# What is the impedance of a 1.25 λ dipole antenna?

I would like to make a dipole antenna. This particular antenna is used for reception only; no transmission is involved. I would like to receive as much power as possible by this antenna as it is used in energy harvesting.

I read that a length of 1.25 λ (i.e. each leg length is 5/8 λ) has more gain. To match it with a load, I need to create a matching circuit. So, what is the impedance of a 1.25 λ length dipole antenna?

• What frequency is this going to operate at? Apr 12, 2015 at 20:37

If you want a closed-form function, Wikipedia gives those:

\begin{align} R &= \frac{Z_m}{2 \pi \sin^2(kL/2)} \Big\{ \gamma + \ln(kL) - \operatorname{Ci}(kL) + \tfrac{1}{2}\sin(kL) \big[\operatorname{Si}(2kL)- 2\operatorname{Si}(kL)\big] \\ &\qquad\qquad\qquad\qquad + \tfrac{1}{2}\cos(kL)\big[ \gamma + \ln(kL/2) + \operatorname{Ci}(2kL) - 2\operatorname{Ci}(kL) \big] \Big\} \\ X &= \frac{Z_m}{ 4 \pi \sin^2(kL/2)} \Big\{ 2 \operatorname{Si}(kL) + \cos(kL)\big[ 2 \operatorname{Si}(kL) - \operatorname{Si}(2kL) \big] \\ &\qquad\qquad\qquad\qquad - \sin(kL)\big[ 2 \operatorname{Ci}(kL) - \operatorname{Ci}(2kL) - \operatorname{Ci}(2ka^2/L) \big] \Big\} \end{align}

I'm not even going to try to explain those equations since they are so hairy. Easier, and more insightful is to see a graph of the equations (from the same Wikipedia article):

In theory, a 5/8 monopole has an impedance of something like $(75-425j)\:\Omega$, and the impedance of a dipole is twice that of the equivalent monopole, so $(150-850j)\:\Omega$. Eyeballing that on the graph looks about right.

However, the thing to note here is that around 1.25λ, the slope of both the real and imaginary components of the impedance is pretty steep, meaning small changes in length make large changes in impedance. Also note that this graph is valid only for a conductor diameter of 0.001λ. Thicker or thinner conductors can also make a significant difference. The sensitivity of the impedance to these parameters explains the variance in the numbers given by various sources.

Consequently, to successfully build this antenna you will need some way to measure the impedance and adjust accordingly.

• I think you meant to say "In theory, a 5/8 monopole has an impedance of something like (75−200j)Ω , and the impedance of a dipole is twice that of the equivalent monopole, so (150−400j)Ω ". The simulation tool, 4nec2, shows 150-j398 with 0.644 mm diameter antenna. It matches with above graph. " Oct 12, 2015 at 3:04
• @ShivaMudide At what frequency did you perform your simulation? Is 0.644 mm equal to 0.001λ at that frequency? Oct 13, 2015 at 0:46
• Freq is 2.44 GHz. I don't think so. Oct 13, 2015 at 14:32
• @ShivaMudide well than explains the discrepancy, then: "Also note that this graph is valid only for a conductor diameter of 0.001λ. Thicker or thinner conductors can also make a significant difference." Oct 13, 2015 at 16:11
• Hi Phil, Can you please see this following link and share your thoughts. ham.stackexchange.com/questions/5397/… Oct 13, 2015 at 23:10

The impedance is double that of a 5/8 wave monopole, which is well studied, there's lots of information out there.

From the ARRL Antenna Compendium:

"The input impedance of a 5/8-wave whip above a ground plane 1/2 wavelength or more in diameter has a resistive component close to 50 ohms, and a capacitive reactance that depends considerably on the whip diameter, typically in the range of 50 to 150 ohms.

The impedance of your 1.25 wavelength dipole will be about 100 -j100 to -j300 Ohms.

Two comments on this choice of antenna:

• higher gain only means more power if you aim the antenna at the source. If you're harvesting in general, rather use a low gain antenna.
• this is a fairly narrow band antenna, so only really useful if you know the frequency of the signal you are harvesting. Its impedance will change quickly with frequency, and the matching inductors change the other way.
• I am keeping receiving antenna closer to the source. I found a website where it will calculate the impedance of a given dipole antenna length. I don't know whether these calculations are correct. I want to cross check with some other website. dryspire.com/dipole-z Apr 11, 2015 at 20:19
• That website calculator looks reasonable, I'm sure it's correct. The problem is that the impedance changes very quickly with length (and frequency). Seems the 100 Ohm point is closer to 1.33 wl, for a 100:1 l/d dipole. Apr 11, 2015 at 21:26
• I found another tool call 4nec2 and it shows a different impedance for 5/4 lambda length dipole antenna. Don't which one to believe. Apr 12, 2015 at 19:59
• All of the tools are giving reasonable answers, within their limitations... what you're finding is that the impedance of long dipole is highly sensitive to the length, diameter, and the modelling algorithm. Try a 0.45 wavelength dipole on each program and you'll find them a lot closer together. Antenna design is about controlling the unexpected effects; the basic electromagnetic design is often trivial. The theoretical impedance of a perfect 5/4 wavelength dipole is only the beginning of complex circuit, sensitive to everything.. Apr 12, 2015 at 20:37
• Most worry-sum is that the website dryspire.com/dipole-z gives -ve imaginary (like in R1-JX1) part whereas 4nec2 gives +ve imaginary (like in R2 + JX2) part when I calculated for 2.4GHz with 1mm diameter. Apr 12, 2015 at 21:08