# Precision voltage divider

Hi guys I'm trying to turn 32 volts into 5v using a voltage divider.

The precision of the device concerns me as because it is fed to an 16-bit analog to digital converter. Could someone provide me a circuit that will precisely step down the voltage.

I tried using resistors but the tolerance of the ones I have are too high....

I do not have any degree of electrical engineering, so keep it as simple as possible.

• Thats actually a good / common way to solve your problem. If you can't calibrate your device afterwards simply use precision resistors (0.01%, 0.1%) – marco-a Mar 27 '16 at 13:13
• There are quite a few tutorials out there that can help you. Generally, you'll find this SE is not a design service, so the expectation is that you try something and come back with a specific questions. See: hackaday.com/2010/12/11/… – user65586 Mar 27 '16 at 13:19
• Are you trying to generate a 5 volt reference for the ADC or trying to convert a 0 to 32 volt signal into a 0 to 5 volt signal? In either case, what kind of accuracy are you trying to achieve? – EM Fields Mar 27 '16 at 14:16
• @EMFields Im trying to get a voltage reading of 3 decimal places also im trying to convert a 0 to 32 volt signal into a 0 to 5 volt signal – Szymon M. Mar 27 '16 at 14:24
• When you say 3 decimal places, are you referring to the 32 volt signal, as in 31.XXX or to the 5 volt signal, as in 4.XXX? In the limit, you'e got a 16 bit DAC so you can represent 32 volts as 65535 488 microvolt steps if you want to go that high. What's your plan? – EM Fields Mar 27 '16 at 15:05

At the input you can use a voltage divider network 1:10, like http://www.caddock.com/Online_catalog/Mrktg_Lit/TypeT912_T914.pdf, then you will have 3.2V from divider at 32v input and you have to amplify with factor 1.5625. So, 1+Rf/Ri = 1.5625; Rf/Ri = 0.5625 you do this with choosing fixed precise resistors, low TCR.

• If the tolerance of his resistors are not enough, how will this circuit help? – pipe Mar 27 '16 at 16:57
• I like this answer as the OP does not specify impedance, etc. of the circuit connected to the divider – johnnymopo Mar 27 '16 at 16:58
• @johnnymopo You're right, the circuit could be much better if he would specify if he has double power supply, and the desired input impedance, so improvemnets are still an option. The basic point is that he will have to find precision resistors, anyway. – Marko Buršič Mar 27 '16 at 17:01

But let's say that zero input gives 0h0007, and 19.9932 volts input gives 0h9a9a. Then a voltage span 19.9932 volts gives an ADC span of 9a9a minus 0007, or 0h9a93, a decimal value of 39,571 counts. A maximum ADC output of ffff will correspond to $$V = 19.9932 (\frac{65535-7}{39571}) = 33.1079\text{ volts}$$