# Solving an equation for semiconductors

hey my lecturer put this example up for an exam tomorrow, could someone please explain how he gets to the 3rd line? is he using factorization?

$$V_{\mathrm{dsb}}=V_{\mathrm{gsb}}-V_{\mathrm t}$$

$$V_{\mathrm{dsb}}=V_{\mathrm{dd}}-(R_{\mathrm d}/2)K_n(V_{\mathrm{gsb}}-V_{\mathrm t})^2$$

$$V_{\mathrm{gsb}}=V_{\mathrm t}+(\sqrt{2K_{\mathrm n}R_{\mathrm d}V_{\mathrm{dd}}+1}-1)/K_{\mathrm n}R_{\mathrm d}$$

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I'm sorry, but your equation is unreadable right now. I'm guessing some of the letters are supposed to be subindices. Are the variables supposed to be $V_{dsb}$, $V_{dd}$, $R_d$, $K_n$, $V_{gsb}$, and $V_t$? –  Arturo Magidin Sep 22 '11 at 13:16
I formatted the equations. It was rather non-trivial to guess what you meant; please check that everything is as it should be. You can right-click on the equations and select "Show Source" to see how to do the formatting so you can do it yourself next time. –  joriki Sep 22 '11 at 13:28

## 1 Answer

The first equation states that the expression being squared in the second equation is $V_{\mathrm{dsb}}$. Thus the second equation becomes

$$V_{\mathrm{dsb}}=V_{\mathrm{dd}}-(R_{\mathrm d}/2)K_nV\;_{\mathrm{dsb}}^2\;.$$

This is a quadratic equation for $V\;_{\mathrm{dsb}}$. Substituting one of its two solutions into the first equation yields the third equation.

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sorry about the formatting, thanks for the help it just wasn't clicking, and thanks for the formatting help –  michael Sep 22 '11 at 14:05