is it just me or is this EE hw impossible?

[Oscar1613]

"A MOSFET with gate dimensions W=2 micrometers, L = 4 micrometers has a gate oxide thickness of 1000 angstroms. Estimate the value of Cgs at Vds = 0V."

the equation for Cgs: Cgs = permittivity*Area/thickness of the oxide. now how the fsck am i supposed to know the permittivity if it's not stated in the problem and there is no table or appendix or any other means of determining it? the real kicker is this is the 4th problem in this hw assignment and 2 of the other 3 are missing vital information as well! :|:|:|:|

 

[konakona]

wish i knew enough to help u out... here is a bump

 

[RaynorWolfcastle]

permittivity is usually constant, since most MOS transistors use silicon oxide. Use e_ox = 3.9 x e_0 ~= 3.45x10^-11 F/m

 

[Martin]

I woudl suggest looking at what values the examples in the book use. Its probably the one they want you to use.

but yes, crappy wording.

 

[silverpig]

Leave it in terms of the permittivity?

 

[JonnyStarks]

Holy hell, what does that even mean?
Is the answer 42?

 

[RaynorWolfcastle]

Originally posted by: silverpig
Leave it in terms of the permittivity?

Damn physicists, EEs want numerical answers, nobody outside of academia wants a very specific tranisistor geometry to be parametrized for permittivity when the material used is nearly always the same. Like I said in my last post, silicon oxide is the standard dielectric.

 

[Oscar1613]

Originally posted by: Martin
I woudl suggest looking at what values the examples in the book use. Its probably the one they want you to use.

but yes, crappy wording.

thats the thing... there are no examples of this type in the book... they just derive the equations and move on to the topics they needed the equations in order to discuss them:roll:

Originally posted by: RaynorWolfcastle
permittivity is usually constant, since most MOS transistors use silicon oxide. Use e_ox = 3.9 x e_0 ~= 3.45x10^-11 F/m

hmm... what about q and epsilon for a pn junction diode?
Cdiode = A(q*epsilon/2 * Na*Nd/(Na+Nd))^.5 * (psi - Vd)^-.5

that problem states everything by q and epsilon, and once again no typical values stated in the text:roll: this problem has the answer in the back of the book so i'll be able to tell you if you're right

 

[her209]

You're right. You're f'ed.

 

[RaynorWolfcastle]

Originally posted by: Oscar1613
hmm... what about q and epsilon for a pn junction diode?
Cdiode = A(q*epsilon/2 * Na*Nd/(Na+Nd))^.5 * (psi - Vd)^-.5

that problem states everything by q and epsilon, and once again no typical values stated in the text:roll: this problem has the answer in the back of the book so i'll be able to tell you if you're right

that really depends on the type of diode, but assuming a garden-variety silicon diode, the permittivity will be the same. I don't recall the other equation, but if I had to guess I'd say that q is the electron charge (1.6x10^-19).

What book are you using btw?

 

[Mo0o]

wow props to you, this seems hard as hell.

 

[George P Burdell]

it is just you

 

[Oscar1613]

well those values didnt give the right answer

i'm using Microelectronic Circuits and Devices by Mark Horenstein

Originally posted by: Mo0o
wow props to you, this seems hard as hell.

most of its not this bad... mostly just applying formulas and concepts, and most of the time you're actually given enough info to do the problem tho :|

 

[Oscar1613]

Originally posted by: hotchilisauce
it is just you

well then how do you solve it oh wise one?:clock:

 

[NutBucket]

Originally posted by: RaynorWolfcastle

Originally posted by: Oscar1613
hmm... what about q and epsilon for a pn junction diode?
Cdiode = A(q*epsilon/2 * Na*Nd/(Na+Nd))^.5 * (psi - Vd)^-.5

that problem states everything by q and epsilon, and once again no typical values stated in the text:roll: this problem has the answer in the back of the book so i'll be able to tell you if you're right

that really depends on the type of diode, but assuming a garden-variety silicon diode, the permittivity will be the same. I don't recall the other equation, but if I had to guess I'd say that q is the electron charge (1.6x10^-19).

What book are you using btw?

For epsilon, assuming its a silicon diode, use 11.7

 

[RaynorWolfcastle]

Originally posted by: Oscar1613
well those values didnt give the right answer

i'm using Microelectronic Circuits and Devices by Mark Horenstein

Originally posted by: Mo0o
wow props to you, this seems hard as hell.

most of its not this bad... mostly just applying formulas and concepts, and most of the time you're actually given enough info to do the problem tho :|

I just busted out my electronics book. Looks like the material used is plain silicon for these diode; try 1.04x10^-12 F/cm <- make sure you convert length appropriately.

 

[white]

isn't it just 11.9 times the permittivity in a vac, which is a constant?

 

[simms]

Yay engineering!

<--- Chem

 

[Oscar1613]

Originally posted by: RaynorWolfcastle

Originally posted by: Oscar1613
well those values didnt give the right answer

i'm using Microelectronic Circuits and Devices by Mark Horenstein

Originally posted by: Mo0o
wow props to you, this seems hard as hell.

most of its not this bad... mostly just applying formulas and concepts, and most of the time you're actually given enough info to do the problem tho :|

I just busted out my electronics book. Looks like the material used is plain silicon for these diode; try 1.04x10^-12 F/cm <- make sure you convert length appropriately.

that did it... thanks! :thumbsup:

 

[MetalMat]

You dont have the solutions manual?

 

[NutBucket]

Originally posted by: white
isn't it just 11.9 times the permittivity in a vac, which is a constant?

Permittivity in free space, epsilon 0 is 8.85x10^-24 F/cm.

 

[NutBucket]

Originally posted by: simms
Yay engineering!

<--- Chem

BOOOO O-Chem!!!!!

 

[MetalMat]

I think I remember a problem like this in electronics exactly, I might go dig up the soloutions manuals to electronics 1and 2.