Microdrive Unit Design

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tofro
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Re: Microdrive Unit Design

Post by tofro »

vanpeebles wrote:This has been a very enjoyable and interesting thread to read :)
...so far. Please keep it going ;)


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Cristian
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Re: Microdrive Unit Design

Post by Cristian »

tcat wrote:Your MDV2_ seems on average a bit slower, is that your star performer? Usually it is the drive mounted as MDV1_ , as MDV2_ picks some noise in the QL case, the QL board is a subject to applying the documented mod for better drive performance.
Thanks for your comments.
So did you apply the famous "mandatory modifications" and you got better performance?
Anyway, concerning my straight-edged mdv: when I mounted them in the QL i decided to swap the units (and the shield). So my current mdv2 was originally the mdv1.
I did't pay much attention on which unit got the best performance. Generally speaking I can say that both drives got evident superior results in comparison with the key-edged ones.
I'll do some tests to find out which one is the best then I let you know.


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Re: Microdrive Unit Design

Post by tcat »

Generally speaking I can say that both drives got evident superior results in comparison with the key-edged ones.
The facts we have learnt about your Iss#5,6 drives:

- all run at the same speed
- all heads have same resistance
- Iss#5 do not have mod applied
- all share same erase current (tested using same QL Iss#6 board)

Cannot comment on inductance, there is a spread of 0.2 measured values.
Cannot comment if your QL board Iss#6 has had MDV mod applied.

Can you comment on motor power leads, if they are laid each from either side of the Iss#5 head board, as shown in RWAPs' pictures?

Tomas


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Cristian
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Re: Microdrive Unit Design

Post by Cristian »

Cristian wrote:concerning my straight-edged mdv: when I mounted them in the QL i decided to swap the units (and the shield). So my current mdv2 was originally the mdv1.
I did't pay much attention on which unit got the best performance. I'll do some tests
After some tests I can say that the best results are averagely obtained with mdv2. I got 238/241 sectors, but frankly I don't think that a so slight speed difference could be sufficient to produce such results.


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Cristian
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Re: Microdrive Unit Design

Post by Cristian »

tcat wrote: Can you comment on motor power leads, if they are laid each from either side of the Iss#5 head board, as shown in RWAPs' pictures?
Yes the wiring seems to correspond. Please see the pictures I posted:
http://qlforum.co.uk/viewtopic.php?f=2& ... =20#p19357


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Re: Microdrive Unit Design

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I don't think that a so slight speed difference could be sufficient to produce such results.
Consider Loop time formula.

Code: Select all

LoopTime[s] = TapeLenght[mm] / ( RollerDiameter[mm] * PI) / MotorSpeed [r/m] * 60
e.g.
5080mm / (6mm *pi) / 2250 * 60 secs/min = 7.19 secs

Where:
2250 - nominal motor speed in [rpm] (revolutions per minute)
5080 - nominal tape length in [mm], 200 inches ~ 5.08 m
6 - roller diameter in [mm]

Consider media Sector count

Code: Select all

SectorCount[n] = LoopTime[s] * 10^6 / SectorTime[us]
e.g.
7.19 * 1.000.000 / 31760 = 226,39 ~ 225 + 5% sectors on media

Where:
31760 - sector time in micro seconds [us] (refer to A.C. Dicken's QL Advanced Guide), this is what QL is tuned to.

This gives nominal sector count ~ 225
This gives nominal loop time ~ 7secs


0.25 secs * 10^6 / 31760 ~ 8 sectors.
1/4 of a second then corresponds to 8 sectors on tape, where each sector is 512 byte long.
225 nominal + 8 sectors ~ 232 could this be close to the sectors you see on your tape?

If the above is at all correct, then media density can be expressed as,
1 byte ~ 40 [us] of tape loop time

Cannot believe that 0.00004 of a second can take one byte, that is 8 bits, can it?

Tomas


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Re: Microdrive Unit Design

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Cannot believe that 0.00004 of a second can take one byte, that is 8 bits, can it?
Hi,
It possibly can. In summary MD parameters:

Code: Select all

Sector time [us]: 31760
Tape length [mm]: 5080 ~ 200"
Loop time [s]: 7,19
Nominal sectors [n]: 225
From the parameters we can calculate, on what tape length one sector will fit.

Code: Select all

SectorLenght[mm] = TapeLenght[mm] / SectorCount[n]
5080 / 225 ~ 22.6mm

One sector will fit into roughly one inch of tape.
How much one byte will take then?

Code: Select all

ByteLenght[mm] = SectorLenght[mm] / BytesPerSector[n]
22.6 / 512 ~ 0.044mm!

If the above correct then 44 um (microns) of tape must accommodate 8 bits, and that is 2*8 FM transitions, we already learnt.
How could this be at all achieved?

Tomas
Last edited by tcat on Mon Nov 13, 2017 8:38 am, edited 3 times in total.


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Pr0f
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Re: Microdrive Unit Design

Post by Pr0f »

Technical docs and hardware shows there are 2 heads, so your calculations need to reflect that 2 heads are in use.


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tofro
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Re: Microdrive Unit Design

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Pr0f wrote:Technical docs and hardware shows there are 2 heads, so your calculations need to reflect that 2 heads are in use.
Erm, no. Probably not. There is no mechanism that ever uses the two tracks on the tape in a parallel way. The tracks are being used for alternate bits, most probably to compensate for the time needed to magnetize and de-magnetize the heads and to space the bits wider apart on the tape to minimize interference.

Tobias


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Re: Microdrive Unit Design

Post by tcat »

Hi,

Yes, that is right!

We have two tracks here of a stereo head (one erase coil is common to both tracks). With a mono head (one track) we would have needed 2x5080mm length of tape, and 2x7.19 secs of loop time to achieve 225 sectors.

I may be wrong, can you please help as to which calculations need correcting with some index of 2?
Thank you.

Tomas


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