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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.

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  • \$\begingroup\$ 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%) \$\endgroup\$ – d3L Mar 27 '16 at 13:13
  • \$\begingroup\$ 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/… \$\endgroup\$ – user65586 Mar 27 '16 at 13:19
  • \$\begingroup\$ 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? \$\endgroup\$ – EM Fields Mar 27 '16 at 14:16
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    \$\begingroup\$ @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 \$\endgroup\$ – Szymon M. Mar 27 '16 at 14:24
  • \$\begingroup\$ 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? \$\endgroup\$ – EM Fields Mar 27 '16 at 15:05
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enter image description here

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.

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  • \$\begingroup\$ If the tolerance of his resistors are not enough, how will this circuit help? \$\endgroup\$ – pipe Mar 27 '16 at 16:57
  • \$\begingroup\$ I like this answer as the OP does not specify impedance, etc. of the circuit connected to the divider \$\endgroup\$ – johnnymopo Mar 27 '16 at 16:58
  • \$\begingroup\$ @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. \$\endgroup\$ – Marko Buršič Mar 27 '16 at 17:01
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In principle, a simple voltage divider, made with standard resistors, will do the job. The important step is to buy, rent or borrow a high-resolution DVM to calibrate it with. A 16-bit ADC will have a resolution of 1 part in 65,536. This establishes that you'll need a 5 1/2 digit meter, which will give you resolution to 1 part in 199,999. Then provide a calibration voltage of about 19.9 volts, but measure it with the meter to get high resolution. Now drive the ADC, and calculate the scale factor for your circuit. Also, just as good practice, apply zero volts and convert in order to find your zero point. Be aware that if the ADC is putting out a solid 0h0000 your reading is not usable for setting the zero point. You don't know what voltage will provide a reading of 0h0001. All you know is that the ADC is saturated.

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 $$V = 19.9932 (\frac{65535-7}{39571}) = 33.1079\text{ volts} $$

You do not state exactly what you're measuring, or why you care about this level of performance, but you do need to learn something about the differences between resolution, precision, and accuracy. Furthermore, you'll need to learn about ideas like thermal drift (your divider resistors will change with changing temperature, and your ADC zero point will also probably change as well). You'll also become acquainted with the concept of noise and noise averaging/lowpass filters. You'll also need to learn about system specification and error budgets - do you REALLY need 16 bits? Will 12 do? What are you doing with the data, and why?

Have fun.

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For a precision +5V reference, buy a low-tolerance 5v zener diode & connect it to the +32V supply through a ~10kOhm resistor.

This will give you a reliable, steady +5V reference voltage (to within the tolerance of the zener) for any draw between 0.001mA-2.0mA.

Less current draw should be fine, but if you need more current, you'll have to select a (possibly lower precision) >5mA rated-current zener & use a lower-value resistor.

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