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Next: Review and Summary Up: Chapter 1: Basic Quantities Previous: Energy Sources

Source and Load

Example I, Maximize Voltage Delivery

Sometimes there is the need to concatenate two circuits in series. As some simple example, the source can be a battery and the load can be just a resistor, or the source is a signal (waveform) generator that can produce various waveforms (sinusoids, saw tooth, square wave, etc.) and the load can be an oscilloscope to display such waveforms. Alternatively, as a more sophisticated example, two voltage amplifiers can be cascaded together so that the output of the first one, the source, is treated as the input of the second one, the load, for the purpose of amplifying some very weak signals. Note that the amplifier circuit is active in the sense it contains a voltage source that depends on the input voltage $v_{in}$ (across the input impedance), e.g., $v_{out}=Av_{in}$, with $A>1$ being the voltage gain.

cascade.gif

To maximize the output voltage, it is important to consider both the input and output impedances (resistances for now) of the circuits. The output impedance $R_{out}$ of the first circuit and the input impedance $R_{in}$ of the second circuit form a voltage divider. For the second circuit to receive maximum voltage from the first one, we want

Example II, Maximize power delivery

powermatch.gif

Sometimes we want to maximize the power delivered from the voltage source to the load:

\begin{displaymath}P_L=I^2 R_L=\frac{V_0^2}{(R_0+R_L)^2} R_L \end{displaymath}

where $R_0$ is the internal resistance of the voltage source. For example, the power amplifier of a stereo system needs deliver the maximum power to the speakers as the load. As the delivered power is a function of the load resistance $R_L$, we let

\begin{displaymath}\frac{d}{dR_L} P_L(R_L)=0
=V_0^2\frac{(R_0+R_L)^2-2R_L(R_0+R_L)}{(R_0+R_L)^4}
=V_0^2\frac{R_0-R_L}{(R_0+R_L)^3}
\end{displaymath}

and get $R_L=R_0$, i.e., when the load is equal to the internal resistance, also called the output resistance, of the voltage source, the power it receives is maximal

\begin{displaymath}P_L=\frac{V_0^2}{(R_0+R_L)^2} R_L\vert _{R_L=R_0}=\frac{V_0^2}{4R_0} \end{displaymath}

In this case, the load current is $I=V_0/2R_0$, and the total power delivered by the voltage source $V_0$ is

\begin{displaymath}P_0=V_0 I=\frac{V_0^2}{2R_0}=2P_L \end{displaymath}

i.e., the internal resistance $R_0$ consumes the same amount of power as the load $R_L$.

maxpower.gif

The efficiency of the circuit is defined as the ratio of the power delivered to the load $R_L$ and the power generated by the source $P_0$:

\begin{displaymath}
\eta=\frac{P_L}{P_0}=\frac{I^2 R_L}{I^2 (R_0+R_L)}=\frac{R_L}{R_0+R_L}
\end{displaymath}

When $R_L=R_0$ and the load receives maximal power, but the efficiency is only $50\%$. Obviously, in order to improve the efficiency, we can increase $R_L$ so that $\eta$ approach 1 when $R_L \gg R_0$. But in this case the power received by the load is no longer maximal. For example, if $R_L=2R_0$, the efficiency becomes:

\begin{displaymath}\eta=\frac{R_L}{R_0+R_L}\bigg\vert _{R_L=2R_0}=\frac{2}{3}>\frac{1}{2} \end{displaymath}

but the power received by the load is less than the maximum power:

\begin{displaymath}P_L=I^2R_L=\frac{V_0^2}{(R_0+R_L)^2}R_L=V_0^2\frac{2R_0}{(R_0+2R_0)^2}
=\frac{2V_0^2}{9R_0} < \frac{V_0^2}{4R_0} \end{displaymath}

In some applications with small power, efficiency can be sacrificed to maximize the load power. For example, in an audio system, it is important for the speaker's impedance to match the output impedance of the power amplification circuit, so that the speaker can receive maximum power.

Example III, Minimize loss in power transmission line:

powerdelivery.gif

In power transmission network, it is more desirable to have high efficiency to avoid wasting energy than delivering maximal power. The power loss along the power transmission line between the power plants and the consumers should be minimized.

We assume the resistance of the power transmission line is $R_T$ and the total load resistance of the power consumers is $R_L$. We also assume the voltage on the consumer's side of the power line is $V_L$.

As the transmission resistance $R_T$ is fixed (minimized) and the power consumption $P_L$ is to be guaranteed, to minimize the power loss $P_T$ along the transmission line we can only increase the voltage $V_L$. For example, when the voltage is increased 10 times, the power loss will be reduced to 1/100.

Summary: The circuits in the three examples above are essentially the same, i.e., they all have a voltage source $V_0$ with an internal resistance $R_0$ (or $R_T$), and a load resistance $R_L$. However, the circuit will be optimized differently according to different requirements:

source_load.gif


next up previous
Next: Review and Summary Up: Chapter 1: Basic Quantities Previous: Energy Sources
Ruye Wang 2014-09-17