Transmission Line Characteristic Impedance

[@usagtv]

In electronics, to distinguish between analog current (i) and SQRT(-1), the convention used among Electrical Engineers is to use the letter "j" to indicate the imaginary term SQRT(-1). Therefore, the formula at 3:30 should be written using the letter "j" instead of the letter "i" to indicate that the characteristic impedance has imaginary terms. The letter "i" is typically used in general mathematics and general physic formulas.

[@Ghost572]

yeah to avoid any confusion with i being used for current

[@bobsimmons5274]

Notayion j actually derives from the unit vectors in vector math, where vectors i, j and k replace the Cartesian axis' X, Y, and Z. Unit vector j is oriented same direction as imaginary axis i in an Argand ( complex plane ) diagram, but j rarely has an arrow overscript, as it SHOULD have, to indicate it expresses a vector. Lots of confusion because of this, very sloppy and careless math notation makes it all much worse. This confusion is very VERY common.

[@saeidesekhavati1518]

Thanks for taking the time to record these tutorials!

[@scottpelletier1370]

Great videos. At ~ 12:27 I think the general value is 6 inches per nanosecond (instead of 6 ns/in)

[@Zachariah-Peterson]

Yes in my haste I inverted the units. I'm a physicist so we use m/s for everything

[@Amos-l6u]

Hi Zach Peterson, thank you for the explanations! But how high does the frequency need to be that skin effect is significant enough that we need to take note of?

[@Zachariah-Peterson]

This is a really great question and it is something I will need to do a video about. It requires some calculation and creating some graphs so I will need to take some time to do that, but I'll make a video about it before I head to Europe.

[@usagtv]

At 7:13, the approximation is assuming that the transmission line is lossless where R and G are zero. A lossless transmission line does not dissipate power, the the signal is not dispersive, and the characteristic impedance behaves like a resistor. A transmission line looks lossless when the frequency used is so low that it acts almost as a DC component to the line. The transmission line length is small with respect to the wavelength of the signal traveling on the transmission line. This is true in ULF and VLF. Lossless is relative to the length of the wire with respect of the wavelength of the signal traveling along the wire.

[@zizogo3417]

Great explanation. Btw, you forgot an omega next to the "i" when you simplified gamma to the lossless case. EE talking here: the epsilon should have a minus sign. Why? so that the loss tangent (Df) is positive. At the end of the day, most EEs care mainly about Dk and Df, and if we define these to be positive, then epsilon must have a negative sign. What is your argument?

[@Zachariah-Peterson]

OMG you're right... count that as the velocity term then!

[@Zachariah-Peterson]

Here's the deal with the negative Df thing: it was brought into RF design from AC power, which incorrectly applies a positive phase angle as a graphically negative angle. I come from an optics/lasers background so I approach this from Euler's identity with the mathematically correct phase definition. This means that a linear transformation forward in time has a negative phase, and a wave traveling along the +x direction accumulates a positive phase that is equivalent to a linear transformation of a wave NEGATIVE in time. This all means that the correct wave equation solution for a forward traveling, forward-in-time wave is A*exp[i(nkx-wt)], where n is the refractive index of the medium. Now, you use the refractive index to determine how the wave behaves during propagation (loss or gain, and dispersion). The refractive index is n = Re(n) + i*Im(n). If you substitute this into the wave equation solution, you'll see that there is an exponentially decaying solution when travelling along the positive x axis. This is attenuation by definition. The dielectric constant is by definition = n^2, so if Im(n) is positive then the imaginary part of the dielectric constant must also be positive. This means that the imaginary part of the dielectric constant MUST be positive in order to describe attenuation. We have gain during propagation only if Im(n) < 0. The loss tangent being positive or negative has nothing to do with anything, it's just a ratio. What matters is that you get attenuation in the forward direction when traveling through an attenuating medium. In order for the negative imaginary part of the dielectric constant to correspond to attenuation, either time flows backwards, or forwards and backwards are the same direction.

[@ehsanbahrani8936]

Thank you ❤

[@lavixl]

If you're looking for smooth copper choose VLP-2 as your copper choice. Also a beautiful well thought out transmission line can goto crap once you meet your connector. SMA SMB, etc.. The artwork time and thought around your connector to transmission line area can greatly improve your over all line loss/ impedance. Some nice calculators are ultracad, Saturn PCB, Polar (cost money) even Rogers website has many calculators. Really nice video I appreciate the way you explain and lay things out. Wheres the link to your channel?

[@TheDigital19]

i recently had a chat with our fabricator - he seemed very anxious about VLP-2 or super smooth copper in general. he told me about how important copper roughness is for proper pcb stack bonding. quite interesting input, i thought :)

[@Zachariah-Peterson]

Thanks for watching! Here's my channel link: kzbin.info/door/EJELy0Puuy47Dpt7zqyX6g

[@KokilaAc]

Hello Zach Peterson, Could you please explain about "is there any calculation to find maximum trace length for pci express or remaining high speed signals".

[@Zachariah-Peterson]

There is a simple way to do it. You need to know the maximum link loss (insertion loss) allowance up to the Nyquist frequency for the bitstream. These values depend on the PCIe generation or on the specific interface, there is no single value for every high speed link. Once you know the total loss at Nyquist, divide this by the loss per unit length and you get the maximum length allowance. It's a good idea to allow 10-20% margin. If you have favorable impedance matching at the input of the link up to Nyquist and there are no other sources of loss at specific frequencies (like stubs), then this will ensure that the length you use will also meet the loss spec at all frequencies below Nyquist.

[@gulafshabhatti9410]

what will be the formula of inductance and capacitance in terms of skin effect?

[@Zachariah-Peterson]

Watch this video from my channel from my IEEE EPEPS presentation in 2020. It explains each of the terms in the characteristic impedance equation and you can see what the inductive and capacitive modifications are due to roughness. The skin effect does not affect the capacitance, but the roughness of the copper does affect the capacitance. kzbin.info/www/bejne/Z2aUaJhrf62pbZY

[@cvillf4694]

Great video

[@AltiumAcademy]

Thanks for the visit

[@PankajKumar-zr3tv]

Sir, how much time it takes to learn about board design?

[@apurvawalunj3407]

Hello Zach, could you please let me know your channel name?

[@Zachariah-Peterson]

You can check out my profile, you'll find my conference presentation videos there.

[@Zachariah-Peterson]

Here's the link: kzbin.info/door/EJELy0Puuy47Dpt7zqyX6g

[@PankajKumar-zr3tv]

Thanks 😘😘😘😘

[@robegatt]

Planes capacitance is negligible, the power and ground planes supply a low inductance path for power and signal.