# Resistive divider accuracy question

Discussion in 'General Electronics Chat' started by tom66, Jul 8, 2010.

1. ### tom66 Thread Starter Senior Member

May 9, 2009
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If I have a resistive divider with two ±0.1% resistors, is the overall output accuracy ±0.2% or ±0.1% or some other value? My senses point towards ±0.2%.

(This is not homework, it's for a power supply project I'm doing...)

Thanks .

2. ### Dx3 Member

Jun 19, 2010
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.2% could happen under worst case conditions.

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3. ### Ghar Active Member

Mar 8, 2010
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4. ### tom66 Thread Starter Senior Member

May 9, 2009
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That makes sense. So when the ratio changes, the accuracy shifts. I don't need too high accuracy, ±0.2% is fine heck ±1% would probably be OK it's only a basic power supply.

5. ### SgtWookie Expert

Jul 17, 2007
22,182
1,728
It's like the old saying - enemies accumulate.

Use metal film resistors, as they are much more stable than carbon or carbon film resistors.

They also create far less noise.

Think of a resistor conducting current as a garden hose nozzle with water flowing through it. Hose nozzles can generate quite a hiss. Note that if you throw that hissing nozzle in a bucket of water, things get very quiet.

Use a small ceramic or metal poly film cap (perhaps 10nF to 100nF) from the junction of the divider to ground; that'll be your "bucket" to keep things nice and quiet.

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6. ### tom66 Thread Starter Senior Member

May 9, 2009
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Of course, SgtWookie, the capacitor decreases the response to load changes, so it's probably not ideal for my application where the load can vary quite suddenly (controlling a motor for example with PWM, the load changes between two levels at a high rate).

7. ### Markd77 Senior Member

Sep 7, 2009
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But of course you wouldn't use a resistive divider to power a heavy load directly. That 0.2% would be right out of the window.

8. ### tom66 Thread Starter Senior Member

May 9, 2009
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No of course not Markd77. I am not powering a load through the divider; I am using the divider as part of the feedback mechanism to the SMPS control logic.

9. ### timrobbins Active Member

Aug 29, 2009
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Save the .1% parts for another more critical project if 1% is way fine!

Similar in logic to the linked article, the .1% part will be better on all accounts than a 1% part for absolute initial accuracy, long term drift and TCR - as long as you keep them within their deratings.

Your smps loop has another main error part - the reference - which is unlikely to be better than 1% over all toleranances either - and then you get onto other external voltage drop issues wrt regulation - doh, it's a fickle path!

Ciao, Tim

Jul 7, 2009
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For you folks that got through basic math in school, a good thing to remember is the total differential. For example, if

$z = f(x, y)$

then

$dz = f_x dx + f_y dy$

Now you've got a function that can do approximate propagation of error for you. You can do fancier stuff using Monte Carlo, Cramer's Theorem, etc., but this one is easy to remember and apply. And if things are too hairy to differentiate analytically, do it quickly with a finite difference.

11. ### tom66 Thread Starter Senior Member

May 9, 2009
2,613
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I'm using a precision TI voltage reference ~ 2.048V ±0.1%, so I wonder, if I have a ±0.1% reference, and a ±0.1-0.2% divider accuracy, what is my overall accuracy? Thanks.

May 11, 2010
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I think the answer is more complicated than 0.2% . . .

I think the accuracy varies based on the ratio of the divider.

Start with the easiest case, when they're equal values. Say you have two resistors that are 1,000 Ohms +/- 0.1%. This means each resistor is from 999 to 1001 Ohms. In the worst case of the error, one will be 999 Ohms and the other will be 1001 Ohms. The divider will output (999)/(999+1001) of the input voltage == 0.4995. The ideal is 0.5, so this ratio is in error by (0.5-0.4995)/0.5 == 0.1%. Voila... this is the answer for the even divider... it is equal to the error of the resistors.

Now let us say that your divider is set up to output a 1/11 ratio. On top is a 1000 Ohm 0.1% resistor, and bottom is 100 Ohm 0.1% resistor.

In the worst error case, they will be off in opposite directions, so the 1000 Ohm will be 999 Ohms. The 100 Ohm will be 100.1 Ohms. Now the ratio is (100.1)/(100.1+999) = 0.0910745... The ideal ratio is 1/11 = 0.0909090909.... The error is 0.182% ... this is bigger than 0.1% but smaller than 0.2%.

Does this make sense?

I am pleased with it because I need to do a 1/2 divider for signal output and now I know it will be the same error as the resistors.

[ As Ghar said ]

13. ### tom66 Thread Starter Senior Member

May 9, 2009
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Thanks for the explaination. But does anyone know the answer to my question about the precision reference and the divider accuracy?

May 11, 2010
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I think this one is simply additive, so if your divider has 0.1% accuracy, and your reference is 0.1% accuracy, your final output is 0.2% accuracy.

Put a small trim pot on the divider if you want better than this!

May 11, 2010
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And one last question -- what is the offset error of your comparator? That is another place where you may have simple error in the circuit.

Another is that if your divider resistors are very large, like 100K and higher, then you may see error due to input bias current on the comparator.

(I assume you're using a simple feedback loop with a comparator to switch on/off the converter.)

16. ### jpanhalt AAC Fanatic!

Jan 18, 2008
5,536
832
If you are asking that from the standpoint of statistical analysis, there are relatively simple formulas for calculating the expected imprecision of a combination, given that the individual imprecisions are defined as mean ± some number of standard deviations (or relative standard deviation = coefficient of variation = %).

When you first posted, I thought that was what you were asking. Your query whether the imprecision would 0.1%, 0.2%, or some other value made it appear you had thought about the statistics. So, I tried to find out what the imprecision term meant in a resistor specification. After a brief search, I could not satisfy myself what exactly it meant. The explanation I found was that it was an assurance that the value was between the limits of the stated percentage. That is, 100% of the components would have a value within those limits. That makes the analysis both harder to believe and easier to do at the same time.

If it is some sort of 100% guarantee, and you want to continue that warranty, then you are left assuming the error of a combination must include the maximum possible adverse deviation for each component, regardless of how unlikely that event may be.

Let me give an example: Assume you have two, 47 ohm resistors, each with an imprecision of ±0.1%. That means each resistor may have a value of 47 ±.047 or any value in the range of 47.047 to 46.953. If the resistors are in series, the combination could have a possible range of 94.094 to 93.906, which has a mean of 94 with an imprecision of ±0.094 (i.e., the imprecision is still ±0.1%). You can do the same calculation for unequal resistors, resistors in parallel, etc. In fact, I found one reference that did the calculation for a divider. I don't have it at hand, but is is easy to do for yourself. My advice would be to use the actual values of each range in the calculation, not percentages.

John

Last edited: Jul 10, 2010
17. ### tom66 Thread Starter Senior Member

May 9, 2009
2,613
213
I did statistics for GCSE, got an A in it, but we were mainly talking about probability. We didn't do much on standard deviation and distributions and the like, but I do understand it well enough...

So I shall try and tackle it here.

My two resistors are $R_1$ = 115k and $R_2$ = 59k ±0.1%. Given a perfect reference $Vref$ = 4.096V.

$\frac{R_2}{R_1 + R_2} V_{in}$ or

$\frac{59}{115 + 59} \times 12$ = 4.068V,
or in reverse
$\frac{4.096}{59} \times (115 + 59)$ = 12.08V.

So the output voltage is nominally 12.08V.

Let's consider the minimum values for all the variables, $R_1$ = 114.885k, $R_2$ = 58.941k, $Vref$ = 4.091904V.

$\frac{4.091904}{58.941} \times ({114.885 + 58.941})$ = 12.0676V (to 4dp).

Let's consider the maximum values for all the variables, $R_1$ = 115.115k, $R_2$ = 59.059k, $Vref$ = 4.100096V.

$\frac{4.100096}{59.059} \times ({115.115 + 59.059})$ = 12.0918V (to 4dp).

The actual range is 0.0124V = 12.4mV, or ±0.1026%.

18. ### jpanhalt AAC Fanatic!

Jan 18, 2008
5,536
832
Thanks for confirming that. I actually started my reply before post #12 and had included some stuff about standard errors (S.E.M.), etc. After an hour or so, I deleted it and decided the analysis was really quite simple based on the assumed definition of the resistor's tolerance limit. I didn't see the intervening posts until after I posted.

In brief, I think your approach using the actual values and ranges is the way to go. I did not check your arithmetic, though.

As a couple of asides:
1) I wish someone from industry would provide a definitive answer for what the precision range means, perhaps with reference to an IEEE standard or similar.
2) When I have tested SMD resistors in a cut tape, the values do not seem to be normally distributed about the labeled value. The values are tightly clustered to one side or the other. I suspect that may mean they are a single lot, and actual manufacturing tolerance for a single lot is pretty tight. In any event, applying statistical tools based on means and standard deviations would not be appropriate if the errors are not normally distributed. In other words, it may be much more likely that both resistors of a pair are low valued, than if the pair were from random lots.

John

19. ### hgmjr Moderator

Jan 28, 2005
9,030
214
This question intrigued me enough to try to solve it for the general case. What follows is my effort to derive the general expression for a two resistor voltage divider in which both resistors have the possibility of being out of tolerance by an amount that ranges from ±Δ ohms.

hgmjr

Last edited: Jul 11, 2010

Jul 7, 2009
1,585
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Tom: unfortunately, nobody has given you a good answer yet. The reason for this is that your question is, in general, an indeterminate problem. This is because the "0.1%" is a meaningless number -- in other words, as John pointed out, there's no agreement on what it means.

It could be a completely "unstatistical" number like manufacturers put on their products (but they get their numbers from detailed statistical analyses). For example, a manufacturer of a voltmeter might say its accuracy is 1%. You'll notice manufacturers never say exactly what this means, but most users think that if they use the instrument while it's within its calibration period and at the environmental specifications set by the manufacturer (e.g., temperature, humidity, line voltage, etc.), then the measurements produced by the instrument can be used to set up a bound ("X% of reading plus Y digits" or somesuch) to apply to the reading. The true value of the measurement will then lie somewhere in the range of (reading) ± bound.

It could be like John says -- 100% of the resistors produced are within this bound. Personally, I'd only feel comfortable with such a statement if I could see the manufacturer's manufacturing process and their statistical distributions. But that's just because I used to be a process engineer in one of my former lives. But if this is true, then you can do worst-case analyses like sage did. If you want to automate this a bit, you can use interval numbers (here's a calculator that can do that for you). If you're only making one or a few, then this can be a good approach, as you'll wind up with bounds you know the answer lies within. However, if you're managing a high-tech process that produces revenue at megadollars per hour, then you'll want better tools to help you estimate producer's and consumer's risks. And your CEO will be on your tail to improve yields, no matter how good they are...

It could be a statistical number like "for a resistance of X, the range X ± (0.1% of X) represents two population standard deviations". Then, assuming e.g. the normal distribution, you would then have a nice stochastic number to deal with.

Or, it could mean something else. The fundamental problem is we don't know.

There are numerous techniques available to deal with these various numbers (and all have various assumptions associated with them). Personally, I like statistical tolerancing best, but it does require good input data with known distributions and the correlation matrix. The reason I like it is because it will give you the best prediction of the resultant distribution. Things like worst-case analyses can give you good bounds, but will make you pessimistically estimate the dispersion. But, if you really get into it, it can be technically a lot of fun -- you get to deal with expensive measurement equipment, hairy integrals you can't evaluate, fancy computer toys, Monte Carlo simulation, and glassy-eyed slack-jawed managers listening to your conclusions...

Here's a simple example. Suppose you want to know the bounds on z = x + y, where x and y are stochastic variables. Suppose you know their distributions are such that their values always lie within a value of ε of their mean values. Then we could predict the bounds that z lies between to be x ± ε + y ± ε = x + y ± 2ε. And this would be a correct statement. However, if you were to plot the probability distribution of z (based on a bunch of measurements of z), its practical limits would be almost certainly be less than the ±2ε bounds given. If the distributions of x and y were approximately normal and they are statistically independent, then the distribution of z will be approximately normal and its standard deviation will be the square root of the sum of the squares of x's and y's standard deviations (if they are correlated, there's another term under the square root sign). Thus, for example, if the uncertainty was 0.1% and this represented one standard deviation of x and y, then z would have a standard deviation of 0.14%, not 0.2% as you would get by worst-case analysis. And, if you measure your resultant distribution, this is what you'll find.

I think the time is ripe for a practical little paper on measurement errors, their estimation, and their propagation. I've added it to my to-do list...

Here's how I'd do a worst-case analysis of your question with the above-referenced calculator. Let

V = the precision voltage source = 2.048 V ± 0.1%
R1 = one resistor = 1 Ω ± 0.1%
R2 = the other resistor = 2 Ω ± 0.1%
Vo = the output voltage

The formula is then

$V_o = \frac{R_1}{R_1 + R_2} V$

Here's a session with the calculator (realize it's an RPN calculator, which may make some of you barf ):

Code ( (Unknown Language)):
1.
2. > 2.048(0.1%) >V
3.  x:  2.0480 (0.1000%)
4. > 1(0.1%) >R1
5.  x:  1.0000 (0.1000%)
6. > 2(0.1%) >R2
7.  x:  2.0000 (0.1000%)
8. > # Now calculate the output voltage Vo
9. > <R1 <R1 <R2 + /     # Calculate R1/(R1 + R2)
10.  x:  0.3333 (0.2000%)
11. > <V *    # Multiply the result by V
12.  x:  0.6827 (0.3000%)  # Our answer
13. > ivc
14.  x: <0.6806, 0.6847>
15. > iva
16.  x:  0.6827 +- 0.0020
17.
I stored the numbers in registers R1, R2, and V only to make the problem (hopefully) a little easier to read for non-RPN users.

The command (ivc) changes the interval number format to show the low and high bounds. I was using the ivb display which showed the center of the interval with the half-width in percent. The iva command shows it with the center ± the half-width.

Let's compare this answer to the formula I gave in a previous post using differentials. The relevant differential is

$dV_0 = ( \frac{V}{R_1+R_2} - \alpha ) dR_1 - \alpha dR_2 + \frac{R_1}{R_1+R_2} dV
$

where

$\alpha = \frac{R_1 V}{(R_1+R_2)^2}$

Plugging in numbers, we get

$dV_o = 0.4551(0.001) - 0.2276(0.002) + \frac{1}{3}(0.002048) = 0.0007$

Note this is about a third of the worst-case estimate and is about 0.1% of the calculated output voltage.

My favorite tool for such tasks is using numpy on the computer, as you can have a believable Monte Carlo answer in short order with a few lines of code -- and you can handle things like bimodal distributions, truncated distributions, etc. I've rambled on enough, so I'll save that for the paper.

Don't trust my calculations, as I didn't go back and do them again.