SMD resistor stray capacitance vs size - Page 1
Forum section: electronics / projects

author: mark03
author_rank: Frequent Contributor
timestamp: September 19, 2018, 01:47:35 am
content: For a wideband photodiode front-end, I'm looking at a fast op-amp (ADA4817) where the desired feedback capacitance will be quite small (<< 1 pF). This has me wondering: How does SMD resistor parasitic capacitance vary with size, i.e. 0402, 0603, ...? If the capacitance is primarily between the metal end caps, you could use the parallel-plate formula to approximate it, but I doubt it is that simple.There is also the possibility of splitting the feedback resistance into a series string, and hoping that the parasitic capacitances also combine in series.Complicating matters somewhat is that the op-amp feedback and inverting-input pins are adjacent and only 0.5 mm apart, so the wiring to a larger part will add more capacitance than it otherwise has to.Basically, I am looking for some sage advice, given that I don't have access to fancy EDA tools (only Kicad).

author: coppercone2
author_rank: Super Contributor
timestamp: September 19, 2018, 02:15:28 am
content: keep in mind that the pad has some capacitance tooI wonder if there is some reason to make a little arch out of capacitors.Don't they account for the parasitic capacitance of a capacitor that is low value when they give you the manufacturing spec though? It's not really a parasitic like inductance, just you have a different structure of capacitor so different transmission mode.I don't think it will be parallel plate because you got dielectric material there, in the long way, but spaced by conductive strips. Turn it sideways in your head, you get like a structure thats likePlate 1 (shorted to) inductance - *dielectric very thin* - plate 2Plate 3 - *dielectric very thin* - inductance (shorted to) plate 4and so on

author: mark03
author_rank: Frequent Contributor
timestamp: September 19, 2018, 02:18:56 am
content: I'm not planning to useany capacitor there. I just want to estimate how much capacitance I'm going to get anyway, with a single resistor, or possibly a string of resistors. And how to keep it small, hopefully in neighborhood of 100 fF.Feedback resistance will be somewhere in the range of 200-500 kohm.

author: coppercone2
author_rank: Super Contributor
timestamp: September 19, 2018, 02:21:11 am
content: i failed

author: Rerouter
author_rank: Super Contributor
timestamp: September 19, 2018, 02:21:36 am
content: Download saturn pcb toolkit. This will help you for most of the layout capacitance. For the actual passive components. That would require knowing the dielectric constant of the smd bulk material. I have personally never seen it. But at a guess the area of the end caps is so small that I cannot imagine it adding up to much.Trace to trace capacitance is generally low as the area is the edge of the traces. Not there width. Just fan out your traces as soon as it leaves your op amp.

author: coppercone2
author_rank: Super Contributor
timestamp: September 19, 2018, 02:37:43 am
content: https://en.wikipedia.org/wiki/Relative_permittivitybetween 1 to around 250,000http://oatao.univ-toulouse.fr/698/1/boulos_698.pdf

author: Cerebus
author_rank: Super Contributor
timestamp: September 19, 2018, 02:52:20 am
content: As a starting point seeFrequency Response of Thin Film Chip Resistorsfrom Vishay.From the first page

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 19, 2018, 02:54:21 am
content: At large values (above ~100 ohms), capacitance is inevitable. The cutoff frequency (the point where C dominates) depends on R. Because C is about constant for a given chip size, of course, while R varies, and F = 1 / (2 π R C).AFAIK, for modest values (100s ohms?), the cutoff is a few GHz.@coppercone2: the substrate is Al2O3, k = 9.8. For a pad-to-pad distance of about half the body length, a chip width of half the body length, and thickness half of that again, the capacitor is approximatelyC = ε A / DC = (9.8 * 8.84 pF/m) * [(L/2) * (L/4)] / (L/2)C = 21.66 L pF/m.For L = 0.4mm, C = 8.66 fF.Seems suspiciously small. This implies a cutoff of 92GHz for a 200 ohm resistor. There are probably geometry and fringing factors that I've skipped, and certainly transmission line effects will take over before frequencies quite that high.Tim

author: mark03
author_rank: Frequent Contributor
timestamp: September 19, 2018, 03:26:57 am
content: Thanks for the Vishay link, Cerebus! It seems that smaller packages are preferred.To be clear, when I said a "wideband" front end I meant about 4 MHz. Phil Hobbs' tutorial explains the need for Cf || Rf for transimpedance amp stability: Cf = 1 / (2*pi*Rf*sqrt(f_t * f_rc)) where f_t is the op-amp GBW (410 MHz for the ADA4817) and f_rc is the corner frequency of Rf and the input capacitance. In my case Cin should be ~ 3-4 pF from the op-amp input and cascode device (a BFG25A/X).A naive application of this formula with Rf=1M yields f_rc = 40kHz, sqrt(f_t * f_c) = 4 MHz, and Cf = .04 pF. My thinking was that this is probably a lower Cf than I can reliably achieve, and therefore I would need to reduce Rf somewhat. But perhaps not. One thing I worry about is that if I leave pads for an actual lumped Cf, in so doing I increase the parasitic Cf. So I should decide now if I need that or notOh, there is also the choice of SOIC or QFN package; presumably QFN would be preferred.

author: DaJMasta
author_rank: Super Contributor
timestamp: September 19, 2018, 04:00:03 am
content: Very small resistors may be tricky if you are using a guard trace to get very low light sensitivity, but with a few hundred k worth of gain that may not be something you're thinking of using. Hamamatsu has a few app notes for photodiode design circuits, and their recommendation for low capacitance was several resistors in series for your feedback path.They also mention lead inductance of the photodiode leads and traces on the board, since these inductive parasitics also have a pretty substantial filtering effect on high impedance parts (and they say specifically chip resistors are preferred to through hole because of this).Though you probably have already considered it, make sure your photodiode is also a high speed low capacitance variety. A lot of the larger area or higher sensitivity photodiodes can have tens of pF of capacitance (or more), and would be entirely unsuitable for the application.

author: David Hess
author_rank: Super Contributor
timestamp: September 20, 2018, 07:20:46 pm
content: Quote from: mark03 on September 19, 2018, 01:47:35 amA series string is commonly used with high resistance values to lower the capacitance. Also cut away any ground plane from sensitive nodes.I am inclined to think that you will need more feedback capacitance for stability reasons. Burr-Brown had a lot of great application notes about photodiode transimpedance amplifiers.

author: mark03
author_rank: Frequent Contributor
timestamp: September 30, 2018, 08:54:09 pm
content: For those curious, the full schematic of my project is posted inthis thread. The resistor in question is R29. My plan is to only place pads for the resistor, but place a very small through-hole on the outside of each pad. If extra capacitance is needed, I can adjust it with a short length of twisted fine wire. The math still says I want < 100 fF.One thing I am still not sure about is whether the feedback resistance needs to be a metal-film type. This is important for the resistors generating bias currents, so as not to create shot noise. But I am not as certain about the feedback resistor. This matters because you can't get metal-film (thin-film) at hundreds of k-ohms in small packages. In fact, it will drive me to 0805!