Figure P 10.41 shows the cross section of an NMOS device...

Figure $\mathrm{P} 10.41$ shows the cross section of an NMOS device that includes source and drain resistances. These resistances take into account the bulk $n^{+}$ semiconductor resistance and the ohmic contact resistance. The current-voltage relations can be generated by replacing $V_{G S}$ by $V_{G}-I_{D} R_{S}$ and $V_{D S}$ by $V_{D}-I_{D}\left(R_{S}+R_{D}\right)$ in the ideal equations. Assume transistor parameters of $V_{T}=1 \mathrm{~V}$ and $K_{n}=1 \mathrm{~mA} / \mathrm{V}^{2} .(a)$ Plot the following curves on the same graph: $I_{D}$ versus $V_{D}$ for $V_{G}=2 \mathrm{~V}$ and $V_{G}=3 \mathrm{~V}$ over the range $0 \leq V_{D} \leq 5 \mathrm{~V}$ for $(i) R_{S}=R_{D}=0$ and $(i i) R_{S}=R_{D}=1 \mathrm{k} \Omega$. (b) Plot the following curves on the same graph: $\sqrt{I_{D}}$ versus $V_{G}$ for $V_{D}=0.1 \mathrm{~V}$ and $V_{D}=5 \mathrm{~V}$ over the range $0 \leq I_{D} \leq 1 \mathrm{~mA}$ for $(i) R_{S}=R_{D}=0$ and $(i i) R_{S}=R_{D}=1 \mathrm{k} \Omega$

Transcript

00:01 Hello students, our question is, so the cross -section of an n -mos device that includes source and drain resistance.

00:14 This resistant deck into account, the bulk and plus semiconductor resistance and plus semiconductor resistance and the omic constant resistance.

00:23 The current voltage relation can be generated by replacing vjs by vg minus idrs.

00:29 So we have to solve it and float the following curve on the same graph.

00:33 Solve this sum values okay so let's see it's the first vg is equal to 3b which is given to us and rd is equals to r s is equal to z hence vds is equal to vd is equal to id r s plus rd so from here we can say that vds is equal to vd and if we say the g s part so this is and this is and this is vgg is equals to vg minus idrs and from here the vg is equals to vg.

01:21 The reason is the rs will be 0 because as you see this rs and rd will be 0 so this term will be the 0 and the vgs is equal to vg.

01:30 So now from here the value of it can be 3 volts which is given in the quotient.

01:37 So let's see it now vgs minus v input so that will be is equal to 3 minus 1 and this is equal to 2 volt and from here for vds is less are equal than vgs minus vg n so from here we can say that vds is lesser are equal than 2 as we know that this value is 2 so this is laser are equal than 2 and the device would be in prior reason and for vds is greater than vgs minus v input the device would be in saturation and in this condition the vds is greater so so here the value of id when it is saturated so this will be k by 2 vgs minus vn and it is square and this is equal to the value of 1 is 1 meter by 2 and 3 minus 1 and square so this is equals to 2 mm.

02:54 This is a value of id.

02:55 If you see the graph, the graph is let's draw it graph so this is the i and in x -axis this is bd and this is the two lines which we got v0 is equal to 3 volt and at v0 is equal to 2 world.

03:12 Let's solve the second part which is r d is equal to this is our second a part okay.

03:18 Rd is equal to rs and that is equal to 1 kilo -oom and rd and r -d is equal to 1 kilo -oom.

03:28 Before it we have put r -d and r -s is equal to 0 but here we are putting a r -d is equal to rs is equal to 1 kilo -m as mentioned in the question and vg is equal to 2 both.

03:42 Hence v -d -s is equal to v -d minus id rs plus r -d.

03:54 And this is equal to vds is equal to vd and vgs is equal to vg minus id rs so this is is equal to 2 minus 1k id and for saturation this will be id saturation is k by 2 the formula will be same okay k by 2 vgs minus v in and whole square and this is is equal to 1 by 2 2 minus here its value is i d as multiply by 10 to the power 3 minus 1 and whole square so this is is is equal to 0 .5 so this is is equal to id is equal to 3 .75 millie ampere are 0 .33 millie ampere as we can say that id saturated is equals to ids okay so and here vjs is equals to 2 minus 1k 3 .8 meter equals to minus 1 .8 volts and then we see vgs we are putting here two values which is like two values of id that's why the here is two value of vgs and here is two value of vgs and here is is 2 minus 1k 0 .33 meter and that is is equal to 1 .67 volt so here is hence id is equal to 0 .33 millamper is considered as gate charge voltage mean to be so positive in case of nmospate okay and was paid for linear reason so vgs minus v n is greater than vds so we have this value 1 .67 minus 1 it is greater than vds.

06:04 So we have 1 .0 .67 is greater than vds.

06:11 And transource voltage vds is equals to vd minus minus id rs plus rd.

06:21 So here when we put the all values, this value will be 1 .33 volt, the value of vd.

06:27 So when we put here all the values from here the value of vd will be 1 .33 when we put this value vds, id, r s and rd...

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