Peter Wheeler: Continuing on, we've covered off Ohm's law equals IR and Kirchhoff's voltage
law, which is the sum of the voltages around a closed loop equals zero.
We've also mentioned the diode, in this case here a silicon diode, which has a 0.7 volt
drop which then in a circuit you can accommodate using Kirchhoff's voltage laws to understand
what are the voltages at other points in the circuit.
Another configuration that we often run into with circuit design is ... We call it a voltage
divider.
We use the voltage divider to create a voltage at some point in a circuit so that we can
establish a bias point, an operating point around which the rest of the circuit works.
Let's go back to our battery design here again.
A voltage divider looks like this.
The essence of it is what is ... This is our volt meter.
What is the voltage at this point that we're trying to set up.
Let's just clean this circuit up so it is truly closed, so the electrons don't have
trouble transferring across through the circuit.
Let's take a voltage.
How about 10 volts.
Let's take two resistors.
How about 8k Ohms and in this case here R2, I could make this 12k Ohms, just for numbers.
Let's call this resistor two and this resistor one.
Okay, so we know that looking at this, 12 and 8 is 20 k.
Again, Ohm's law, I equals V over R. If we have 10 volts divided by 20,000 Ohms, what
we have is .5 milliamps of current in this circuit.
Now, if that is the case then what is the voltage here?
Well let's write some equations down.
We know that V source is equal to I times R1 plus I times R2.
Okay, the voltage that drops across the entire circuit is that.
We could've constructed Kirchhoff's laws and equal to zero and take the 10 across like
that, but basically that's the numbers.
Let's just clean that up a little bit.
That's I times brackets R1 plus R2.
In this case here, let me isolate my “I”, my current, it's equal to V source so we're
taking this from multiplying to this side where you would dividing over R1 plus R2.
Okay.
Now, what is this voltage here which we're trying to establish?
Basically, it is the voltage at R2 is equal to the current through here, same current
flows through this entire circuit, is the current through R2 times R2.
That's that.
We have this equation.
We have I here and we have I here.
Why don't I substitute this I for this I?
What I've got is VR2 equals VS over R1 plus R2 times R2.
This is the classic ... Well, it's like saying this is the total pie.
I'm taking a piece of that pie.
This is the formula that people use all the time, rather than calculating currents, resistances,
voltage drops, subtract the voltage drop across here from the source.
You can just use this formula.
VR2 equals the source voltage times this particular resistor R2 divided by R1 plus R2.
Please keep in mind, brackets are important because if you don't use them on your calculator
you will probably make an error in the order.
Remember BEDMAS, brackets, exponents, multiply, divide, add, subtract, in the order of your
calculations.
Again, just voltage divider and this is going to be used all the way up to your final semester
in your designs.
It's simply this resistance over the total resistance times the source voltage will give
the voltage across that particular point.
Thanks.