ (9).

substituting (9) into (6b) yields; \$ C_1 \frac{d}{dt} V_{R1} + \frac{V_{R1}}{R_{1}} + \frac{V_{in} + V_{R1} - V_{be}}{R_{in}} + gm V_{be} = 0 \$ (10)

or if you want the solution in the laplace/s -domain; \$ s C_1 V_{R1} + \frac{V_{R1}}{R_{1}} + \frac{V_{in} + V_{R1} - V_{be}}{R_{in}} + gm V_{be} = 0 \$ (10b).

Comment; Please do not remove your previous picture etc, but add the new one as an edit!