Peter Wheeler: We went through a circuit where we applied Ohm's Law, calculating voltage,
and current, and ... Well, we were given the resistance.
That's fine.
There was gentleman then, Gustav Robert Kirchhoff, who looked at what Ohm did, and he added his
interpretation to some of the circuit analysis that we're talking about today.
What he said was, let's go back to our circuit.
We had that three-resistor circuit.
Let me come down here.
All right.
We had 12 volts, we had one 1k Ohm, we had 2k Ohm, and we had 3k Ohm.
As we remember, across here, we had 2 volts, because we had 2 milliamps going through the
circuit.
Something you may notice that I do, is I do identify polarities when I draw my circuits.
Why?
When current enters a component, a passive component like a resistor, there is a plus
side, a high side, and a low side to it.
Identifying the high sides and low sides becomes helpful in some other circuit types of calculations
we do.
In this case here, let's just go back to the numbers we had.
We had 2 volts across here.
We had 4 volts across here, and here, we had 6 volts.
What Kirchhoff said was first of all, you need a closed loop, so that we have current
flowing.
In a closed loop circuit, we can note that the sum of the voltages around the loop equal
zero.
Let me just use some abbreviated terminology.
The sum, sigma, of the vdrops around the closed loop equals zero.
Let's see if that's true.
Here, we have R1, the voltage across R1, was 2 volts.
If I'm following the current through, so I would say plus, plus the voltage here, which
was 4 volts, plus the voltage here, which is 6 volts, minus, see?
Minus the voltage here, which was 12 volts, equals to zero.
Does that add up?
This is 6, 4, 10 is 12 volts equals other sign, reverse the sign, 12 volts.
Of course, they're equal.
That's what an equation is.
This is an interesting situation with circuits, that you can figure out then, using Ohm's
Law to maybe find a current, what the voltages are at a certain point in a particular circuit.
Let's look at this again.
I'm going to give you another circuit, and we'll go around the circuit, using this Kirchhoff
Law of Voltages, KVL is sometimes how we abbreviate.
We'll analyze the voltage drops around the circuit, and see where we stand.
We talked about Ohm, George Simon.
We've talked about Gustav Robert Kirchhoff.
I want to recall that voltage equals current times resistance, from Mr. Ohm, and the sum
of the voltage drops around a closed-loop circuit, where current is flowing, is equal
to zero, also known as Kirchhoff's Voltage Laws.
Let's take an example of Kirchhoff's Voltage Laws, where we may have to do a little bit
of work.
I'm going to use a circuit that's, let's say, a resistor like this.
I'm going to introduce a device called a diode, and another resistor in the circuit.
Again, I want a reference point, which is my ground.
Let's use 12 volts again, that's fine, plus is minus here.
This is going to be plus/minus, this will be plus/minus, as the current goes around
the circuit, plus/minus.
The current will go this way in the circuit.
I'm showing that as I. Let's try 4,000 Ohms for this one, 5,000 Ohms
for this one.
Now, this device, which may be new to you, maybe not, is a diode.
A diode is two semiconductor pieces of material, P&N junctions, generally.
The nature of the energy levels in the two materials is such that it's as if you had
a very small, reverse-bias battery in this diode, like this.
That particular battery is fighting this flow.
If it's a silicon diode, that battery voltage is 0.7 volts.
You'll come into this kind of configuration in your later design courses, but right now,
let's just say there's a 0.7 volt drop across a diode, when current flows through it.
Let's use Kirchhoff's Voltage Laws.
Just going to go back here.
This is resistor 1, this is resistor 2.
This is diode 1, this is our source voltage.
If we look at this, the voltage ... He said the sum of the voltages around a closed loop
is equal to zero.
We should have the voltage from resistor 1, plus the 0.7, remember plus/minus, plus/minus,
minus.
Plus the 0.7 volts, plus the voltage drop across resistor 2, minus the 12 volts, will
equal zero.
Actually, you know what?
I'm just going to change the numbers, just for calculation purposes.
That's okay.
It's not a big deal.
What we'll do is, we're going to change this to, let me just use 7 volts.
It'll make the math a little easier.
This is a 7-volt source now.
Again, VR1, diode drop, VR2, minus the source voltage, 7 volts, equal zero.
Let's just collect terms.
Here, we have VR1, VR2, and on the other side, there will be plus 7, minus 0.7, 6.3 volts.
What is “I” in this circuit?
Back to Ohm's Law, I is equal to V over R. In this case here, we have 6.3 volts, and
the total resistance is 4k plus 5k, which is 9,000 Ohms.
Anybody?
0.7 volts.
Correct?
Is my math good?
Looks pretty good there.
We have 0.7 volts ... I'm sorry, you didn't correct me.
It's 0.7, we're talking about current.
0.7 amps, 700 milliamps.
0.7 milliamps.
All right, reasonable.
That's 0.7 times 10 to the minus 3.
That's good.
Now, we have a current flow.
According to Kirchhoff then, we have this voltage, which happens to be I times R1, plus
the 0.7, plus this I times R2.
It was equal to 7 volts here.
I times R1 plus R2 is equal to 6.3 volts, as we had there.
If you do this math, which we did over there, we came up to 0.7 milliamps.
What is the voltage here, at this particular point?
Let's solve this particular equation here.
The voltage is I times R1.
R1 is .7 times that, so it should be 2.8 volts.
R2 should be 5 times 0.7, which is 3.5 volts.
Does that equal 6.3?
Let's see.
2.8, 5, 6.3 volts.
Kirchhoff is correct.
The sum of the voltage drops, the 2.8, the 0.7, and the 3.5, minus the 7, equal zero.
Or, if we rearrange the equation, we now have our numbers.
What's the significance of all that?
Let's say that you did want to know, at this point, if I put a volt meter here, what is
that voltage, let's see.
We dropped 2.8 volts here, so that means 7 minus 2.8, which is 4.2 volts.
You will come across a number of circuits which will have some kind of configuration
here, where you have a resistance, you have a voltage drop, due to a semiconductor device,
and you will have another resistor.
You need to figure out what is the voltage at that particular point?
Or, what is the current through that particular resistor in this particular circuit.
Kirchhoff is your solution to that.