Max Power Transfer
Introduction: This article reviews Thevenin and Norton Equivalent and the theory to find the maximum power delivered to a load circuit.
The Essentials
To this point, we know about Thevenin Equivalent Circuits when a circuit has a set of terminals that we are interested in. One important feature of these circuits is calculating the maximum power transferred to the load.
For the purpose of the topic, we'll represent the resistance of some load as \( R_L \). We are curious to know what \( R_L \) value will ensure a maximum power transfer in our Thevenin circuit. After some calculus, which can be found in the textbook, we learn that \( R_{Th} = R_L \) in order for maximum power transfer to happen.
Figure 1. shows a schematic of a maximum power transfer Thevenin circuit.
Figure 1: Max Power Transfer Circuit
To calculate the maximum power in the circuit, we are given this equation.
Example
Let's do an example problem to solidify this concept.
Figure 2: Max Power Transfer Example
Let's say for this problem we are interested in 2 things: the value of \( R_L \) that results in maximum power and the maximum power that can be delivered to \( R_L \).
Now, we recognize that to achieve maximum power, we need to find \( R_{Th} \). That is simply done by combining resistors in the circuit to make an equivalent resistance. So for this circuit, we can combine the parallel resistors to get our \( R_{Th} \), which will be 20.59 \( \Omega \).
The reason why this is is because we need to see the circuit from the perspective of the load. From the perspective of the load, the two resistors are in parallel. Since they are in parallel, we calculate the two resistors as if they are in parallel and not in series.
Now we can calculate the maximum power transferred to the resistor. We also need to find the value of \( V_{Th} \), which is done by finding the voltage applied across the terminal. This can be done by voltage division. The voltage applied across \( R_L \) is the same as the voltage applied to the 35 \( \Omega \) resistor.
\[ V_{Th} = 20\cdot \frac{35}{85} = 8.23 \, \text{V}\]
After solving for the Thevenin voltage, we can then solve for maximum power transfer. Using equation 1, we can find the power.
\[ p_{max} = \frac{V_{Th}^2}{4R_L} = \frac{8.23^2}{4 \cdot 20.59} = 0.822 \, \text{W}\]
One other detail that may be important is to know what percentage of power supplied by the source reaches \( R_L \). Let's first find the voltage across the \( R_L \)
\[ V_{ab} = \frac{V_{th}}{2R_L} \cdot R_L = \frac{8.23}{2 \cdot 20.59} \cdot 20.59 = 4.12 \, \text{V}\]
Now let's find the current through \( R_L \)
\[ i_s = \frac{V_{in}-V_{ab}}{50} = \frac{11.77}{50} = 235 \, \text{mA} \]
Now, to find the max power delivered by the source is the voltage multiplied by the current through it.
\[ p_s = -i_s \cdot V_{in|} = -(0.235) \cdot (20) = 4.7 \, \text{W}\]
Now we can get a percentage of how much power is being transferred to \( R_L \).
\[ \frac{p_{max}}{p_s} \cdot 100 = \frac{0.822}{4.7} \cdot 100 = 17.489\, \%\]
So in order for the circuit to deliver its maximum power, \( R_L = 20.59 \), \( p_{max} = 0.822 \, \text{W} \), and power percentage delivered from the power source to the load is 17.489 \( \% \).
Practice
Problem 1.)
Figure 3: Problem 1
Using Figure 3., find a) the value of \( R_L \), b) the maximum power transfer to \( R_L \), and c) the percentage of power from the source being transferred to \( R_L \).
Problem 2.)
Figure 4: Problem 2
Using Figure 4., find a) the value of \( R_L \), b) the maximum power transfer to \( R_L \), and c) the percentage of power from the source being transferred to \( R_L \).
Problem 3.)
Figure 5: Problem 3
Using Figure 5., find a) the value of \( R_L \) and b) the maximum power transfer to \( R_L \).
Solutions:
Problem 1 Solution.) \( R_L = 3.55 \, \text{k}\Omega \), \( p_{max} = 52.4 \, \text{mW} \), and \( p_{\%} = 14.69 \, \% \). Let's first find the value of \( R_L \). Remember, find \( R_L \) from the perspective of the load.
\[ R_L = R_{Th} = \frac{6500\cdot7800}{6500+7800} = 3.55 \, \text{k}\Omega\]
Now to calculate the maximum power using equation 1. We also need to find \( V_{Th} \).
\[ V_{Th} = \frac{6500}{6500\cdot7800}\cdot60 = 27.273 \, \text{V}\]
\[ p_{max} = \frac{27.273^2}{4\cdot3550} = 52.4 \, \text{mW}\]
Now let's calculate the total power delivered from the source. To do this we need to find the current through the 7.8 k\( \Omega \). We do this by finding the voltage drop across that resistor and then divide by the resistance. Using voltage division, we find the voltage after the resistor to be 13.64 V.
Now let's find the current:
\[ i_s = \frac{60-13.64}{7800} = 5.944 \, \text{mA}\]
\[ p_s = 60 \cdot 5.944 \cdot 10^{-3} = 356.6 \, \text{mW}\]
Now with the power of the source known, we can find the percentage of the power delivered to the load resistance.
\[ p_{\%} = \frac{52.4}{356.5} \cdot 100 = 14.69 \, \% \]
Problem 2 Solution.) \( R_L = 991 \, \Omega \), \( p_{max} = 0.327 \, \text{W} \), and \( p_{\%} = 49.92 \% \). First, let's solve for \( R_{Th} \), which is also \( R_L \).
\[ R_L = R_{Th} = \frac{10000 \cdot 1100}{10000+1100} = 991 \Omega\]
We need to calculate the max power transfer. To do that, we need to find the Thevenin voltage.
\[ V_{Th} = 40 \cdot \frac{10000}{11100} = 36.036 \, \text{V} \]
With this we can calculate the max power transfer using equation 1.
\[ p_{max} = \frac{36^2}{4\cdot991} = 0.327 \, \text{W}\]
Now let's calculate the percentage of power going from the source to the load resistance.
\[ V_{ab} = \frac{36.036}{2 \cdot 991} \cdot 991 = 18.018 \, \text{V}\]
Now that we know the voltage across the terminal, we can find the current through the 1.1k\( \Omega \) resistor.
\[ i_s = \frac{36-18.018}{1100} = 16.35 \, \text{mA}\]
With the current through the source and voltage across, we can find th power supplied.
\[ p_s = 16.35 \cdot 10^{-3} \cdot 40 = 0.655 \, \text{W}\]
Now we can calculate the percentage of power supplied to \( R_L \) to the power supplied by the source.
\[ p_{\%} = \frac{0.327}{0.655} \cdot 100 = 49.94 \, \%\]
Now this power efficiency is much better. While the source doesn't supply a lot of power, the circuit does utilize what it is given by delivering a near ideal amount of power to the load resistor.
Problem 3 Solution.) \( R_L = 5.586 \, \text{k}\Omega \), \( p_{max} = 0.254 \, \text{W} \), and \( p_{\%} = 22.9 \, \% \). Let's solve for \( R_L \)
\[ R_L = R_{Th} = \frac{15000\cdot8900}{15000+8900} = 5.585 \, \text{k}\Omega\]
With that, let's solve for the maximum power transfer through \( R_L \).
\[ V_{Th} = 120\cdot\frac{15000}{15000+8900} = 75.31 \, \text{V}\]
\[ p_{max} = \frac{75.31^2}{4\cdot5586} = 0.254 \, \text{W}\]
Now let's solve for the power through the source.
\[ V_{ab} = 120\cdot\frac{4070}{4070+8900} = 37.66 \, \text{V}\]
Now to solve for the current through the source.
\[ i_s = \frac{120 - 37.66}{8900} = 9.25 \, \text{mA}\]
Now we can solve for the power supplied by the voltage source.
\[ p_s = 120 \cdot 9.25 \cdot 10^{-3} = 1.11 W\]
Now for the power percentage delivered to \( R_L \).
\[ p_{\%} = \frac{0.254}{1.11} \cdot 100 = 22.9 \%\]