View Full Version : How do you match impedance when driving 2 video devices from the same Propeller

Dennis Ferron
02-02-2007, 12:21 PM
In my suitcase computer, I have a video feed going to an internal LCD TV display, and an RCA video jack on the outside of the case as well. I would like to be able to connect an external TV and the internal TV and drive them both at the same time from the same video source. Do I need to account for a difference in impedance caused by adding an extra load?

What if I use an op-amp as a voltage buffer? (That is, configured as a voltage follower with it's (-) input connected to the output and (+) input connected to the video source.) Since the op-amp has a high-impedance input, do I need to put a 75 ohm resistor to ground across the (+) input?

I realize it will have to be a fast op-amp to keep up with the 3.58 MHz color signal. How high does it's bandwidth need to be in order to not skew the colors?

Or perhaps I should simply change the values of the video output resistors to make a 37.5 ohm impedance? What would those values be?

02-02-2007, 01:17 PM
The maximum shift of a color phase is 1/16th of a clock (315/88mhz).

The plus side of NTSC is so long as the shift is consistant, the color should never be distorted. There is a portion of the front porch that establishes the baseline phase, everything after that is compared to it. As far as distortion, so long as you can get a clean 3.58mhz output, nothing terrible should happen. Rounding the color is fine, but levels need to be correct. Sometimes rounding actually improves the color, depending on the way the television handles phase sense. If you wind up blurring too much, you will wind up desaturating the color.

It might just be worth it to use two different video outputs. It only takes 3 pins to drive a TV output, unless you're using broadcast mode.

Phil Pilgrim (PhiPi)
02-02-2007, 01:50 PM

The usual way this is done is to daisy-chain the two monitors from one output, with only the one at the end being terminated with 75 ohms to ground. This is probably not a convenient thing to do in your situation, though, unless the suitcase monitor has loop-through connectors and a switch to disconnect the termination. So an alternative would be to use separate output buffers. The attached circuit shows how this might be done with transistors. I'd start with DAC resistors maybe twice the value as those used in the Demo Board. Then adjust R2 to get a 1V peak-to-peak signal at the summing junction. Next adjust R1 so that the sync tips at the summing junction are riding at about 0.7V. You will have to go back and forth with these two resistance values until both conditions are met. (I'm sure there's a way to calculate these, but the transistor base currents into the 1K emitter resistors complicate things a bit. So it's probably faster just to tweak it until it's right.)

The transistor buffers are emitter followers whose output impedance is dominated by the 75-ohm collector resistors. What's nice about this configuration is that it will provide a 1V P-P signal at 75 ohms impedance no matter how (or whether) the other end of the cable is terminated. (Of course, a long, unterminated cable will be rife with reflections, with visible consequences.)

Note: I haven't actually tried this exact circuit, but have had success with similar, single-buffer circuits. Be sure to use the specified transistors: 2N4401. These are fast enough not to screw up the chroma, as a garden-variety 2N2222 would. If you do try it, I'd be interested to hear how it works out.


Dennis Ferron
02-04-2007, 01:22 AM
Thanks. It might be a while before I have an actual need for it, but when I do try it I'll let you know.

I've used similar emitter-follower configurations to make high power linear regulated power supplies with 2N3055s. In those cases, the .6 volt base to emitter voltage drop means the emitter will keep the output voltage at .6 volts below whatever the base voltage is. In the circuit you gave me, does the base to emitter .6 volt drop have any effect? I'm guessing that since a bipolar transistor is current-controlled, then as long as the b-e junction remains forward biased, the real controlling factor is the amount of current through the 75 ohm resistor?

Dennis Ferron
02-04-2007, 01:27 AM
asterick: Thanks for pointing that out about the front porch determining the color phase so that a shift is not a problem as long as it's constant. That's good to know.

Phil Pilgrim (PhiPi)
02-04-2007, 01:46 AM

Yes, the B-E drop has an effect. That's why I suggested adjusting the sync tips to 0.7 volts above ground at the summing junction (base), so they don't get clipped off.