Continue to Site

# Need help: nmos operating question

Status
Not open for further replies.

#### Fei

##### Newbie level 6
Hi,

I am confused with the nmos operating method.

The circuit is just a "R + nmos". S and B of nmos are tied to ground, Drain is connected to the Resistor, the other terminal of R is floating. Vg change from H to L at 5ns.
Netlist is like this:
r1 d1 d 100 \$[RP]
m1 d g s s nmos L=0.35u W=40u
vg g 0 pwl(0n 3.3 5n 3.3 5.2n 0)
vs s 0 0

Could you explain why the drain voltage is droped?
Why is the i(vs) so big?

Thanks.

capacitive coupling between gate and drain.
You have a floating cap (Cgd).
Next charge in the channel will get distributed eually between drain and source as MOS was ON so hopfully it look to be in linear region.

### Fei

Points: 2

I can understand the effect caused by Cgd. But I can't understand the effect of the charge in channel.

My understanding is that when Vg is lower, the charge in channel will go back to the depletion region.

Can you explain more? thanks.

basically, you should consider that the MOSFETs used are not ideal. Many equations should make sense to us, who only look at the ideal transistor or regards it with 1 level of non ideality. (that is the capacitors that you were talking about here). But when you consider that HSpice simulates your circuit very very accurately then thing are not that strange. (this is what we mean by saying level = 49; 49 levels of accuracy)

Things can be teken into two parts: (switching effect and charges in the channel)
1. At beginning, Vd=Vs=0, then Vg starts to decrease. Befoer Vg hit the Vth, channel is still there, which means the drain and source are connected by a on resistance (on resistance values changes). since Vg changes dramatically at this period, it make drain voltage changes as well. High freq signal pass through cap. This is one part of the current. The other part of the current is from the discharge of the Cgs.
2. Charges in the channel. In my opinion, they are still flow to decrease the drain-body depletion region. Since the drain voltage become negative, which almost make the drain-body diode forward biased. It needs charges to fill the depletion region. That's the place where the channel charges go.

Thanks alot gusy

Status
Not open for further replies.