# dividing voltage of shunt Vref Lm4040

the whole thing i'm trying to do for about 3 mounth's is measuring IR-level with high accuracy, i tried many newbie circuits with huge noises, unstable results etc..
then i focused on ADS1110, an ADC chip which has an internal 2.048V vref so i can't use any value bigger that Vref for measurement. in other side, i'm using a voltage divider to generating voltage changes in different IR levels. not matter how much photodiode is sensentive and what is maximum and minimum resistance of it, the importance is just steady results for ever... so an steady power supply is the key, and i just found LM4040CIZ-2.5 available for myself (and don't know any other steady supply) which can provide 15mA maximum current.
but that is 2.5 not 2 in voltage! then i decide to use this formula to calculate R1 :
and this one for R2 & R3 (considering an absolute maximum current drawing or passing trough Lm4040) :
also i moved photodiode before the divide line in serie with R2 (instead R3) here is the circuit :

simulate this circuit – Schematic created using CircuitLab

i cant validate this method, im not crowling for a negative answer (80% i guess my circuit not works) i just trying to find a way to get an steady result from photodiode and ADS1110

TLDR : i read some things about Op Amps with a few understanding, im not sure if they are undependent voltage stability? here is the reference

# EDIT:

Answer to comments, my imagine of steady is something near to a Big 1.5v Battery which i used many times to measuring IR level but unfortunately the whole of the big batteries goes empty then and new batteries are different from each other in millivolts,

• Please clarify what you mean by "steady". How much variation, as a percentage of the supply voltage, can you allow? What is the operating temperature range of your circuit? Commented Jul 18, 2018 at 11:25
• @ElliotAlderson tnx, Edited. Commented Jul 18, 2018 at 11:30

A photodiode produces a photo-current proportional to the incident light power hitting its active area but, it has to be the correct wavelength light (for the photodiode) or values measured will be highly confusing.

The circuit you have shown with a photo-diode built into a potential divider sat around a voltage reference is missing the point. You should consider using a transresistance amplifier to convert the photo-diode current to a voltage that then feeds into what measurement system you have. In simple terms it is one of these: -

So, for the TEMD1000 specified above, the data sheet will show you what current output you get for a given light incident power: -

Take into account the photo-diodes active area to convert incident power density to actual power received and you get photo-current and that photocurrent is amplified by the op-amp and resistor to give you the required output voltage per uW you require for your ADC input.

• thanks in advance mr andy, am i allowed to use LM4040 as power supply for Op-Amp and connect its -2.5 to GND ? is that output voltage relative to Resistor and Capacitor which was placed between output and input? Commented Jul 18, 2018 at 12:26
• I’m unsure why you want to do this. An opamp will run from various supply voltage ranges and they don’t need to be regulated. If you want to run from a single supply you need to bias the grounded pins shown in the diagram to a mid rail voltage. Commented Jul 18, 2018 at 12:41
• today i bought two LF356 op-Amp's, then i started reading more, now every thing is slightly more clear for me BUT virtual ground stays my problem please guide me, i just have op-amp, resistors, caps, a 5V supply.. few components! the first method for measuring IR, voltage divider was not horrible like Op-amps in TIA mode, forgive me if i'm begging knowledge. Commented Jul 19, 2018 at 22:14
• Will an LF356 work from a single 5 volt supply. Am on android and can’t check. Commented Jul 19, 2018 at 23:11
• They can’t convert a smaller voltage to a higher voltage or a positive voltage to a negative voltage so, if all you have is 5 volts then these won’t help. A dc to dc converter will help. Commented Jul 20, 2018 at 22:33