Voltage Divider Circuit

Question

We were doing an experiment on how to get the battery reading. we will be using a resistive voltage divider with a LM741 op-amp and we were wondering what resistor values is the best to use for the r1 and r2 given that our battery source is around 8v.

thank you.

@Soph: what does the circuit look like? E.g. is the op-amp used as a buffer (high input impedance unity gain voltage buffer)? — Peter Mortensen, Oct 10 '10 at 10:22
Why are you still using an LM741? Running it from a single battery source will not work well; it is limited to V- + 1.5V to V+ - 1.5V, i.e. 1.5V to 6.5V with an 8V battery. Use something like an LM358. — Thomas O, Oct 10 '10 at 10:22
educated guesses could be made (how does 10K grab you?), but it's difficult to say anything definitive about resistor values w/o seeing the circuit. — JustJeff, Oct 10 '10 at 11:10
soph, is there something different that you needed then my answer, if you expand I will expand on my answer. Thomas, if it is a buffer for half the battery voltage it will not need to go much towards the limits you noted. — Kortuk, Oct 11 '10 at 03:01

score 4 · Answer 1 · answered Oct 10 '10 at 16:16

Choose what precision you want. Lets say .1V(100mV).

Now you look up your op-amps leakage current. Lets say it is 1.5uA.

For a voltage drop of .01(10mV) from 1.5 uA(.0015mA) you need a resistance of 10kOhms.

If I stay below 10kOhms I cannot get more than a 10mV voltage drop from it. You now know what you upper limit is for the resistance of your divider.

If you need higher precision, use an op-amp with lower leakage current or use a lower resistance. In all reality with a 1.5uA leakage you will probably not know the difference with a 100K resistance, but I always plan for worst case and then test it.

Let me know if there is anything I can do to make this more clear.

Voltage Divider Circuit

1 Answers1