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(e) An n-type In$_{0.53}$Ga$_{0.47}$As epitaxial layer doped at 10$^{16}$ cm$^{-3}$ is to be used as a channel in a FET operating at 300 K. A decision is to be made whether the JFET or MESFET technology is to be used for the device. In the JFET technology a p$^{+}$ region can be made with a doping of 5$\times$10$^{17}$ cm$^{-3}$. In the MESFET technology a Schottky barrier with a height of 0.4 V is available. The materials parameters are: $D_p$ = 20 cm$^2$/s, $D_n$ = 50 cm$^2$/s, $L_p$ = 5$\mu$m, $L_n$ = 5$\mu$m, $n_i$ = 2.37 10$^{11}$ cm$^{-3}$, $A^*$ = 5 A cm$^{-2}$ K$^{-2}$. (i) Calculate the reverse saturation current for JFET. (ii) Calculate the reverse saturation current for the MESFET. (iii) Considering gate isolation issues, which technology should be used?

          (e) An n-type In$_{0.53}$Ga$_{0.47}$As epitaxial layer doped at 10$^{16}$ cm$^{-3}$ is to be used as a channel in a FET operating at 300 K. A decision is to be made whether the JFET or MESFET technology is to be used for the device. In the JFET technology a p$^{+}$ region can be made with a doping of 5$\times$10$^{17}$ cm$^{-3}$. In the MESFET technology a Schottky barrier with a height of 0.4 V is available.
The materials parameters are:
$D_p$ = 20 cm$^2$/s, $D_n$ = 50 cm$^2$/s, $L_p$ = 5$\mu$m, $L_n$ = 5$\mu$m, $n_i$ = 2.37 10$^{11}$ cm$^{-3}$,
$A^*$ = 5 A cm$^{-2}$ K$^{-2}$.
(i) Calculate the reverse saturation current for JFET.
(ii) Calculate the reverse saturation current for the MESFET.
(iii) Considering gate isolation issues, which technology should be used?
        
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(e) An n-type In0.53Ga0.47As epitaxial layer doped at 10^16 cm^-3 is to be used as a channel in a FET operating at 300 K. A decision is to be made whether the JFET or MESFET technology is to be used for the device. In the JFET technology a p^+ region can be made with a doping of 5×10^17 cm^-3. In the MESFET technology a Schottky barrier with a height of 0.4 V is available.
The materials parameters are:
Dp = 20 cm^2/s, Dn = 50 cm^2/s, Lp = 5μm, Ln = 5μm, ni = 2.37 10^11 cm^-3,
A^* = 5 A cm^-2 K^-2.
(i) Calculate the reverse saturation current for JFET.
(ii) Calculate the reverse saturation current for the MESFET.
(iii) Considering gate isolation issues, which technology should be used?

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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(e) An n-type In_(0.53)Ga_(0.47) As epitaxial layer doped at 10^(16)cm^(-3) is to be used as a channel in a FET operating at 300K. A decision is to be made whether the JFET or MESFET technology is to be used for the device. In the JFET technology a p^(+) region can be made with a doping of 5\\times 10^(17)cm^(-3). In the MESFET technology a Schottky barrier with a height of 0.4V is available.\nThe materials parameters are:\nD_(p)=20c(m^(2))/(s),D_(n)=50c(m^(2))/(s),L_(p)=5\\mu m,L_(n)=5\\mu m,n_(i)=2.3710^(11)cm^(-3),\nA^(**)=5Acm^(-2)K^(-2)\n(i) Calculate the reverse saturation current for JFET.\n[2]\n(ii) Calculate the reverse saturation current for the MESFET.\n1\n(iii) Considering gate isolation issues, which technology should be used?\n[1] (e) An n-type Ino.53Gao.47As epitaxial layer doped at 1016 cm-3 is to be used as a channel in a FET operating at 300 K. A decision is to be made whether the JFET or MESFET technology is to be used for the device. In the JFET technology a p region can be made with a doping of 51017 cm-3. In the MESFET technology a Schottky barrier with a height of 0.4 V is available. The materials parameters are: Dp=20cm2/s,Dn=50cm2/s,Lp=5um,Ln=5um,ni=2.371011cm3 A*=5Acm-2K2 (i) Calculate the reverse saturation current for JFET. (ii) Calculate the reverse saturation current for the MESFET (iii) Considering gate isolation issues, which technology should be used? [2] [1] [2]
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Transcript

-
00:01 Hi, this is a question based on jfet.
00:04 So, here we have the first statement is a true statement because the gate source junction is always reversed by us.
00:15 So, vgs must be negative.
00:16 The second statement is also true statement because we have the drift current or drift current has the value idss into 1 minus vgs by vp whole square.
00:28 Then the third statement is a false statement.
00:33 So, we have the value of gain is different.
00:37 So, now we have the fourth statement is a true statement because they obey square law.
00:44 So, then we have the fifth statement.
00:47 So, fifth statement is again a true statement.
00:50 Then we have sixth statement is also true statement and seventh statement is a true statement that all mosfets are subjected to damage from the electrostatic discharge...
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