Question: Why is it that the produced clock frequency is 6.144 MHz, but internally it (8085 processor) uses only 3.072 MHz. Also what leads to the specific value of 6.144 in a clock. 

I found an answer at yahoo.... <http://answers.yahoo.com/question/index?qid=20080810090119AAurr2i>

but I must admit I didn't still get it it well. Could any of you guys throw out a few lines on this, please?

Answer: 
At the time this chip was designed, people used as few transistors as possible in the CPU, to make them small enough to fit on the available chips.

I suspect that practically every "register" (both programmer-visible instruction-set registers and also internal microarchitecture latches) in a CPU of that era stored data in a transparent [gated D latch](http://en.wikipedia.org/wiki/gated_D_latch) or something similar.
Nowadays, there's plenty of transistors on a chip, so it's simpler to use full master-slave D flip-flops, even though they use twice as many transistors.

Many instructions take data from some register A, combine it with some other data with the ALU, and store the result back in register A.
That is pretty easy to do if register A is implemented with a full master-slave D flip-flop.

But if register A is a transparent gated D latch, you need non-overlapping clocks.
You use a pulse on one clock to store some intermediate result somewhere (while register A holds its output constant), and then a pulse on another clock to load register A with the new value (while the intermediate register holds its output constant).

This requires a 2-phase clock.
The easiest way to make a non-overlapping 2-phase clock (in those days when transistors were scarce) was a little external circuit that takes an input clock and divides it by two.

As time went on, people figured out how to pack more and more transistors onto an IC.
So people designing CPUs integrated more and more of the stuff around the CPU in a full computer system onto the CPU chip.

Reading between the lines of the [Wikipedia clock signal article](http://en.wikipedia.org/wiki/clock_signal),
I get the impression that the people who designed the 8085 and the 6502 and other chips of that era had just a little more room than the previous generation of integrated CPUs,
and they decided the best use of that room was to put that little external circuit on-chip.
But they kept all the registers the same gated D latch as before.

So that's why the clock frequency is divided by two. You can think of the first external clock pulse generating a pulse on the phase\_one internal clock signal to update that intermediate result register, and the second pulse from the external clock generating a pulse on the phase\_two internal clock signal to update the programmer-visible register.
Internally, the core of the 8085A requires a two-phase clock. The internal logic that derives the two clock phases also divides the input clock by two. As previously stated, the reason for using a 6.144MHz input clock is for baud-rate purposes, the chip will run just fine at 6MHz. The chip is actually rated at 3MHz requiring a 6MHz crystal, but runs happily with a 6.144MHz giving easier baud rate generation (An Uart could be clocked with either the 6.144MHz from an Oscillator driving the 8085 or at 3.072MHz from the 8085's CLK output providing many usable Baud rates). I still use these archaic chips to perform special functions in some of my robots. I clock the Uarts with their own oscillator and I clock the 8085A's with a 6.4MHz oscillator, which runs the chip at 3.2MHz. The 3.2MHz divides down nicely to provide the 40KHz clock for my ultrasonic transducers. It makes more sense to use more modern I.C. devices in my 'bots, but I have a ton of old 8085's, Z80's, 63C09 and 63C09E's, 68B09 and 68B09E's, etc. that I really enjoy playing with.