How clock gating reduces power dissipation

As discussed in clock gating - basics, enable signal coming in data path is transferred into clock path in order to save dynamic power. But the question is exactly how is this power saved. In this post, we will discuss the same. 

A flip-flop implemented as a standard cell mostly has two internal inverters to generate clk' and clk_delay signals. So, even if the flip-flop input is kept constant, there is still toggling of data at these inverters, thereby dissipating dynamic power. In addition to this, there is internal power dissipation inside flip-flop due to charging and discharging of transistors' gates repetitively because of clock toggling, but this component is not a significant factor compared to dynamic power of inverters. Figure 2 below shows the internal structure of flip-flop, which has two latches in master-slave configuration and two inverters in clock path.

Figure 2: Flip-flop internal structure

Every clock cycle, these two inverters toggle regardless of flip-flop output toggling. However, implementation of clock gating will prohibit the toggling of these inverters when data is not toggling. Let us assume that a latch-based ICG is inserted. Thus, a mux in data path is replaced by an ICG in clock path. But there is a difference here. If there are, say 1000, flip-flops with same enable signal, there will be a common ICG inserted for these. Thus, instead of now 2000 inverters (inside 1000 flops) toggling when flip-flop output will be constant, we have only 2 inverters inside ICG consuming dynamic power. This is how dynamic power is saved. However, if only 1 flop had been clock gated in this manner, there would not have been any dynamic power saving, instead we have an ICG instead of a latch, it may result in overall loss in terms of area and power.

Whether there is any net saving is governed by how many flips-flops have been clock gated using a single ICG.

Also, as discussed, many muxes in data path with same enable are replaced by an ICG in clock path. Thus, there are advantages in terms of area and leakage power too, in addition to dynamic power.

Design query : How can we construct a 101 non overlapping counter using only combinational circuit for 32 bit input for example on considering 10101001 i want an output as 1 since there is only 1 101 non overlapping sequence

Solution: The design in question is a combinatorial design with 32-bit input (given) and a 4-bit output as shown in figure 1 below. How the output is 4-bit is a bit tricky. For this, we have to understand the problem. We have to count the number of non-overlapping "101" sequences in 32-bit input. Thus, "10101" counts as only a single occurrence, and "101101" counts as two occurrences. So, the maximum number of such patterns will occur when "101101" is repeated, which comes out to be 10 in a 32-bit number. 10 can be represented by a 4-bit number.

Figure 1: Design representation

One of the solutions, of course is to make a truth-table and then find a solution using logic equation solving. But the number of combinations possible here is huge and practically impossible to find a solution. So, we need to follow a modular approach here.

We can divide the problem into two parts, detecting the required pattern and then counting how many patterns actually were detected. We are introducing an intermediate 32-bit output, each bit (Nth bit) detecting if the pattern was found with Nth bit of input as the middle symbol of pattern. To detect non-overlapping "101" pattern, we need to look into 2 bits on each side, thereby making a combinational logic comprising of 7 bits. There will be special cases for terminal bits (here bit-0, bit-1, bit-30 and bit-31) where we know that there are less than 2 bits on one side. So, we need to have special logic for these bits.

The overall combinational logic will look like as shown in figure below. Int-N (Nth bit of intermediate output) is a resultant of (Bit-N-3 to Bit-N+3). The number of 1's in the intermediate output will tell how many patterns were detected, which, as discussed earlier, will be maximum 10.

Figure 2: Block diagram representation of design

Let us, now, proceed for a generic logic for Nth intermediate output. As discussed, Nth output (On) depends upon 7 bits, 3 bits on the up and 3 on down. The Nth bit of output will show "1"only for following cases: X1101XX & 00101XX. As 10101XX will be detected as "1" for N+2 bit of output. If we denote the variables involved as G,F,E,D,C,A, then the output expression becomes

On = FED'C + G'F'ED'C

On = ED'C (F+G')

 For bit 31, the upper two bits do not exist. 101XX is the expression for getting output as 1. Thus, the equation, on a similar note, can be expressed as:

O31 = ED'C

For bit 30, the expression comes out to be X101XX. So, O30 also has same logic as O31.

For O1 and O0, we get the same expression as we get for On. The logic diagram for obtaining intermediate outputs is shown in figure 3 below:

Figure 3: Circuit with logic for intermediate output

Now, we have got the intermediate outputs showing which bits have the specified pattern detected. The number of "1"in the final output is our final answer. So, we need a special circuit to count the number of 1's combinationally in a bus as shown in figure 3 above. "Combinationally count number of 1's in a bus" explains how we can do this.

This post is in response to a query posted on our "post your query" page. In case you want to have an answer to your query, you can post a comment. We will try our best to answer.

Design query :: Combinationally count number of 1's in a 32-bit bus

Solution: The design in question is a combinational design with 32-bit input and 6-bit output as there can be maximum 32 1's and 32 stands "100000" in binary. Making a truth-table or K-map for this problem is not practical, so we have to take a modular approach. Let us divide the problem into detecting number of 1's among 4 bits and then adding the resulting numbers together providing the total count.

Let us first create a truth-table converting the number of 1's in a 4-bit stream into a 2-bit number. The resulting truth table is shown in figure 1.

Figure 1: Truth table for 4-bit count 1's circuit

Solving the above for O2, O1 and O0 using K-maps, we get the expressions as shown in figures 2, 3 and 4 below.

Figure 2: Expression for O2

Figure 3: Expression for O1

Figure 4: Expression for O0

Thus, we have 8 instances, each counting the number of ones pertaining to respective 4 bits. The next thing we need is to add these 8 three-bit numbers to obtain the resultant total number of 1's in the 32-bit number we got. For this, we can again follow modular approach to add two numbers at a time until we are left with a single number. The block diagram of the complete solution is shown below in figure 5.

Figure 5: Complete block diagram of counting number of 1's