[[SOLVED]] a question about resistance (i did think before posting)

so i did some calculation and it turns out i need a resistance of 450.21ohm . the thing is , all i have is E24 series resistors that have 5% tolerance . i am going to use a combination of 430 and 20 ohm resistors or 330 and 120ohm resistors . the thing is , i want to get resistors that have the closest value possible to 450.21 . and they are cheap so i can buy alot of them . the thing is , how can i test them to figure the most suitable ?

thank you

How close do you need to be?

Since you are using 5% resistors, do you have a way of measuring the resistors with more accuracy than that? Or, with whatever accuracy you need?

If you have a way of accurately measuring resistance, you can select two values that add closest to 450.21 and wire them in series.

If you can’t measure the resistors accurately, you can’t be guaranteed of being any closer than 5%.

A 25 Ohm 50 Ohm pot would give you 5% adjustability. A 10-turn pot would probably give you the precision-adjustment you need. (Actually, you’ll need about 50 Ohms for a little more than +/-5%, and remember that a pot in series will add to the resistance, so the other 1 or 2 resistors should add to less than 450 Ohms.)

Selecting resistors (or other components) usually isn’t good engineering practice because it’s difficult to reproduce the circuit. You end-up scrapping lots of parts and you may buy 100 parts and not find any usable parts.

Please explain why you think you need a resistor whose value is accurate to 0.002%.

For probably 99% of common circuits, component value tolerances of 5% are entirely acceptable.

In theory, you can improve your accuracy by using more resistors. This only works if the errors in each resistor are random and not systematic.

It works like this. A 100 ohm resistor with a 5% standard deviation would have a resistance of 100 +/- 5 ohms. If you put two 50 ohms resistors in series you would wind up with a resistance of 100 +/- 3.535 ohms. This is because when independent random variables are added their variances (i.e., the square of the standard deviations or sigmas) add. The resulting standard deviation is then the square root of the sum of the squares. So the standard deviation increases as the square root of N where N is the number of resistors. But, the resistor values diminish as 1/N. The result is the error is diminished by sqrt(N) divided by N, or 1/sqrt(N).

So, if you put 100 of your 5% resistors in series you would have a .5% resistor.

Needed a 9M but closest I had was 820k, scraped the 820k with a knife 'till it measured 9M and “dressed the wound” with clear nail polish.
I know, its rude crude and unconventional but sometimes you gotta do what you gotta do. It worked!

PC250048.png

-Jremington :

thank you ,

i am assembling a DAC , though to achieve the needed quality i need some really good values .
so the first input goes to a standard R2R resistor ladder (1K and two 1Ks) so i can get a sig between 0 and 5v

then comes a simple voltage divider to change the scale to 0 - 0.714v
calculations indicate that i need a 75Ohm resistor ( i have good ones)
and the 450.21 one

-DVDdoug :

thank you very much , i do not have a tool that can measure at a scale of 0.1 or below , just anything superior to 1ohm … i think it’s really hard to get the exact value using a pot .

  • Charliesixpack :

very very interesting thank you very much :slight_smile:

  • Outsider : though the scale here is really tight , scraping might get you to be further away from the correct value , though that’s a good thing to know :slight_smile: thank you very much

charliesixpack:
In theory, you can improve your accuracy by using more resistors. This only works if the errors in each resistor are random and not systematic.

It works like this. A 100 ohm resistor with a 5% standard deviation would have a resistance of 100 +/- 5 ohms. If you put two 50 ohms resistors in series you would wind up with a resistance of 100 +/- 3.535 ohms. This is because when independent random variables are added their variances (i.e., the square of the standard deviations or sigmas) add. The resulting standard deviation is then the square root of the sum of the squares. So the standard deviation increases as the square root of N where N is the number of resistors. But, the resistor values diminish as 1/N. The result is the error is diminished by sqrt(N) divided by N, or 1/sqrt(N).

So, if you put 100 of your 5% resistors in series you would have a .5% resistor.

Since resistors are made in a factory and packaged right on the line, isn't it more likely that errors will be systematic? And wouldn't that make it worse?

i am assembling a DAC ,

Just buy one and have done, you don't stand a chance of getting the values you need.

But you still have not explained why you would think it appropriate to calibrate your circuit in that manner. Are you using 0.5% resistors for the ladder? If not, you are talking nonsense for a start.

Heard of a variable resistor?

32cbde53b0bb813b34ce1473d2a74dafde7c2f1b.png

I think that is pulling your leg.

calculations indicate that i need a 75Ohm resistor ( i have good ones)
and the 450.21 one

Are these "good ones" also accurate to 0.002%? Otherwise, you are wasting your time.

ChrisTenone:
Since resistors are made in a factory and packaged right on the line, isn’t it more likely that errors will be systematic? And wouldn’t that make it worse?

Interesting question. See the attached pdf for a bunch of 1% resistors. The data came from here.

ResistorDistribution.pdf (49 KB)

Heard of a variable resistor?

Oh sure, thats what the scraped resistor replaced, a 1M 10 turn trimmer ruined by Orville Overtorque, but when faced with a breakdown at 2 AM and no spare parts you have to get creative.

ChrisTenone:
Since resistors are made in a factory and packaged right on the line, isn't it more likely that errors will be systematic?

I've noticed, if the resistors are from the same batch, the resistors tend to be all above the target resistance or all under the target resistance. The error does not appear to be a random scattering around the target resistance.

That is because resistors come off the line and are sorted into percentage bands. Only the highest tolerance resistors will be scattered around the value. Others will have a hole in the distribution where the tighter tolerance values have been removed.

The report cited by charliesixpack suggests that the distribution is gaussian, at least in the long run. However it hints at clusters of off center batches.

@Grumpy_Mike, I have never seen resistor manufacture. Do they sort the tolerances as part of the assembly line? A separate QC step would seem not to be cost effective.

No they make them and then test them. That is why over the years that 1% resistors have become increasingly cheaper as they get better at making resistors. Also they can get too many 1% and have to sell them as 5%.

Grumpy_Mike:
Just buy one and have done, you don't stand a chance of getting the values you need.

i am using one , an 8bit to 0-5v analog , then i need to divide that voltage to get a 0-0.714v interval . the only issue is the 450.21 resistor .

thank you though Mike

Oh and Mike ... what if i use one of those resistors with changing resistance i don't know what they are called in English but what if i put one of those in then follow the output with a multimeter once ti get's to 0.7v i block it at that position .. is that possible Mike ?

variable resistor, potentiometer, trimmer.
Setting a level is a good use for a trimmer, especially a multi-turn part, can dial a voltage right in:

CrossRoads:
variable resistor, potentiometer, trimmer.
Setting a level is a good use for a trimmer, especially a multi-turn part, can dial a voltage right in:
http://www.digikey.com/product-detail/en/PV36W103C01B00/490-2875-ND/666502

can i use it as the needed resistor ?