Block 07 — Power-relevant figures

Power and Characteristics in Diagrams

The simulation in Abbildung 2 shows the following:

• The circuit with linear voltage source ($U_0$ and $R_\rm i$), and a resistive load $R_\rm L$.
• A simulated wattmeter, where the ammeter is implemented by a measuring resistor $R_\rm S$ (English: shunt) and a voltage measurement for $U_\rm S$. The power is then: $P_\rm L = {{1}\over{R_\rm S}}\cdot U_\rm S \cdot U_\rm L$.
• in the oscilloscope section (below).
  • On the left is the power $P_\rm L$ plotted against time in a graph.
  • On the right is the already-known current-voltage diagram of the current values.
• The slider load resistance $R_\rm L$, with which the value of the load resistance $R_\rm L$ can be changed.

Now try to vary the value of the load resistance $R_\rm L$ (slider) in the simulation so that the maximum power is achieved. Which resistance value is set?

Abbildung 3 shows three diagrams:

• Diagram top: current-voltage diagram of a linear voltage source.
• Diagram in the middle: source power $P_\rm S$ and consumer power $P_\rm L$ versus delivered voltage $U_\rm L$.
• Diagram below: Reference quantities over delivered voltage $U_\rm L$.

The two powers are defined as follows:

• source power: $\, \, \large{ P_\rm S = U_0 \cdot I_\rm L} $
• consumer power: $\large{ P_\rm L = U_\rm L \cdot I_\rm L} $

1. Both power $P_\rm S$ and $P_\rm L$ are equal to 0 without current flow. The source power becomes maximum, at maximum current flow, that is when the load resistance $R_\rm L=0$. In this case, all the power flows out through the internal resistor. The efficiency drops to $0~\%$. This is the case, for example, with a battery shorted by a wire.
2. If the load resistance becomes just as large as the internal resistance $R_\rm L=R_\rm i$, the result is a voltage divider where the load voltage becomes just half the open circuit voltage: $U_\rm L = {{1}\over{2}}\cdot U_{\rm OC}$. On the other hand, the current is also half the short-circuit current $I_\rm L=I_{\rm SC}$, since the resistance at the ideal voltage source is twice that in the short-circuit case.
3. If the load resistance becomes high impedance $R_{\rm L}\rightarrow\infty$, less and less current flows, but more and more voltage drops across the load. Thus, the efficiency increases and approaches $100~\%$ for $R_{\rm L}\rightarrow\infty$.

The whole context can be investigated in this Simulation with a resistor or this one with a variable load.

The Efficiency

To understand the lower diagram in Abbildung 3, the definition equations of the two reference quantities shall be described here again:

The efficiency $\eta$ describes the delivered power (consumer power) concerning the supplied power (power of the ideal source): \begin{align*} \eta = {{P_{\rm out}}\over{P_{\rm in}}} = {{R_{\rm L}\cdot I_{\rm L}^2}\over{(R_{\rm L}+R_{\rm i}) \cdot I_{\rm L}^2}} \quad \rightarrow \quad \boxed{ \eta = {{R_{\rm L}} \over {R_{\rm L}+R_{\rm i}}} } \end{align*}

Once we want to get the relative maximum power out of a system (so maximum power related to the input power) the efficiency should go towards $\eta \rightarrow 100\%$. This situation close to (1.) in Abbildung 3.

Application:

1. In power engineering $\eta \rightarrow 100\%$ is often desired: We want the maximum power output with the lowest losses at the internal resistance of the source. Thus, the internal resistance of the source should be low compared to the load $R_{\rm L} \gg R_{\rm i} $.

The utilization rate $\varepsilon$ describes the delivered power $P_{\rm out}$ concerning the maximum possible power $P_{\rm in, max}$ of the ideal source. Here, the currently supplied power is not assumed (as in the case of efficiency), but the best possible power of the ideal source, i.e. in the short-circuit case: \begin{align*} \varepsilon = {{P_{\rm out}}\over{P_{\rm in, max}}} = {{R_{\rm L} \cdot I_{\rm L}^2}\over{{U_0^2}\over{R_i}}} = {{R_{\rm L}\cdot R_{\rm i} \cdot I_{\rm L}^2}\over {U_0^2}} = {{R_{\rm L}\cdot R_{\rm i} \cdot \left({{U_0}\over {R_{\rm L}+R_{\rm i}}}\right)^2} \over {U_0^2}} \quad \rightarrow \quad \boxed{\varepsilon = {{R_{\rm L}\cdot R_{\rm i}} \over {(R_{\rm L}+R_{\rm i})^2}} = {{R_{\rm L}} \over {(R_{\rm L}+R_{\rm i})}}\cdot {{R_{\rm i}} \over {(R_{\rm L}+R_{\rm i})}}} \end{align*}

The Utilization Rate

In other applications, the absolute maximum power has to be taken from the source, without consideration of the losses via the internal resistance. This corresponds to the situation (2.) in Abbildung 3. For this purpose, the internal resistance of the source and the load are matched. This case is called impedance matching (the impedance is up to for DC circuits equal to the resistance). The utilization rate here becomes maximum: $\varepsilon = 25~\%$.

Application:

1. In communications engineering the impedance matching of the source (the antenna) and the load (the signal-acquiring microcontroller) uses resistors, capacitors, and inductors. There, we want to get the maximum power out of an antenna. For this purpose, the internal resistance of the source (e.g., a receiver) and the load (e.g., the downstream evaluation) are matched. An example can be seen in this application note for near field communication.
2. Furthermore, also for photovoltaic cells one wants to get the maximum power out. In this case, the concept is often called Maximum Power Point Tracking (MPPT)

The impedance matching/power matching is also here explained in a German video.

Power-flow chains (series stages)

The usable (= outgoing) $P_{\rm O}$ power of a real system is always smaller than the supplied (incoming) power $P_{\rm I}$. This is due to the fact, that there are additional losses in reality.
The difference is called power loss $P_{\rm loss}$. It is thus valid:

$P_{\rm I} = P_{\rm O} + P_{\rm loss}$

Instead of the power loss $P_{\rm loss}$, the efficiency $\eta$ is often given:

$\boxed{\eta = {{P_{\rm O}}\over{P_{\rm I}}}\overset{!}{<} 1}$

For cascaded conversions (cf. Abbildung 4), the overall efficiency is the product of stage efficiencies:

$\boxed{\eta = {{P_{\rm O}}\over{P_{\rm I}}} = {\not{P_{1}}\over{P_{\rm I}}}\cdot {\not{P_{2}}\over \not{P_{1}}}\cdot {{P_{\rm O}}\over \not{P_{2}}} = \eta_1 \cdot \eta_2 \cdot \eta_3}$

Exercises

Exercise E3.3.3 Power of two pole components

Two heater resistors (both with $R_\rm L = 0.5 ~\Omega$) shall be supplied with two lithium-ion-batteries (both with $U_{\rm S} = 3.3 ~\rm V$, $R_{\rm i} = 0.1 ~\Omega$).

1. What are the possible ways to connect these components?

Solution

2. Which circuit can provide the maximum power $P_{\rm L ~max}$ at the loads?

Solution

At the maximum utilization rate $\varepsilon = 0.25$ the maximum power $P_{\rm L ~max}$ can be achieved.
The utilization rate is given as: \begin{align*} \varepsilon &= {{P_{\rm out}}\over{P_{\rm in, max}}} \\ &= {{R_{\rm L}\cdot R_{\rm i}} \over {(R_{\rm L}+R_{\rm i})^2}} \\ \end{align*}

As near the resulting equivalent internal resistance approaches the resulting equivalent load resistance, as higher the utilization rate $\varepsilon$ will be.
Therefore, a series configuration of the batteries ($2 R_{\rm i} = 0.2~\Omega$) and a parallel configuration of the load (${{1}\over{2}} R_{\rm L}= 0.25~\Omega$) will have the highest output.

Result

The following configuration has the maximum output power.

3. What is the value of the maximum power $P_{\rm L ~max}$?

Solution

The maximum utilization rate is: \begin{align*} \varepsilon &= {{{{1}\over{2}} R_{\rm L} \cdot 2 R_{\rm i} } \over { ({{1}\over{2}} R_{\rm L} + 2 R_{\rm i} )^2}} \\ &= { {0.25 ~\Omega \cdot 0.2 ~\Omega } \over { ( 0.25 ~\Omega + 0.2 ~\Omega )^2}} \\ &= 24.6~\% \end{align*}

Therefore, the maximum power is: \begin{align*} \varepsilon &= {{P_{\rm out}}\over{P_{\rm in, max}}} \\ \rightarrow P_{\rm out} &= \varepsilon \cdot P_{\rm in, max} \\ &= \varepsilon \cdot {{U_s^2}\over{R_{\rm i}}} \\ &= 24.6~\% \cdot {{(3.3~\rm V)^2}\over{0.1~\Omega}} \\ \end{align*}

Result

\begin{align*} P_{\rm out} = 26.8 W \end{align*}

4. Which circuit has the highest efficiency?

Solution

The highest efficiency $\eta$ is given when the output power compared to the input power is minimal.
A parallel configuration of the batteries (${{1}\over{2}} R_{\rm i} = 0.05~\Omega$) and a series configuration of the load ($2 R_{\rm L}= 1.0~\Omega$) will have the highest efficiency.

Result

5. What is the value of the highest efficiency?

Solution

The efficiency $\eta$ is given as: \begin{align*} \eta &= { {2 R_{\rm L} }\over{ 2 R_{\rm L}+ {{1}\over{2}} R_{\rm i} }} \\ &= { { 1.0~\Omega }\over{ 1.0~\Omega + 0.05~\Omega }} \end{align*}

Result

\begin{align*} \eta = 95.2~\% \end{align*}

Detailed Comparison

Summary

Common pitfalls checklist

1. Forgetting units in intermediate results (always write $x = \text{number} \times \text{unit}$).
2. Mixing up goals: high $\eta$ vs. high $P_{\rm L}$ lead to different $R_{\rm L}$.
3. Using ideal source formulas for a real source (always include $R_{\rm i}$).
4. Ignoring the sign convention when interpreting $P=U\cdot I$ (source vs. load).