Hello,

I'm new to this forum, and I need your help. I have been trying to figure out how to solve this question, but I honestly don't even know where to begin or what kind formula to use. I know the rules state that I must come up with something first. I have asked my classmates for help, searched online, but I can't find an answer.

An LDR has a resistor of 2.7 KΩ at the maximum light intensity to be measured and an (almost) infinite resistance in total darkness. You want to read this sensor with a voltage divider circuit consisting of a voltage source of 9.0 V and a series resistor, so the output voltage fits in the range of ADC 0.0-5.0 V.

What is the correct value of the series resistor?

 

First, sketch up your basis circuit consisting of a Vs = 9V supply, two resistors configured as a voltage divider (call the R1 and R2) and your ADC (specifically, indicate which node(s) your ADC is measuring the voltage at).

Then work out the relationship between Vs, R1, and R2 and the voltage seen by the ADC (call it Vadc).

 

Firstly, draw a voltage divider circuit with a source voltage of 9V.
What resistor R in series with 2.7kΩ will result in an output voltage of 5V?

 

Firstly, draw a voltage divider circuit with a source voltage of 9V.
What resistor R in series with 2.7kΩ will result in an output voltage of 5V?

Depends on which resistor the voltage is taken across. This matters since if it is done the wrong way the circuit will end up producing a MINIMUM voltage of 5V into the ADC.

 

Depends on which resistor the voltage is taken across. This matters since if it is done the wrong way the circuit will end up producing a MINIMUM voltage of 5V into the ADC.

I left that as an exercise for the TS to figure out.

 

Ah. I was trying to do the same. Apologies. We can still let the TS figure out how to deal with the issue.

 

Hi,

I was thinking since there is only one series resistor allowed then it would have to connect to the 9v supply. That way the voltage divider output can be limited to a range of 0 to 5v.

This would not be my preferred method either but we are only allowed one resistor here.