Alright so you're working on a project,
and you want to measure a battery. Say like, this 9V.
Well if you're using something like an Arduino, such as this one, you can't simplly take and put the 9V directly in.
You'll blow the input! [buzzer]
So what can you do?
Well you can use two resistors as a voltage divider. I can use a voltage divider.
Which is what we'll talk about on this AddOhms video.
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Let’s start with a quick Ohm Law’s overview, using this resistor.
Adding a battery creates a very simple circuit.
If we give the resistor 10 ohms of resistance, (the Omega Symbol means Ohms), and make
the battery 10 volts, Ohm’s Law tells us how much current will flow through the circuit.
According to Mr. Georg (Yorg) Ohm, current is equal to voltage divided by resistance.
In this case that’s the battery’s 10 volts is divided by the resistor’s 10ohms, giving
us 1 amp of current.
So….
What happens when we add another 10ohm resistor and call it R2?
This creates a "Series Circuit."
Think about this circuit from Electron’s Point of View.
We’ve just doubled the amount of resistance all of the electrons leaving the battery sees.
Now our current calculation becomes 10V divided by 20ohms, for a total current of 500mA.
What about the voltage of R1 and R2?
They can’t both be 10V [buzzer] and here’s why.
To find the voltage drop of each, we can use Ohm’s Law.
Let’s look at R1 first.
We know it is 10ohms with this 500mA of current flowing through it.
Ohm’s law says the voltage drop will be resistance times the current, which comes
out to be 5V.
Since R2 has the same 500mA of current flowing through it,
it's also going to have a voltage drop of 5V too.
See, each electron has to go from the battery, through R1 and then through R2, back to the
battery.
Notice that the voltage drops of the components add up to the original source voltage.
In a series circuit, components have the same current flowing through them, while their
voltage drops add up.
Okay.
New scenario.
What if we change the value of R2 to something else?
Like say, 100 ohms?
Since we are changing 1 element of our circuit, we are going to start over.
Let’s start with total circuit current, which is now 10V divided by R1 plus R2, which
is now 110ohms.
Giving a current of about 91mA.
Voltage of R1 is 10ohms times 91mA which is 0.91V or 910mV.
Voltage of R2 is 100ohms times 91mA which is 9.1V.
See what has happened?
By making R2 10 times bigger than R1, the voltage drop of R2 is also 10 times bigger
than R1!
We’ve divided the voltage of our battery.
There is a problem with using these resistor values.
We've been focusing on voltage, let's take a look at the current.
At 10 volts and 91mA, we are wasting 0.91 watts — almost 1W — of power.
It would make much more sense to use values like 10K or 100K for a divider, which reduces
that wasted power.
Want a faster way to calculate a voltage divider?
Check this out.
Vout = R2 divided R1 plus R2, times Vin.
No need to calculate the currents.
For example, let’s say you have a 9V battery, with a 47kOhm resistor and a 6.8kOhm resistor.
It’s okay, I’ll wait while you calculate it. [Clock Ticking]
[James Laughs] Just kidding.
Vout is 6.8kOhm divided 47k plus 6.8k times 9V, for an output of 1.13 volts.
Okay, so we've taken a look at what resistor values go into a voltage divider.
Using this magic formula, you should be able to calculate those values easily.
If you have any questions about voltage dividers, or suggestions about future topics.
Feel free to leave a comment.
In the meantime, follow and subscribe to AddOhms to know when new tutorials are released.
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