Thevenin Equivalent Circuits - Why Are They Used?
What Does A TEC Explain?
What Is A TEC?
Finding A TEC
You are at: Elements - Sources - Thevinin EquivalentsWhat Is A Norton Equivalent Circuit?The Maximum Power TheoremProblems
Whenever you need to predict how something is going to behave you don't need to analyze things down to the lowest possible level. For example, when current flows in a resistor, you don't need to know what happens to every atom in the resistor. That ability to describe what happens to a large number of atoms in the resistor by using a macromodel for the resistor is convenient. Electrical engineers often think at different levels of complexity.
What do you want to know about Thevenin and Norton equivalent circuits.
When you want to use different components you often need to use macromodels of the components you use. You do that because you don't always need to have totally detailed knowledge of what goes on inside the component.
Thevenin Equivalent Circuits - or TECs - are macromodels that are used to model electrical sources. Those sources are as diverse as batteries, stereo amplifiers and microwave transmitters. In this lesson we will develop TEC models of sources and learn how to use them in larger circuits.
A Thevenin Equivalent Circuit is used to explain some of the things that happen when you use sources. One good example is what happens if you start a car with your headlights on. If you have ever done that you probably noticed that the lights of the car dim when you start the car with them on.
Obviously, something happens that causes the voltage across the headlights to change. The battery is not an ideal voltage source since that dimming occurs whenever the starter motor draws a lot of current from the battery. If the battery were an ideal source, the voltage would never change and the headlights would never dim. In this lesson, we will examine Thevenin Equivalent Circuit - TEC - models for nonideal sources. Those models can be used to predict the sort of voltage decrease that the battery exhibits under heavy load current.
Does a TEC explain this phenomenon? Actually, nothing can explain this phenomenon because it doesn't exist. It violates the laws of physics - i.e. light travels in a straight line. However, although the light beam doesn't droop, the voltage of a nonideal source, like a battery, often droops when current is drawn - the car lights don't droop! They just get dimmer!
The Thevenin Equivalent Circuit is an electrical model composed of two components shown below.
Another important feature of the TEC is that is can explain a drooping terminal voltage. That's inherent in the electrical model itself.
We can write equations that describe the behavior of the TEC when it interacts with other components. First, let us define some variables. If we have a load attached to the terminals some load current will flow. We'll define a load current and a terminal voltage for the TEC as shown below.
Now, write the equation for the circuit. Notice that there will be a voltage across the internal resistance if load current flows.
Using the expression for the terminal voltage, we can get the plot of terminal voltage against load current.
The first interesting property is that the source, V_{o}, is the voltage that is measured when no load is attached. That's called the open circuit voltage. If you just attach a voltmeter to the output terminals - and didn't attach anything else, the voltmeter woud read the value of V_{o}.
Another interesting property is that the there is a definite limit to how much current this source can supply to a load. If we short the terminals - something you can do in your mind, but not often in practice - the current that will flow is given by:
Short circuit current = V_{o}/R_{o}= I_{o}.
Notice the interesting relationship between open circuit votlage, V_{o}, internal resistance, R_{o}, and the short circuit current, I_{o}. It looks like Ohm's Law but it really isn't! It doesn't relate current through a resistor to voltage across a resistor.
P1 A
source has an open circuit voltage of 24 volts and an internal resistance
of 600 W.
Determine the voltage at the terminals of this source when a 1200 W
load resistor is attached to the source.
Q1 If you attach a resistor to a source with a TEC, will the voltage drop below the open-circuit voltage or can it rise above the open-circuit voltage?
P2
A source has a terminal voltage of 10 v when 20 A is drawn from the source.
The open circuit voltage is 12 v. What is the internal resistance
of the source?
The TEC is a useful way of reducing complexity. If you have a complex circuit interacting with other circuits you don't want to look at all of the details of what is taking place inside the complex circuit - at least you don't often want that.
To illustrate some of the power of the TEC we'll get the TEC for a voltage divider.
The voltage divider isn't a very complex circuit, but it is more complex than the TEC. It can be represented with a TEC, and wherever you find a voltage divider you can replace it with the equivalent TEC. We need to establish values and/or formulas for two items in the voltage divider. Basically, we need two of the following:
Next, focus on theShort Circuit Current. Short circuit current, I_{o} is the current that would flow if the output of the circuit is short circuited. We can imagine shorting the Voltage Divider and shorting the TEC. Wires that short them are shown below and the short circuit current, I_{sc}, is indicated.
For the TEC you should be able to see quickly that the short circuit current is just
V_{o} / R_{o} . For the voltage divider it's not any more difficult. There the short circuit current is just V_{s} / R_{b}. We can use both of these observations to compute the internal resistance of the voltage divider TEC.
Calculate the Short Circuit Current.
I_{o} = V_{o} / R_{o}
I_{o} = V_{s} / R_{b}
Combining we have
Continuing, we find:
R_{o}= V_{o} / (V_{s} / R_{b} ) = [ V_{s} ( R_{a}/( R_{b} + R_{a}))] / (V_{s} / R_{b} )
= R_{a}R_{b} /( R_{b} + R_{a})
So, we have the internal resistance, and it looks like the parallel formula. (But it really isn't!)
At this point you can compute both the open circuit voltge and the internal resistance for the voltage divider, so you have a TEC for the voltage divider. You can use that TEC any place you find the voltage divider. It doesn't matter where you find the voltage divider, you can replace it with its TEC, just like you can replace two parallel resistors by their parallel equivalent.
P3 A thirty (30) volt source experiences a .5 volt drop when a load drawing 0.9 amperes is attached to the source. Determine the Thevenin equivalent model of the source.
First determine the open circuit voltage, V_{o}, in this TEC representation.
P4 Next,
determine the internal resistance, R_{o}, in the TEC representation.
P5 A source has a Thevenin equivalent circuit with:
Q2 You have a battery with an open-circuit voltage of 12.6 volts, and an internal resistance of .05 W. Is there a .05 W resistor inside the battery?
P6 When 200 amperes is drawn from Bob White's car battery (when he uses the starter), the headlights dim and the voltage across the battery drops from 12 volts to 10 volts. Determine the Thevenin equivalent model of Bob's battery.
First determine the open circuit voltage, V_{o}, in this TEC representation.
P7 Next, determine the internal resistance, R_{o}, in the TEC representation.
In electrical engineering there is a recurring duality. Voltage and current are duals, and where we find one variable appearing in an expression or a theorem, we usually find the other appearing in a dual expression or theorem. The dual of the Thevenin equivalent model is the Norton equivalent model, a sample of which is given in the figure below.
The relation between terminal voltage and load current for this circuit is interesting because it is indistinguishable from that of the Thevenin Circuit, and the two are electrically equivalent. Write KirchHoff's Current Law for the Norton circuit with a load attached. The result is:
V_{terminal}/R_{o} + I_{load} = I_{o}
So, we find:
If we identify R_{o}I_{o} with the V_{o} that appears in the Thevenin model, this is, as claimed, equivalent to the Thevenin model.
Note the following relations between the Thevenin and Norton equivalents.
P8 A thirty (30) volt source experiences a .5 volt drop when a load drawing 0.9 amperes is attached to the source. Determine the Thevenin equivalent model of the source.
First determine the short circuit voltage, I_{o}, in this NEC representation.
P9 Next,
determine the internal resistance, R_{o}, in the NEC representation.
Whenever we have a source of electrical energy we might pose the question of what circumstances produce maximum load power.
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:
The maximum power for positive load resistance occurs when:
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:
P_{max} = (V_{o} )^{2}/(4R_{o})
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!
The load power is given by:
P_{load} = V_{load}I_{load} = (V_{o} - I_{load}R_{o})I_{load}.
Now, differentiate with respect to I_{load}.
dP_{load}/I_{load} = V_{o} - 2I_{load}R_{o}.
Now solve for the load current that maximizes load power, and we get:
I_{load} = V_{o}/2R_{o}
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 - V_{load}I_{load}- 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.
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
A DC source has an open circuit voltage of 9.6 volts and an internal resistance of 8.0 ohms.
P10 What
resistance will dissipate maximum power when attached to the source?
P11 How much power will be dissipated for the resistance that dissipates maximum power?
P12 A source has a Norton equivalent model with a 200 milliampere source and a 3.2 ohm internal resistance.What load resistance dissipates maximum power?
P13
What is the load voltage and load power for the load resistance found above?
One common circuit is a voltage divider. Here is a voltage divider.
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:
Now, if your expert genes are still working you will recognize a combination in the Norton circuit.
Now, the two resistors, R_{a}and R_{b}, are in parallel, and they can be combined to give the circuit below.
Notice the equivalent circuit has the following properties.
In this voltage divider,
P15 Next,
determine the internal resistance, R_{o}, in the TEC representation.
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.
What If You Need To Know About Things Inside The Source?
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!