Vo = Open Circuit Voltage in the Thevenin model, and,
Io is the short circuit current in the Norton model.
This is not Ohm's Law because the voltage and the current are not for a resistor, but are - in fact - the parameters in two different models.
Both circuits satisfy:
Vterminal = RoIo - RoIload = Vo - RoIload
Whenever we have a source of electrical energy we might pose the question of what circumstances produce maximum load power.
Teenagers solve a closely related problem whenever they are able to determine the speaker connections that produce the maximum amount of noise from a stereo system.
Engineers processing radar signals solve the same problem when they extract the maximum power from a low power signal received through an antenna.
In both cases we can think of the system being represented by a Thevenin (or Norton) equivalent circuit with a load connected. The TEC will determine how much power can be extracted from the source.
Consider a load connected to a source, and consider that the source is well represented by a Thevenin equivalent model. Then the situation is the one shown in the figure below.
The question we want to answer is:
What load will dissipate the maximum amount of power when connected to the given source?
We proceed by deriving an analytic expression for the power dissipated by a resistor (load) attached to the source.
First calculate the voltage across the load resistor, calling that voltage Vload.
Vload = Vo Rload/(Ro+ Rload)
Then the power dissipated in Rload is:
(Vload)2 / Rload
= (Vo)2 Rload/(Ro+ Rload)2
For an extremum (maximum or minimum) we differentiate the load power with respect to Rload, and set that derivative to zero. The derivative is:
(Vo)2(Ro- Rload)/(Ro+ Rload)
The maximum power for positive load resistance occurs when:
Rload = Ro
When the load resistance is equal to the internal resistance of the source, the load voltage is exactly half of the open circuit voltage, so the maximum power delivered to the load is:
Pmax = (Vo )2/(4Ro)
Whenever a load resistance is equal to the internal resistance of a source, the two (source and resistance) are said to be matched.
Above we have derived the maximum power theorem as it is usually presented. However, it gives the impression that maximum power is delivered only when a resistive load equal to the internal impedance is attached. That is not the case!
Maximum power is delivered to a load when it drops the terminal voltage to half the open circuit voltage and the load current is half the short circuit current.
However, that is not quite what we proved, so let us look at a more general situation.
Consider the following situation.
You have a source with a relationship between terminal voltage and load current as shown in the graph below.
The load power is given by:
Pload = VloadIload = (Vo - IloadRo)Iload.
Now, differentiate with respect to Iload.
dPload/Iload = Vo - 2IloadRo.
Now solve for the load current that maximizes load power, and we get:
Iload = Vo/2Ro
That's half the short circuit current, and if you put that value back into the equation for terminal voltage, the terminal voltage at max power is half of the open circuit voltage.
To give another perspective, consider that the power to the load - VloadIload- can be thought of as the area of a rectangle at a point on the load curve for the source. At any point on the plot of terminal voltage against load current, the product of the terminal voltage and the load current is the power delivered. That's also the area of a rectangle with a height equal to the terminal voltage and a width equal to the load current. Thus, we can look at the rectangle under the load curve for a series of points and the point with the rectangle with the largest area is the point which delivers the most power to the load.
Click the double arrow green button below to see how that rectangle area varies as the point moves along the load line.
Click the single arrow green button to step the action one step.
Click the single arrow green button pointing left to back up the action once it is started.
Estimate the point at which the area of the rectangle is largest - remembering that the area of the triangle is the power delivered to the load.
Can you see that there is a max?
Not all circuits have a straight line relationship between load current and load voltage. A solar cell might have the load curve shown below.
To try to determine the maximum power that can be extracted from this solar cell is more difficult than when the load curve is a straight line. However, some of the ideas we have used so far may also be used here. For example, the power supplied by this source may still be viewed as the area of a rectangle at a point on the curve. You can check that below.
Here's the solar cell load curve again. Here you can click to see three different operatin point, and you can determine which operating point provides the most load power. Remember, the power supplied will be the product of load voltage and load current, so it is the area under the rectangle that shows. Click each button to see a point on the load line and a rectangle whose area is the power supplied when the solar cell at that point.
Here are a few points to note
Clearly, the max power point occurs near the rounded corner at the top right.
When the max power is delivered, the load voltage is well above half of the open circuit voltage, and the current is nearly the short circuit current.
Q3 Which point provides the most power?
A DC source has an open circuit voltage of 9.6 volts and an internal resistance of 8.0 ohms.
Using TECs In Circuit Analysis
One common circuit is a voltage divider. Here is a voltage divider.
Now consider the following:
A voltage divider is a linear circuit,
Therefore, a voltage divider has a Thevenin Equivalent Circuit.
The unresolved question is how to determine the TEC.
In the process we can learn how to use Thevenin and Norton equivalents to simplify circuits, including source conversions.
In another lesson we discuss how one aspect of being an expert involves the ability to see larger chunks in a situation. The advice there can be applied to circuits with voltage dividers, Thevenin equivalents and Norton equivalents.
Consider the voltage divider circuit shown below. We have considered this circuit earlier in this lesson. Now that we know about TECs, we can see that there is a TEC inside this voltage divider. Click the button below to outline that TEC.
Now, any time we have a Thevenin equivalent circuit we can replace it with a Norton equivalent. That's what we will do below. Recall:
The short circuit current = Open circuit voltage/Internal resistance.
The internal resistance is the same for the Thevenin and the Norton circuits.
You can click the button to initate the transformation from a Thevenin Equivalent to a Norton Equivalent. (And you can click it again to reset back to a TEC.)
Now, if your expert genes are still working you will recognize a combination in the Norton circuit. Q4 Do you see two resistors in series or in parallel?
Now, you can combine the two resistors in the combination. If you do that, you should get the circuit below.
Now, the two resistors, Raand Rb, are in parallel, and they can be combined to give the circuit below.
This is the Norton equivalent circuit for the voltage divider.
If you want the Thevenin equivalent circuit, you can convert the Norton equivalent to a TEC by computing the open circuit voltage.
The open circuit voltage is given by:
Vo = (Vin/Ra) (RaRb/(Ra + Rb))
The result is the TEC shown below.
Notice the equivalent circuit has the following properties.
The open circuit voltage is the voltage we get from the voltage divider formula.
The internal resistance has a formula for a parallel resistance even though the two resistors would appear to be in series in the original voltage divider.
That's all there is to it.
If you can't calculate the TEC, there is always the possibility of measuring the TEC. To measure the TEC you can often measure the open circuit voltage pretty easily. Measuring the internal resistance is a different problem. To do that you almost certainly have to attach a load to the source to cause the terminal voltage to drop. That might be a problem with a source like the plug on the wall, for example. You might not want to draw enough current from the wall plug to drop the voltage enought to measure the difference. Spots in the wiring inside the wall could get hot - possibly enough to start a fire.
Sometimes you need to know about variables inside a source. For example, in the voltage divider, you might want to know how much power is consumed in the resistors in the voltage divider. You can't tell anything about that from the TEC. You have to go back to the actual circuit in that case.
The TEC can't do everything. What it does is enable you to use a simple model to predict how a real source - no matter how complex - interacts with the rest of the world. But it doesn't tell you about what goes on inside the source itself!
