Quite often, the values you need for voltage regulators, like the venerable LM317 and its ilk, don’t work out to anything you have in your parts bin. What to do?
One of the really nice things about SMD resistors is that you can stack them up without much effort. That parallels their value, so you can only make the final value smaller than any of the stacked resistors, but we can work with that.
The schematic shows part of a multi-voltage power supply for the EPROM programmer I mentioned there. Normally you use a 240-Ω resistor between the Output and Adjust terminals, but anything in that range will work fine. Alas, when I went to the parts bin, that’s the value I didn’t have any of.
But, having recently acquired an assortment of 60-some-odd 1% chip resistors, 100 to the bag, I had enough raw material to make it work. In fact, the values in the schematic reflect the parts on hand, which is how it sometimes happens.
A pair of 499-Ω resistors in parallel gives you 249.5 Ω, close enough to 240 (and shown as 250 because that’s only 0.2% off). Plug that value and the desired voltages into the LM317 equation to find the other resistors:
12.5 V = 1.25 * (1 + R / 250)
R = 250 * ((12.5 / 1.25) – 1) = 2250 Ω
If you happen to have something close to that in your parts heap, great. I didn’t, and a stock 2200 Ω 5% resistor would produce 12.25 V; a bit lower than I wanted.
This pic shows the solution: stack some SMD resistors to get the right value.
To make this work easily, you need a calculator that has a reciprocal (1/x or x-1) key. My ancient HP-48 does that, natch, but both of the Official School Calculators our daughter uses has 1/x as a shifted function. Your mileage will certainly vary.
Anyhow, the reciprocal of the resistance of two parallel resistors, RA and RB, is the sum of their reciprocals. Got that?
1/R = 1/RA + 1/RB
If you know the total resistance R that you want and one of the resistors RA, then you find the other resistor RB thusly:
1/RB = 1/R – 1/RA
In order to get 2250 Ω, I started with the next higher value in the assortment, 2740 Ω, and turned the crank:
1/RB = 1/2250 – 1/2740 = 79.4809E-6
RB = 1/79.4809E-6 = 12.58 kΩ
As it happens, the assortment didn’t have that value, either, but it did have 15 kΩ. The parallel resistance of 2740 and 15 k is:
2317 = 1 / (1/2740 + 1/15000)
So turn the crank one more time to find the third resistor RC:
1/RC = 1/2250 – 1/2317 = 12.814E-6
RC = 78.04 kΩ
Well, that isn’t one of the values I have either, but I do have 82.5 kΩ. The parallel value of those three resistors is:
1/R = 1/2740 + 1/15000 + 1/82500 = 443.75E-6
R = 2254 Ω
Which is 0.1% off the desired value. Close enough.
Actually, it won’t be nearly that close, because the 2740 Ω resistor can be off by 27 Ω either way. If you really care, measure the actual values and feed those into the equations. If, of course, you can measure resistors better than 1% and you don’t care about temperature effects and suchlike.
This is appropriate for one-off projects and prototypes, not production runs, but it’s a handy trick to keep in mind. If you want to be fancy, you can lay the circuit board out with parallel resistor tracks and make it look like you knew what you were doing all along…
The Smell of Molten Projects in the Morning 
We often do this with capacitors as well, since a space-sensitive layout might only have a single pad and we need a lot more capacitance than is available in that package. I have on occasion made a 3×3 cube of SMT caps that have gotten soldered onto the top of a single existing cap to give us 10x the amount the single cap on the board provided. It works best, I’ve found, to solder this up somewhere else, by adding one cap at a time to the growing cube, and once both sides are soldered, then add it onto the top of the cap on the board, so everything doesn’t fall all over.
Haven’t been that desperate yet, but I can see it from here.
I suppose it would be considered wanking to build a one-cubic cap block using higher melting point solder, then dial back your precisely calibrated iron to tack the block to the board with eutectic solder.
Oh, right, we can’t use eutectic tin-lead solder these days… I keep forgetting.
In the jewelry trade (my background) there are at least five grades of commonly-available solder in a spread of temp ranges, for exactly this situation. When you design a piece of jewelry, you mentally assemble it hierarchically, using the-next-lower-temp solder as you move down the hierarchy. There have been times I’ve wanted that for electronics.
Thankfully, since I’m in R&D, we have thus far managed to scoff at the idea of lead-free solder.
By the way, we also use the stacked-resistor plan for when we need to put a 1 watt current sense resistor in a 1206 space. I’m really not convinced that’s a great idea, given the reduced cooling the bottom resistor sees, but we’ve even done that on applications evaluation boards in somewhat-mass-production, and apparently none has burst into flame. Yet.
apparently none has burst into flame. Yet.
That you know of …
*sound of helicopters approaching*
Might be interesting to check the color on the underside of the board below that stack o’ resistors…
I actually wrote a program to build up resistors of desired value from series or parallel combinations of existing ones. It has internal tables for E24, E96, and even the peculiar series available in a Radio Snack resistor assortment. Took me far longer than longhanding ’em with a calculator whenever I need to, but it was fun.
That’s the spirit: nothing exceeds like excess!
I don’t have any complete sets, so I make do with what I have, plus the contents of a jumbled bag of surplus SMD resistors. It’s pretty much ad-hoc; for resistors I don’t care about, finding one in the jumble is quick and easy… for the rest, I start with the bags of known values and work it out.
Bonus: I don’t have to deploy the workbench laptop…
Does anyone know if IPC-J-001 standard (or any standard) cover the topic of stacking up resistors? Is there any recommended way of seldering/assembling the stacked up resistors?
Thanks a lot,
Muhammet
I think stacking resistors is suitable only for hand-built prototype boards, because the top resistor prevents cleaning and inspection of the underlying solder bonds.
A production board should provide separate pads for each trim resistor so (at least) one could be soldered as part of the normal assembly process. A trim resistor could then be added by hand to the adjacent pads using well-qualified techniques.
But I could well be wrong, as I’ve seen some pretty strange things on circuit boards…