Vdd min = Vt,n + |Vt,p| ?

Status
Not open for further replies.
K

kweijun

Newbie level 3
Joined
Sep 12, 2005
Messages
3
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Location
Penang, Malaysia
Activity points
1,304
I've read this in a CMOS book. It stated that the minimum supply voltage, Vdd min, in order for a CMOS inverter to work is equal to nMOS threshold voltage, Vt,n + pMOS threshold voltage, Vt,p.

The question is,
Since only one of the MOS in a CMOS inverter will conduct at one time, the Vdd min should be ≥ Vt,n right?
 

There is no strict cut off point. Instead the current versus voltage dependence will transit into the exponential domain. If you simulate the selfoscillation of a ringoscillator over VDD you will get a linear dependence above your threshold, below you will get a exponential. This region is called subthreshold.
 

In CMOS one NMOS and one PMOS transistor connected in such a way that,
when NMOS turn-on PMOS must turned off,
and vice versa.
suppose when input voltage equal to Vtn-0.01V, then NMOS is turned off.
and PMOS must be turned on.
Thus VCC must (VCC>=|Vtp| +Vtn).

Regards,
Davood.
 

If the supply voltage is less than (Vtn+|Vtp|), there is a undetermined logic output.
e.g. when the Vdd=1V, input voltage=0.8V, Vtn=0.7 & |Vtp|=0.7.
 

Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…