Resistor dissipation vs. pulse current

You can break open one resistor and weigh the actual resistance element. Using that and a basic guess at the heat capacity will give you a worst-case estimate of the rate of temperature rise during the pulse. If that is a relatively small temperature rise, and the continuous dissipated power is within the rating of the part, then you are probably OK. If the temp rise is substantial within a single pulse, and the pulses are in rapid succession, then it gets more complicated.

Jon

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I measured 80K/W hot-spot temp for a 1206 on one of my boards, a multilayer with a plane 12 mils below the surface.

John

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Maybe Boron Nitride? It's got some wicked thermal conductivity. I have no idea about the cost.

George H.

In 10s of milliseconds, mounting won't matter much.

John

Aluminum nitride is affordable, and very cool stuff. It conducts heat about as well as aluminum alloys.

John

Why? I still have thousands of unmarked Germanium diodes for that. :)

Still not as exciting as exploding electrolytics. :)

On the plus side, a metal film resistor makes a dandy and very inexpensive fuse for certain kinds of transients (eg. induced current from nearby lightning strikes).

It doesn't have to be a circular cylinder, but the calculation doesn't strictly apply to serpentine structures because the heating isn't as uniform.

I'm using a pulse-rated resistor in my little downhole laser gizmo,

R6 Res 100 pulse 0805 Vishay CRCW0805100RJNEAHP $0.14 Digikey

In the 0805 size, it's rated to handle 20W for 1 ms. THe bigger ones can do 200 W for 1 ms or 1 kW for ~50 us.

See

Cheers

Phil Hobbs

One interesting number, for a conductor, is degsK/W per ohm. Most metals run around 180,000. Brass is about twice that, for some reason.

John

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Yeah, 180,000 is the 1 ohm =3D~ 10^5 K/W number (That I first wrote as

George H.

Copper is the clear winner for wiring stuff on cold plates, e.g. cooled CCDs. Its thermal conductivity is very high, but its electrical conductivity is so much higher than other metals that it still wins. I have a table of that in my online thermal control chapter (first edition),

Of course you have to use tiny tiny wires, which used to be hard. Nowadays you just use 1/2 oz Cu on flex, with 4/4 mil rules.

FFC connectors rule.

Cheers

Phil Hobbs

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Hi Phil, Thanks for the second edition draft of thermal control. (something to read at lunch.) What do you mean by "Copper is the clear winner..."? Are you talking about the ratio of thermal to electrical resistance?

George H.

Cool, no pun etc. I'll add that to my collection of thermal design notes.

I'm reminded of a laser pointing problem that needs sub-arc-second stability, and your point about thermal bending. Masses of metal do lowpass filter thermal noise, but a tenth of an arc-second is a pretty small angle. Some sort of active (piezo?) or detector-end signal-processing compensation might be possible... a sort of a spin on your LNC?

John

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Right. Folks used to use brass or manganin or constantan wire, just to have something strong enough to handle, but copper wins, and with flex circuits, it's very nearly trivial.

Cheers

Phil Hobbs

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You can do amazingly well with a split detector--shot noise limited pointing stabilities of around 1 part in 10**8 of the beam diameter per hertz. If you do a good job of subtracting the currents, and servo around zero, you get noise cancellation for free, so you can get down to sub-picometer stability even if the laser has significant residual intensity noise (RIN).

Of course you'll probably be limited by things like the temperature coefficient of small etalon fringes well before you get to that level, which will change the beam shape a bit, but the centroid will be _stable_.

Cheers

Phil Hobbs

Doesn't the thermal conductivity of the substrate dominate?

I suppose with a two layer flex and a slow twist you could get a more-or-less twisted pair.

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Hmm, Well I thought the whole point of the Wiedemann Franz law was that the ratio is pretty much the same for all conductors (at a given temperature).

Here's some data from Guy White's "Experimental Techniques in Low- Temperature Physics" Temperature =3D 295 K material resitivity thermal conductity Product 10^-8 ohm-meter W/(meter-K) aluminum 2.74 237 649 Be 3.25 200 650 Bismuth 120 10 1200 Cadmium 7.3 97 708 copper 1.68 403 677 Lead 21 35 735 Tin 11 67 737 SS 304 71 15 1065 Cu/Zn (70/30) 7.2 120 864 manganin 48 22 1056

Well dang, The elements do look better than the alloys. I went looking for data on Phosphor-Bronze which is what I've used for cryogenic wiring, but I got discouraged with all the different units. (I did find a nice paper from Nasa though.) ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.../

Copper on flex it is then.

George H.

Kapton HN is quoted as 0.12 W/m/K, copper is 250-400 W/m/K, and it's the ratio of the cross sectional areas that matters. So if you print the flex on 6-mil flex, on 4/4 rules with 4 mils keepout at the edge, you

You can do a cross-hatched ground plane on the bottom, which can work fine, or you can use two-layer flex and fake it with vias. Half ounce copper is nominally 18 um, so for a 10 cm length of N-conductor flex, the copper conductance is

G_Cu = N*(300 W/m/K)(18 um)(100 um)/(10 cm) = 5.4 uW/K

and the substrate conductance is

G_sub = (N+1)(0.12 W/m/K)(150 um)(200 um)/(10 cm) = 36 nW/K

There's a factor of 16-2/3 in cross-sectional area, but a factor of

By comparison, 40 AWG is 80 um in diameter (3.1 mil), so

G_#40Cu = (300 W/m/K)(pi/4)(80 um)**2/(10 cm) = 15 uW/K

and

G_#40Constantan = (20 W/m/K)(pi/4)(80 um)**2/(10 cm) = 1 uW/K.

not counting the insulation. However, the electrical resisivity is 0.5 microohm-meters as opposed to 0.017 for copper, i.e. even with that factor of 3 larger area, your current capacity is down by a factor of 10.

If 'twere me, I'd rather try quarter-ounce copper and a serpentine pattern than mess around with a bunch of #40 wires with the same performance, and keep the higher current capacity.

You can fan the flex out at the ends to fit an 0.5 mm pitch FFC connector for durability. It's great stuff, flex.

Cheers

Phil Hobbs

sqrt(10) of course.

Cheers

Phil Hobbs