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Could you explain why when bulk voltage is negative (depletion region charge increases) as shown in the picture the threshold voltage Vth increases? enter image description here

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The source to body voltage has an direct effect on the threshold of the MOSFET. The relation between the two is given by the Shichman-Hodges model as

enter image description here

So, as the body voltage increases (positive), the effective threshold voltage of the MOSFET decreases.Hence for the same gate to source voltage a greater inversion occurs in the channel.

To put it in simple terms,

The negative voltage you apply at the body actually pushes more minority carriers in the substrate towards the channel area which is relatively at a higher voltage because of the applied gate voltage.

edit:

Since the minority carriers in the substrate move towards the channel region, they recombine with the holes to result in more depletion layer charge across the channel.

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  • \$\begingroup\$ Thank you. In this case S is grounded. So the more negative B is the more Vsb and the larger Vth as shown in the formula. However, I don't understand why large Vsb causes large Vth physically. \$\endgroup\$
    – emnha
    Nov 17 '15 at 9:35
  • \$\begingroup\$ check my edited answer and do mark the answer as complete if your doubt is cleared. \$\endgroup\$ Nov 17 '15 at 12:55
  • \$\begingroup\$ Hi, could you explain why more depletion layer charge across the channel will increase threshold voltage physically? \$\endgroup\$
    – emnha
    Nov 17 '15 at 13:59
  • \$\begingroup\$ Its because we more recombination of the charge carriers that occur at device level, i dont think there aint any other simple explanation for this and also what actually do you mean by PHYSICALLY??? \$\endgroup\$ Nov 17 '15 at 15:58
  • \$\begingroup\$ Hi, I meant that when depletion is enlarged why Vt increases. Actually, I would like to understand it based on physics not just the formula above. Why larger depletion region make it hard for the device to turn on? \$\endgroup\$
    – emnha
    Nov 18 '15 at 1:26
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Presence of the depletion region means more(fixed) charges in our would-be channel region. Now before the inversion channel is actually formed, the oxide AND the depletion layer act like two of capacitors in series - both layers having different dielectric properties, with the gate and the substrate acting as the plates sort of. Before Vgs is sufficiently large to form the inversion channel(i.e.,before reaching our Vt), the gate electrode has to exactly "mirror" the charge that is present in the substrate - the whole point of a CAPACITOR.

Hence, the more charge you have on the substrate, the more charge you need to put on the gate electrode - which implies an application of a greater Vgs. Hence the presence of the greater Vt.

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