If gate (NMOS) is connected to Ground how is current able to flow

[PalmTreeSD]

I am confused about this example circuit (attached) where the Gate it connected to Ground.
I thought in order for current to flow in a MOSFET (i.e. NMOS) a positive gate voltage (Vg) AND Vds (drain to source voltage) needs to be applied. If Vg = 0, how is a channel created in order to allow current to flow in this circuit if this is not a depletion-type MOSFET?
I only have a guess as to how it is possible for current to be able to flow.

Thanks in advance

 

[gnurf]

Looks like a typo, the FET in your attached figure is a PMOS.

 

[gnurf]

Err, I was about to link you to this wiki page

http://en.wikipedia.org/wiki/MOSFET#Circuit_symbols

when I realized the symbol in your assignment very much resembles what they call a N-channel MOSFET enh (no bulk) on that wiki page. I thought arrow pointing out always meant PMOS, but apparently not ... ?

(At least my advice is cheap.)

 

[Phrak]

Why don't you add 5 volts to all three nodes--it will change nothing but voltage offsets, and see how it looks.

 

[PaulS1950]

Um, the grounded gate is 5 volts positive as compared to Vss (-5v)
If you think of it as:
Vdd = 10v
Vgate = 5v
Vss = 0v
does that explain it better?

Paul, the 60 year old student

 

[rajeshpamula]

It is Ok. The transistor is NMOS and it is operating in triode region. For transistor to conduct in strong inversion, the condition is gate to source voltage has to be greater than threshold voltage. I said that it is in linear region because Vd < Vg - Vt. Now using the MOS equation is triode region, the problem can be solved.

 

[lostinxlation]

It is because Vss is negative.. To turn on the nMOS, gate-source or gate-drain voltage must be more than the threshold. In this case, you have more than threshold voltage between the gate and Vss, therefore, nMOS turns on.