Boot up time

[Lauri Kettunen]

You learn pretty quickly to pick and choose when to reboot for battery swaps.

Removed

 

[Obin Olson]

C'mon guys it's a 4k digital raw motion picture camera. not a DV cam!! let it take it's time booting! give em a break!

 

[Brent J. Craig]

Give me a call. 17 seconds is way too slow.

Agreed!

 

[Brent J. Craig]

I would hope that once Red is running a stable firmware they could get the boot times way down by simply loading a preset operating system image.

We already know the little Red computer writes 28 Megabytes per second to its CF cards. How many megs does it theoretically need to read when booting up? In the 60 seconds it takes now, it should be able to read more than 1.68 Gigabytes. How big is the Red OS? I would hope it's a lot smaller than a gig and a half! Some Linux distributions are less than 32 Megs. Adding in all the Red magic and drivers for the multiple Red displays will obviously increase that, and simply reading from red's internal flash memory should be much faster than writing to external CF cards.

Perhaps someone more familiar with Red's internal workings can explain why the process is more involved than simply copying a couple of hundred megs into RAM off of some fast internal flash.

 

[Steven Caesare]

I would hope that once Red is running a stable firmware they could get the boot times way down by simply loading a preset operating system image.

We already know the little Red computer writes 28 Megabytes per second to its CF cards. How many megs does it theoretically need to read when booting up? In the 60 seconds it takes now, it should be able to read more than 1.68 Gigabytes. How big is the Red OS? I would hope it's a lot smaller than a gig and a half! Some Linux distributions are less than 32 Megs. Adding in all the Red magic and drivers for the multiple Red displays will obviously increase that, and simply reading from red's internal flash memory should be much faster than writing to external CF cards.

Perhaps someone more familiar with Red's internal workings can explain why the process is more involved than simply copying a couple of hundred megs into RAM off of some fast internal flash.

Much OS boot time is spent interrogating hardware, and waiting for it to initialize.

If you have a static hardware configuration in a known state, there's opportunity for optimization here as well.

-Steve

 

[Leif Thomas]

There's a cheap and trivial workaround to avoid booting when swapping the battery; One needs a Y-connected power cord with a diode in one branch.

I like the idea, but why do you need a diode on at least one branch? As you recommend, you will first attach a small battery to the diode-branch but then attach a full charged one again to the not-diode-branch.

I just don't know the function of the diode.

 

[Kevin Halverson]

A diode in only one branch is a foolish idea. It would only prevent excessive current flow if the diode isolated battery exhibited higher unloaded Voltage, but if the conditions were reversed, no protection would exist at all.

A pair of isolation diodes are necessary to prevent one battery from becoming the load to the other battery. Considering the very low source impedance that a battery of this capacity exhibits, connecting them directly in parallel (or using a single isolation diode) could easily result in very high peak currents and could damage the batteries. Explosions or fire are also a possibility, so you need something control the current flow, diodes are the obvious choice.

However, considering that the current consumption of the camera is near 8 Amps, this would cause a conventional diode (Si) to exhibit a forward drop of near 1 Volt. This drop along with the current consumption would force the diode to dissipate 8W which would necessitate a moderate sized heat sink and a stud mounted diode. A Schottky dual diode in a common cathode configuration wouldn't fair that much much better as their forward drop would still be about 700 mV in the same configuration and this is still over 5 Watts to dissipate.

The best approach is an intelligent set of MOS-FET switches. If one choose the correct devices and implemented them in an appropriate topology, the forward drop could be gotten down to around 100 mV which would reduce the Pd (power dissipation) to only .8 Watts. This is a manageable thermal load.

 

[Lauri Kettunen]

A diode in only one branch is a foolish idea. It would only prevent excessive current flow if the diode isolated battery exhibited higher unloaded Voltage, but if the conditions were reversed, no protection would exist at all. ... A pair of isolation diodes are necessary to prevent one battery from becoming the load to the other battery. ... The best approach is an intelligent set of MOS-FET switches.

Of course, fully agree with you -a too hasty thought from my side. A pair of diodes is not that great idea as it will cause losses. MOSFET? Why not, but it's not a everyman-build-it-yourself approach.