By Linda E. M. Brackenbury
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Extra info for Design of VLSI Systems — A Practical Introduction
3 V. Thus Vcts < V gs- Vte and so T1 operates in the resistive region. A where (WI/Lt) is the aspect ratio of transistor Tl. T2 is on and V ds = Vgs. Hence the inequality V ds > Vgs - Vte is true and T2 is saturated. The gate-source voltage of T2 is VP- Vout and since Vout is low, the body effect can be neglected. 4 L2 2 L2 where (W 2 /L 2 ) is the aspect ratio ofT2. Ids ofTl and T2 are equated to obtain an inverter ratio k for the circuit It is usual to split the inverter ratio obtained between transistors T1 and T2 in order to prevent T1 from occupying a much larger area than T2.
The maximum current through T1 occurs when the gate-source voltage is a maximum (5 V) and the device is saturated. 3) (W1) (5-1) . maxtmum/ds ofTl =30 - 2 Lt 2 (W1) 1J,A = 240 - Lt. 14. Cout arises because of the capacitance of the gate of T2, the source of T2 and the drain of Tl. In addition, Cout includes the capacitance of the connections between Vout and these transistor terminals. Thus Cout is proportional to the silicon area necessary to implement these features. 1 pF. 3 V. If the input now changes instantaneously from high to low, then T1 turns off.
Thus for a high and low level input, Vout = Vin and V gs of T4 is approximately 0 V. An inverting superbuffer is obtained by exchanging the connections from T1 and T2 to T3 and T4. 28 NMOS non-inverting superbuffer The advantage of the superbuffer lies in its action when it switches state. 3 V, turning T3 off. T4 is on and supplies current to the load capacitance on Vout• causing Vout to rise. Initially, the gate-source voltage of T4 rises to 5 V and thus T4 can supply more current than the conventional depletion load such as T2 where the gate is connected to the source.