Dynamic gates burn more power because of the windows 7 any product keys ! associated clocks.
What happens in this region is that electrons now travel through inversion channel, until the channel ends, and then it travels through depletion region to cross the gate to drain junction.
Lets first see, what does the delay of a MOS transistor depend upon, in a simplified model.
At least one end channel formed, so pmos is definitely not in cut-off.
This multiplier a is our fanout.It means a large current would from from VDD to VSS, which is also called crowbar current or rush-through current.Another thing to notice is that for this range of input voltage Vin, both transistors are.That is the driver inverter output, or the end of resistance.Dynamic gates use nmos or pmos logic.Again it will depend on what the initial voltage is at node out.You can see that top end is going to be cut-off as gate to top end voltage difference is zero which is less than Vth.For nmos device, source is tied to ground and if gate voltage is less than Vth we know gate to source junction is cut-off.While in the pre-charge state, NOR input A goes high.
For an equivalent cmos NOR gate, there would be pull up tree made up of p-mos devices.
Sum up such R and C product for all nodes.
Given the finite rise time of input signals it is inevitable to avoid this region, but one should take care to minimize this region.
Lets assume the input capacitance of first inverter is C as shown in figure with unit width.In reality graph still follows similar contour even when you improve inverter delay model to be very accurate.Total delay along the chain D Total inverters along the chain * Delay of each inverter.This gives rise to increased electron scattering.Following is a fairly popular graph of I v/s V curves of a MOS transistor.Inverter with input at VSS We shall discuss the answer in next post.At high voltage mobility determines the drain current where as at lower voltages threshold voltage dominates the darin current.