[Solved] Purely Inductive Load on Transmission Line [Smith Chart] - Page 1
Forum section: electronics / beginners

author: unassailable
author_rank: Newbie
timestamp: February 09, 2017, 04:14:44 pm
content: Heads up, university student here probing the internet for answers. This could get messy.I can't find an example online to compare with and I've read over many of the related smith chart questions on the blog but this is a pretty specific case I need help with.Question is to plot various loads on a smith chart at 300 MHz. Characteristic impedance (\$Z_0\$) is 50. Plotting open, short, matched, resistor and \$|\Gamma| \angle \theta \$ was easy. Inductor is throwing me off.Given:50 \$\mu H \$Then:\[Z_L = 2 \pi(300 \times 10^6)(50 \times 10^{-6}) = 94200 \; \Omega\]I'm hung up on normalizing this by dividing by \$Z_0\$;\[Z_n = \frac{94200}{50} = 1884 \; \Omega \equiv 0+1884j\]I have two conclusions to choose from:- this purely inductive load at the given frequency isequivalentnearly equivalent to an open circuitor-my logic is flawedAny guidance would be appreciated, thank you.edit: solved

author: Vtile
author_rank: Super Contributor
timestamp: February 09, 2017, 05:07:29 pm
content: Seems to my eyes 1880 ohms worth of impedance, but I have zero hours of hf transmission line theory under my belt. So don't quote me to your doctoral thesis.Needed to double calculate (with calculator, I only do it by hand if you give me salary):Assuming z0: 50j0 and ZL: 0j94200 and your equation is correct.

author: Vtile
author_rank: Super Contributor
timestamp: February 09, 2017, 06:30:12 pm
content: I don't really understand where you get the assumption that there is any sort of open circuit?Xl = L*s( or w) frequenzy dependaple resistance goes up when frequenzy goes up.Xc = 1/C*s frequenzy dependaple resistance goes down when frequenzy goes up.

author: PartialDischarge
author_rank: Super Contributor
timestamp: February 09, 2017, 06:46:19 pm
content: Quote from: unassailable on February 09, 2017, 04:14:44 pmNot exactly, everything is relative to the frequency you are working with and the inductor value in question. In this case that inductive load at 300mhz is a near open circuit.But a certain inductor + a length of transmission line could present at the source as either an inductive or capacitive load depending on the wavelength.In RF and microwaves everything is relative to the wavelength, keep in mind that.

author: unassailable
author_rank: Newbie
timestamp: February 09, 2017, 06:58:12 pm
content: Quote from: Vtile on February 09, 2017, 06:30:12 pmGood comment. Using asmith chart:- The impedance of C/L has no real part (=0), which means the plot of a C/L is on the outermost circumference where \$\dfrac{R}{Z_0} = 0\$- To plot the imaginary point along this circumference, the normalized reactance \$\dfrac{X_L}{Z_0}\$ is used. Starting at \$0j \equiv 180^\circ\$ and moving clockwise shows increasing levels of reactance.- At \$0^\circ\$ the reactance is infinite (open), the next smallest reactance level is \$50j\$.Plotting my calculated normalized impedance (\$1884j\$) on a smith chart is infinitesimally close to an open circuit, which is where I am getting that notion from

author: unassailable
author_rank: Newbie
timestamp: February 09, 2017, 07:02:19 pm
content: Quote from: MasterTech on February 09, 2017, 06:46:19 pmOkay, so since the problem states that lead length can be neglected, I guess the correct response would be to plot it near an open circuit.

author: KD0CAC John
author_rank: Frequent Contributor
timestamp: February 09, 2017, 07:09:05 pm
content: I suck at the math part , so not sure what your asking - because seems to me your getting stuck with the math and loosing sight of transmission line theory ?A few more items I see [ with limited vision , but do a lot ham radio , tower / antenna work ] your mentioned of " Plotting open, short, matched , resistor " to me that is referring to calibrating a spectrum analyzer or VNA before testing ?Also just in case , in your quote " matched , resistor " again to me match is saying 50 ohms and using the term resistor may be misplace , dealing with RF resistance varies with frequency and is referred to as impedance .Then mentioning Smith Chart , if it is 50 ohms - the center point on a Smith Chart - the center or the horizontal line and not enough info to determine anything else , like reactance , wether inductive - negative reactance below horizontal impedance line , or capacitive positive reactance above the horizontal impedance line .If I'm on the right track and you are at a university , try to get into a lab with someone that has some experience that can demenstrate a VNA in Smith Chart mode , or a signal analyzer in Smith Chart mode and get a little feel for it .

author: unassailable
author_rank: Newbie
timestamp: February 09, 2017, 07:14:54 pm
content: Quote from: KD0CAC John on February 09, 2017, 07:09:05 pmFunny you mention that, this is actually a prelab assignment for our network analyzer lab tomorrow morning. And I normally would consult with my classmates but there's a blizzard today, so that's out of the question.

author: rfeecs
author_rank: Frequent Contributor
timestamp: February 09, 2017, 07:17:50 pm
content: It is just because your inductor is so big. Try a 50nH inductor and see what you get. A small inductor will be close to a short, a large inductor will be close to an open.

author: unassailable
author_rank: Newbie
timestamp: February 09, 2017, 07:22:41 pm
content: Quote from: rfeecs on February 09, 2017, 07:17:50 pmYup, 50 nH places the normalized impedance around 1.88.I think you all have addressed my concerns. The relationship between the inductor, characteristic impedance and the frequency being used here is causing the normalized impedance to be near an open.edit: amended relationship statement, see comment below.

author: Vtile
author_rank: Super Contributor
timestamp: February 09, 2017, 07:36:17 pm
content: Quote from: unassailable on February 09, 2017, 07:22:41 pmNope, the difference between Z0 and ZL, if I start to get on hold what all this fuzz is about.But you know better.Edit: Oh, my brains just did got some bruises. I'll stay on area what radioheads call DC <10MHz and lines below 50km@50Hz. So the waves stay where they should and I can still live in fallacy with the centralised network models.

author: MrAl
author_rank: Super Contributor
timestamp: February 14, 2017, 09:11:57 am
content: Hi,I see you noted that this issue had been 'solved' where your main concern seemed to be trying to figure out why a certain inductive load at the end of the transmission line looked like an open circuit. Your conclusion seems to be that it looked like an open circuit because the frequency was high relative to the inductive reactance at that frequency, which is reasonable.However, being a self proclaimed university student you may want to look into the transmission line equations which are a set of partial differential equations that describe transmission lines. You can use them to calculate just about anything you want to know about a transmission line and how it works with just about any load you can think of. It will be especially easier because you are using a sine wave as a test signal. So these equations are very very interesting and definitely worth a look if you are already studying this stuff anyway.Good luck to you in your future endeavors.