I need to
determine the video signal level. Apparently VGA uses 0.7V
p-p, so I might need to add a resistor in-line to reduce the voltage if
it's any higher.
Or more properly a voltage divider.
Most VGA monitors terminate the input siganls into 75 ohms -- basically
there's a 75 ohm resistor from the input signal to ground. You can use
that as the lower resistor of the potential divider.
Unfortunately
I don't have access to an oscilloscope.=20
Too bad, as the peak-to-peak amplitude of the signal is really what you'=
d=20
need to know here.
Could I just arrange for a full-screen white
image to be shown, measure=
=20
the voltage using a normal DC multimeter, and
multiply up the value
shown to account for the proportion of each scanline corresponding to
sync/flyback?
Not likely.
The first thin to do is to see what supplies the converter board takes.
If it takes some high voltage (say around 40V), this could well be for a
video output stage on that board, which will complicate things. Look out
for any signs of a step-up voltage converter on the board too.
Assuming there's just a 5V supply, I think it's safe to assume the video
output siganl has an amplitude of less than that.
A tirck I've used in the past to find the peak value of a signal in the
absense of a suitable 'scope (in my case, the signal was a narrow pulse,
so I'd hae needed a good storage 'scope), is to use an analogue
comparator chip.
Get a suitably fat one that covers the voltage range you're looking at
and power it approriately. Connect one input to the signal under test
(the video output signal here), the othre to a mutil-turn potentiometer
conencted across a stable voltage source. Monitor the output of the
comparator with a logic proe, and turn up the pot' until the output just
stops pulsing. Now the voltage on the pot' slider is just above the peak
voalue of the signal. And since that voltage is a steady DC one, it can
be measured with a DMM.
-tony