Question: Below figure represents an approximate equivalent circuit of a field effect transistor amplifier. Using KCL find the output voltage \$v_2\$ in terms of input voltage \$v_1\$.

My try: I redrew the circuit as:

Here, \$i_1=\frac{v_1}{r+r_g}\$, \$i_2=\frac{v_2}{r_d}\$, \$i_3=\frac{v_2}{r_l}\$ and \$v_g=\frac{v_1r_g}{r+r_g}\$. I applied the KCL at node x, $$i_1+\alpha v_g+i_2+i_3=0\\ \implies \frac{v_1}{r+r_g}+\frac{\alpha v_1r_g}{r+r_g}+\frac{v_2}{r_d}+\frac{v_2}{r_l}=0\\ \implies \frac{v_1}{r+r_g}(1+\alpha r_g)=-v_2\left(\frac1r_d+\frac1r_l\right)\\ \implies \frac{v_1}{r+r_g}(1+\alpha r_g)=-v_2\frac{r_l+r_d}{r_lr_d}\\ \implies v_2=-\frac{v_1r_lr_d}{(r+r_g)(r_l+r_d)}(1+\alpha r_g)$$ But, in my book the answer is: $$v_2=-\frac{\alpha r_gv_1r_dr_l}{(r_g+r)(r_d+r_l)}$$ Please check my mistake.

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    \$\begingroup\$ Hint: all current flows in loops. Where is the return for \$ i_1 \$? Does that help you give an exact value for \$ i_1 \$? \$\endgroup\$ – Transistor Aug 2 at 18:22
  • \$\begingroup\$ You can solve this problem by considering the two "parts" of the circuit as a voltage divider and a current divider, respectively. \$\endgroup\$ – Carl Aug 2 at 18:28
  • \$\begingroup\$ This circuit is not a good approximation for a MOSFET.MOSFET V/I diagram is a curve it is not linear. \$\endgroup\$ – Helena Wells Aug 3 at 0:16
  • \$\begingroup\$ @Transistor I got it. Thank you very much😊 \$\endgroup\$ – Ankita Pal Aug 3 at 0:53

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