Dave, you missed a point that may increase the insight of the viewers: You can divide a large square in smaller squares and consider this as a series parallel circuit. Say that the end to end resistance of the small square is 1 Ohm. Then a two times as large square is made up by two strips of two smaller squares in series. The series resistance of two squares is 2 Ohm. And these series circuits are placed side by side, thus in parallel, what make a total resistance of 1 Ohm. Thus the same value as the smaller square.
meettechniek That's what I figured out after some minutes, because I was one of the confused people in his last video. But Dave's approach is more general and mathematically easier.
meettechniek 'Infinite' S/P network would be a good analogy for explaining sheet resistance. Then again, be careful you don't nerd-snipe yourself trying to calculate this guy: xkcd.com/356/
meettechniek Exactly.This could have been a 15 second fun fact. "Double the series resistance and double the number of resistances connected in parallel and bingo, net resistance remains same."
It makes perfect intuitive sense. Dave, I would have mentioned that you can visualize sheet resistance by imagining a *matrix* of infinitestimally small, identical resistors, connected along the width and the length of the sheet. The wider (and the thicker) the sheet, the more resistors in parallel, the longer the more of these sets in series. If you increase or decrease the size of the sheet, but keep the *aspect ratio* (squares remain squares 2x1 rectangles remain 2x1), you are adding or removing resistors *proportionally*. For example: 8 x 1 ohm resistors in parallel, in series with 3 more (4 total) sets of 8 parallel resistors, make 0.5 ohms (0.125 * 4). So do 4 x 1 ohms in parallel, in series with another 4! (0.25 * 2 = 0.5)
Dave, I really enjoy your fundamental videos. As a student studying mech engineering I don't have as much access to electronics education as I would like. Keep up the great work!
Yup, I was confused when you said ohms per square in the resistor network video. Thanks for making this. Sheet resistance makes perfect sense once you think about it.
When I saw the last video I was taken aback for a few seconds, then I thought of a square divided into 4 is the same as 4 resistors in series-parallel which we all know gives the same total resistance as the individual resistors. Took me 20 seconds to work it through in my head.
Dave, Hello from St. Louis, Missouri. Great vlog post, I'm a beginning electrical engineer (year 2), I can appreciate the value this information brings to the viewer. How a chip resistor works and how sheet resistance applies to it. I use many 0603 and 0805 chip resistors in my PCB layout for work. Now I know how they are construction and the reasoning behind the size of the solder footprint. Thumbs up!
Great video. I really enjoy this channel. I'm sure others will have made a similar point, but the easy way to show the truth of unitless square fact (in my world of elastomeric adhesives and sealants, we generally talk about surface resistivity but it's kind of a 6 or half dozen distinction) is to show that you have some specific resistance per square unit. Then, place two square units in series and, as expected, the resistance doubles. But you're left with a 2:1 aspect ratio rectangle so the linear units haven't really disappeared yet. Then, take two of these rectangles and place them alongside each other in parallel and you're left with square twice as wide as the original where, if the original film resistance or surface resistivity is R, the 2:1 rectangle is R+R = 2R and so the paralleled rectangles are 1 / (1/2R + 1/2R) = 2R/2 = R. The a part that I find more interesting is the formulation of electrically conductive rubbers like that foam you used with the meter. The usual way to get there is to load an elastomeric material with something conductive. I would wager you have a urethane foam bearing a considerable amount of carbon black pigment. Normally, in formulating elastomers, the goal of a filler is to reinforce the binding polymer. So ideally, the filler load would essentially dress every filler particle in an exceedingly thin polymer coating. Then, when it cures, the fillers are bound but separated by the polymer and the filler can lend the maximum reinforcing strength to the material's lattice structure. With a conductive formulation, though, the amount of filler is actually a bit higher. It makes the working viscosity of the uncured material generally higher, necessitating more plasticizer in the system to allow for flow and extrusion of the uncured product. As the system cures, the polymer shrinks a bit and the plasticizer migrates to the surface and eventually leaches out (effectively promoting additional shrinkage). This allows the filler particles come into physical contact with one another, providing countless tiny "traces" to pass current. In addition to providing better contact between your trace boards, the pressure you exert also helps force contact through those fillers. So the bouncing readings on the scope were quite literally measuring the jitter in your arms as you subconsciously adjusted your muscle flex to try to hold everything consistently tight. Kind of amazing when you think about, eh? :D
A good way of thinking of it is like this: As you increase size of square, you would think you are increasing resistance (due to longer path) but you are also increasing the parallel conductors which lowers the resistance across the distance, so this cancels out the increase of resistance from the size increase, keeping things constant. Any increase in conduction path (which increases resistance) also incurs extra conduction paths (which lowers resistance).
A suggestion for a better way to control the resistance: cut the foam longer than a square, and put it face up on your bench with the shorter length of the foam between them. Ideally, you would put another blank pcb of that width in between them, but upside down (assuming single sided copper-clad). Measure the area of your foam, then put a weight proportional to the area on the foam, making sure the load is evenly distributed, to get a consistent contact pressure.
Dave's "paddles" have a sheet resistance as well. That resistance is in series with the sheet resistance that is being measured and is figured into hls measurement - all from a single point of contact (so his logic of having to connect to the entire edge is in error). It is just a measurement of series and parallel resistance.
At first, this concept seemed totally impossible, but after I considered the actual current path a bit further, it came to be perfectly reasonable. Really interesting, how your mind can find things that make perfect logical sense seem unintuitive for totally illogical reasons. So, thanks for the video Dave, it's always good to have something simple, yet not immediately obvious explained by someone who knows their stuff as well as you, keeps me on my toes :)
Chris Tate EEVblog Also, Dave, please, keep on making these FF videos. They may not get as many views, but they've been a huge help in mine (and I'm sure many other's) understanding of basic electronics concepts. If my uni had tutors like you, I get the feeling I'd learn an awful lot more in classes. Peace!
Nice video! I think the counter-intuitive part comes from the fact that when you think higher current, you think thicker traces, but the reason for that is not to lower resistance (at least when comparing thin/thick traces with the same number of "squares"), it's to raise the power dissipation capacity of the trace. Instinctively you'd think you need to lower P=I^2R, but that stays the same for two equal square traces ;)
As a practical example people should watch episode #317 where Dave is doing some experiments on PCB tinning and current capacity. Here he is performing a 4 wire resistance measurement on a strip of veroboard (1oz). He kindly provides the length and width of the copper strip (358x4.2mm respectively). Do a quick calculation based on the 0.5mΩ for copper sheet and we get the calculated value of 42.6 mΩ for 85.23 squares. Pretty close to the measured value of 51 mΩ in episode #317.
You mentioned square furlongs. When I was I high school (AKA secondary), I thought about trying out for the track team. A friend said they weren't able to time furlongs per fortnight... :>))
Thanks for the explanation. After the thin film video I was thinking of ways to demonstrate this and I came up with the same thing, except I was going to thread a wire through the end of the foam. Also considered making some silicon rubber sheets doped with graphite. You saved me the trouble.
Dave, it should be noted that while it might be a good approximation it still an approximation. Not every shape's resistance can be satisfactorily calculated by a sum of squares as current density may not be uniform over the whole square.
vihai It’s not a “sum” over area, but a sum in finite element sense, which happens to be a simple arithmetic sum when the shape is rectangular and the mesh (subdividing squares) are symmetric along the length. For any other shape the finite element approximation takes care of no uniform current density. Given how relatively easy it is to compute, I wonder why the feature is not standard in expensive EDA packages :/
Good video, but you should say how they trim the the thin film network : They start to reduce the area (w x T) (coarse : vertical cut) aim to have smaller square ; and they reduce the network resistance (fine : horizontal cut) ; at the end you have a equivalent model of 3 resistances in star, with one with a disconnected terminal. (may induce some noise)
i was hoping you would cut the foam using similar patterns as those seen in the trimmed resistor to demonstrate how it changes the resistance, but i can see how that would have been difficult with the improvised experimental setup.
I think one simple way of explaining it is that even though the distance (length or series resistance) is increasing between the two sides of the square sheets, the width (parallel resistance) is also changing which equally counteracts any increase in length.
I really would have liked to see you make a cut in the foam to demonstrate the laser trimming and then counting the blocks in the cut version of the foam. You'd of course need some sort of insulator to place in the cut.
russdill Mere “counting” the blocks is insufficient unless you’d use that with a shape formula derived experimentally or theoretically. For a general solution you need a finite element grid.
What I took from this is the thickness of the giant "trace" and the length of the trace remain a constant ratio Like paralleling 2 10k resistors and putting that in series with 2 paralleled 10k resistors will be 10k
It's simple enough to understand just using wire as an example. If you double the length of a wire, the resistance doubles. But if you add a second wire of the same diameter, the resistance halves. One 10cm wire is the same resistance as two 20cm wires, for a given conductor size.
Love the Fundamentals Friday videos - please do more of them! But I have to say Dave... you have that beautiful 7.5 digit multimeter - you surely could have measured the resistance of the copper clad board instead of the foam. I suppose getting the contacts right would be tricky though.
Or would be noise because it’s one square, so the lowest possible resistance of copper in that thickness. That is not something even the best handheld meters could reliably measure
It makes perfect sense. If you compare, say, a 2 in. square with a 4 in. one, the 2 in. square would have twice the resistance as the 4 in, due to half the width, but then, half the resistance, because it's half as long, so the two cancel out.
The Sheet resistance formula is common sense. The easiest way to explain it is if you see it like a resistor: if you double the length of a piece of copper, it will have double the resistance (2 resistors in series), buy if you double the width you will have 2 resistors in parallel, so the resistance will be half and so the resulting resistance will always be the same.
Makes perfect sense, just visualize a sheet as a bank of resistors connected in parallel, then add resistors as you lengthen or widen it. Of course you must count squares in series, not squares in parallel, squares in parallel are fractional squares. So a 2x1 rectangle measured lengthwise becomes 2 squares, but measured edgewise becomes 0.5 squares.
An intuitive way to grasp the concept is the following: Think of the atoms/molecules of the material as tiny resistors of the same value that are all connected in a grid. We know from electronics 101, that if you put two resistors (of the same value) in series, their total resistance doubles. If you put them in parallel, their resistance halves. You can easily reproduce this experiment with four 10k Ohm resistors connected in a grid. The overall resistance is - lo and behold - 10k Ohms. Now go, grab some resistors and measure it :)
Amen... this is what I was expecting to see. We all know those formulae for series/parallel resistances, so I was expecting them to be used to explain this.
I think the confusion is coming about because people are thinking of it in terms of a wire where cross sectional area increases by the square of the change in dimension, but because in a sheet, the thickness remains the same, so the cross section increases linearly with the change in dimension.
When I was first grasping the concept, I would think of standard series and parallel resistance math, where the length is series and width is parallel. Pretend each unit, like sqr cm is just a resister of a certain value, the length would be series, so say each sqr cm is 10ohm, if length is 10cm, each 10cm wide strip would be 100ohm, but then the width would be parallel math, you have 10 100ohm resistor in parallel, so again you total resistance is 10ohm...It is cool for changed orientation too, say you have a sheet that is twice as long as it is wide, the resistance across it lengthwise would be twice that of just the square, as you shown, but if turned it 90deg and measured widthwise, it would be 1/2 the square resistance, because you have twice the amount in parallel.....Hope that does not confuse more people.
Well, that explains the square laser cuts we saw when you did the teardown of the thin film resistor networks... but what about the curved cuts? I think you even said in that video that there was a reason for that. I'm just curious - can you explain them, too?
Probably we can say that by increasing the size of the sheet, we increase the distance between terminals - so the resistance has to rise, but we also increase paralelism of currents between terminals - so resistance drops :)
I have to say I did the math while I was watching Your video on the Resistors - 'cause at first it sounded strange. So the math was not new to me. But Dave, was there really no other way for a practical measurement? You could have done this with two pieces of copper heating pipe as the probes. In reference to the copper board the resistance of the pipes would have been to low to really change the result. And You could have shown the incredible Resistance measurement capabilities of the Keithley... Or at least You could have used any solid material. A square of salami schould have been better evidence than the spongy foam, were the resistance, as You said, was mainly determined by ''the force''.
I think what is happening is that there are more paths for electrons with larger squares, but more material to go through, so they cancel out the resistance. Granted you have the one side the same, the other getting longer, the resistance goes up...
Awesome Dave, love this format and love the explanation! Could have been interesting to try this with liquids, like a pan of salt water. Then contact unreliability wouldn't be an issue.
At 8:50 into the video you show how the laser cut works. However the first smaller square is not contacting all of the right edge of the bigger square to the left. Why is it that the squares can be added at the top of the laser cut but we must probe along the full edge to measure properly? Also I don't think the squares under the laser cut are added as the net current in and out of the area under the laser cut is 0A therefore this area would be an open circuit.
Maybe you could have given another geometrical explanation too (explaining the same thing in various manners to help various type of learners catch the concept): if I double a square in the direction of the current, I double the resistance, if I double a square across from the direction of the current, I halve the resistance. So if I double both directions at once I have not changed the resistance. One mildly interesting factoid, is that now you have an easy tool to control the admissible power dissipation of your resistor: by scaling it, you scale its dissipation without changing the resistor value. Plus when you trim it, you can make it bigger so that it is more sensitive to trimming.
I'm surprised you didn't use the analogy to wire thickness. Same formula, much easier to understand and probably everyone is aware that a wire with smaller diameter/area and longer length will have larger resistance than thicker, shorter wire.
Dave, my non-engineer mind has a much easier time thinking of sheet resistance as a massive array of resistors wired both in parallel and in series. The more resisters you put in the parallel, the lower the resistance; and the more you put in series, the higher the resistance. Therefore, it doesn't matter how large you make the array (if you keep it square), the resistance will be the same.
Well this is funny thing that I really realized after I did part time job as chip layouter. 1um wide 100um long trace on chip is actually almost the same resistance as 1mm wide 10cm long trace on PCB...
Great video. I have two questions. I hope I'm not commenting to late to get an answer from some of the smart people here. 1, When you cut the L-shape with a laser, you create more squares but what happens with the area under the L? Is that counted as well? 2, In the video with the precision resistors, there was a curved cut in one place, how do you apply the squares there? A curved line can be approximated by an infinite amount of squares which will be a bit of a pain to calculate. Do you give up the concept of squares there and just calculate the area (conductor volume) instead?
Make an analogy to resistors being put in series as the trace gets longer… but also in parallel when the trace gets wider. The series resistance is cancelled by parallel resistance. So as long as the aspect ratio of all traces (boards) is the same, then the resistance remains constant.
Is sheet resistance like having an almost infinite number of same-value resistors, all stacked in series and parallel? (_same value_ being due to the thickness of the sheet, and the other two dimensions corresponding to the _square_ because there are 2 dimensions [^2]).
So how does trimming (laser cuts, etc) factor into it? Does it just break up the larger square into more smaller squares, which increases the resistance?
Scott Lawrence yep. If you look at the resistor video and imagine the trimming ladders as squares, you can see if you break a connection, you're adding squares. If you look at the L shaped laser trim drawn on the board, you can see that the thin section the current travels through is now made of more smaller squares because the width of the conducted path is smaller.
This correlation to square is quite interesting. Resistance imagined as some sort of material obstacle that scales with size fails. Electric energy is really hard to grasp, no wonder so many theories are out there and many representations of it but all of them fail in some way.
I don't really think it's counter intuitive, we have all learned in school that the resistance of length of material depends on it's length and it's thickness. The Ohms per square is just a consequence of that rule. It's just a simple formula.
Eviltech according to your formula length and thickness it's very counter intuitive since the formula explained use the width, length and thickness. and only double the length = double resistance. double width and length = same resistivity.
I was expecting tessellations of the smaller board onto the larger and then using the resistors in series and parallel equations e.g. for a board 3 times larger in length & width than your original (r) then using the resistor parallel/series formulae, you'd get (⅓r)*3=r or 3r*⅓=r (depending on whether you calculated the parallel or series resistances first)
You had the three square sheet in serial, and got 15k ohms, would it be safe to assume then, that the three square piece turned 90 degrees to get it in parallel would give us about 1.6k ohms?
This is very good but what about non-ideal properties, does the current avoid the middle or edges of the metal, even if not externally influenced by static/magnetic fields?
Great video, how many college lectures are using your videos in there lectures. Wondered if took a copper clad board and stripped out the middle leaving two tracks for the high conductivity area then silk screening some sort of conductive paint across the board that gives more resistance than copper. Then you could take a dremel to it and carve out paths across painted area to physically demonstrate how the intricate cuts affect the ohms per square.
At about timestamp 16:30 with the rectangle, what if you 1) measured it along the LONG edges? ALSO, What if you 2) measured a stack of three squares compared to ONE (would it be three in parallel)?
I would like to see you explore the path of electricity when the contact is indeed at a point instead of the the entire edge. I know that's more physics than engineering but I think many people would be interested. Thanks.
Ok, now let us add a T-Line and do some Smith charts. By the time we have learned a fair amount in engineering then we would be too old to enjoy it. This is an example of another day where another video has proven I do have the information but can't put it together on the fly and hence still don't know fair enough. yet. My question for Dave is: Does it mean I won't be a good enough engineer?
smellysam Yes. But the interface between foam and a conductor is very tricky and interfacing foam to copper usually ends up with copper slowly oxidizing or outright corroding (some older foams turned out to be corrosive over decades). It’s OK for a demo or an experiment but in a product you’d need way more engineering to qualify the change in foam properties over time and load cycles, changes in interface properties over the same, durability of the top plate should it be exposed to non-uniform loads (like people stepping on it), and so on. For homemade load sensors, the conductive black 3M anti static bags and similar materials are great: their life and properties are much better than those of foam, because they are a void-free solid. Linnstrument is made that way - look it up.
Hi Dave ! Really a very very interesting info - though it's a little hard to grasp, but the math is there to help you - right ? BUT ,,, I think - maybe I remember wrong - the current is ONLY running in a thin layer on a round wire - IF that's true, then "a wire" can be interpreted as a "folded squar to a tube" and we can use the same "thinking" - maybe I was too quick there ! PI cm of wire with Ø1mm (the "squar" will be equal to the circumference !) should have the same resistance as 10 *PI cm of wire with Ø10mm
Keld Sørensen You're thinking of the skin effect, but afaik, that only applies to AC. Although it doesn't matter, because your thinking is still true with DC, I think. A 1 mm diameter wire of length 10 cm should have the same resistance as a 10 mm diameter wire of length 100 cm. Correct us if we're wrong, Dave, maybe in a followup vid?
Daedalus Young No, its not correct. For R to stay constant L/A needs to stay constant. If you take 10 times the length, you need 10 times the area. But by multiplying the diameter by 10 you get 100 times the area!
Baurophon Ah of course, that's right, but you could calculate how much bigger the diameter needs to be to be 10 times larger than the smaller wire, then it would work?
Baurophon That is useful to know, if you've got a certain length of wire and you want to keep the resistance over that distance to a certain maximum. Dave's showed those probes with a certain resistance before, I can see how this theory applies to those.
Also, wouldn't you be pretty much be doing the same thing by running a thin wire along the edge of the resistive foam and connecting your probe to the end? And then for that matter, just probing the corner? The spread of the electrons due to those sheets are probably so miniscule it probably didn't make any noticeable difference
If the resistance of a sheet is the same regardless of the size, as long as the shape remains the same, what's the advantage in having a larger sheet? Why not make it minuscule?
WarpRulez I would say a bigger sheet gives finer control for adjusting the resistance. If you have a wonky laser with a precision of +/- 1 mm, then that will be a huge error on a 5 mm^2 sheet, but much smaller on a 50 mm^2 sheet.
WarpRulez Larger currents = more power dissipation required to keep the tracks at lets say 10 degrees C above ambient. So you can have the same number of "squares" for two different tracks, one tiny and one large, which will have the same resistance, and therefore the current will be the same given an applied voltage, but the tiny trace will heat up considerably and probably melt.
Dave, you missed a point that may increase the insight of the viewers:
You can divide a large square in smaller squares and consider this as a series parallel circuit. Say that the end to end resistance of the small square is 1 Ohm. Then a two times as large square is made up by two strips of two smaller squares in series. The series resistance of two squares is 2 Ohm. And these series circuits are placed side by side, thus in parallel, what make a total resistance of 1 Ohm. Thus the same value as the smaller square.
meettechniek That's what I figured out after some minutes, because I was one of the confused people in his last video. But Dave's approach is more general and mathematically easier.
meettechniek 'Infinite' S/P network would be a good analogy for explaining sheet resistance. Then again, be careful you don't nerd-snipe yourself trying to calculate this guy: xkcd.com/356/
meettechniek Exactly.This could have been a 15 second fun fact. "Double the series resistance and double the number of resistances connected in parallel and bingo, net resistance remains same."
@@lhxperimental BINGO
At only 18 minutes, this has to be one of the shortest videos I've ever seen on this channel ;)
soviut There is ~2 min video. Few videos back.
Valtra103 Heh, the book? Even that was pretty long to reveal a 3rd edition ;)
Valtra103 We have the same icon! :O
It makes perfect intuitive sense.
Dave, I would have mentioned that you can visualize sheet resistance by imagining a *matrix* of infinitestimally small, identical resistors, connected along the width and the length of the sheet. The wider (and the thicker) the sheet, the more resistors in parallel, the longer the more of these sets in series.
If you increase or decrease the size of the sheet, but keep the *aspect ratio* (squares remain squares 2x1 rectangles remain 2x1), you are adding or removing resistors *proportionally*.
For example: 8 x 1 ohm resistors in parallel, in series with 3 more (4 total) sets of 8 parallel resistors, make 0.5 ohms (0.125 * 4).
So do 4 x 1 ohms in parallel, in series with another 4! (0.25 * 2 = 0.5)
This kind of videos are better i think, more informative than just the teardowns.
Dave, I really enjoy your fundamental videos. As a student studying mech engineering I don't have as much access to electronics education as I would like. Keep up the great work!
Yup, I was confused when you said ohms per square in the resistor network video. Thanks for making this. Sheet resistance makes perfect sense once you think about it.
When I saw the last video I was taken aback for a few seconds, then I thought of a square divided into 4 is the same as 4 resistors in series-parallel which we all know gives the same total resistance as the individual resistors. Took me 20 seconds to work it through in my head.
Dave,
Hello from St. Louis, Missouri. Great vlog post, I'm a beginning electrical engineer (year 2), I can appreciate the value this information brings to the viewer. How a chip resistor works and how sheet resistance applies to it. I use many 0603 and 0805 chip resistors in my PCB layout for work. Now I know how they are construction and the reasoning behind the size of the solder footprint. Thumbs up!
Great video. I really enjoy this channel.
I'm sure others will have made a similar point, but the easy way to show the truth of unitless square fact (in my world of elastomeric adhesives and sealants, we generally talk about surface resistivity but it's kind of a 6 or half dozen distinction) is to show that you have some specific resistance per square unit. Then, place two square units in series and, as expected, the resistance doubles. But you're left with a 2:1 aspect ratio rectangle so the linear units haven't really disappeared yet. Then, take two of these rectangles and place them alongside each other in parallel and you're left with square twice as wide as the original where, if the original film resistance or surface resistivity is R, the 2:1 rectangle is R+R = 2R and so the paralleled rectangles are 1 / (1/2R + 1/2R) = 2R/2 = R.
The a part that I find more interesting is the formulation of electrically conductive rubbers like that foam you used with the meter. The usual way to get there is to load an elastomeric material with something conductive. I would wager you have a urethane foam bearing a considerable amount of carbon black pigment. Normally, in formulating elastomers, the goal of a filler is to reinforce the binding polymer. So ideally, the filler load would essentially dress every filler particle in an exceedingly thin polymer coating. Then, when it cures, the fillers are bound but separated by the polymer and the filler can lend the maximum reinforcing strength to the material's lattice structure. With a conductive formulation, though, the amount of filler is actually a bit higher. It makes the working viscosity of the uncured material generally higher, necessitating more plasticizer in the system to allow for flow and extrusion of the uncured product. As the system cures, the polymer shrinks a bit and the plasticizer migrates to the surface and eventually leaches out (effectively promoting additional shrinkage). This allows the filler particles come into physical contact with one another, providing countless tiny "traces" to pass current. In addition to providing better contact between your trace boards, the pressure you exert also helps force contact through those fillers. So the bouncing readings on the scope were quite literally measuring the jitter in your arms as you subconsciously adjusted your muscle flex to try to hold everything consistently tight. Kind of amazing when you think about, eh? :D
Your excitement about this stuff is awesome and contagious 😊
Love it! You should make more fundamental videos!
Bravo Bravo! Another great demonstration from Mr. Jones. Thank you!!!
I'm loving these fundamentals Friday videos. Very informative and useful to a beginner hobbyist.
A good way of thinking of it is like this: As you increase size of square, you would think you are increasing resistance (due to longer path) but you are also increasing the parallel conductors which lowers the resistance across the distance, so this cancels out the increase of resistance from the size increase, keeping things constant.
Any increase in conduction path (which increases resistance) also incurs extra conduction paths (which lowers resistance).
A suggestion for a better way to control the resistance: cut the foam longer than a square, and put it face up on your bench with the shorter length of the foam between them. Ideally, you would put another blank pcb of that width in between them, but upside down (assuming single sided copper-clad). Measure the area of your foam, then put a weight proportional to the area on the foam, making sure the load is evenly distributed, to get a consistent contact pressure.
dave you have a great value on earth!
Dave's "paddles" have a sheet resistance as well. That resistance is in series with the sheet resistance that is being measured and is figured into hls measurement - all from a single point of contact (so his logic of having to connect to the entire edge is in error). It is just a measurement of series and parallel resistance.
At first, this concept seemed totally impossible, but after I considered the actual current path a bit further, it came to be perfectly reasonable. Really interesting, how your mind can find things that make perfect logical sense seem unintuitive for totally illogical reasons. So, thanks for the video Dave, it's always good to have something simple, yet not immediately obvious explained by someone who knows their stuff as well as you, keeps me on my toes :)
Chris Tate EEVblog Also, Dave, please, keep on making these FF videos. They may not get as many views, but they've been a huge help in mine (and I'm sure many other's) understanding of basic electronics concepts. If my uni had tutors like you, I get the feeling I'd learn an awful lot more in classes. Peace!
Nice video! I think the counter-intuitive part comes from the fact that when you think higher current, you think thicker traces, but the reason for that is not to lower resistance (at least when comparing thin/thick traces with the same number of "squares"), it's to raise the power dissipation capacity of the trace. Instinctively you'd think you need to lower P=I^2R, but that stays the same for two equal square traces ;)
As a practical example people should watch episode #317 where Dave is doing some experiments on PCB tinning and current capacity. Here he is performing a 4 wire resistance measurement on a strip of veroboard (1oz). He kindly provides the length and width of the copper strip (358x4.2mm respectively). Do a quick calculation based on the 0.5mΩ for copper sheet and we get the calculated value of 42.6 mΩ for 85.23 squares. Pretty close to the measured value of 51 mΩ in episode #317.
the trick to understanding electronics is making the connection
brandon carter groan. shameful! :D
brandon carter Ohm I see watt you did there.. This guy has potential..
hank bizzo only if he keeps his material current.sorry, I couldn't resistant.
brandon carter Oh you have the power to make me chuckle. No negative feedback for you!
That is mean. Chris Gray
eevblog is the best!!!!
20 years in electronics -- never considered that. Thanks, Dave!
Great video as usual, Dave! I always learn something interesting from you.
Oh nice, You are part of my lecture. Say "hallo, TU Berlin"
You mentioned square furlongs. When I was I high school (AKA secondary), I thought about trying out for the track team. A friend said they weren't able to time furlongs per fortnight... :>))
Dave i give you a big thumbs up.
Always look forward to your videos. Thanks Dave.
Thanks for the explanation. After the thin film video I was thinking of ways to demonstrate this and I came up with the same thing, except I was going to thread a wire through the end of the foam. Also considered making some silicon rubber sheets doped with graphite. You saved me the trouble.
Awesome video. Nice concept of explaining and very nice idea with the foam!
Equations just don't lie...
Dave, it should be noted that while it might be a good approximation it still an approximation. Not every shape's resistance can be satisfactorily calculated by a sum of squares as current density may not be uniform over the whole square.
vihai It’s not a “sum” over area, but a sum in finite element sense, which happens to be a simple arithmetic sum when the shape is rectangular and the mesh (subdividing squares) are symmetric along the length. For any other shape the finite element approximation takes care of no uniform current density. Given how relatively easy it is to compute, I wonder why the feature is not standard in expensive EDA packages :/
Good video, but you should say how they trim the the thin film network :
They start to reduce the area (w x T) (coarse : vertical cut) aim to have smaller square ; and they reduce the network resistance (fine : horizontal cut) ; at the end you have a equivalent model of 3 resistances in star, with one with a disconnected terminal. (may induce some noise)
I love to watch your great fundamential videos!
Great Video. Would really like a similar video on inductance - Calculating trace length and the pit falls/tips + the theory
Thanks for explaining this Dave.
i was hoping you would cut the foam using similar patterns as those seen in the trimmed resistor to demonstrate how it changes the resistance, but i can see how that would have been difficult with the improvised experimental setup.
I think one simple way of explaining it is that even though the distance (length or series resistance) is increasing between the two sides of the square sheets, the width (parallel resistance) is also changing which equally counteracts any increase in length.
I really would have liked to see you make a cut in the foam to demonstrate the laser trimming and then counting the blocks in the cut version of the foam. You'd of course need some sort of insulator to place in the cut.
russdill Mere “counting” the blocks is insufficient unless you’d use that with a shape formula derived experimentally or theoretically. For a general solution you need a finite element grid.
This makes perfect sense, and series/parallel resistor networks work exactly the same way...
What I took from this is the thickness of the giant "trace" and the length of the trace remain a constant ratio
Like paralleling 2 10k resistors and putting that in series with 2 paralleled 10k resistors will be 10k
It's simple enough to understand just using wire as an example. If you double the length of a wire, the resistance doubles. But if you add a second wire of the same diameter, the resistance halves. One 10cm wire is the same resistance as two 20cm wires, for a given conductor size.
Great video backed by tests!
Something new to learn again. Thanks for sharing.
You said square 76 times in this video.
Pedram I don't whether to applaud you for counting, or suggest you get a life :->
Pedram EEVblog Drinking game!
Pedram That was easy. Impressive would be: how many times Dave says "we're in like Flynn" in all of his videos!
Frédéric Dutrey My mates and I have a drinking game for Dave's videos.
Pedram Thumbs-up for simply not saying "porn" this time
Thanks for this extremely enlightening video, Dave. I also understand how a 50 ohm trace work, just 10000 squares!
Very interesting, thanks Dave.
Love the Fundamentals Friday videos - please do more of them!
But I have to say Dave... you have that beautiful 7.5 digit multimeter - you surely could have measured the resistance of the copper clad board instead of the foam. I suppose getting the contacts right would be tricky though.
Or would be noise because it’s one square, so the lowest possible resistance of copper in that thickness. That is not something even the best handheld meters could reliably measure
thanks dave fantastic explanation !
It makes perfect sense. If you compare, say, a 2 in. square with a 4 in. one, the 2 in. square would have twice the resistance as the 4 in, due to half the width, but then, half the resistance, because it's half as long, so the two cancel out.
I love your tutorial videos. Keep up the good work!!!!! Cheers from Italy
The ohm/square is used in many fields, including microelectronics. At first it does sound weird, but it's ok after a while, and it's quite useful.
thank you for the video lesson
Great video Dave
The Sheet resistance formula is common sense. The easiest way to explain it is if you see it like a resistor: if you double the length of a piece of copper, it will have double the resistance (2 resistors in series), buy if you double the width you will have 2 resistors in parallel, so the resistance will be half and so the resulting resistance will always be the same.
Interesting video. Never knew about sheet resistance.
Great stuff as always Dave. Love your work.
Makes perfect sense, just visualize a sheet as a bank of resistors connected in parallel, then add resistors as you lengthen or widen it.
Of course you must count squares in series, not squares in parallel, squares in parallel are fractional squares. So a 2x1 rectangle measured lengthwise becomes 2 squares, but measured edgewise becomes 0.5 squares.
An intuitive way to grasp the concept is the following: Think of the atoms/molecules of the material as tiny resistors of the same value that are all connected in a grid. We know from electronics 101, that if you put two resistors (of the same value) in series, their total resistance doubles. If you put them in parallel, their resistance halves.
You can easily reproduce this experiment with four 10k Ohm resistors connected in a grid. The overall resistance is - lo and behold - 10k Ohms.
Now go, grab some resistors and measure it :)
Amen... this is what I was expecting to see. We all know those formulae for series/parallel resistances, so I was expecting them to be used to explain this.
Excellent explanation! :D
I think the confusion is coming about because people are thinking of it in terms of a wire where cross sectional area increases by the square of the change in dimension, but because in a sheet, the thickness remains the same, so the cross section increases linearly with the change in dimension.
When I was first grasping the concept, I would think of standard series and parallel resistance math, where the length is series and width is parallel. Pretend each unit, like sqr cm is just a resister of a certain value, the length would be series, so say each sqr cm is 10ohm, if length is 10cm, each 10cm wide strip would be 100ohm, but then the width would be parallel math, you have 10 100ohm resistor in parallel, so again you total resistance is 10ohm...It is cool for changed orientation too, say you have a sheet that is twice as long as it is wide, the resistance across it lengthwise would be twice that of just the square, as you shown, but if turned it 90deg and measured widthwise, it would be 1/2 the square resistance, because you have twice the amount in parallel.....Hope that does not confuse more people.
Well, that explains the square laser cuts we saw when you did the teardown of the thin film resistor networks... but what about the curved cuts? I think you even said in that video that there was a reason for that. I'm just curious - can you explain them, too?
Probably we can say that by increasing the size of the sheet, we increase the distance between terminals - so the resistance has to rise, but we also increase paralelism of currents between terminals - so resistance drops :)
I have to say I did the math while I was watching Your video on the Resistors - 'cause at first it sounded strange. So the math was not new to me.
But Dave, was there really no other way for a practical measurement? You could have done this with two pieces of copper heating pipe as the probes. In reference to the copper board the resistance of the pipes would have been to low to really change the result. And You could have shown the incredible Resistance measurement capabilities of the Keithley... Or at least You could have used any solid material. A square of salami schould have been better evidence than the spongy foam, were the resistance, as You said, was mainly determined by ''the force''.
That was rather interesting. Makes me wonder about other similar stuff I don't know yet.
I think what is happening is that there are more paths for electrons with larger squares, but more material to go through, so they cancel out the resistance. Granted you have the one side the same, the other getting longer, the resistance goes up...
Awesome Dave, love this format and love the explanation! Could have been interesting to try this with liquids, like a pan of salt water. Then contact unreliability wouldn't be an issue.
Dave, I believe you !!!
Anyway, thank you very much for your lessons !
At 8:50 into the video you show how the laser cut works. However the first smaller square is not contacting all of the right edge of the bigger square to the left. Why is it that the squares can be added at the top of the laser cut but we must probe along the full edge to measure properly?
Also I don't think the squares under the laser cut are added as the net current in and out of the area under the laser cut is 0A therefore this area would be an open circuit.
Maybe you could have given another geometrical explanation too (explaining the same thing in various manners to help various type of learners catch the concept): if I double a square in the direction of the current, I double the resistance, if I double a square across from the direction of the current, I halve the resistance. So if I double both directions at once I have not changed the resistance. One mildly interesting factoid, is that now you have an easy tool to control the admissible power dissipation of your resistor: by scaling it, you scale its dissipation without changing the resistor value. Plus when you trim it, you can make it bigger so that it is more sensitive to trimming.
dave, try passing 1A through the pcb squares, and use your 7.5 digit bench meters to measure the voltage drop.
I'm surprised you didn't use the analogy to wire thickness. Same formula, much easier to understand and probably everyone is aware that a wire with smaller diameter/area and longer length will have larger resistance than thicker, shorter wire.
Love these videos! Thanks Dave! :)
cheers Dave, that was cool
Dave, my non-engineer mind has a much easier time thinking of sheet resistance as a massive array of resistors wired both in parallel and in series.
The more resisters you put in the parallel, the lower the resistance; and the more you put in series, the higher the resistance. Therefore, it doesn't matter how large you make the array (if you keep it square), the resistance will be the same.
Well this is funny thing that I really realized after I did part time job as chip layouter. 1um wide 100um long trace on chip is actually almost the same resistance as 1mm wide 10cm long trace on PCB...
Great video. I have two questions. I hope I'm not commenting to late to get an answer from some of the smart people here.
1, When you cut the L-shape with a laser, you create more squares but what happens with the area under the L? Is that counted as well?
2, In the video with the precision resistors, there was a curved cut in one place, how do you apply the squares there? A curved line can be approximated by an infinite amount of squares which will be a bit of a pain to calculate. Do you give up the concept of squares there and just calculate the area (conductor volume) instead?
Make an analogy to resistors being put in series as the trace gets longer… but also in parallel when the trace gets wider. The series resistance is cancelled by parallel resistance. So as long as the aspect ratio of all traces (boards) is the same, then the resistance remains constant.
Is sheet resistance like having an almost infinite number of same-value resistors, all stacked in series and parallel? (_same value_ being due to the thickness of the sheet, and the other two dimensions corresponding to the _square_ because there are 2 dimensions [^2]).
Amazing, thanks for making great wideo!
So how does trimming (laser cuts, etc) factor into it? Does it just break up the larger square into more smaller squares, which increases the resistance?
Scott Lawrence yep. If you look at the resistor video and imagine the trimming ladders as squares, you can see if you break a connection, you're adding squares. If you look at the L shaped laser trim drawn on the board, you can see that the thin section the current travels through is now made of more smaller squares because the width of the conducted path is smaller.
Question----1) Are you trying to show that the black foam sheet conducts too?
2) irrespective of its shape
This correlation to square is quite interesting. Resistance imagined as some sort of material obstacle that scales with size fails. Electric energy is really hard to grasp, no wonder so many theories are out there and many representations of it but all of them fail in some way.
I don't really think it's counter intuitive, we have all learned in school that the resistance of length of material depends on it's length and it's thickness. The Ohms per square is just a consequence of that rule. It's just a simple formula.
Eviltech Perhaps a touch counter intuitive as Dave just showed pieces of material of differing lengths that have the same resistance.
Eviltech according to your formula length and thickness it's very counter intuitive since the formula explained use the width, length and thickness. and only double the length = double resistance. double width and length = same resistivity.
I was expecting tessellations of the smaller board onto the larger and then using the resistors in series and parallel equations
e.g. for a board 3 times larger in length & width than your original (r) then using the resistor parallel/series formulae, you'd get (⅓r)*3=r or 3r*⅓=r (depending on whether you calculated the parallel or series resistances first)
You had the three square sheet in serial, and got 15k ohms, would it be safe to assume then, that the three square piece turned 90 degrees to get it in parallel would give us about 1.6k ohms?
Yes.
This is very good but what about non-ideal properties, does the current avoid the middle or edges of the metal, even if not externally influenced by static/magnetic fields?
Great video, how many college lectures are using your videos in there lectures.
Wondered if took a copper clad board and stripped out the middle leaving two tracks for the high conductivity area then silk screening some sort of conductive paint across the board that gives more resistance than copper. Then you could take a dremel to it and carve out paths across painted area to physically demonstrate how the intricate cuts affect the ohms per square.
Thank you for the film but,
What about the thickness? How dose it affect the Sheet Resistance?
Gabi Saada It scales inverse with thickness. E.g. 3x the thickness yields 1/3rd sheet resistance.
This seems quite simple to me.
Yes, this is ;)
At about timestamp 16:30 with the rectangle, what if you
1) measured it along the LONG edges?
ALSO, What if you
2) measured a stack of three squares compared to ONE (would it be three in parallel)?
I would like to see you explore the path of electricity when the contact is indeed at a point instead of the the entire edge. I know that's more physics than engineering but I think many people would be interested. Thanks.
Hey can you make a video about sheet capacitance or sheet impedance in general
Ok, now let us add a T-Line and do some Smith charts. By the time we have learned a fair amount in engineering then we would be too old to enjoy it. This is an example of another day where another video has proven I do have the information but can't put it together on the fly and hence still don't know fair enough. yet. My question for Dave is: Does it mean I won't be a good enough engineer?
Could this be used to make giant DIY foam load sensors?
(And thanks for the FFriday, really appreciate these)
smellysam Yes. But the interface between foam and a conductor is very tricky and interfacing foam to copper usually ends up with copper slowly oxidizing or outright corroding (some older foams turned out to be corrosive over decades). It’s OK for a demo or an experiment but in a product you’d need way more engineering to qualify the change in foam properties over time and load cycles, changes in interface properties over the same, durability of the top plate should it be exposed to non-uniform loads (like people stepping on it), and so on.
For homemade load sensors, the conductive black 3M anti static bags and similar materials are great: their life and properties are much better than those of foam, because they are a void-free solid. Linnstrument is made that way - look it up.
What is the resistance if contact points are two adjacent edges? Same resistance?
Hi Dave !
Really a very very interesting info - though it's a little hard to grasp, but the math is there to help you - right ?
BUT ,,,
I think - maybe I remember wrong - the current is ONLY running in a thin layer on a round wire - IF that's true, then "a wire" can be interpreted as a "folded squar to a tube" and we can use the same "thinking" - maybe I was too quick there !
PI cm of wire with Ø1mm (the "squar" will be equal to the circumference !)
should have the same resistance as
10 *PI cm of wire with Ø10mm
Keld Sørensen You're thinking of the skin effect, but afaik, that only applies to AC. Although it doesn't matter, because your thinking is still true with DC, I think. A 1 mm diameter wire of length 10 cm should have the same resistance as a 10 mm diameter wire of length 100 cm. Correct us if we're wrong, Dave, maybe in a followup vid?
Daedalus Young No, its not correct. For R to stay constant L/A needs to stay constant. If you take 10 times the length, you need 10 times the area. But by multiplying the diameter by 10 you get 100 times the area!
Baurophon Ah of course, that's right, but you could calculate how much bigger the diameter needs to be to be 10 times larger than the smaller wire, then it would work?
Daedalus Young Then it schould work! A 1mm wire of length 10cm should have the same resistance as a 10mm wire of 10m.
Baurophon That is useful to know, if you've got a certain length of wire and you want to keep the resistance over that distance to a certain maximum. Dave's showed those probes with a certain resistance before, I can see how this theory applies to those.
Also, wouldn't you be pretty much be doing the same thing by running a thin wire along the edge of the resistive foam and connecting your probe to the end? And then for that matter, just probing the corner? The spread of the electrons due to those sheets are probably so miniscule it probably didn't make any noticeable difference
Next FF should be Sheet current capacity per square and thermal calculations
I must have missed it. At what point do you insert the black magic onto the sheet to make the resistances the same??
elimenohpee182 Only magic smoke!
If the resistance of a sheet is the same regardless of the size, as long as the shape remains the same, what's the advantage in having a larger sheet? Why not make it minuscule?
WarpRulez I would say the bigger sheet can carry greater current.
WarpRulez I would say a bigger sheet gives finer control for adjusting the resistance. If you have a wonky laser with a precision of +/- 1 mm, then that will be a huge error on a 5 mm^2 sheet, but much smaller on a 50 mm^2 sheet.
Daedalus Young
Wouldn't it be cheaper to buy a more accurate laser than make millions of large sheets?
WarpRulez bigger current (but its more easy to make it more thick). Maybe heat dissipation and ability for more accurate trimming?
WarpRulez Larger currents = more power dissipation required to keep the tracks at lets say 10 degrees C above ambient. So you can have the same number of "squares" for two different tracks, one tiny and one large, which will have the same resistance, and therefore the current will be the same given an applied voltage, but the tiny trace will heat up considerably and probably melt.