# Voltage at antenna?

Discussion in 'General Technical Questions and Answers' started by ZL1UZM, Apr 8, 2010.

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1. ### ZL1UZMHam MemberQRZ Page

I'd like to be able to calculate the voltage at the antenna. Do you have a formula?
What, for instance, is the voltage at a 1/4 wave mobile whip at 2m and 50 watts power?
TIA

2. ### AG3YGuest

It is easy enough to get in the ballpark with the old EIR forumula if you are considering just the feedpoint and a theoretically perfect 1:1 match. However, out at the tip of the antennna, the voltage gets extremely high as the impedance approaches infinity !

I will defer this one to the antenna experts, I am only saying that since an antenna is NOT a fixed impedance along its length, the voltage/current relationship is going to vary all over the place. Look in an older Amateur's Handbook and you should be able to see diagrams of what I am talking about. They show the voltage as a sine wave with the highest points on the ends, and the current as a curve with the highest point at the feedpoint.

3. ### W8JIHam MemberQRZ Page

The voltage across the base insulator is easy to estimate. A 1/4 wave uniform cross section radiator has about 35 ohms resistance at resonance.

This means to radiate 50 watts, it would have sqrt of 35*50 = 41.83 volts RMS or 60 volts peak. You would use RMS voltage for heat or work and 60 volts for arcing or insulation.

That voltage would decrease with losses in the system, or increase if you were operating it off resonance.

If you are using it to pick the voltage rating of a capacitor it would probably not apply. The voltage across a capacitor would be the current through the capcitor times the reactance times 1.414 plus a safety factor.

4. ### ZL1UZMHam MemberQRZ Page

Thanks for your reply. I only know the R in the EIR formula. If I knew the I, I could work out the E. Do you know the amps?

5. ### K9STHPlatinum SubscriberVolunteer ModeratorPlatinum SubscriberQRZ Page

UZM:

P= (I^2)R

Therefore, P/R = I^2

or, I = sqrt P/R

If the power is 50 watts and the impedance is 35 ohms then use the following:

50/35 = I^2
1.4286 = I^2
1.1952 = I

Glen, K9STH

6. ### ZL1UZMHam MemberQRZ Page

Thanks Glen! Now I remember the formula. It's been 20 years ago when I learned this.
How can I work out the voltages along the antenna? I am mainly interested at the tip of the antenna and the halfway point.
I need to know this for safety reasons.

7. ### WB2WIKPlatinum SubscriberPlatinum SubscriberQRZ Page

At the tip of the antenna the RF voltage is much higher (but power is virtually nothing).

The voltage is even higher if the tip is a "point" as opposed to a "ball." This is one reason that most whip antennas have a ball (metal sphere) at the tip.

This is difficult to measure but *might* be estimated if you make some assumptions. If the impedance at the tip of the antenna is 10K Ohms, use the same formula to derive voltage: Sqrt (P*R) = 707Vrms (for 50W of applied power at the base) = 1000Vpk.

The "assumption" is the 10K Ohms. This actually varies with relative humidity, the shape of the tip, etc.

I've seen commercial use formulas that use 5K, 10K and 20K for this value.

8. ### ZL1UZMHam MemberQRZ Page

Thanks a lot WIK, great!!!

9. ### W8JIHam MemberQRZ Page

Remember when you are dealing with voltages and arcing, use peak voltages.

When you are dealing with heating or power, use RMS values.

The antenna is a transmission line. Assuming it is uniform cross section, the surge impedance of that single conductor transmission line determines the voltage along the antenna including the open end.

If the conductor diameter is such that surge impedance is 600 ohms, and you had 35 ohms at the feedpoint, the peak impedance would be 600^2/35 = 10286 ohms. Assuming negligable losses the RMS voltage at the tip would be
about 454 volts RMS or 641 volts peak.

The voltage halfway up should be about 70% of the open tip voltage.

While I made some assumptions, this will be pretty close for most typical conductor diameters.

Those are RMS voltages, not peak.

It's a whole lot more difficult to be a little more accurate.