# Does increasing the frequency of an antenna require energy?

1. Mar 22, 2015

### Rhannmah

I'm under the impression that if you increase the frequency of a circuit coupled to an antenna, for the same amount of photons emitted, the frequency of those photons will increase therefore having them carry out more energy total, thus increasing the amount of energy required to power the circuit. Am I wrong in thinking this? How does this relate to the amplitude of the electromagnetic wave?

Are there any counter-examples to this in radio or other telecommunication circuits where increasing the frequency does not require extra energy? If so, how is this reconciled with the fact that higher frequency photons carry more energy?

2. Mar 22, 2015

### rootone

Increasing the frequency in itself requires additional energy to power the circuit.
The amplitude of the signal produced is not directly altered as a result, at least that should be the case for something like a simple multivibrator oscillator circuit producing square waves.
However you could design a circuit in a way so that the amplitude is made to to decrease as frequency increases.
Thus the circuit could be made to consume a constant amount of power regardless of what frequency it is operating at.
I don't know of any practical application of this.

Last edited: Mar 22, 2015
3. Mar 23, 2015

### tech99

If you increase the frequency, the total energy stays the same but the size of each photon increases. Fewer, but larger, packets.

4. Mar 23, 2015

### rootone

In which case my reply was wrong then, at least in part anyway.
I do know that in practice higher frequency circuits generally do need more power,
but given what you said, the reason for that probably has more do with their tendency to lose more power as heat.
That in turn is probably due to physical characteristics of semiconductors and such.

5. Mar 23, 2015

### davenn

That is just so wrong

6. Mar 23, 2015

### zoki85

Photons? What the heck is that? In my radio engineering handbook I have nowhere that mentioned

7. Mar 23, 2015

### Staff: Mentor

There is no fixed relation between frequency and power. You can operate an antenna at every power you like (assuming it does not melt down), independent of the frequency.

Photons are a bad model for the emission of antennas.

That does not make sense at all.

8. Mar 26, 2015

### Rhannmah

Being unsatisfied with the answers given here (it didn't seem quite right), I did a lot more research and yes, radio photons are workable, they are just hard to research because everything at room temperature emits radio photons saturating the environment with a ton of noise. That said, I found excerpts from this textbook asking the reader exactly what I've been trying to figure out on page 853 (stop to think 25.5):

"Two FM radio stations emit radio waves at frequencies of 90.5 MHz and 107.9 MHz. Each station emits the same total power. If you think of the radio waves as photons, which station emits the largest number of photons per second?"

Answer http://www.cabrillo.edu/~cfigueroa/2B/2Bprob_sets/2B_book_problems/25_problems.pdf [Broken] (stop to think 25.5, bottom of the document).

As I thought, the lower frequency radio photons carry less energy, therefore for equal power, more photons are emitted from the antenna.

Does this seem correct?

Last edited by a moderator: May 7, 2017
9. Mar 26, 2015

### davenn

Please don't use that term, there's no such thing as radio photons ... photons are photons

10. Mar 27, 2015

### Staff: Mentor

Correct.

@davenn: Where is the fundamental difference between speaking of "10 GeV photons" and "radio photons"? The former is used everywhere in high-energy physics, as the reference frame (earth) is clear.

11. Mar 27, 2015

### Baluncore

Wave particle duality can be a nuisance here, but only because of the language used. The OP question does not specify the electronic circuit involved or which parameters are maintained constant as frequency is changed.

If we consider light emitting diodes, the number of photons is proportional to the current while the energy of each photon is proportional to the voltage. Power is clearly the product of current and voltage.

If the same current is maintained, then increasing the frequency of the photons is equivalent to reducing the photon wavelength and so increasing the energy. That is achieved by changing the semiconductor material to increase the voltage. V*I = W must therefore increase because I is maintained constant while V is changed.

Exactly the same reasoning can be applied at Radio Frequencies.

12. Mar 31, 2015

### sophiecentaur

E = hf for all photons of EM at all frequencies. That is correct in principle.
There is, however, very little point in talking about photons where RF signals are concerned because of the ridiculously low energies associated with RF photons, compared with the Energy levels in electronic components. QM is not always the best way to approach problems.

13. Mar 31, 2015

### Rhannmah

Well if you have, for example, a radio station broadcasting with a circuit operating an antenna at a frequency of 50 Mhz and a 2nd one broadcasting at 100 Mhz, would the 2nd one require twice as much energy as the 1st one if they are both emitting the same amount of photons? If so, it would seem like a concern when building an emitter.

Does this situation even happen in real life or other factors are in play, here?

14. Mar 31, 2015

### Staff: Mentor

How would that be relevant in any way?

15. Mar 31, 2015

### davenn

As others have said, you really do need to stop thinking about this as the emission of photons and just stick to EM wave

Considering your 50MHz and 100MHz antennas. Lets for a moment make them both 1/2 wave dipoles

The 100MHz dipole is only 1/2 the length of the 50MHz one and if say 100Watts of RF was applied to each, then less energy is going to be wasted
heating the metal elements of the 100MHz dipole ( just a consideration )

D

16. Mar 31, 2015

### Baluncore

The specification of transmitters is based on power and frequency. EM energy radiates from the antenna as waves of photons, not as individual photons. The efficiency of operation is photon independent since the radiated energy is measured in power per unit area. The number of photons emitted cannot be counted because they are so small and there are so many travelling together.

We can do the numbers; Photon energy in electron volts is, eV = 1239.84 / (wavelength in nm). That is also the forward voltage of a Light Emitting Diode. A 10 MHz signal has a wavelength of 30 metre. So the energy of a 30 metre wavelength photon is 41.328x10-9 eV which is equivalent to 6.621x10-27 joule.

We measure the frequency as only 10,000,000. cycles per second, but at 10MHz there are about 1 / 6.621x10-27 = 151.x1024 photons emitted per second per watt.
Those photons are much much too small and many, too many to count.

17. Apr 1, 2015

### sophiecentaur

If you choose to think of photons as little bullets and an antenna shooting them out in all directions, you are in a hiding to nothing. The fact that energy is transferred in distinct lumps ( which is how photons are often best viewed) then fine; you can come up with an answer about how many per second are emitted. But what relevance has that number? It's certainly not what QM is all about and it does nothing for your understanding of EM and Antenna theory. Don't propagate the nonsense that so often gets told you in School. PF can put you straight on this topic.

18. Apr 1, 2015

### sophiecentaur

The one with the same number of high energy photons would need to be transmitting more power. Is that a problem?
E = hf
As I wrote before.

19. Apr 1, 2015

### f95toli

Just to add what has been written above: When you DO need to think about individual radio frequency photons things get very complicated very quickly. For example,. if most states of light do not contain a fixed number of photons (the states that do are so called Fock states or number states) and it is actually very difficult to generate a "stream" ofindividual photons (this is one of the things I work on, albeit at microwave frequencies). A typical source of radiation at radio frequencies will emit something more like coherent or perhaps thermal radiation.
Hence, whereas it sometimes make sense to talk about the average number of photons in an electrical circuit (if you are working at very low powers and very low temperatures) you can still not say anything about how changing the power changes the number of emitted photons; simply because that number is not fixed..