How can we generate a pulse for every edge of the incoming pulse

It is a very common requirement to detect either positive edge, negative edge or both edges of a signal. And the circuit that can detect and generate a single cycle pulse is quite simple. In this post, we will discuss how we can detect positive edge, negative edge and both edges of a signal.

• Detect positive edge of a signal: Positive edge of a signal means that the current state of the signal is "1" and previous state is "0". And a pulse signal means that the output of the circuit is "1" for one cycle. So, we need a circuit which generates "1" as output when present state is "1" and previous state is "0". It generates "0" as output otherwise.

Thus, output = D(n-2)' & D(n-1)

The required implementation is shown in figure 1 below:

Figure 1: Detection of positive edge of signal

• Detect negative edge of a signal: Negative edge of signal means that the current state of the signal is "0" and previous state is "0". So, we need a circuit which generates "1" as output when present state is "0" and previous state is "1". It generates "0" as output otherwise. 

Thus, Neg_edge_detect = D(n-2) & D(n-1)'

 The required implementation is shown in figure 2 below:

Figure 2: Detection of negative edge of signal

• Detecting both positive and negative edges of the signal: Simply doing an "OR" operation of Pos_edge_detect and Neg_edge_detect signal will produce an output which is a single cycle pulse for any of the edge of incoming signal. The requirement for consecutive edges of incoming signal is to be at least 2 cycles apart otherwise, the output will not be pulse, but will be a continuous signal.

Any_edge_detect = Pos_edge_detect + Neg_edge_detect

The required implementation is shown in figure 3 below:

Figure 3: Detection of both positive and negative edges of signal

The technique we discussed here delays the output by two cycles. Can you think of any other way to detect the edges of a signal which is more efficient?

Why is body connected to ground for all nmos and not to VDD

To prevent latch-up in CMOS, the body-source and body-drain diodes should not be forward biased; i.e, body terminal should be at same or lesser voltage than source terminal (for an NMOS; for a PMOS, it should be at higher voltage than source). This condition will be satisfied if we connect all the nmos bodies to their respective sources. But we see that all the body terminals are connected to a common ground.

This is due to the reason that all the nmos transistors share a common substrate, and a substrate can only be biased to one voltage. Although it introduces body effect and makes transistors slower and deviate from ideal mos current equation, there is no other way.

One could achieve different body voltage for all nmos transistors by putting all transistors in different wells, but that would mean a tremendous penalty in terms of area as there needs to be minimum size and separation that needs to be maintained which is huge in comparison to transistor sizes. This is the reason why body is connected to ground for all NMOS.

Similarly, body of all PMOS transitors is connected to a common terminal VDD.