Voltage divider (worst case)

[TsAmE]

A voltage divider with 2 equal 100Ω resistors divides the voltage of a 9V battery, that has an output that from 7-9V depending on its charge state. What are the limits on its output voltage (which is 4.5V) if the resistors have a ±10% tolerance?

I understand voltage division and applying the formula: Vout = Vin * R2 / (R1 + R2), but I am confused with this question.

How would you set it out? Would you say Vin = 7V and both resistors = 100 - 10% for the lower limit? And Vin = 9V and both resistors = 100 + 10% for the upper limit?

 

[keith1200rs]

To get the worst case you need to increase one resistor by 10% and decrease the other. Then do the opposite.

Keith

 

[franciscorc]

Hi
Assuming the resistor connected to the power supply is R1 and the other is R2.
Independent of the power supply, always you will have the maximum voltage in the center tape when R2>R1 and the lowest value when R1>R2.A simple way to see that without maths (most important) is - Imagine that R2 have a very small value (close to ZERO Ohms).What is the voltage in the center tape? It is almost zero or ground.

If you need further information be confortable to ask me any time.

Best Regards

Francisco

 

[kjineesh]

Use a Resistor Array..!

 

[keith1200rs]

Use a Resistor Array..!

I don't think that answers the original question. Also, if the original question was "how do I make an accurate voltage divider" the answer would probably be "buy high tolerance resistors". They are readily available without using resistor arrays. I would think it is probably more difficult to find a 10% resistor!

Keith.