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I'm studying about 8085 microprocessor and found that it works on 3 Mhz and takes for instance 4 clock cycles for opcode fetch, 3 clock cycles for IO/M read or write.

So now, If I somehow overclock 8085 to work at 10 Mhz then will the number of clock cycles get reduced ?
Or the number of cycles will remain same but the effective time for 3 or 4 complete cycles will get reduced ?

Shivam Aggarwal
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1 Answers1

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Overclocking reduces the length of a cycle. Instructions still take the same number of cycles.

The clock cycle is a heart beat that is used to synchronise actions across a circuit. Different parts act simultaneously, passing data at appropriate moments based on the clock. For example, a four-cycle instruction for a fictional architecture might be performed as:

  1. load accumulator with value of register while issuing memory read cycle;
  2. add low four bits of fetched value to accumulator;
  3. add high four bits of fetched value to accumulator;
  4. move value from accumulator back to register.

You can't cut a cycle from that without changing the architecture somewhere — simply providing a faster clock can't make the ALU suddenly work in 8-bit quantities and, even if it did, the register to which its final value goes wouldn't expect to receive a value earlier than the fourth cycle.

Simply changing the clock rate doesn't change the layout or overriding logic of the circuit so it can't change the number of cycles it takes different parts to do things or the relative times at which they expect other parts to have things done.

Tommy
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  • Thanks. It's the perfect answer to the question. Just one thing I'd like to ask you is, What if I overclock the cpu to such a extent that the clock rate reduces by soo much that the time isn't just enough to even complete one task, what will happen in that case ? Thanks. – Shivam Aggarwal Feb 24 '16 at 16:28
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    Once you overclock beyond the tolerances that your particular chip can handle (they're mass produced to tolerances; it's luck of the draw — maybe you got lucky), different parts within the chip will begin to act erratically, usually with the fast result of a complete breakdown of predictable function. One of the normal effects of overclocking is a significant increase in heat, which is often the thing that first starts to cause negative effects so you can often go quite a long way by applying absurd cooling. Which, for an 8085, is more or less anything active. – Tommy Feb 24 '16 at 16:50
  • Thanks. I agree with the fact that the cpu will start getting more and more hotter but apart from that, 'technically' what must have be going on inside the system along with the data, registers, memory and data lines. Please reply. – Shivam Aggarwal Feb 24 '16 at 17:02
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    We're running well up against the limits of my knowledge; however things like transistors work by storing and releasing charge, which is a physical process, and therefore have a minimum switching time. If you get to the point where you're expecting a transistor to have switched before it physically can have done so then it's going to be in an inconsistent state. So maybe that value you thought you stored somewhere isn't actually stored yet. In which case maybe you're not adding the thing the instruction asked you to add. Etc. – Tommy Feb 24 '16 at 17:15
  • Got it bro. It brilliantly answered my question and all my further doubts. Thanks alot :D . – Shivam Aggarwal Feb 24 '16 at 17:32
  • Good answer. By the way, you may actually see more cycles for a given task, since you can't overclock all the components. For example if you're memory bound, you'll still see the same memory latency, which is now *more* core clocks. – Leeor Feb 27 '16 at 20:17