Electrical – Setting the minimum value of a potentiometer using a resistor

The main point is current.

Take a look at this circuit. Hover your mouse pointer over the ground symbol and you'll see that the current is 25 mA. Now take a look at this circuit and you'll see that the output current is \$ 2.5 \mbox{ } \mu A \$.

Now let's see how the circuits behave under load. Here's the first circuit with load. As you can see, is a 2.38 mA current going through the load resistor on the right and the voltage on it is no longer the expected 2.5 V but instead 2.38 V (because the two bottom resistors are in parallel). If we take a look at the second circuit here, we'll see that now the top resistor drops around whole 5 V while the two bottom resistors have voltage of 4.99 mV. That is because the resistor ratio have been changed here. Since the two bottom resistors are in parallel now, and we have one resistor with significantly larger resistance than the other, their combined resistance is negligible compared to the resistance of just the bottom right resistor (you can check that using parallel resistor formulas). So now the voltage output is significantly different from the 2.5 V we get in case of no-load condition.

Now let's take a look at opposite situation: Two small resistors in voltage divider and one large as load here. Again the combined resistance of the two lower resistors is smaller than the resistance of the smaller resistor of the two. In this case however this doesn't make a big impact on the voltage seen by the load. It still has the voltage of 2.5 V and everything is fine so far.

So the point is when determining the resistance of the resistors, we should take into account the input resistance of the load and the two voltage divider resistors should be as small as possible.

On the other hand, let's compare the current going through the divider in the circuit with large resistors on the divider and the circuit with small resistors on the divider. As you can see, the large resistors have current of just \$2.5 \mbox{ }\mu A\$ going through them and the small resistors have current of 25 mA. The point here is that the current is wasted by the voltage divider and if this was for example part of a battery operated device, it would have a negative impact on the battery life. So the resistors should be as large as possible in order to lower the wasted current.

This gives us two opposite requirements of having as small as possible resistors to get better voltage regulation at the output and as large as possible resistors to get as small as possible wasted current. So to get the correct value, we should see which voltage we need on the load, how precise it needs to be and get the load's input resistance and based on that calculate the size of the resistors we need to get to have the load with acceptable voltage. Then we need to experiment with higher voltage divider resistor values and see how the voltage will be affected by them and find the point where we can't have greater voltage variation depending on the input resistance. At that point, we (in general) have good choice of voltage divider resistors.

Another point that needs to be considered is the power rating of the resistors. This goes in favor of resistors with larger resistance because resistors with lower resistance will dissipate more power and heat up more. That means that they will need to be larger (and usually more expensive) than resistors with larger resistance.

In practice, once you do a number of voltage dividers, you'll see few popular values for the voltage divider resistors. Many people just pick one of them and don't bother too much with calculations, unless there is a problem with the choice. For example for smaller loads, you can pick resistors in the \$100 \mbox{ } k \Omega\$ range while for bigger loads you can use \$10 \mbox{ } k \Omega\$ or even \$1 \mbox{ } k \Omega\$ resistors, if you have enough current to spare.

A simple battery backup IC will do what you want.

Essentially, a comparator samples two voltage inputs, the main power, and a backup power supply (Batteries). Based on that, two cascading inverters paired with two p-channel fets power the circuit at Vo, while two n-channel fets are used as open-drain outputs. Pull them up with a resistor, and they are the perfect interrupt source for your arduino.

The specific one shown is the ICL7673, but it's maximum current is 38mA. It's designed for rtc and ram with a coincell, but an extra pair of P-channels can be used to increase the current.

There are other battery backup or power switches which will do higher current builtin.