# How do I compute $y_{12}$ in this circuit?

Consider the circuit below from Chapter 3.3 of Analysis and Design of Analog Integrated Circuits by Gray, Hurst, Meyer, and Lewis (the genesis of this question is my trying to understand whether this amplifier is bilateral or unilateral). Note that $$\i_o\$$ seems to be placed oddly within the amplifier. I will leave that as drawn and call $$\i_2 \$$ the current draw into the amplifier as a whole, as per the convention with y-parameter calculations. I'll likewise call $$\i_i =: i_1\$$, $$\v_i =: v_1\$$, $$\v_o =: v_2\$$.

To find $$\y_{12} := \frac{i_1}{v_2} \rvert_{v_1 = 0}\$$ I'll set $$\v_1 = 0\$$. From this I obtain by KCL at Node 1 that $$\frac{v_e}{r_e || r_b} = g_mv_e \,\,\,\,\,\,\,\, [1]$$ By KCL at the output node we find that $$g_mv_e + i_2 = v_2/R_c.$$

Further, we obviously have $$i_1 = -v_e/r_e.$$ Now I can't seem to get rid of $$\i_2\$$ in this calculation. It appears that if I do take $$\v_e = 0\$$ then I get a consistent solution (and then $$\y_{12} = 0\$$), but I can't convince myself of why I should have to choose that rather than prove it. Any help would be greatly appreciated.

*Edit: Should the argument be that since $$\frac{1/g_m}{r_e || r_b} =\frac{r_e+r_b}{g_mr_er_b} = \frac{\alpha_0/g_m+r_b}{\alpha_0r_b} = \frac{1/g_m + r_b/\alpha_0}{r_b} = 1/\alpha_0 + \frac{1}{g_mr_b} > 1 + \frac{1}{g_mr_b} \neq 1$$ so that equation $$\[1]\$$ above demands $$\v_e = 0\$$. Is there an easier way to see this argument?

• Well for this circuit I1 will be 0A. Because we have V1 is 0V, the Ve will also be 0V. Thus, our VCCS will be off (V2 cannot turn on VCCS). To turn it on you need to add a resistor in parallel with VCCS. Now with this extra resistor (ro), the VCCSC will be turn on by V2 voltage.
– G36
Commented Nov 24, 2023 at 21:03

• Thanks for your answer :) I'm not quite sure it answers as my question though, as I'm looking for a rigorous demonstration that, indeed, $i_1 = 0$ in the circuit above. I get intuitively that $y_{12} = 0$ but want to show it explicitly.