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Can someone take a look at this datasheet and tell me if I'm missing something?

Okay... I'm trying to create a circuit that interrputs a video signal to a monitor. (Right now I'm using an old arcade monitor and arcade game to test). The part I chose was the TI OPA3692 tripple video amp with disable. Here's the datasheet:

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Basically all I did was tie -Vs to GND, +Vs to +5V, attach the RGB signals from the game to +IN A/B/C respectively, and the OUT A/B/C signals to the monitor. I left the -INA/B/C signals floating so that the amps would provide a +1 gain as specified in the datasheet. When the disable pin is high (deasserted) the video looks normal, but when I ground it the video dims but is still very noticeable. According to the datasheet, disabling the amps puts them in a high-impeadance state. I even threw some 100k pulldowns in the outputs to the monitor but no difference. Is there something I'm missing here? Input voltage levels off? Can't think of a reason why the disable function would not work.... Thanks for any help guys.

-Adam Courchesne

Reply to
ajcrm125
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That Disable signal has a bar over it, meaning that the disable is active when low.

If your monitor has pretty high input impedance, then the signal is jumping across the + to - input and going out the 402 ohm feedback resistor. Try putting something like a 1k load across the output.

Normal video signals are 75 ohm impedance, so the 75 ohm input load resistors, gain of 2 and 75 ohm series output resistors are typical, with the monitor having another 75 ohms to ground at the monitor end of the cable. But I suspect your monitor is a low bandwidth type without the terminating resistors.

Reply to
John Popelish

Right,... that's why I'm grounding it to try to disable the amps and put them in a high impeadance state.

Jumping accross the + to - input? Never even crossed my mind. Is that possible? I'm totally a digital guy so my analog experience is very slim. I thought that the inputs to an op amp were supposed to have a virtually infinite resistance between them. (One of the only things I recall from my analog course) :-)

Nevertheless I'll give it a shot.... thanks!

Reply to
ajcrm125

There is a non zero capacitance across the two input pins. With a reasonable output load, this current path is small.

Reply to
John Popelish

Ahhhhh.. gotcha. Good point. I just had a revalation BTW... I think what I'll do is ground the - inputs so the amp's gain is 2. Then I'll put a series resistor (as suggested) accross the output pins, bringing the output back down to proper elevels. That way, it will act as an amp with gain of 1 when active and when in high impeadance, the internal resistors will be in series to GND and should pull down the output for me. Does that sound like it would work?

Reply to
ajcrm125

That is its optimum configuration. Don't forget the terminating resistor between the + input and ground, to absorb the incoming signal, so it doesn't bounce back and forth between the source and the amplifier. That will cause thin ghosts at any high contrast vertical edge. The monitor should also terminate its signals the same way.

Reply to
John Popelish

Good point. Thanks... I looked at the monitor datasheet and schematic and it states (and shows in the schematic) a 5.6k terminating resistor at the monitor. Seems high but what do I know....

Reply to
ajcrm125

A modern, high resolution monitor could not get away with that kind of terminator. This reminds me of the old black and white monitor (just a stripped down black and white television) on my Radio Shack TRS-80.

Reply to
John Popelish

Yeah these monitors are really ooooooold school. :-)

Reply to
ajcrm125

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