# How to attenuate a (seemingly) overdriven VGA signal

I own an 8 port Belkin OmiPro2 KVM switch, which is a device used to connect multiple computers to a single display/mouse/keyboard. I've owned this KVM for years and it has exhibited the same symptom since I purchased it, so this is not a new behavior. Basically, when I connect this KVM to a CRT with a standard VGA cable the display looks perfect, but when I connect the KVM to a flat panel display the resulting video looks hot or overdriven. The colors (especially whites) are so intense they overwhelm surrounding colors/text and severely wash out the display.

I have seen this behavior with multiple flat panels, so I think I can safely rule out a problem with a single flat panel and focus on the Belkin.

My suspicion is that the manufacturer purposefully over-drives the VGA video signal with the assumption that the switch may be rack mounted and have to overcome whatever line loss is introduced when routing the signal 10 or 20 feet to the connected display.

In the past, I placed three 100ohm resistors directly on the Red/Green/Blue lines of the VGA with some reasonable success in my output. This had the effect of reducing the over-driven video but introduced 'ghosting' into the output. I suspect this is some sort of impedance matching or frequency-response type problem, but this is definitely outside my normal areas of expertise...

However, feeling like I was on to something, I have googled for how to properly attenuate a VGA signal but with no luck. Perhaps I am using the wrong terminology, or barking up the wrong tree completely.

Can anyone offer any suggestions or a proper circuit that I could try to achieve a better/cleaner/sharper result?

• Can you adjust the image in the LCD monitor into an acceptable form by manipulating the monitor's contrast & brightness adjustments? These are usually located in the on-screen display menu. Domenix has provided a very good hardware solution, but you may be able to cure your problem without hardware mods. Do you have access to a SMPTE133 test pattern? These can be downloaded from the web. This is the correct test pattern to use while making the contrast & brightness adjustments. Lacking that, any image with a wide black to white content will do. – FiddyOhm Oct 22 '16 at 23:23
• FiddyOhm, thanks for the ideas. I did indeed play with the brightness and contrast and anything else I could find on both LCD monitors with no success. I think Domenix answer is going to be exactly what I needed. Thanks! – Geo... Oct 23 '16 at 1:00
• If you don't get the results you need from Domenix's solutions (which I agree are completely valid), post another question about this. There is a certain procedure for adjusting contrast and brightness of VGA signals so the resulting image has correct black-to-white (and color) gradation. There is something odd here; I have worked with both CRT and LCD monitors extensively and have not found there to be a categorical difference in the way they handle VGA signals. – FiddyOhm Oct 23 '16 at 13:46
• FiddyOhm, I completely agree that something odd is happening, but I am almost certain the signal from the Belkin is being over-driven (possibly a defect in the Belkin). - But I disagree with your assertion there is no categorical difference between CRT's and LCD's. In my experience CRT's are more forgiving of out-of-tolerance inputs and are far more likely to handle older 15Khz video (which most LCD's completely fail to sync). - I will be sure to post my results here, thanks again! – Geo... Oct 24 '16 at 14:20
• BTW, in the on-screen adjustment menu of your LCD monitor(s) is there a button labeled "Auto Gain" or "Auto Adjust"? – FiddyOhm Oct 25 '16 at 10:41

Yes, you caused impedance mismatch when you connected the $100\;\Omega$ resistor in series with the RGB signals.

Most likely your output signal is larger than $0.7\;V$, which is the maximum specified signal voltage for the RGB lines. Your solution for this situation would be an L-pad attenuator on the three signal lines. You don't need a more complex passive attenuator for this application, because signal travels only in one direction.

$\hskip2cm$

Your source ($R_s$) has an impedance, in this case its $75\;\Omega$. The load has the same impedance (the display). If you insert only one resistor in series with the signal line, then the impedances will be mismatched, signal reflection will occur, hence the ghosting.

The best would be if you could open a fully white screen on one of the computers and measure the voltage on the RGB lines with an oscilloscope, and if its larger than $0.7\;V$ then design the attenuator accordingly. For example, if your output signal is about $0.9\;V$, then you can calculate the attenuation in decibel with this:

$\hskip6cm$ $20\cdot\log\cfrac{0.9\;V}{0.7\;V}=2.1829\;dB$

If you want to calculate the resistance values from the dB value, it will look like this:

$\hskip5cm$ $R_1=Z_s\cdot\left(\cfrac{10^\cfrac{2.1829\;dB}{20}-1}{10^\cfrac{2.1829\;dB}{20}}\right)$

$\hskip5cm$ $R_2=Z_s\cdot\left(\cfrac{1}{10^\cfrac{2.1829\;dB}{20}-1}\right)$

It turns out, $R_1=16.\dot6\;\Omega$, and $R_2=262.5\;\Omega$.
The closest available resistor values are $16\;\Omega$ and $270\;\Omega$. If you still experience ghosting problems with these values, then you should use trimmer potentiometers. The power doesn't exceed $10\;mW$ in this application so it should be safe to control the signal itself.

Here's the schematic of the solution:

Sources:

• domenix, thanks for the awesome response. I actually did try sending a white screen to my o-scope, but my scope is ancient (and I don't think it works right anymore), so the results were suspect (I think the signals were reading ~1v and not 0.7v but I dismissed the observation as flakyness). Your example of working out the dB and then back to the resistor values is very helpful. To my limited knowledge, the schematic looks similar to 3 small voltage dividers... but there was no way I was going to come up with the math on my own. This is a great learning experience; Thanks! – Geo... Oct 23 '16 at 0:58
• You're welcome. Yes, they are voltage dividers. L-pad is just a fancy name for it, and because it looks kinda like an L. Just to expand on my answer, the voltage divider equation on the site I linked is not the best, its only applicable for open-circuit output voltage. To get a reliable result, you have to calculate the output voltage while the load is connected. R2 and RL are parallel, so you have to calculate that, and the input voltage is shared between the calculated resistance and R1. Just to make sure the calculations are correct. – domenix Oct 23 '16 at 1:16
• Trimmer potentiometers ..... at VHF? – rackandboneman Feb 27 '17 at 14:05