4 bit shift and add BCD multiplier

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arjunarikeri

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Hi,
As a part of my project I need to make a 4 bit BCD shift and add multiplier.
My idea is to multiply the inputs(which are less than 9) as binary and then convert then to BCD.
Both the above steps have standard circuit designs. However, the emphasis of our instructor is on reducing hardware and overall efficiency. Can anyone suggest anything better please.

Thanks in advance,
bye,
 

Only to move into the 21st century, and use a micro-controller, efficiency increased, and hardware vastly reduced.

This is presumably just a theoretical question though, and not intended as a practical design?.
 
for a 4 bit BCD you don't need to do anything to the numbers before you multiply them (for numbers less than or equal to 9 HEX = BCD). Since the largest output is 9*9 you can eliminate the calculation stages on your multiplier for larger numbers. You then just put a Hex to BCD converter on the output.
 
Thanks to all of you for the replies.
I think haven't explained the problem porperly. This is what I paln to do on my assignment

1) Use three 4-bit shifters registers and three 4-bit full adders to get the 7-bit binary product by repeated addition.

2) Now I need to convert this 7-bit bianry product to BCD, but I'll be given only 4-bit BCD converters. I have a circuit which uses 3 such 4-bit converters and, 3 full adders to convert a 7-bit binary to BCD.

But all this is pretty brute force. Plese give me suggestions to improve my circuit, the components being pretty much the same.

I have listed all the things the instructer is willing to give

1) standard 74xx series gates
2) 74151 MUXs
3) 7483 full adders
4) 7442 BCD to decimal decoders
5) 7476 flip flops
6) 74191 counters
7) 7490 decade counter
8) 7495 shift registers
9) ic 555
10)8-bit DAC08 data converter

The general instruction is to avoid fancy components and minimise the circuit restricting ourselves to these componrnts as far as possible.Some other may be given if found resonable(but again not very fancy ones)
 
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