# bogatin

's profile**prof**

Biography has not been added

## bogatin

's contributions- 06.22.2012
- How to recover a pulse signal with a large capacitance load
- the root cause of this problem is too high an output impedance of your driver. Just use a 50 Ohm driver and your pulses will look pristine at the end of even long cables.

- 07.19.2016
- What is the resonant frequency of a cavity?: Rule of Thumb #30
- excellent example!

- 07.19.2016
- What is the resonant frequency of a cavity?: Rule of Thumb #30
- The cavity resonance will be about 160 MHz. The DC blocking caps on the board will shift the resonance to higher frequency, but its their inductance that determines how much shift you get. You need a surprising number ot have any real impact on the resonant frequency.

- 07.19.2016
- What is the resonant frequency of a cavity?: Rule of Thumb #30
- There will be resonances between each end and even diagonals. However, the longest dimension will give the lowest frequency resonance.

- 02.19.2014
- Total inductance in the return path: Rule of Thumb #7
- I_Drppa I cannot tell from this plot what the simulation was set up for, what is being plotted or what type of field solver the Maxwell D tool is. I assume it is a quasi static tool and is probably similar to Q3D, in which case it is calculting the B field lines from just the current in the wire, assuming ideal sources and sinks on each end. However, it depends on your boundary conditions on the ends of the wire. Do you have ideal sources and sinks? If what you are plotting is the B field at a fixed distance from the center of the wire, along the length of the wire, then you might want to extend the length to beyond the edges of the wire to see the B field drop off. You also might want to use more mesh elements- increase the number of adaptive iterations. I can see the finite size of your mesh in the last B field density plot. You maybe seeing some resolution issues in the B field. It is good for you to question what you are seeing. I always recommend to start a problem for which you know the answer if you can't get from the tool what you expect, do not proceed with the tool. Figure out why it is not as expected. This discussion should probably continue offline if you are interested.

- 03.13.2015
- What is the ringing period on an unterminated line?: Rule of Thumb #26
- This is a really good question, one which come up for discussion during my AmpHour interview. The highest data rate at which you need to worry about the SI effects depends on which problem will cause the product to break first. Unfortunately, the answer is often, "it depends." The reflection noise from unterminated transmission lines is just one of the problems, and often not the first one to cause a problem. When the reflections happen during just the early part of the unit interval, sometimes they don't affect the details within the mask. Likewise when your UI is 80 nsec, a length skew problem will never appear. But, there are three other problems which can kill a 16 MHz arduino porject. Any of these problems may result in a simple 16 MHz Arduino Uno to not work if the layout is screwed up. 1) the short rise times means you can get large ground bounce. This will happen in the cheap packages, and when signals cross gaps in the return planes. This is very common in 2 layer, low cost boads. Even the 16 MHz Uno will show this problem. WIth die shrinks, the rise time decreases and you may encounter ground bounce problems with a cheaper chip you didn't see in the last gen chip. 2) if you have to go thru EMC certification, the short rise time or ground bounce problems may also cause EMC failures. You would only see this problme when doing EMC testing, though. 3) Many arduino projects are now IoT with wiFi or Zigbee rf interfaces adjacent to noisy digital I/O. Shielding the noisy digitial from the senitive rf receiver will become a significant issue in future IoT designs. Just paying attention to a few SI principles will push these problems into much higher frequencies.

- 02.19.2014
- Total inductance in the return path: Rule of Thumb #7
- I_droppa- If you look at just the partial circuit consisting of a short section of round wire, and calculate the external B field density around the wire, only from the current in the wire- with an ideal source at one end and sink at the other, the field density will drop off near the ends. This is related to the partial self inductance. This situation is physical not real, but can be calculated mathematically. If now you take an infinitely long rod and calculate the B field density around it, which can be done just solving Ampere's law, the B field density is seen to be unifrom up and down the line. If you take a cut from this infintiely long wire and show the wire and the field distribution, it looks like a uniform B field. Whether they say explcitly that their plot is for an infintiely long wire, that's what they are doing. BTW, both links you offered did not work, so i could not see the original figures.

- 11.19.2013
- Rules of Thumb #0: Using Rules of Thumb Wisely
- A fair point about QM adding a level of uncertainity to the predictable nature of the world. But, on a macoscopic level, I have a lot of confidence about how well science can predict the real world.

- 09.04.2014
- Estimating wire & loop inductance: Rule of Thumb #15
- I agree, it is confusing. The key principle is that k = L21/(*sqrt(L1 * L2). This is the fundamental relatiionship. The inductance elements come firt, then we get k. It is only in the special case when L1 = L2 AND the two loops are on top of each other, that k = 1. In all other cases, k will be less than 1. You do not enginer a k, you engineer an L21. If the two loops are very different, and L2 is 10% L1, for example, and the 2 loops are overlapping as much as possible, then at best, L21 ~ L2 and k ~ 1/sqrt(10) ~ 0.3

- 09.04.2014
- Estimating wire & loop inductance: Rule of Thumb #15
- Thanks for taking the time to post a question. The loop mutual inductance between two loops is only indirectly about the loop self inductances. Where ever you saw that loop mutual inductance is M = sqrt(L1 * L2), is flat out wrong. There is a coupling term, in SPICE, which describes the mutual inductor element between two loop inductors. In SPICE language, the dimensionless term which decribes the k element, in which you have identified the two self loop inductances, L1 and L2 which have an actual loop mutual inductance between them of L21, is k = L21/(sqrt(L1 * L2). Sometimes this is written that L21 (your term, M) = k x sqrt(L1 * L2). Note the use of the term k. This is always less than or equal to 1. Only oif the two loops are literally on top of each other and identical is k = 1. The rest of the time k is small, typically on the order of 0.01 to 0.1 depending on the geometry. And, the loop mutual inductance, L21, between two inductors, with loop self inductances L1 and L2, will ALWAYS be less than either L1 or L2. Hope this helps.

## Almost Done

Please confirm the information below before signing in.

{* #socialRegistrationForm *} {* firstName *} {* lastName *} {* displayName *} {* emailAddress *} {* addressCountry *} {* companyName *} {* ednembJobfunction *} {* jobFunctionOther *} {* ednembIndustry *} {* industryOther *}