How to make a precision 1:10 resistor bridge with any resistor value...

The E series of resistors do not allow to make a 1:10 divider bridge with e.g. 2 R22 values. You never will find a pair of values corresponding to an exact required ratio of 1:9.

Here is a trick to make a precision 1:10 divisor bridge with any resistor value that you may have in quantities in a drawer. You need 6 identical resistors:

The upper part of the bridge consist of 3 resistors in series. The lower part of the bridge consist of 3 resistors in parallel.

So you will get te exact ratio 3*r / (r/3) which is 1/9. When all the resistors are from the same batch, you will get an excellent precision, much better than the given individual precision of the resistors.

Enjoy!

I have such a circuit where I use 9.1M,1M. It's not exact but 10.1:1. That's only off by 1%.

aarg: I have such a circuit where I use 9.1M,1M. It's not exact but 10.1:1. That's only off by 1%.

If you use 6 pieces of 3,3MΩ resistors from the same batch you can get 0,1% and even better...

So, what do the engineers in the test equipment business do? They design the divider as close as possible, taking all the tolerances into account, using fixed resistors and then add a trimpot to calibrate it.

Exact never will be but close can be achieved. A precision resistor is a resistor that has a tolerance level as low as 0.001%. This means a precision resistor will only vary 0.001% from its nominal value. You can place them in series, Parallel combinations to get what you want. 2 ppm/°C resistors are on the high end of the better end of scale. As you pass current through them they will change temperature and hence there value. You will need to design your circuit to compensate for this. These can be found relatively easily on line, and custom values in the range you want are available form many sources. Next thing for you to consider these will not be exact but can you measure the error?

Actually, I can also make a divider with three E12 series resistors, [R1=6.8+2.2=9]/[R2=1.0] for what it’s worth at 5% accuracy.

aarg: Actually, I can also make a divider with three E12 series resistors, [R1=6.8+2.2=9]/[R2=1.0] for what it's worth at 5% accuracy.

Three resistors only. I call that the winner! :sunglasses:

I have never been in the situation that I needed a 1:10 exact ratio, and that it had to be that exact ratio. I've often enough dealt with a voltage divider ratio that can't be made exactly with standard values. No problem. Get something close (either too high or too low, depending on the application), and go with that.

Normally it's a case of "I don't care what the actual value is, as long as I know what that value is". Looking for a 1:10 ratio, but if it ends up being 1:9.87 that's just fine. After that it's just a matter of maths. Being able to use just two resistors instead of six is much more important than whatever number I have to use in my maths. The maths have to be done anyway.

Having a precision 10:1 resistor divider seems to be not worth the effort to me.

With say 3.3V on the input, the 10:1 tap might well be 0.33V, but then you go and read it with a 10 bit AD ( 0-1023) and maybe use a Vref of 1.1V, so to get the real voltage you have to some maths anyway and probably floating point.

What you can do is arrange the resistor values so that you get close to some even value (1,2,5,10) of mV for each AD bit. Then the maths are simple.

srnet: What you can do is arrange the resistor values so that you get close to some even value (1,2,5,10) of mV for each AD bit. Then the maths are simple.

Only if you can be sure of your ADC reference voltage. The built-in 1.1V has a 10% tolerance, and Vcc is not exactly 5.0V (it's closer to 4.5V when on USB power), let alone guaranteed to be stable... In most situations I don't even bother calculating the voltage from the reading, I don't even remember having this requirement at all. In many cases it cancels out, in other cases I just don't care enough about the actual voltage so anyway work with the ADC reading itself.

These days 1% E48 resistors are commonplace, 0.1% E96 are not that much more expensive, but that doesn't get round the problem that stocking 1000's of resistor values is a pain.

For a 1:9 ratio with minimum E12 resistor counts I'd suggest:

1k+1k : 18k 1k : 18k || 18k 1k : 2k2+6k8 (E6 series only) 1k+330 : 12k - error only 0.25% 1k5 : 12k + 1k5 1k5+1k5 : 27k

If E24 values are added 3k0 : 27k

(1k || 10k) : 10k gives 100/11 ~ 9.09. Far away but E1 only. (I am using E2 set - only multiples of 1k and 3k3; it is enough in nearly any situation and still lot of work when the child mix them together.)

There isn't E1 or E2, it starts at E3 https://en.wikipedia.org/wiki/E_series_of_preferred_numbers

There's also the Renard series, only seen these days for capacitor voltage ratings I think. https://en.wikipedia.org/wiki/Renard_series

MarkT: There isn't E1 or E2, it starts at E3

There is not. But from the "definition" from the Wiki page:

where the number after the 'E' designates the quantity of value "steps" in each series

(per decade they mean) you can reduce those to "E1" as 1R, 10R, ... which is enough for most of digital electronics or add 3R3, 33R, ... for finer work to get 2 steps per decade ("E2"). I rarely NEED something else and it is still so many resistors.

wvmarle: and Vcc is not exactly 5.0V (it's closer to 4.5V when on USB power), let alone guaranteed to be stable...

Oh yes! That is too frequently forgotten. That is also what makes 3,3V systems somewhat better, the 3,3 the ADC refers to is somewhat stabler than a "5V" that it almost never is.

MarkT: ...but that doesn't get round the problem that stocking 1000's of resistor values is a pain. For a 1:9 ratio with minimum E12 resistor counts...

You just gave the best arguments: The point is NOT to use less resistors, but to get the precise divider with ANY value you just have got available. Additionally if they are from the same batch, they are likely to have the same value far closer than the rated precision.

herbschwarz:
So, what do the engineers in the test equipment
business do? They design the divider as close as
possible, taking all the tolerances into account,
using fixed resistors and then add a trimpot to
calibrate it.

Oh! you believe that your multimeter is full of trimpots?
:smiley:

Beside that: 4 resistors do cost nothing (and far less than a trimpot).
Adjusting a trimpot on a test bench is today exorbitant expensive, you will barely find them in current electronics.

Most high-end test equipment use bespoke precision laser-trimmed resistor arrays, which are available to very accurate specifications if you pay enough. Metal film or foil resistors with low-tempco and 0.01% tolerance are another approach. Precision resistors are not cheap, but note that these days even 0.5% tolerance is not really precision in this sense. Increasingly test-equipment does its calibration in software, routing an internally generated reference to measure and resistances, saving money on precision resistors (they still need to have low tempco though).

MarkT: Increasingly test-equipment does its calibration in software, routing an internally generated reference to measure and resistances, saving money on precision resistors (they still need to have low tempco though).

Here the matter is to make a chain of op-amps that have each exactly a gain of 10.

RIN67630: Oh! you believe that your multimeter is full of trimpots? :D

I do, and a web search for schematics proved that most of them have trimpots for calibration. I can't say that they are "full of them", though. :)