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Drain current in the linear region is given by \$I_D=k[(V_{gs}-Vt)V_{ds}-.5V_{ds}^{2}]\$.

Now transcoductance \$g_m=\frac{\mathrm{d}I_D}{\mathrm{d}V_{gs}}=kV_{ds}\$, so \$g_m\$ should be constant parallel to the x-axis, but no such option is given. Is there anything I am missing?

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  • \$\begingroup\$ Use the formula: \$ g_m = \frac{2I_{D}}{V_{GS}-V_{tn}} \$ ... then use that \$ I_D \$ formula \$\endgroup\$
    – user103380
    Oct 7, 2017 at 6:18

2 Answers 2

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In the sub-threshold region of operation the drain current increases exponentially with the gate voltage:

$$I_D\propto \exp\left(\frac{qV_G}{kT}\right)$$

This equation says that at very small \$V_G\$ the drain current is almost zero, hence D as well as C cannot be true.(C because it changes linearly with Vg). As drain current reaches Vt it rapidly grows up and approaches a maximum. Then it starts decreasing until a lower value. So the right answer is A.

Also the interesting thing is that a MOS device operating in weak inversion has a transconductance similar to that of a BJT operating in active region, because $$g_m=\frac{I_D}{\zeta V_T }.$$

It's tempting to use a MOS device working in weak inversion for high gain applications, but since it requires large device width or very low drain currents it will limit the speed of these circuits.

One typical use of transistors working in weak inversion region is in very low power applications. Of course, sub-threshold conduction can result in power dissipation or loss of information.

The equation for the drain current in weak inversion can be written more accurately as $$I_D=A.\exp\left(\frac{V_{gs}-V_T}{nV_T}\right),$$

where \$A\$ is a constant which depends on the MOS characteristics. Calculating \$ \frac{dI_D}{dV_{gs}}\$ gives \$g_m=\frac{I_D}{nV_T}.\$ This equation for \$g_m\$ obtained for the MOS operating in the weak inversion suggests that \$g_m\$ increases linearly with the drain current, and that it's independent of the gate voltage, $V_G\$. Note that the slope is quite steep and can be modeled as given in Fig. A. As you see it increases almost linearly until it exits the weak inversion area. At this time the transistor enters the moderate region where the above ratio is no longer true and is given by

$$ \frac{g_m}{I_D}=\frac{2}{V_{ov}},$$ where \$V_{ov}=V_{gs}-V_T.\$ Now as \$V_{gs}\$ increases the ratio falls as a function of \$ \frac{1}{V_{gs}}\$. This is why you see a decaying behavior after it reaches a maximum point.

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  • \$\begingroup\$ Why did you use the current equation in subthreshold? question is saying its in linear region.is the current equation i wrote in the linear region wrong? \$\endgroup\$
    – Rohit
    Oct 7, 2017 at 10:38
  • \$\begingroup\$ Your equation is true only for Vgs>Vth. It doesn't predict what happens at Vgs less than Vth. The question in fact combines both subthreshold and linear regions to conclude the overall behavior. \$\endgroup\$
    – dirac16
    Oct 7, 2017 at 10:51
  • \$\begingroup\$ according to the answer you have given its start increasing because of exponential nature of current its showing below threshold voltage..but why its decreasing after that? \$\endgroup\$
    – Rohit
    Oct 11, 2017 at 11:57
  • \$\begingroup\$ @Rohit I updated the post. Hope it now clarifies things for you. \$\endgroup\$
    – dirac16
    Oct 11, 2017 at 14:23
  • \$\begingroup\$ Hii,Thanks for undating...just one doubt i have after that i will accept it.:-).As far as i know the weak inversion region is at the point of thresold voltage also that graph is decresing after the thresold voltage,it that right? \$\endgroup\$
    – Rohit
    Oct 11, 2017 at 15:47
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Ideally, ๐‘”๐‘š should follow slope of Id-Vg curve. At higher Vgs, large transverse electric fields from gate to channel will pull the carriers more closer to oxide-channel interface, which will cause an effective mobility degradation, hence ๐‘”๐‘š falls again.

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