DMA transfer rate

Question

An 8-bit DMA device is operating is cycle stealing mode (single transfer mode). Each DMA cycle is of 6 clock states and DMA clock is 2MHz. Intermediate CPU machine cycle takes 2 microsecond, determine the DMA data transfer rate.

Comments

isn't there a concept like this???

8 bit dma meaning 2^8 bytes to be transferred

is it in burst mode???

Answer 1

Each DMA cycle is of 6 clock states with clock Frequency is 2MHz. = 6 $\times \frac{1}{2MHz}$ = 3$\mu sec$

An 8-bit DMA device is operating is cycle stealing mode (single transfer mode) = so 8 bits are transfer in one transfer

But CPU machine cycle takes 2 microsecond,  after one transfer of 3 micro sec CPU take bus control and do some work ,

then again After 8 bits are sent so to transfer 8 bits in 3[Dma] + 2[CPU ] = 5 $\mu sec$ after that next 8 bit sent.

So transfer time in terms of 8 Bits =

 5 $\mu sec$ = 1byte

1sec = $\frac{1}{5\mu sec}$ = 200000 bytes 

Hence the DMA transfer rate = 200 kBps

Comments

because of this word cycle stealing mode (single transfer mode), In this mode, 1 Bytes is transferred only.

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sorry, but i know this...in single machine cycle it will only wait for 1 word(1 byte) to be ready...

but what if machine cycle is 2μsec and device is transmitting at the rate of 1 byte/6μsec...in this case dma has to wait for 3 machine cycles(6μsec)..and hence,

transfer time for 8bits(1 byte)= 3[Dma] + 6[CPU ]=9μsec...

correct me if i am missing something..thankyou..

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isn't there a concept like this???
8 bit dma meaning 2^8 bytes of data to be send..
is it in burst mode?????

Answer 2

2MHz = 2 x 10^6 cycles/s DMA cycle = 6/(2 x 10^6) = 3 mic s (10^-6 s) Intermediate CPU cycle is 2 mic s So, every 5 mic s, we have a 8 bit transfer - 1Byte In 1s => 10^6/5 = 200 x 1000 B

Comments

I am not getting...what is given in the question.

@anirudh sir...please explain.

Answer 3

what i think is that they are asking the DMA transfer time. so nowhere we are going to use the cpu cycle .

dma is a 8 bit . " and its one cycle contain 6 states" i think this is also not needed.. 

time taken will be 1/frequency = 0.5 * 10^-6 seconds.

as it is a 8 bit device 6 cycle will output 8 bit.

8 bit = (0.5 * 10^-6 seconds) * 6.

now in one second i can get 8/ (0.5 * 10^-6)*6 bits /second

= 2.66 Mbps

plz let me know the answer. 

Comments

in question i missed its "6 clock state"
and answer given is 200 kb/s

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i think 2.66 is the answer have u got the solution plz post it.

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i dont have solution!