# 1/4 wave resonantor equiv to a series LC circuit and voltage magnification?

1. Aug 19, 2011

### hobbs125

I recently read that a 1/4 wave resonator is equivalent to a series lc circuit but with a much higher q.

In a typical resonant LC series circuit voltage magnifications exists, does the same thing occur in a 1/4 resonator since it is equivalent to a series LC circuit?

2. Aug 19, 2011

### yungman

I don't think so, The voltage amplification exist in the discrete LC only because you can measure the junction between the L and C. The 1/4 wave resonator has the equivalent impedance of the LC in series, there is no junction in it to be measure.

From my understanding, it is the impedance that equal to the LC circuits, that at frequency at $\;\frac {\lambda} 4 \;$ the quarter wave line acts like an shorted circuit. Also the quarter wave line acts like either an inductor or a capacitor below and above the center frequency resp. according to the Smith Chart just like the LC tank circuit.

The formula is:

$$Z_{in}=R_0\frac {Z_L+jR_0 \tan\beta l}{R_0+jZ_L\tan\beta l}$$

So when you have an open end quarter line:

$$Z_{in}|_{Z_L=\infty}=R_0\frac{Z_L}{jZ_L tan \beta l}= -jR_0 \frac 1 {\tan \beta l}\;\hbox { for }Z_L \;\hbox {>> } R_0$$

So when $l=\frac {\lambda} 4 \;\Rightarrow \tan \beta l = \tan(\frac {2\pi}{\lambda}\frac {\lambda}{4})=\tan \frac {\pi} 2 =\infty\;\Rightarrow Z_{in}=0$

You can also so if length is smaller than quarter wave, the input impedance is -ve which indicate it is capacitance and in length is larger than quarter wave and smaller than half wave, it is +ve and it act like an inductance. This is just like a series LC resonance circuit.

Last edited: Aug 19, 2011
3. Aug 19, 2011

### yungman

I have to point out it is not usually practical to do a quarter wave open line resonator like this except using it as distributed capacitance when the length is much smaller than quarter wave ( more like no more than 1/10 wave length). Too many parasitic elements and you likely not getting a good approximation. We use shorted quarter wave transmission line more to simulate a parallel LC circuit which is short circuit at other frequency and behave like an open circuit and the center frequency. In RF design, this is commonly used as power feed for the transistor or amp. because at the operating frequency, it act like an open circuit and out of the way of the rest of the matching network that is so critical in microwave design.

4. Aug 19, 2011

### vk6kro

Yes, it does produce voltage magnification.

A quarter wave length of wire placed above a metallic sheet (such as the body of a car) has an impedance of about 39 ohms and can be driven by a radio transmitter at the resonant frequency.

The top end of the quarter wave of wire develops a much higher voltage than the drive voltage. Even low powered transmitters can produce voltages of hundreds of volts at the top of a quarter wave antenna.

Touching the top of such an antenna can produce painful (but usually not dangerous) RF burns. Transmitters of 100 watts or more can produce corona effects where you can see a blue glow near the tip of the antenna in darkness.

5. Aug 19, 2011

### yungman

Do you have any link that talk about this, In rf transmission lines design on pcb, I never heard of this. But then we pretty much avoid open end tx line close to quarter wave as I said before, we make it a point not to have any open line more than 1/10 wave length to simulate a cap and ensure it really behave like a cap.

6. Aug 19, 2011

### vk6kro

I'll have a look, although this is more like textbook stuff.

You will find it in practically any textbook dealing with antennas.

There are plenty of such images here:

7. Aug 19, 2011

### yungman

Thanks, I'll look into this. I am just starting to study antennas.

8. Aug 20, 2011

### Averagesupernova

If transmission lines can act as impedance transformers doesn't it make perfect sense that they also have to transform voltage?

9. Aug 20, 2011

### yungman

I don't think so. It is well known that at the open end of a transmission line, you get double the voltage. The standing wave at any point on the line regardless of length vary between 0 to 2 of the input voltage. That is the reason I post the first post, We call quarter wave tx line as resonator also. I still have to find time to read up how open end antenna work, how is it different from transmission lines like stripline, microstrip or coax. With those, for sure you don't get high voltage out of it. Just double.

From the formula I provided, you can see the reason you say the line behave like cap or inductor only because of the + or - j. The impedance behave like the tank circuit, but not the voltage. This is in any of the basic RF text books. This is nothing more than the standing wave pattern of an open end transmission line.

10. Aug 20, 2011

### skeptic2

First to answer the OP's question, in some aspects, yes it does act like a series resonant LC circuit with a much higher Q. That doesn't mean that every aspect of one will have an equivalent in the other.

Regarding the question about transmission lines acting as transformers and being able to transform voltages above twice the input voltage the answer is yes. To transform one resistance to another resistance, the transmission line is given an impedance equal to the geometric mean of the two resistances. Geometric mean = √ R1 * R2. Since resistances can be transformed in ratios greater than 1:4, voltages also will be transformed to greater than 1:2 ratios. In fact that's one way of getting more than a few watts out of a mobile transmitter operating on 12V into a 50 ohm antenna without using a DC/DC converter.

Edit: I forgot to mention that the transmission line in the example of impedance matching, must be an electrical 1/4 wavelength long.

Last edited: Aug 20, 2011
11. Aug 20, 2011

### yungman

You look at it in phasor form, the Max is only forward phasor + reverse phasor and max can only be additive. As I said, look at the standing wave pattern, never go beyond twice.

Yes, resistance can be from a short to open circuit, but not the voltage to more than twice. Please provide the formula if you think otherwise.

Last edited: Aug 20, 2011
12. Aug 20, 2011

### skeptic2

I checked a few online references but couldn't find anything appropriate. The only way I can explain it is through use of a Smith Chart. http://sss-mag.com/smith.html

I believe you are assuming the transmission line is the same impedance as the voltage source. As such you are right, the voltage cannot exceed twice the source voltage. However if the impedance of the transmission line is the geometric mean between the real parts of the two impedances, then it is possible for the voltage to exceed twice the source voltage.

13. Aug 20, 2011

### vk6kro

In a quarter wave of transmission line, the voltage step-up (or down) can be a lot more than twice.

Take a 100 ohm line, one quarter wavelength long (allowing for velocity factor).

Put a 1000 ohm load on it. At the other end, it will now look like a 10 ohm resistor.
(Zin = Zline^2 / Zload ie 100 * 100 / 1000 = 10 ohms).

So, if you put 100 watts into the line there will be 31.6 volts across the input.
( 31.6^2 /10 ohms) =100 watts.

Assuming no losses, there will be 100 watts delivered to the 1000 ohm resistor, or 316 volts.
(316^2 /1000 =100 watts).

This is 10 times the input voltage.

14. Aug 20, 2011

### Averagesupernova

I guess I haven't assumed anything as far as source impedance goes. I just want to point out what seems so obvious to me and that is that transforming impedances regarless of the method also transforms voltage. You cannot have one without the other. VK6KRO's last post is most certainly spot on correct as far as I know.

15. Aug 20, 2011

### yungman

I need to see the reference of the assertion. I worked with transmission line quite a bit and I really study study transmission lines. What we are talking about is nothing more than standing wave with open end termination. There are so many basic rf books talking about this. All you can get is the max when the incident and the reflected voltage phasor in phase and add to give you double the voltage.

As I said, I cannot argue about antenna as I am just studying it right now. And antenna is not perfect tx line so I don't want to say one way or another. But I read a lot of books and I designed transmission line filters, matching network and couple line circuit, you don't get high voltage out of it.

I am not saying I am expert, I am still learning. But I think I have a lot of books to back me up on this. I typed the formula exactly from the book on my post. It is not about the impedance go from 0 to infinity. This is not like a ideal current source that if you have open circuit, then you get very high voltage. This is about phasor and if you have the reflection coef $\;\Gamma \;$ equal to 1 in open circuit, your reflection voltage is only equal to the incident voltage.

Remember the very basic formula:

$$\Gamma =\frac {Z_L-Z_0}{Z_L+Z_0}\;\hbox {, if }\; Z_L=\infty \;\Rightarrow \Gamma = 1$$

Last edited: Aug 20, 2011
16. Aug 20, 2011

### yungman

If that is the logic, all I need is to put a small section of tx line and use a relay to snap in a very low impedance voltage source ( say a good low loss cap charged up to say 10V). This create a step function that contain broad frequency components, then one of the frequency will have the quarter wave equal to the length of the tx line and you should get very high voltage out of it. This will eliminate the ignition coil in cars, you get sparks out of a transmission line with a low impedance step generator.

My understanding is you can inject in power, but if you an not having a perfect match on the line ( which depend on the termination of the line where in this case is open), reflected power just shoot back into the generator and the result is you cannot put in the power that you want. I think you are assuming that you inject the power and the tx line is going to take in all the power. But in real life, the power just got reflected back if the the transmission line system is not a perfect match and you only get double the voltage at the open end. This is how I understand it. I would like to see some material otherwise.

Last edited: Aug 20, 2011
17. Aug 20, 2011

### skeptic2

yungman, when you were designing transmission line filters and matching networks, what frequencies did you work with? When you were designing matching networks, I presume you obtained good matches in which nearly all the power was transferred to the load with a low VSWR or return loss. If the load impedance was higher than the source impedance, how could the power supplied by the source be dissipated in the load unless the voltage was higher at the load? Neither vk6kro nor I am talking about reflection from the load. In our posts we are talking about perfectly matched sources and loads. A transmission line can act as a transformer to transform impedances and voltages just as do matching networks or transformers.

18. Aug 20, 2011

### vk6kro

Above is an excerpt from the ARRL Antenna Handbook 1988 edition .

Or, you could check Terman's book "Electronic and Radio Engineering" 1955 ed, P120. This is regarded as a Bible by most of us.

This is an unusually useful technique and it would be covered in almost any book on transmission lines or antennas. Your local library or technical bookshop would easily provide such a book.

19. Aug 21, 2011

### yungman

Yeh, you two are right, I was watching tv and it just dawn on me!!!! I remember now that I actually design using quarter wave line to transform from low impedance to high impedance and have perfect matching for max power transfer. The voltage got to be more than double. I was so sure I was right that I was so dead wrong.

Sorry, it's been over 6 year since I was working and I have not been using it, things start leaking out the wrong end, blame it on old age!!!