andrewb
Well-known member
Thanks for that. It's gonna take a bit to digest completely, but if I'm grasping this correctly; it almost sounds as if the DC could infer relative position on the screen. I'd think it would have to travel into negative to get to the other half of the screen on whichever axis. Or does 0 vdc represent the opposite extreme of deflection?
0V represents the center of the screen (for each axis).
Positive voltage deflects the beam up. Negative deflects it down. (Or right/left, for the X axis). The game board outputs these XY voltages, and the monitor amplifies them into huge currents, which it sends through the yoke, to deflect the beam, and draw on the tube face.
The issue is that for whatever graphic you're drawing, unless it's a perfectly symmetric shape, you're going to spend more time on one side of zero than the other. So there's some constantly changing DC imbalance as the monitor is drawing, but it's usually within +/-1V (ish).
When there's a problem with the game board, the game board can output a stuck X or Y value, which is on the order of 8 to 10V. That will force the beam off the edge of the screen and keep it there, for seconds (i.e., hundreds of times longer than it was designed to). And that's when your deflection system nukes itself, as it was only designed to have the beam out there for 1/1000 of a second while it's drawing something, and not out there permanently (which draws so much power that the transistors basically melt down.)

