# How to create precision reference 2.5 voltage

Can I create a precision reference voltage of 2.5V using 7805 regulator and ordinary resistors(10% error)/diodes or do I need to rely on specific reference voltage generator?

• How precise is "precision"? What load are you connecting to it? Commented Mar 15, 2015 at 14:42
• A 7805 wouldn't be classified as precision therefore using it is pointless. Commented Mar 15, 2015 at 14:42
• @Andyaka It'd be better labelling them as "7805-ish" Commented Mar 15, 2015 at 14:43
• You can buy 2.5 V voltage references? Commented Mar 15, 2015 at 14:53
• A 16-bit ADC has a precision of approx $2^{-16}\ = 0.000015$ or 0.0015% . A 7805 type regulator can probably get you to 1% or 2% of accuracy. A TL431 type reference -- 0.5% accuracy. A search for precision reference can get you lower with corresponding increase in cost. Consider the system as a whole, do you really need 0.0015% (16-bit) of precision? What level of accuracy do you need (I think it is safe to say that you probably won't get anywhere near 0.0015% of accuracy)? Commented Mar 15, 2015 at 20:11

A precision voltage reference is everything that a 7805 and resistive divider isn't.

Precision isn't all about how accurate the voltage is, but how accurate the voltage remains over time.

An precision voltage reference:

• Has a precise voltage - the 7805 isn't very precise at all.
• Is low noise - By comparison the 7805 has massive amounts of ripple.
• Has a low temperature coefficient - 10% resistors drift massively in resistance at different temperatures.

So if you want precision then a 7805 and some resistors are really not the way to go.

Yes, you can. It all depends on how much error you can tolerate. Since you didn't quantify "precision", it is a meaninless term and can only be ignored.

Obviously the error of a reference derived from a 7805 regulator is all the errors added up. Find the worst case output voltage range of the 7805, the worst case resistances of your resistor divider, and from those calculate the worst case range of your nominal 2.5 V reference. Keep in mind that the worst case for each of these is over the full temperature range you want your device to operate over.

Also consider the loading on a resistor divider. The output impedance of that will be the parallel combination of the two resistors. For example, if you use two 10 kΩ resistors, then the output impedance will be 5 kΩ. If that is feeding something that, for example, has 100 kΩ input impedance, then that adds another 120 mV drop to your reference. Even if the output was a perfect 2.5 V unloaded, you would only have 2.38 V when in use. Lower resistances in the divider decrease this effect at the expense of more current.

However, the easiest way to get a 2.5 V reference is to use a 2.5 V reference chip. These are intended for the application, and will have much better accuracy than a 7805 followed by a divider with two off the shelf resistors.

Although it's not designed as such, a 7805 (the L version rather less so) contains a bandgap reference and a divider made from matched resistors so its actually not a horrible reference provided you don't draw much current from it and the source is reasonably well filtered- for many purposes it's actually okay, especially if you calibrate out the initial inaccuracy (which may eliminate other system errors). It's not a 3ppm/degree C precision reference but plenty good enough for measurement of humidity, ambient comfort temperature and similar, so it really depends on your requirements.

Edit: here is the typical temperature stability of the TI 78M05

/Edit

The problem with a divider (and this would apply with a precision 5V reference as well) is that many ADCs have a requirement for a low source impedance so the divider could affect accuracy, sometimes in subtle ways.

For modest accuracy, a TL431 shunt reference may be a better choice, or any number of precision series or shunt 2.5V references.

• @MattYoung The "precision" (TI's words, not mine) TL431 is not much better, though it's specified with the box method which does not tell you the slope. In the range around room temperature it's typically 100ppm-ish.. More than adequate for many applications. Of course that all goes sideways if you also try to draw a lot of current from it rather than spending another 5 cents for a second device. Commented Mar 15, 2015 at 16:15

If you want a precision reference voltage of 2.5V, you are better off buying a purpose-built AD584 precision voltage reference. You can pick one up packaged on a circuit board with filtering capacitors and jumpers and even a "calibration certificate" (in quotes because it is not officially traceable to a reference standard, and in many cases appears not to be even an actual measurement of the device) from Amazon for under $20. Some of the units come with the AD584LH, which is specified as ±2.5 mV at 25ºC (0.1%) and 10ppm/ºC deviation from there. Lots more real-world information available here courtesy of Mark Hennessy. • That is a very expensive and large and complicated solution to an easy problem. Precision voltage references in a two or three pin package cost cents and need nothing more than an external, non-precision resistor! Commented Jan 3 at 3:26 • @ConstantineA.B. Would you please be more specific? Better yet, post your solution as an answer. The cheapest precision voltage references (0.1% or better, temperature compensated) I know of are around$7 + shipping in single quantities, and require a couple of capacitors for best performance. The referenced packaged product can be had for ~$10 including shipping if you look hard enough. – Sara Commented Jan 3 at 4:05 • Well, that's your definition of "precision", and it is not close to the point op started from (3%-5%), not is it close to the point they assumed they would get (1/2¹⁶). Commented Jan 3 at 4:32 • But sure, if you need a fixed 2.5 V, better than 0.1% including temperature error, you get to spend like AUD 4: digikey.com.au/short/41tf47zmnot, not$10. The need for external capacitors is kind of inherent to any control loop, so your device isn't better - it's just that it's inherently slower (and the datasheet is a bit more old-fashioned in letting the user calculate that themselves) Commented Jan 3 at 4:40