Advertisements

Archive for July 23rd, 2009

Trimming Voltage Regulators by Stacking SMD Chips

LM317 Regulator (Partial) Schematic

LM317 Regulator (Partial) Schematic

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.

Three sets of stacked chip resistors

Three sets of stacked chip resistors

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…

Advertisements

9 Comments