I decided to open up a VGA cable which started to produce only red / blue / green colored image. I thought that inside I will find something like shown on this scheme:

enter image description here

Instead I found that my VGA cable is a more complex one. Allso the wire colors do not mach with the ones in the scheme above:

enter image description here

It has 11 wires instead of just 6. My cable has yellow and cyan cables for V-sync and H-sync but then there are these colours (white, gray, brown, black, dark red and bright red) which I don't know where to solder. I need an advice.

  • \$\begingroup\$ Can't you open the other connector, or use a multimeter to check which wire is connected to which pin? \$\endgroup\$ Commented Dec 9, 2013 at 8:04
  • \$\begingroup\$ Unfortunately no. The other side of this quite long cable is plugged onto a projector on the ceiling. Is it possible that all unknown wires are GND? \$\endgroup\$
    – 71GA
    Commented Dec 9, 2013 at 8:14
  • \$\begingroup\$ en.wikipedia.org/wiki/VGA_connector \$\endgroup\$
    – jippie
    Commented Dec 9, 2013 at 8:42
  • 2
    \$\begingroup\$ Where is the old connector? Can you open it up to check the wiring? \$\endgroup\$
    – jippie
    Commented Dec 9, 2013 at 8:48
  • 3
    \$\begingroup\$ WIRE COLOR IS NEVER A GOOD INDICATOR OF SIGNAL PURPOSE. You need access to the connector to find out which is what. \$\endgroup\$
    – Passerby
    Commented Dec 9, 2013 at 10:25

2 Answers 2


You actually have 14 wires if you count the ground return for each colour, it may use one of those on pin 5 (GND) as in your diagram to complete the set of 15. The other pins are often used for communications. For example from VGA connector on Wikipedia some of the ones your diagram showns as not connected are:

  • Pin 4 - ID2/RES - formerly Monitor ID bit 2, reserved since E-DDC
  • Pin 9 - KEY/PWR - formerly key, now +5V DC
  • Pin 11 - ID0/RES - formerly Monitor ID bit 0, reserved since E-DDC
  • Pin 12 - ID1/SDA - formerly Monitor ID bit 1, I²C data since DDC2
  • Pin 15 - ID3/SCL - formerly Monitor ID bit 3, I²C clock since DDC2

The reference to DDC refers to the Display Data Channel that the host can use to query the monitor for its capabilities such as supported resolutions and refresh frequencies. So while I believe in many cases they could be left disconnected with the loss of that functionality they won't all be additional ground signals as you mentioned in a comment.

If you're feeling experimental (and there are no guarantees this won't cause damage) because I'm not sure if the colours are 100% standard I'd try the following:

  • Connect the RGB center wires to pins 1/2/3 with the shields going to pins 6/7/8 respectively.

  • Short pin 8 to pin 5 to get the additional ground.

  • Connect the other wires as per your diagram leaving unknown ones disconnected

Now see if you can get an image on the projector by forcing a resolution you know that it supports because you won't have the DDC channel. If that works OK and it's not a problem leave it at that. Otherwise you could probably identify the remaining pins by measuring between ground and the DDC2 pins 12 / 15 using a scope during initial negotiation to check for the square clock versus data. The 5V to go to pin 9 should be easy enough to spot and is presumably present the whole time.


Nearly all modern PC that have a VGA connector use the same 15 pin VGA connector that the original IBM VGA card used.

There are at least four versions of the VGA connector, which are the three-row 15 pin DE-15 (also called mini sub D15) in original and DDC2 pinouts, a less featureful and far less common 9-pin VGA, and a Mini-VGA used for laptops. The image and below table are the newer 15-pin VGA VESA DDC2 connector pinout.

enter image description here

  • \$\begingroup\$ Nearly all modern PC graphics cards do not have a VGA connector at all. \$\endgroup\$ Commented Dec 9, 2013 at 12:09
  • \$\begingroup\$ @PeteKirkham - Nearly all modern computers who has the connector has that type of VGA. \$\endgroup\$
    – Butzke
    Commented Dec 9, 2013 at 12:11
  • \$\begingroup\$ If that's what you meant, correct the answer, don't add a comment. \$\endgroup\$ Commented Dec 9, 2013 at 12:18

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