When building or designing an electronic circuit, we may need to simplify the circuit into an equivalent or analogous circuit consisting of only a few essential components. To achieve that, we can apply Norton’s theorem or Thevenin’s theorem to the circuit. Where the former simplifies the circuit into an equivalent current source and an equivalent parallel resistor, the latter simplifies it into an equivalent voltage source and an equivalent series resistor.
Both Norton’s and Thevenin’s theorems are considered dual to each other, as the equivalent circuit for one theorem can be converted to the other. The equivalent circuit is like a black box, where the internal workings are disregarded, combined into one, and only the input/output or functions matter.
In this article, we will go over Thevenin’s theorem first, then Norton’s, before looking at how to convert their equivalent circuit. But first, let’s learn about the voltage and current divider rules, as these will become useful shortly.
Voltage and Current Divider Rules
To simplify a circuit, we should first understand two important ideas, which are Ohm’s law and voltage-current dividers, as these are the basis of the two aforementioned theorems. We already have an article on Ohm’s law, so let’s start with the voltage divider.
Voltage Divider
The idea behind the voltage divider is very simple: it distributes the input voltage to each component based on its resistance. In the circuit below, two resistors are connected to the Vin (voltage source), and the Vout is pulled from between the two resistors. We could say that Vout is the same as the voltage across R2.
We know that the voltage across R1 and R2 cannot exceed Vin, so to find the voltage across R2 (Vout), we can use the following formula:
Vout = (R2/R1+R2)Vin
The formula determines the ratio of resistance between R1 and R2, then multiplies it by Vin to get the exact amount of voltage across R2. For example, if Vin = 12V, R1 = 2kΩ, and R2 = 4kΩ, Vout will be as follows:
Vout = (2kΩ/2kΩ+4kΩ)12V = ⅓*12V = 4V
Current Divider
How a current divider works is also straightforward. A current divider essentially splits the input current among its components. The resistance of each component determines the split ratio. This is because electrical current, like water, will follow the path of least resistance. The basic circuit below illustrates a current source and three resistors in parallel. How do we find the current IX?
Basic Current Divider Circuit
To find the value of IX, we can use the following formula:
IX = (RT/RX+RT)IT
RT is the total resistance of all other parallel resistors, whereas IT is the current from the source. RX and IX represent the resistance and current of the resistor under consideration, respectively. Note that RT is derived from resistors in parallel, so we need to calculate it using the parallel circuit rule. In this example, RT = R1 ‖ R2.
For example, suppose that RX = 30kΩ, RT = 12kΩ, and IT = 0.5A, the calculation for IX would have been as follows:
IX = (12kΩ/30kΩ+12kΩ)0.5A
= (12/42)0.5A
IX ≈ 0.143A
Additionally, the current divider is based on Kirchhoff’s current law, which says that the current entering a node is always equal to the current leaving it. In the circuit above, the current that flows through R1 combined with the current that goes to RT is always equal to the current source. In other words, the current that does not flow through R1 will flow to the rest of the resistors and be distributed among them using the same formula.
Note: you may learn more about current divider here
Thevenin’s Theorem
Applying Thevenin’s theorem yields an equivalent circuit consisting of a voltage source Vth and a single resistor Rth in series.
It depicts the circuit from the perspective of a load connected to its output. Let’s walk through the process of converting the example circuit below into an equivalent Thevenin circuit.
This example circuit is pretty straightforward. It consists of a 20V voltage source connecting to three resistors arranged in a T-shape. The resistance values of these resistors are 20Ω, 30Ω, and 10Ω for R1, R2, and R3, respectively.
The first step is to determine the Thevenin equivalent voltage Vth for this circuit. Vth represents the voltage measured between output A and ground. Currently, R1 and R2 function as a voltage divider, and R3 is the output. Therefore, we can use the voltage divider formula from earlier:
Vth = (R2/R1+R2)Vs
Vth = (20/50)20V
Vth = 8V
You may notice that R3 is not part of the calculation; this is because to find VTH, we assume the output is an open circuit, which means that there is no current over R3, and as a result, no voltage as well. So the voltage at both ends of R3 is equal to 8V.
The next step is to find the Thevenin equivalent resistance, or Rth. We begin by replacing the voltage source Vs with a straight connection. This is because Rth represents the resistance observed by a hypothetical load connected across outputs A and B of the circuit. Or if the circuit under consideration contains a current source, we will have to replace it with an open circuit.
Now we calculate the Rth of the circuit. Notice that R1 and R2 are effectively in parallel. This means that the overall resistance of this circuit is equal to R3 plus R1 and R2 in parallel.
Rth = R3 + R1 ‖ R2
Rth = 10Ω + (20Ω*30Ω/20Ω+30Ω)
Rth = 10Ω + 12Ω = 22Ω
After calculating both Vth and Rth, we can now construct the Thevenin equivalent circuit. As shown above, the circuit includes an 8V Thevenin voltage source and a 22Ω Thevenin resistor in series with the output A. From the perspective of a load, this circuit is no different from the original circuit we started with.
Norton’s Theorem
Applying Norton’s theorem will result in an equivalent circuit with a current source, Ino, and one resistor, Rno, parallel to the current source.
However, the resulting Norton circuit does not contradict Thevenin’s theorem; it is just another way to view and describe a circuit. Let’s use the same example circuit we previously used for Thevenin’s theorem and try to convert it into a Norton equivalent circuit.
We will begin with the same circuit as before, and now we need to determine the Norton equivalent current, or INO, of this circuit.
The Ino represents the current flowing from output A to B in the case of a closed circuit; in other words, the output current. However, we must first determine the total current flowing out of the voltage source, IT. We will simply use Ohm’s law:
IT = VS / RT
IT = 20V / 20Ω + 30Ω ‖ 10Ω
= 20V / 27.5Ω
IT ≈ 0.727A
VS is the voltage from the source, and RT is the total resistance from the perspective of IT. Note that the resulting IT has a very long decimal number, so for the sake of simplicity, we will round it up to the closest three digits.
Now that we know the approximate total current, we can calculate the Ino. Notice that the circuit performs the same function as a current divider, where part of the total current flows through R2, while another part flows through R3. Therefore, we can apply the current divider formula as follows:
Ino = (R2 / R2 + R3) * It
Ino = (30Ω / 40Ω) * 0.727A
Ino ≈ 0.545A
The Norton equivalent resistance, or Rno, is the circuit’s total resistance observed from the load. It is identical to the Thevenin equivalent resistance; this means that we can apply the same process we did previously. Start by replacing the voltage source with a straight connection—or, in the case of a current source, with an open circuit—then use the following formula:
Rno = R3 + R1 ‖ R2
Rno = 10Ω + (20Ω*30Ω/20Ω+30Ω) = 22Ω
We can now assemble the final Norton equivalent circuit using the Ino and Rno we learned. The circuit consisted of a Norton equivalent current source Ino (0.545A), and a parallel Norton equivalent resistor Rno (22Ω). Although it differs from Thevenin’s equivalent circuit, both the Norton and the Thevenin equivalent circuits retain the output characteristics of the original circuit. As a result, we can create a Thevenin circuit based on the values of a Norton circuit and vice versa.
Conversion of Equivalent Circuits
To convert the equivalent circuit between the two theorems, we simply do so following these guidelines:
Rth = Rno
Vth = Ino * Rno
Ino = Vth / Rth
But of course, we would also have to change the position of the equivalent resistor.
Conclusion
We thought that Norton’s theorem and Thevenin’s theorem would only be applicable in basic-level electronics class exams; you might be thinking the same as well. However, in reality, when doing electronics, even at the high level, we still cannot avoid these basic class exam principles.
Knowing these basic principles can be very helpful; for instance, when designing an amplifier circuit, we usually have to calculate the input and output impedances in order to match them. Learning about these things is quite necessary. We try to give examples or introduce them in an easy way.
We hope that this article is useful for you. In the future, we will try to introduce more basic electrical principles. If possible, we will try to make it clearer and easier to understand. Unfortunately, our old meter is broken; we are currently looking for a solution. Thank you for following.
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