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Give the clock out frequency and state time T of 8085 when the crystal frequencies is 5MHz?
Wiki User
∙ 14y ago
The clock out frequency of an 8085 is one half the crystal frequency. The period of one T cycle is the inverse of the clock frequency.
At a crystal frequency of 5MHz, the clock is 2.5MHz, and T is 400 ns.
Wiki User
∙ 14y ago
Wiki User
∙ 14y agoThe clock out frequency of an 8085 is one half the crystal frequency. The period of one T cycle is the inverse of the clock frequency.
At a crystal frequency of 6.144MHz, the clock is 3.072MHz, and T is 326 ns.
Wiki User
∙ 13y agoAt a crystal frequency of 6.144MHz, the clock out frequency of an 8085 is one half that, or 3.072MHz. The state time T is the inverse of clock out, so T is 326ns.
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How can you determine the duration of one T-state?
The duration of one T-state is one divided by the clock frequency, or two divided by the crystal frequency. This is for the 8085.
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At a clock frequency of 5 MHz (10 MHz crystal) the 8085 has a clock period of 200 ns. An instruction using 18 cycles would use 3.6 us. (Microseconds)This is for the case with no wait states. Each wait state adds 200 ns. Since an 18 cycle instruction has 5 memory accesses, one wait state per access would add 1 us to the execution time.
If the clock frequency is 5 MHz how much time is required to execute an instruction of 18 T states?
At a clock frequency of 5 MHz (10 MHz crystal) the 8085 has a clock period of 200 ns. An instruction using 18 cycles would use 3.6 us. (Microseconds)This is for the case with no wait states. Each wait state adds 200 ns. Since an 18 cycle instruction has 5 memory accesses, one wait state per access would add 1 us to the execution time.
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