Electronic – Little change in Verilog module results in high change in power consumption (Synopsys Design Compiler)

clock-speeddigital-logicpowersynthesis

I am comparing two Verilog designs:

  • Design (1): A top module that is driven by a clock running at 50MHz, which is the main system clock.

  • Design (2): The same top module as in Design (1) with one difference, i.e., the main system clock runs at 400MHz, but I use a divide-by-8 (/8) clock divider to obtain the same clock frequency as Design (1).

Intuitively, the two designs should have comparable power consumptions. The primary reason behind this intuition is that the two top modules of the two designs are running at the same clock frequency of 50MHz. However, it is expected that Design (2) consumes slightly more power, due to using the overhead clock divider.

However, after synthesizing the two designs using Synopsys Design Compiler, the power reports of the two designs indicate that Design (1) consumes 63.4 micro Watts, whereas Design (2) consumes 500 micro Watts, i.e., Design (2) consumes almost 8X the power consumed by Design (1). Please note that the reported power figures refer to the total power consumption, i.e., the sum of all power categories.

In an effort to understand the reason behind this 8X higher power, I suspected that the /8 clock divider is responsible. However, after synthesizing the divider alone (all by itself) with a master clock frequency of 400MHz, its total power consumption turns out to be 35.4 micro Watts only. The numbers don't seem to add up somehow, no?

Anyway, I hope someone can explain to me the source of this 8X higher power, please? Thanks in advance.

PS: I guess the question does not need my Verilog sources and synthesis scripts, right? Regardless, please let me know if these are needed, and I will update the question with more details.

Best Answer

I have seen that some synthesizers are intelligent enough to optimize logic expressions.

If the source code for example contains:

(A or B) and (A or C)

... they will actually produce the following logic on the FPGA:

A or (B and C)

If your synthesizer did so it may happen that not only the /8 divider will "see" the 400 MHz signal.

A hypothetical example:

B might be a 400 MHz input signal and C might be another signal which behaves in a way that the expression B and C might result in a 50 MHz signal. The duty cycle of B and C would not be 50% in this case but the duty cycle is not that important for the power consumption of CMOS circuits.

If you have the following code (here pseudo-code, not Verilog):

Z = B and C
Y1 = X1 or Z
Y2 = X2 or Z
...
Y100 = X100 or Z

... you might think that only one logic cell receives a 400 MHz signal: The one that generates the signal Z.

However the optimizer might optimize the logic the following way:

Y1 = X1 or (B and C)
Y2 = X2 or (B and C)
...
Y100 = X100 or (B and C)

... which means that all logic cells are driven with 400 MHz.

For many FPGA types the power consumption of a cell mainly depends on the fastest input signal of that cell while the other input signals have less impact on the power consumption.

Please also note that some FPGAs have "global clock lines" which are inputs to all logic cells used (maybe even if the signal is unused). Feeding a high-frequency signal into such a line would be a bad idea.

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