# Ohms & coax cabling miscalculation?

I measured the resistance for 15 meters RG6 cable, which showed 3 ohms on my multimeter ! I've got 0.7 ohms resistance on the multimeter itself when touching the negative and positive. So 3 ohms - 0.7 ohms makes 2.3 ohms absolute value for 15 meters RG6. Now for 300 meters / 15 meters = 20 times longer. 20 times the ohms would be = 2.3 * 20 = 46 ohms for 300 meters cable OR 92 ohms for 600 meters. How dare these people grade this cable RG6 if it should be 75 ohms per 1000 meters ?

Is this correct what I am calculating or I do mistake somewhere?

Edit: I was touching the copper only on both sides.

Edit2: I'm testing this because : I learned that some people have tested wifi over coax in the past at 2.4 Ghz, 22 dBm and with RG6 cable successfully got RSSI of -9 at the 64th meter of the cable. Nobody explains what happens after the 64th meter..

I have 2W 2.4 Ghz amplifier. So I wanted to make rough calculation and see how far I can reach with the same cable and get decent RSSI. If you look at the charts for attenuation for RG6 it says ~34 dB loss @ 2.4 Ghz at 100m, which makes no sense of what I found as Ohm resistance and calculations. This ~34 dB loss seems way too much because if 2W are 33 dBm this means at 100m the signal will be totally gone ? What am I missing?

I went to the simple path that this amplifier should be in fact nothing more than Direct Current on the RF out. My calculations show 500-550mW loss each 100m of RG6 without considering Ghz but DC. At 300m I will be left only with 100-200mW (which is 20-21 dBm) and possibly still decent RSSI.

What your calculation show ? I guess you'll take different path of calculations. How far 2W @ 2.4 Ghz and RG6 ?

Thanks !

First: if your multimeter reads 0.7 ohms with the probes shorted, it's not very trustworthy, and the rest of the calculation should be ignored. (Try putting a new battery in). The correct DCR for RG6 should be about 2 ohms per 100m.

Second: You're measuring DC resistance, which is different from the characteristic impedance (75 ohms) and also completely different from the RF loss (so many dB per meter).

Third:

If you look at the charts for attenuation for RG6 it says ~34 dB loss @ 2.4 Ghz at 100m

Yes, that seems about right. The datasheet I have here says 32dB/100m at 2.2GHz and 36dB/100m at 2.5GHz. Close enough, considering variation between different manufacturers.

because if 2W are 33 dBm this means at 100m the signal will be totally gone ? What am I missing?

No, at 100m the signal will be 33dBm - 34dB = -1dBm, which is 0.8mW. "Totally gone" isn't 0dBm, it's -∞ dBm.

I went to the simple path that this amplifier should be in fact nothing more than Direct Current on the RF out. My calculations show 500-550mW loss each 100m of RG6 without considering Ghz but DC. At 300m I will be left only with 100-200mW (which is 20-21 dBm) and possibly still decent RSSI.

Unfortunately the difference between DC and RF can't be ignored. Skin effect and dielectric loss don't exist for DC, but they do exist for RF. The higher the frequency, the higher the loss.

RG6 is simply not a good choice for WiFi. 35dB/100m is too much loss, and since you're trying to use 75-ohm coax in a 50-ohm system, the loss will actually be even higher than that.

For comparison, LMR-400 has a loss of 22dB per 100m at 2.4GHz, LMR-600 is 14.3dB, 1/2" hardline is about 12dB, and 7/8" hardline is about 6.5dB. Microwave frequencies are not easy.

• This does not answers how the other testers achieved 64 meters with RG6 if the attenuation chart is correct. Nor it answers the approx. calc for 2W. Do you even know the price for LMR-400 per meter ? Nobody considers LMR-400 longer than 50 meters because of that.. The loss for 50 to 75 ohms connection is just 5% May 3, 2023 at 20:54
• @Svetoslav what the professionals do when working with microwave is figure out how to make the cable run much shorter than 50 meters. But if they can't do that, they buy something even more expensive than LMR-400, because there's no point paying for cable and installing it if what comes out the other end isn't even usable. May 3, 2023 at 21:05
• "figure out how to make the cable run much shorter than 50 meters". Sometimes that is just not possible. I have already assigned rules and that's not considered. "The correct DCR for RG6 should be about 2 ohms per 100m." - from where did you got this information ? Because if my multimeter is wrong then the DC calculations will show something different. May 3, 2023 at 21:12
• @Svetoslav "The loss for 50 to 75 ohms connection is just 5%" — The penalty for an SWR mismatch isn't a fixed number, it depends on what your cable loss is to start with. In this case it adds about 0.35dB per 100m (which is not huge, but it's more than 5%). May 3, 2023 at 21:13
• @Svetoslav DC resistance is on the datasheet too. If you're measuring pin-to-pin or shield-to-shield use the "DCR" numbers, if you're measuring pin-to-shield (with the far end shorted) use the "loop resistance" number. The datasheet I have says 6.5 ohm/1000ft, which is 2.1 ohm/100m. Other brands may differ slightly. May 3, 2023 at 21:15

You measured DC resistance correctly, however, resistance is not the same as the 75ohm characteristic impedance specified for the cable. Impedance includes both DC and AC components.

Also, cable (and transmission line) impedance is not specified over unit length. It is the impedance which a signal sees as it traverses down the cable, not over the entire length.

Resistance is important for total cable loss, which is specified over unit length. So, given a certain impedance, a larger wire inside a coax will have lower loss than the same length with a thinner wire.

• I have edit. Please check :) May 3, 2023 at 19:44

Your multimeter is measuring DC resistance - it applies a DC signal, and measures the resistance value. This, by its nature, ignores the reactive (inductance and capacitance) components of the impedance[1].

RG6 is specified to have a characteristic impedance[2] of 75 Ω for AC signals. The correct way to measure this involves an AC signal, not a DC signal - for example, from a VNA; you cut a length that's approximately $$n/8$$ wavelengths, where $$n$$ is an odd integer (because both quarter and half wavelength transforms result in very high impedances from a transmission line in this procedure, which makes your result less accurate).

Using a calibrated VNA, measure the impedance with the cable left open circuit, to get $$Z_{open}$$. Repeat your measurement with the centre conductor shorted to the shield, to get $$Z_{short}$$. The characteristic impedance of your coaxial cable is $$\sqrt{Z_{short}Z_{open}}$$; this method falls out from the Telegrapher's Equations.

You can also determine the characteristic impedance if you have the ability to measure inductance and capacitance. Measure the inductance and capacitance of a wire, to get the reactive components of the impedance; the characteristic impedance will be $$Z_0 = \sqrt{\frac{L}{C}}$$ where $$Z_0$$ is the characteristic impedance, $$L$$ is the inductance, and $$C$$ is the capacitance.

[1] Impedance is the combination of resistance, capacitance and inductance into a single figure; from the point of view of AC analysis, capacitance and inductance are both just forms of reactance with opposite signs, and by turning them into imaginary components of a complex number, and treating voltage and current as complex numbers, we get the mathematics of complex number calculations to work to give us the correct answers for AC analysis as we get when we use reals for DC analysis.

[2] Characteristic impedance is not resistance - it's the impedance the transmission line would present to AC signals if it were infinitely long.

• I have edit. Please check :) May 3, 2023 at 19:44