# Voltae Divider Vout current- sense check

Referring to the attached circuit diagram, which shows an LDR-based voltage divider and its illumination circuit - am I correct in assuming that the current passed to the output will be limited by the total resistance offered by SR-LDR1 and SR-R2?

For example, I have measured the resistance of the LDR when fully illuminated as 300 ohm, therefore the output current using a 10k resistor at SR-R2 will be

5/(300+10,000) = 0.48 mA fully illuminated

That is I do not take into account the resistance of the resistor SR-R1 or the potentiometer that come after the connection point?

I have found that this works well for triggering a one-shot 555 timer when the light is cut off(with or without SR-R2), will be trying it out on the Arduino later but want to make sure I don’t blow one of the analogue pins (or worse!), though hopefully there will still be enough current for the Analogue pin to sense.

(I have an excel spreadsheet used to calculate the voltage output using measured values from the LDR and a variety of resistances for SR-R1 but I don’t think this is needed for this question)

Thank you for your time - Richard.

Assuming that R2 is connected to an Arduino analogue input then virtually no current is drawn by the pin.
I would remove R2 altogether.
The pot is not acting as a pot just a 10K resistor because the wiper is not connected anywhere. If you want it to do something connect the wiper to one of the ends.

Grumpy_Mike:
Assuming that R2 is connected to an Arduino analogue input then virtually no current is drawn by the pin.
I would remove R2 altogether.
The pot is not acting as a pot just a 10K resistor because the wiper is not connected anywhere. If you want it to do something connect the wiper to one of the ends.

Thanks for that Mike - slight glitch on the circuit diagram! Revised digram attached to original post.

Removing R2 altogether would allow a draw of 16mA by whatever the circuit is connected to - maybe I'm being overcautious, or have you just helped my learning curve by telling me that the current draw will be defined by what the Arduino wants, not by what I allow the circuit's maximum to be?

maybe I’m being overcautious

Yes. And a bit simplistic. The maximum current you can get in the output is defined by the value of the LDR. But even then, at the maximum current, the potential divider action is totally trashed.

As a rule of thumb never take more than one tenth of the current out of the center tap of a potential divider than is flowing down the divider. As I said the actual current draw is determined by the load you put on the center tap. If that is an Arduino input then the technical term for the current drawn is “bugger all”.