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Sinusoidal Inductor on PCB

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Jun 28, 2011
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I'm a first year student in engineering school and i have to design a sinusoidal inductor on a FR4.I haven't found subjects about this,only formula about spiral inductor.I thought i could use a microstrip line and bend it on PCB but it's wrong :roll: Frequency:1Ghz or more

I'm supposed to find a link between the inductance and the dimension of the sinusoid but I'm stuck at the begenning: I can't find a theorical model nor a way to include mutual inductance between the line...

Does somebody has an idea ? I'm trying to figure out of Fasthenry work to realise the shape in 3D.

Many Thanks!

Thanks you i didn't know the english name for this line.Did you have any idea what would be the inductance of the quarter circle ? because meaner line have right-angled corner.So i'm lost when i have to find the mutual inductance.

Many Thanks!

No sorry my english is bad. I try to do something like this:

I have no idea how to work with circular corner.

I would like to solve my problem like this: **broken link removed** But I need to find the self inductance of the corner and mutual inductance between 2 corners.
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I finally manage to realise something.I used ADS to realise my line:


Then I use the simulation tool to get all the parameters S and I convert them to Y with matlab and solve L:



I want L=68nH and I try to change values such as w or the length but I always get a value of L near 360nH. I have no idea how to solve this problem.Maybe i made something wrong with ADS but I have never used this software before.

Many Thanks for helping!

You might be missing a basic concept. Inductors act, pretty much, like an inductance. At 1 MHz, a lumped 100 nh inductor looks like a 100 nh inductor. At 100 MHz, a lumped 100 nh inductor looks like a 100 nh lumped inductor. It is only at very high frequencies, where the length or width of your lumped inductor starts becoming a significant fraction of a wavelength that the lumped inductor stops acting like an inductor.

One reason for this is the energy from an inductor is confined in a small space.

If you try to approximate a lumped inductor by using a transmission line, you will quickly find that:
1) is only approximates a fixed value inductor over a limited frequency range. Over a broader range, the apparent inductance value starts to change. THis is a problem in things like lowpass filters, which need the inductance value to stay ~ fixed over a broad frequency range.
2) at some frequencies, where the approximation transmission line starts getting to be a significant fraction of a wavelength, the "inductance" can look like a capacitance, or a short circuit, or an open circuit. In fact, if you study transmission line theory, you will see that the actual impedance of your transmission line changes as a hyperbolic tangent function! i.e. not linearly.
3) that structure you analyzed in ads uses long lengths of 50 ohm lines. If you study the telegrapher's equation, you know that a transmission line looks like distributed series inductances and shunt distributed capacitances. The ration of L to C governs the impedance, by Z =square root (L/C). So, by that reasoning, if you are trying to approximate an inductance with transmission lines, use HIGH IMPEDANCE ones and it will work better. In your case, make the lines 100 ohms, and they will look more like an inductance.√√
And maybe you should do momentum analysis.

Thank you all.
@BigBoss : It was a really stupid mistake thanks ! But it hasn’t solved my problems ^^

@biff44 : Thank to you I realize that I did understand a lot of part in my Transmission Line Lesson… In the beginning I wanted to use a T-model and solved it to find L and C but I didn’t find the impedance Z.When you said in 3 “make the lines 100 ohms” what is the meaning? I set ports to 100 ohms but I also need to set all lines impedance to 100ohm? I tried to use Line calc to find the correct length but I can impose Z and W for a strip line.

@tony_lth: Thanks for the advice I tried to use momentum analysis. Finally it seems to work:


I made few modifications on the size:
W=0.5mm; Circle radius:1mm; Length L1/L5=2.6mm,Length L2/L3/L4=5mm
And I get and inductance of –77.35nH (I don’t know why but I have a – with the previous equation…).I take PCBlines16_lay.slm as substrate. I think I fail using the port:
I set port 1 to Single Mode with 100 ohm impedance and Port 2 to Ground reference (I think it is a mistake).

Then I plot S and Z0 .I have 2 columns for Z0 (1 complex: 113.105+i*0.215 and the other: 50) I make the resolution with Z0 complex and I get -77.35nH.Is it wrong?

Many Thanks Again!

Use LineCal, set the Z0=100, and you can get the width for 100 ohms, the set the width of the trace as your calculation.

You can set another schematic with a lumped inductor with your required value, then compare the S params,

Thank for this idea! I should have thought to it before!
So I used line calc and get the width for 100 ohm: 0.6417 mm for my substrate.
Then I made a schematic inductor of 68nHand I try to compare the 2 S parameters.
For simulation I use 50 ohm port in Momentum for the line and in Schematics for the lumped one(I think I should have use 100 ohm?)


I modify the lengths of the different parts. Finally I succeed in equalizing the imaginary parts but not the real parts. The determinant of the S matrix is the same for theirs real parts but not for their imaginary parts: lumped inductor is twice smaller.

Is it still something correct? Or am I completely wrong again? ^^

Many Thanks!

I reviewed the thread and had serious difficulties to understand your exact intention.
- You said, you want to implement an inductor, not a transmission line? That means, that it has inductive beahviour in the frequency range of interest.
- But you are inplementing a microstrip transmission line, involving a substrate with a ground plane
Consequently, the implemented component behaves much more like a transmission line (respectively a cascaded LC circuit) than a lumped inductor. Furthermore the transmission line is a two port (four terminal) device, while an inductor woul be a two terminal device.

Thanks, I will try to explain clearly:
My boss worked on a project years ago where they had an “inductor on a PCB”: There was something with the shape I draw and next to it was written “L=68nH”. He told me that it was working like an inductor but it had the look of micro strip line. He wants me to find a way to recreate this “inductor” because the company who did that thing does not respond to his mail.

That why I tried to use a micro strip line but it seems clearly that I cannot denied the capacitance between the line on the PCB and the ground plane. I think we have 4 ports (2 on the top of the PCB and 2 on the ground plane). So I’m trying to find the values of L and C in the telegrapher equation and also trying to have a lossless transmission line. I think this is the only solution. Anyone agree? Or somebody has another idea?

O.K., if the structure has a ground plane, it can't be appropriately described as an inductor for GHz frequencies. S parameters can describe it, of course. But the question if it can be treated as an inductor respectively which of its parameters are significant refers to the embedding circuit or in other words, the externally connected impedances.

If you have a distributed (two port) circuit and want to measure it's inductance, you have at least two possible measurement circuits:
- short one port and measure the input impedance of the other (by measuring S11)
- perform a two port measurement and derive a series inductance from S21

In general, both results will be different.

you keep responding to quesitons but do not give any additional information. FINALLY you tell us you want 68 nh. What frequency? What board area do you have available? Is there a ground plane in the board, and if so how many layers down? Do you want a series inductor or a shunt inductor to ground? Why are you not looking at spiral inductors--just being stubborn?

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