Control Unit

Question

Design a vertical micro programmed control unit to generate 40 signals. Out of first 35 those only 3 signals can be active at a time.  And remaining 5, anyone can be active anytime. The micro instruction of the control unit stores control signal information along with 3-bit MUX select and 12 bits address field.  The size of the control memory required is?

Answer 1

To design a microprogrammed control unit, we need to determine the number of bits required to represent the microinstructions and calculate the size of the control memory.

Given information:

1. 40 signals are generated.
2. Out of the first 35 signals, only 3 can be active at a time.
3. The microinstructions include control signal information, a 3-bit MUX select, and a 12-bit address field.

Let's calculate the number of bits required for each microinstruction:

• For the 35 signals where only 3 can be active at a time, we need ⌈log⁡2(353)⌉⌈log2​(335​)⌉ bits for the control signal information.
• The 3-bit MUX select requires 3 bits.
• The 12-bit address field requires 12 bits.

So, the total number of bits per microinstruction is ⌈log⁡2(353)⌉+3+12⌈log2​(335​)⌉+3+12.

Now, we can calculate the size of the control memory:

Control Memory Size=Number of Microinstructions×Bits per MicroinstructionControl Memory Size=Number of Microinstructions×Bits per Microinstruction

Since there are 40 signals, and we want to generate 40 microinstructions, the number of microinstructions is 40.

Control Memory Size=40×(⌈log⁡2(353)⌉+3+12)Control Memory Size=40×(⌈log2​(335​)⌉+3+12)

You can use this formula to calculate the size of the control memory required for the given microprogrammed control unit.

Comments

How $\left \lceil log_{2}(353) \right \rceil$ and $\left \lceil log_{2}(335) \right \rceil$ are calculated there?? Can you please explain.