Gert's VGA - 888 possible?

[Snoop05]

RPi B+ and 2 have 40pin (26 usable GPIO).
6bpp (666) is using 6 pins for every color (RGB) + 2 pins for horizontal and vertical sync, totaling 20 pins used.
8bpp (888) will use 8 pins for every color, totaling 26 pins.

What value resistors will be needed for 7th and 8th pins? Starting at 16k Ohms, every pin half of previous, so i guess 7th will be 250 Ohm and 8th 125 Ohm. Or not?

Is that possible? I think everybody would like full 24 bit color depth on their analog output (me too).

Documentation say's that PCB is designed for 666, so just add another 6 lanes.

Then just use "dpi_output_format=7" and it should work

I'm mainly asking about needed resistors and connecting to DE15.

If its possible, i will try to make it and post final "product" here.

https://github.com/fenlogic/vga666/blob ... manual.pdf
http://www.cpmspectrepi.webspace.virgin ... apter.html

[mahjongg]

According to Gerts documentation 888 seems possible, but with simple 1% resistors you won't get the necessary accuracy, and for 256 different levels you will need far better than 1% resistors, for 666 you have just 64 levels, so 1% accuracy is enough.

also the extra resistor shouldn't be halve of the last one, that would mean doubling the drive level, and will probably overload the GPIO current. Instead all current resistors must be doubled, and shift one position, then the new resistor (for the LSB bit) should be double that of the largest value used.

I wonder what gert will think of this idea.

[Snoop05]

Since he used 120 Ohm for sync, 125 should work

And what tolerance do you think it need? ±0.5% ?

[toxibunny]

Get a load of 5% resistors and measure them. you're bound to find one that has the exact value you need - depending on the amount of resistors you have, and the accuracy+precision of your meter.

An interesting maths problem would be to calculate how many resistors you'd need to get in order to have a >50% chance of having the right ones...

[plugwash]

The trouble with the "measure a load till you find one that matches" technique is that resistors form the same batch are likely to have very similar values. Plus you will need a very good meter.

8 bit means 256 steps so at minimum you would want resistors accurate to 1 part in 256 (0.4%) of the ideal value. This deviation needs to consider both selection innacuracy and tolerance inaccuracy.

Now an 8 bit resistive DAC isn't entirely unreasonable. 0.1% resistors are not terribly expensive nowadays. and if you go for a R-2R ladder design you only need two different values (and there are standard values that differ by a factor of exactly 2).

[toxibunny]

aha. good information, thanks!

[rotwang]

The trouble with the "measure a load till you find one that matches" technique is that resistors form the same batch are likely to have very similar values. Plus you will need a very good meter.

8 bit means 256 steps so at minimum you would want resistors accurate to 1 part in 256 (0.4%) of the ideal value. This deviation needs to consider both selection innacuracy and tolerance inaccuracy.

Now an 8 bit resistive DAC isn't entirely unreasonable. 0.1% resistors are not terribly expensive nowadays. and if you go for a R-2R ladder design you only need two different values (and there are standard values that differ by a factor of exactly 2).

My solution to this was always to use 3 resistors for each leg of the ladder (either 2 in series, or 2 in parallel), this leaves you free to choose values. Buy your resistors by the yard on a bandolier and you'd be surprised just how tight the match between pairs of resistors is.

[Gert van Loo]

In all the posts below nobody seems to have mentioned an Interesting fact: Not all the resistors have to 0.4%.
The big ones have to be accurate as their deviation should be smaller then the value of the smallest resistor.