0
\$\begingroup\$

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\$.

enter image description here

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?

\$\endgroup\$
1
  • \$\begingroup\$ 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. \$\endgroup\$
    – G36
    Nov 24, 2023 at 21:03

2 Answers 2

0
\$\begingroup\$

Some guidance

Y12 tells how output voltage affects the input current. This circuit has Y12=0 because you'll find that Ii=Vi/input impedance. You need only to make a normal circuit analysis equation and you'll see that the output voltage doesn't at all exist in the formula for Ii. The input impedance will be an expression of Re, Rb and Gm. Most of us probably would say Y12=0 because there's no feedback from the output to the input.

\$\endgroup\$
2
  • \$\begingroup\$ 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. \$\endgroup\$
    – EE18
    Nov 24, 2023 at 20:28
  • \$\begingroup\$ Writing the proper circuit analysis equation for Vi is the rigorous proof for what I said. No other rigorous proof exists for it, no matter how hard you want and search it. When you get Ii = Vi divided by an expression of Rb, Re and Gm you should see that adding Y12 * Vo is possible for all Vo values only if Y12=zero. Nothing is more rigorous. Formally you can subtract equation Ii=Vi/Zin from equation Ii=Vi/Zin + (Y12)Vo. BTW Ii is zero only if Vi =0. \$\endgroup\$
    – unawriter
    Nov 24, 2023 at 20:54
0
\$\begingroup\$

In every good text on the analysis of electricity networks, there is at least one chapter relating to quadripoles. Y parameters are also considered shorted parameters.

enter image description here

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.