Wednesday, June 13, 2012

Design a Sequential Multiplier


On the previous tutorial, I was presented how the Register Transfer Level (RTL) designs methodology is successful to design a GreatestCommon Divisor (GCD) core engine.

Here again the same method I would like to apply it into sequential multiplier unit.  In the book “Fundamentals of Digital Logic with VHDL Design” by Stephen Brown & Zvonko Vranesic, they also provide one example to design this sequential multiplier. However they name it as Shift-and-Add Multiplier.


Let’ me start with design specification.
We are going to design a multiplier to multiply two an 8-bit unsigned numbers to produces 16-bit product. For example 8-bit Data A multiply with 8-bit Data B and produce 16-bit Product.

Algorithmic modelling 
In this step, we can translate the design specification to produce the behavioural model of the multiplier. This model is expressed in terms of an algorithm as shown in Figure 1, and this should be completed with IO Block diagram of the top-level system as shown in Figure 2.


Figure 1: Algorithm in Psedo-Code for sequential Multiplier for 8-bit input (n=8)


Figure 2: Top-Level of Sequential Multiplier

RTL Modelling
The RTL Model is first provided in the form of an ASM flowchart as shown in Figure 3. Then from the ASM flowchart, we can construct the RTL control sequence table in the form of an RTL Code is derived as shown in Table 1.


Figure 3: the ASM Flowchart of Sequential Multiplier


Table 1: The RTL-CS Table

Datapath


Figure 4 shows the datapath circuit for the sequential multiplier. The datapath consist of two shift registers, namely shift-left register for data A and shift-right register for data B. Other components are adder, multiplexer and register to store the Product (result).


Figure 4: Datapath of Sequential Multiplier

Meanwhile Figure 5 shows the top level datapath using Verilog code.


Figure 5: A Verilog Datapath code

Control Unit
The Verilog program in figure 6 shows how the control unit is constructing using Moore Model. There are three separate blocks, Next-State Logic block, State Register block and Output Logic block.



Figure 6: Verilog Code for Control Unit Sequential Multiplier

Waveform Simulation
Figure 7 show example waveform simulation to perform multiplication of hexadecimal number of FFh multiply with FFh. The product should be FE01h. In decimal number represents 255 x 255 = 65025. Inputs A and B are define as FF hexadecimal respectively. The sequential multiplier will be start the processing the data according to control unit. You can verify the state movement by State_Y output. Signal output valid will be asserted in state S3 to indicate the process of multiplication is finish and the valid result (output R) should be taken at the same clock. The state resume again at state S0.  As you can see for the worst case (FFh xFFh) the result should be ready around 18 clock cycles compare if I want to multiply 3 x 2 which is take only 5 cycles as shown in Figure 8.


Figure 7: Output Waveform Simulation for test vector FFh x FFh



Figure 8: Output Waveform Simulation for test vector 3h x 2h

Can you estimate the total clock cycles required if I would like to multiply Data A =00h with Data B= FFh ? It’s there are redundant clock have been waste? Can you do some improvement on this design? Please provide your feedback here!













8 comments:

  1. i'm also from UTM..final year student..i want to ask you...how to design CU for mealy and moore.. i don't clearly understand about it..

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  2. I know about FSM(Finite State Machines) for moore and mealy. What is CU

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  3. The CU is Control Unit. I'm using Moore model to implement CU... Nazrin

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    Replies
    1. Hi can you please send the data-path instance modules ? Would be helpful. Thanks

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  4. hey nice information for us,thanks for sharing the nice thoughts and ideas of the FPGA Design and this information really helpful for us.I have to sure bookmark this blog.

    FPGA Development Service

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  5. Hello, nice job you've done here, it helped me a lot!
    However, I have 2 questions. At the Output Waveform Simulation, at the right of State_Y, why does it go from state 2 to state 1 and then 3? Why doesn't it skip state 1 and go directly to state 3?
    The algorithm says: S2: if(zero) Y=S3;

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  6. Hi good job, can i have your email or contact number as i have question to ask you personally.thanks

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    Replies
    1. im learning too about proteus, lets together!

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