MOS Inverter

MOS INVERTER

3.1 General circuit of an nMOS inverter

Voltage Transfer Characteristic (VTC)

3.2. Voltage transfer characteristic

* In VTC for every low input voltage levels, the output voltage Vout is equal to the high value of VOH (output high voltage).

* In the above case, the driver nMOS transistor is in cut-off and hence, does not conduct any current.

* As the input voltage Vin increases, the driver transistor starts conducting a certain drain current, and the output voltage eventually starts decreasing.

* There are two circuital voltage points on this curve, where the slope of the Vout(Vin) characteristic becomes equal to -1 i.e.,

dVout/dVin =-1

* The smaller input voltage value satisfying this condition is called the input low voltage VIL, and the larger input voltage satisfying this condition is called the input high voltage VIH.

* The output voltage continues to drop and reaches a value of VOL (output low voltage) when the input voltage is equal to VOH.

* The inverter threshold voltage Vth, is defined as the point where Vin = Vout on the VTC.

* The functional definition for these critical voltages are given below:

o VOH: Maximum output voltage when the output level is logic”1”.

o VOL: Minimum output voltage when the output level is logic”0”.

o VIL: Maximum input voltage which can be interpreted as logic”0”.

o VIH: Minimum input voltage which can be interpreted as logic”1”.

Noise Immunity and Noise Margins

* The noise immunity of the circuit increases with noise margin (NM).

* Two noise margins will be defined: the noise margin for low signal levels (NML) and the noise margin for high signal levels (NMH).

NML = VIL – VOL

NMH = VOH - VIH

Type of Inverters

1. Resistive load inverter

2. nMOS load inverter

3. cMOS inverter

Resistive Load Inverter

3.3 Resistive load inverter circuit

I. Operation region of the driver transistor in the resistive load

II. Current equation

IL = ID

IL = (VDD – Vout)/RL

ID (cut off) = 0

ID (linear) = [2(Vin – VTo)Vout - V²out] Kn / 2

ID (saturation) = [Vin – VTo] ² Kn/2

III. Calculation of VOH

Vin = VOL (given)

Vout = VOH (given)

Region of operation

Check Vin and Vout

Vin <>out

Therefore, cut off region (from operation region table)

Current equation

IL = ID (cut off)

(VDD-Vout)/RL = 0

VDD = Vout

Therefore, VOH = VDD

IV. Calculation of VOL

Vin = VOH (given)

Vout = VOL (given)

Region of operation

Check Vin & VTo

VOH > VTO

Vout <>in - VTO

Therefore, linear region (from operation region table)

Current equation

IL = ID (linear)

(VDD – Vout)/RL = [2(Vin – VTo)Vout - V²out] Kn / 2

on solving we get

VOL = VDD – VTO + (1/KnRL) – [(VDD – VTO + 1/KnRL)² – (2VDD/KnRL)]½