circuit for remainder for divide by 64

Please provide a circuit that can provide the remainder when any 16 bit input to the circuit is divided by 64.

It's a bit strange when an experienced Edaboard member makes his questions sound like a first year's student homework problem...

Isn't the answer a simple part select (Verilog) respectively slice (VHDL)? You can also desribe it with a modulo operator.

sun_ray said:

Please provide a circuit that can provide the remainder when any 16 bit input to the circuit is divided by 64.

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please correct me if iam wrong
copy the dividend into MSB of 32 bit register , right shift 4 times you will get quotient [31:16] and fractional remainder [15:0]

FvM
Will that be synthesizable if modulo operator is used?

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dipin

Can you please explain how is it bringing remainder?

Regards

Will that be synthesizable if modulo operator is used?

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modulo 2^n simpy synthesises as a shift operation, other numbers infer a divider core, if supported by the respective tool.

dipin

Can you please explain how is it bringing remainder?

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take number 22 ,it sbinary equals 0000 0000 0001 0110
copy it to MSB of 32 bit register 0000 0000 0001 0110 0000 0000 0000 0000

then 2^4 =16 ,so shift 4 times 0000 0000 0000 0001 0110 0000 0000 0000

22/16= 1.375
quotient(msb)=0000 0000 0000 0001 =1
(lsb) remainder(in fraction)=0110 0000 0000 0000 = 2^-2+2^-3
=.25+.125
=.375

the output fractional accurcy is depends upon the number of fractional bits we are using
regards

Hi,

Just use bit 0 ...5 of the 16 bit input.

Or use "AND 0x003F".

Klaus