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escher
Posts: **113**

I'm constructing an 8-bit VGA DAC (RRRGGGBB), and found this Stack Overflow answer going into some solid detail on implementation.

If you look at the bottom of the answer (the "edit"), he makes the claim that...

"I believe you can see in the image above that you don't want 400 mV at the input of the other side of the transmission line. You want 200 mV there. The 75 Ω series termination resistance and the 75 Ω parallel termination resistance forms a voltage divider with a ratio of 2:1. So it will expect half the voltage, not the entire voltage."

I'm not quite understanding his logic here, and I'm not seeing how the correct voltages are reaching the monitor. Can anyone give me a clue?

If you look at the bottom of the answer (the "edit"), he makes the claim that...

"I believe you can see in the image above that you don't want 400 mV at the input of the other side of the transmission line. You want 200 mV there. The 75 Ω series termination resistance and the 75 Ω parallel termination resistance forms a voltage divider with a ratio of 2:1. So it will expect half the voltage, not the entire voltage."

I'm not quite understanding his logic here, and I'm not seeing how the correct voltages are reaching the monitor. Can anyone give me a clue?

## Comments

6 Commentssorted by Date Added Votes1,2530Vote UpVote DownJim

1,9290Vote UpVote DownAt the source end of your connection, install a series resistor of 75ohm.

At the destination end, install a shunt termination resistance of 75 ohms, this consumes current but stops signal bouncing/ghosting.

When doing a rgb dac, you're are adjusting the source 75ohm to a higher value on the fly as to create analog values within range to the destination side.

It's simple resistor voltage divider stuff.

A mcu gpio have 50-100ohm already, so take that in to account.

1,6960Vote UpVote DownVideo signals are wideband, so they also have to be impedance matched so the response is flat and particularly to avoid

phase distortion/reflections which are visually bad (ghosting etc). Thus the source, destination and cable all have to be

a resistive 75 ohms.

1130Vote UpVote DownThat makes perfect sense, thanks! However, for the circuit in question, I'm not seeing how the proper voltage levels are being reached.

1,6960Vote UpVote Downcan be generated in one of several ways:

Generate twice the voltage, feed through a series 75 ohms before sending out.

Generate the signal as a current, feed to a 75 ohm resistor to ground and also to the output. The

current should be twice that needed at the load.

With an R-2R ladder you can always load the ladder to any given impedance with a shunt resistor since

the R-2R itself is constant impedance, so can be loaded without affecting its linearity. This is one of the

reasons its much better for DAC use that a 1-2-4-8-16R network which can only work into a high-Z load.

If you load the R-2R ladder to 75 ohms and find the voltage is then too low, an opamp buffer is the way

to go. Lowering the value of R allows a higher output voltage for a given load impedance, but accuracy

will tend to suffer as the logic outputs become loaded.

1130Vote UpVote DownAn R-2R network can be simplified to a Thevenin source R = R, V = 3.3 * (2n -1) / 2n. Add on a parallel shunt and now you've got Rth = R//Rshunt, V = 3.3 * (2n -1) / 2n Rshunt / (Rshunt + R). Solve for Rth = 75, V = 1.4 (because a 75 ohm - 75 ohm 2:1 voltage divider is created between the DAC and monitor).

Solving the system can be done by hand or via Wolfram

I appreciate the inputs everyone but at the end of the day it's linear algebra and elementary circuit theory to the rescue.