Understanding voltage dividers

[OldWorldBlues]

Hi! I'm working on a project in which a small microcontroller with a max. analog input of 5v takes readings from multiple sensors. I would like to make this input window a bit more versatile, and turned to voltage dividers as a solution. 'Problem is, I'm having a bit of trouble understanding how to work with them. I saw the above diagram while researching the topic, but I'm not sure about exactly how it works. I read that R1 creates a voltage drop, but so does R2. I assume that this means that since the voltage from R2 to ground will be lower than from R1 to ground, the voltage between R1 and R2 will be proportional to the input and such. But then why not just use one resistor? Thanks for any help I can get :)

 

[sophiecentaur]

But then why not just use one resistor?

That's the number one question that people ask. "Why not keep it really simple? "
BUT. If you just put a single resistor in series with the battery + terminal, the voltmeter will pass no current (an ideal voltmeter has infinite resistance) and it will read the same as the battery Volts, whatever value of resistor you use. Using a Voltage Divider circuit, you can get any voltage V_out that you want, between zero and V_in according to the Ratio of the two resistors.
This link shows the maths behind it and even gives you a calculator to use. If you are starting on a career of circuit projects, this formula is one of the first things you need to get a grasp of.

 

[Windadct]

Ideally - V out is only a high impedance, sensing circuit, so no current flows and thus does not impact the reading. ( that is part of it why this can work)

The R1, R2 thing makes it a little confusing but R1 + R2 is constanct, the value of the potentiometer. As a signal is applied Vin, let's say it has a max value of 24 V, you want V out to not exceed 5V. You adjust the potentiometer to limit Vout to 5 V. ( and then you do not adjust it for that application).

Example if this was a 10K potentiometer, and the Vin is a 24V max signal. You want to divide the voltage into two parts.

10K/24V = R2/5V - we would adjust the pot to 2,083 ohms, (R2 ( Term 2 to 3) .

This would be the same as using 2 fixed resistors 2,083 on the bottom, and 7,917 on the top. Since getting exact fixed resistor values is very expensive, using an adjustable resistor is pretty common.

 

[Asymptotic]

A fixed voltage divider uses two resistors in series. V_IN is dropped across both of them, and the divided voltage is dropped across the 'bottom' (R2) resistor. For instance, if R1 = 7K, and R2 = 3K (10K total resistance), and 10V is dropped across the pair, then 3V will be across resistor R2 (measured from the R1/R2 resistor connection point, and circuit common), and 7V will be across R1..

Your drawing shows a potentiometer, which can be thought of as a variable voltage divider where the connection points labeled '3' and '1' are a fixed resistance (here, connected to circuit common, and the voltage source), and '2' (the wiper) would be equivalent to the connection point between R1 and R2 in a fixed divider. When the wiper is adjusted all the way down to terminal '3', zero voltage is between common and the wiper, when the wiper is all the way up to terminal '1', then full voltage is at the wiper. Wiper voltage follows the physical position set between the two extremes of travel. For instance, if the wiper is set exactly between terminal 1 and 3, then wiper voltage from terminal 2 to terminal '3' is half of the voltage between terminal 1 and 3.

 

[OldWorldBlues]

That's the number one question that people ask. "Why not keep it really simple? "
BUT. If you just put a single resistor in series with the battery + terminal, the voltmeter will pass no current (an ideal voltmeter has infinite resistance) and it will read the same as the battery Volts, whatever value of resistor you use. Using a Voltage Divider circuit, you can get any voltage V_out that you want, between zero and V_in according to the Ratio of the two resistors.
This link shows the maths behind it and even gives you a calculator to use. If you are starting on a career of circuit projects, this formula is one of the first things you need to get a grasp of.

Hmm, that's sort of what I was thinking. If you use only one resistor, a voltmeter will just read the potential between the positive and negative. But if you have a resistor between the positive cable of your voltmeter and ground, it will make current flow?

 

[sophiecentaur]

Hmm, that's sort of what I was thinking. If you use only one resistor, a voltmeter will just read the potential between the positive and negative. But if you have a resistor between the positive cable of your voltmeter and ground, it will make current flow?

Yes. Current will flow (I=V/R). This holds for an ideal battery (Voltage Source).
I suggest you look at the link I posted earlier and actually do some calculations. This circuit stuff is very number dependent and arm waving doesn't get one very far. A few numerical examples can work wonders.