The horizontal and vertical synchronization signals are TTL compatible, so there’s no problem there. For the R, G, and B lines (0–2 V/75 W), a 150-W series resistor does an adequate 5-V to 2-V/75-W adaptation, thanks to the high power capacity of the PIC outputs (20 mA/line, 100 mA total for ports A and B).
Impedance is, like resistance, an opposition to current flow. Resistance affects both AC and DC signals, while impedance affects only AC signals.
The monitor is designed to have a 75 ohm impedance. This means that, as far as you (as an engineer) are concerned, the monitor looks like three 75 ohm resistors
to ground.
Also, don't use W as a unit of measurement for resistance -- internationally, it means Watts, which is a measure of power. Instead, in the absence of using the Omega character, use the European schematic notation for resistance, like this:
Code: Select all
4700 ohms ==> 4.7KOhms (American standard) ==> 4K7R (European Standard)
1500000 ohms ==> 1.5MOhms ==> 1M5R
2003 ohms ==> 2003 ohms ==> 2K003R
75 ohms ==> 75 ohms ==> 75R (R indicates resistance)
Note several things occur: an internationally recognized letter serves as the decimal point, which not only tells you how big the resistance is, but it also eliminates ambiguity between using "." and "," (3.14159 in America is 3,14159 in many European countries). Also, the last "R" tells you it's resistance.
2V seems
awfully large to me. I would not generate a signal that powerful. But let me explain to you how their logic works.
Remember that the monitor presents a 75 ohm impedance to the driving circuit. Therefore, it appears to be a 75 ohm resistor to ground (remember, everything in electronics ultimately is referenced from ground). Therefore, the 150R resistor in series with the red, green, or blue signal line will give a total resistance of 225R, from PIC chip to ground. The PIC will therefore supply (5V/225R)=22.22mA of current (22m22A??? I've not yet seen any particular standard for identifying currents or voltages similar to that for resistances. Can someone clarify?). Since all currents in a serial circuit must be equal, we can compute the voltage drop across the 75R resistance: (0.02222A)(75R) = 1.666V. Thus, when the PIC's output voltage is high, the monitor will see 1.666V at its input. When low, obviously, it'll see 0V (or substantially close to it to not matter).
Now, I think that driving the VGA port with 1.6V is awfully dangerous. I prefer 0.7V. So how would I go about designing the series resistor value?
The first step is to determine what current needs to flow through the 75R resistor to develop 0.7V across it. This is done by evaluating 0.7V/75R = 9.33mA. Now that we have this figure, we know we need to drop 4.3V in the series resistor (since 4.3V in the series resistor + 0.7V in the 75R resistor = 5V being supplied to it); therefore, we know the series resistor (from Ohm's law) is 4.3V/0.00933A = 460R.
So, am I going to blow up my monitor?!
When driving the input with anything larger than 1V, I think you very much could do that. That seems dangerously high to me.
Even if it doesn't blow up the monitor, it could the monitor's video pre-amps to overdrive, creating what we call "splatter" in the ham radio community. In ham radio, splatter shows up as audio that sounds, unmistakably, like water splattering in your headphones (not EXACTLY like it, but figuratively). On the monitor, splatter will likely show up as annoying "ringing" or "ghosting" effects on the screen. It may even affect the color correctness of surrounding pixels.
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- Presumably, I could put an R-2R ladder in front of this. Or, figure out how to use the DAC (per other posts), and I'd have multiple values?
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