Does this fix seem reasonable and safe? (link to first post with solution...my specific hardware is addressed in this post.)

I have the exact same problem. Asus A8N-VM CSM and a Dell 1907FP monitor that's not recognized on the DVI-D port.

I'd hate to hose my motherboard and/or my monitor.

A pull up is a perfectly reasonable and safe fix. (As an EE) I wouldn't hesitate to try it.

If I understand things correctly (looks like I do: http://en.wikipedia.org/wiki/Display_Data_Channel ), the DDC stuff is I2C based in which case it should have a pull up on both clock and data lines (SCL and SDA in I2C speak).

I'm not sure about the post mentioning a pull down though. I'd personally ignore that and put a 4.7k pull up on that line. Go for it

edit: and you really can't damage anywith

Cool. I hope to try this sometime today then.



You've obviously never seen me wield a soldering iron! I think I can do this one though.

Thanks!

Hi,

Shonky's right, you can use a pull-up without damage.

The I2C bus allows up to 3 mA sink which at 5V is around 1.66k Ohms. This low of pull-up value would only be needed for a very capacitive load.

The amount of capacitance is determined by cable type and cable length. Coaxial cables typically have 20 pF to 30 pF per foot. Twisted Pair cables have less (12 pF to 15 pF).

The specification also specifies a maximum rise and fall time requirement based on data rate. The edge rate is degraded by the system capacitance. The capacitance degrades (rounds-off) the edges and reduces the amplitude of the signal causing errors on the I2C bidirectional bus. It can also cause loss of the first transition bit if the capacitance is very high.

For a short length of interconnect cable <10ft (around 200 pF) the 4.7k resistor should work fine. It will still preserve reasonable rise and fall time and signal level. At lengths greater than 10 ft, you might need a lower value of resistor (3.3k, 2.7k, or 2.2k) for up to 400 pF of cable capacitance. The 400 pF value is the maximum allowed in the specification, but then, there is the real world... In any case, don't go less that 1.66k per the specification.

Remember that there may be a destination termination in the monitor. This may need to be considered as a parallel resistance.

I also would not bother with a pull-down resistor at lower data rates.

Ross

Thanks for the confirmation (although most of what I understood was in the first and last lines).

Hi,

Engineers are nerds, everybody knows that. I was just trying to explain how the value was arrived at and provide you enough information to help you tailor the resistance value to your application based on cable type and length.

Ross

Thanks. It's a standard 6ft. cable, so 4.7 kOhm it is.

Does it matter whether it's a 1/4 watt or 1/2 watt resistor? I'm getting ready to do this now, so if I don't hear anything, I'll use the 1/4 watt.

1/4 watt is heaps. 5V into 4.7k is only ~1mA so at most it will dissipate only ~5mW.

It worked! Thanks for the help.

Can you imagine? Working without dual monitors? Oh the pain!