Comments on: Puzzle #5 – Binary-Gray

By: Puzzle #5 - Binary-Gray counters - solution « Adventures in ASIC Digital Design

Thu, 05 Jul 2007 13:50:38 +0000

[...] Layout Considerations Puzzle #5 – Binary-Gray counters - solution July 5th, 2007 The binary-Gray puzzle from last week generated some flow of comments and emails. Basically, the important point to notice [...]

By: Nir Dahan

Nir Dahan — Tue, 03 Jul 2007 09:52:27 +0000

correct solution!
in this specific case the binary counter is cheaper regarding power consumption.
This puzzle was inspired by a “true story” btw, maybe I will write about it sometime…

By: Sriram S

Sriram S — Mon, 02 Jul 2007 17:51:48 +0000

Easy enough …

Gray counter: Toggles 1-bit and there are 2^n counts to go through. So total power for gray = 3P * (2^n)

Binary counter: Bit-0 toggles every count. Bit-1 toggles half-rate, bit-2 toggles quarter-rate and so on. So total power = total toggles * P

total toggles = (2^n)*(1 + (1/2) + (1/4) + … + (1/(2^n-1)))
total toggles = 2^n * ( 1 * (1 – 1/(2^n))/(1/2) ) (using formula for geometric progression sum).

total toggles = 2*(2^n – 1)

total_binary_power = 2P * (2^n – 1)
total_gray_power = 3P * 2^n

So total_binary_power is ALWAYS smaller than total gray power regardless of value of n!

By: Santhosh K.R

Santhosh K.R — Fri, 29 Jun 2007 04:24:56 +0000

4 bit binary code will take 30P pwer because

for counter from 0000 to 1111 it takes (1P+3P+7P+15P)

and counter resets from 1111 to 0000 so 4P

so totally 26P+4P

By: nirdahan

nirdahan — Thu, 28 Jun 2007 14:54:34 +0000

I think you are way off regarding the amount of toggling in a binary counter. n! grows by far faster (it actually explodes) than 2*2^n = 2^(n+1).
You are on the right track – try to identify precisely how much does a binary counter toggle through a single cycle. You could basically list the values for 1 to 6 bits, it will give you an idea if not the general rule exactly.
If you are into sum of a series calculation you could get the precise info by doing some maths…

By: ASHWATH BL

ASHWATH BL — Thu, 28 Jun 2007 09:00:48 +0000

In binary encoding, for ‘n’ bit wide counter, number of toggles will be around 2*2^n+(n!) [Approximately]. So it is slightly more than two times the gray counter togglings. Given the Gray counter consuming three times more power – Gray counter will always consume one third more power than binary counter[approximate].

By: nirdahan

nirdahan — Wed, 27 Jun 2007 21:32:19 +0000

Your end result is right but the numbers are wrong.
The 4 bit Gray counter will have by definition 2^4 bits toggle and thus will consume 3*16=48 units of power.

The Binary counter will toggle 30 times through a complete counting cycle and thus will consume 30 units of power.

can you prove why, given the assumptions in the puzzle, the Gray counter will always consume more power? regardless of the number of bits!

By: Santhosh K,R

Santhosh K,R — Wed, 27 Jun 2007 19:12:00 +0000

Gray Counter consumes more power.
Assume 4 bit counter.

Gray Counter takes (3P+6P+12P+24P) = 45P

Binary Counter takes(1P+3P+7P+15P) = 26P

You can list the Truth Table for both counters and list down, how many times each bit toggles.

By: nirdahan

nirdahan — Thu, 14 Jun 2007 07:20:30 +0000

The answer is irrelevant of n …

By: edoardo

edoardo — Thu, 14 Jun 2007 05:47:43 +0000

Didn’t you forget to tell what n is?
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