3.10 Jfet

There is a lot more that we could do with field effect devices, but it is probably time to move on to new topics. For one finalpoint however, we might just look at something called the JFET, or junction field effect transistor. The JFET structure lookslike . It consists of a piece of p-type silicon, into which two n-type regions have beendiffused. However, instead of being both on the same surface, as with a MOSFET, the two regions are opposite one another oneither side of the crystal. In cross-section, the JFET looks like . We also show the biasing here.

The two n-regions are connected together, and are reverse biased with respect to the p-type substrate. A second battery, $V_{\mathrm{ds}}$ is used to pull current out of the source, by applying a negative voltage between the drain and thesource. The reverse biased n-p junctions creates a depletion region which extends into the p-type material through which theholes travel as they go from source to drain (a channel?). By adjusting the value of $V_{\mathrm{gs}}$ , one can make the depletion region smaller or larger, thus increasing or decreasing the draincurrent.

The observant student will also note that the polarity of the $V_{\mathrm{ds}}$ battery makes it so that there is more reverse bias across the p-n junctions at the drain end of thechannel than at the source end. Thus, a more accurate depiction of the JFET would be what is shown in . When the drain/source voltage gets large enough, the two depletion regions will join together, and, justas with the MOSFET, the channel pinches off, as shown in .

Surprising as it may seem, when you work out the equations which describe how the depletion region extends with $V_{\mathrm{gs}}$ and how the pinch-off mechanism changes $I_D$ , you end up with behavior, and equations, which are quite similar to those of a depletion-modeMOSFET.

Using JFETs is a little more cumbersome than a normal MOSFET. You must make sure that the gate-substrate junctionalways remains reverse biased, and since the JFET can only be a depletion-mode device, you have to have a voltage on the gate ifyou want to turn the transistor off. The JFET does have one advantage over the MOSFET however. A while back we calculated the value for $C_{\mathrm{ox}}$ the oxide capacitance and found that it was on the order of $10^{-7}\frac{F}{\mathrm{cm}^2}$ . A typical MOSFET gate might be 1μm long by 20μm wide, and so it would have a gate area of 20 $\mathrm{\mu m}^2$ or $210^{-7}$ $\mathrm{cm}^2$ . Thus, the total gate capacitance is only about $10^{-14}F$ .