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.

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

• The output impedance $R_{out}$ of the first circuit to be small, so that little voltage drop across it will be caused even if the second circuit draws a large current due to its small input impedance;
• The input impedance $R_{in}$ of the second circuit to be large, so that it draws little current from the first circuit, causing small voltage drop across its output impedance.

Example II, Maximize power delivery

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

$$P_L=I^2 R_L=\frac{V_0^2}{(R_0+R_L)^2} R_L$$

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

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

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

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

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

$$P_0=V_0 I=\frac{V_0^2}{2R_0}=2P_L$$

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

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 $P_0$:

$$\eta=\frac{P_L}{P_0}=\frac{I^2 R_L}{I^2 (R_0+R_L)} =\frac{R_L}{R_0+R_L}\bigg\vert _{R_L=R_0}=\frac{1}{2}$$

We see that when $R_L=R_0$, the load receives maximal power but the efficiency is only $50\%$. The efficiency will approach 1 when $R_L \gg R_0$, but in this case the power received by the load is not maximal. For example, if $R_L=2R_0$, the efficiency becomes:

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

but the power received by the load is

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

which is less than the maximum power.

In power network, it is more desirable to have high efficiency to avoid wasting energy than delivering maximal power. But in communication network and other 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:

The power loss along the power transmission line between the power plant and the consumers should be minimized. Assume the resistance of the power line is $R_T$ and the consumer load is $R_L$. Also assume the voltage on the consumer's side of the power line is $V_L$.

• Power consumption by the load: $P_L=V_L I$, i.e., $I=P_L/V_L$
• Power dissipation along the transmission line: $P_T=R_T I^2=R_T P_L^2/V_L^2$

For fixed transmission resistance $R_T$ and power consumption $P_L$ by the consumer, the power loss along the transmission line $P_T$ can be minimized only by increasing the voltage $V_L$. For example, when the voltage is doubled, the power loss is reduced four times.

Summary: The circuits in the three examples above are essentially the same, 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:

• Maximize voltage delivery (or minimize load effect): minimize output resistance $R_0$, maximize input resistance $R_L$.
• Maximize power delivery ($50\%$ of total power): match the output and input resistances $R_L=R_0$
• Minimize power dissipation in transmission: increase output voltage $V_L$.

Example 0 (Homework)

A realistic voltage source (e.g., a battery) can be modeled as an ideal voltage source $V_s$ in series with an internal resistance $R_s$. Ideally, the voltage $V_s$ can be obtained by measuring the open-circuit voltage $V_{oc}$ with a voltmeter

$$V_s=V_{oc}$$

while the internal resistance $R_s$ can be obtained as the ratio of the open-circuit voltage $V_{oc}$ to the shirt-circuit current $I_{sc}$, which can be measured by an ammeter:

$$R_s=\frac{V_{oc}}{I_{sc}}$$

However, in reality, any voltmeter has an internal resistance $r_v$ in parallel with the meter, any ammeter has an internal resistance $r_a$ in series with the meter. For better measurement accuracy, should $r_v$ be small or large, how about $r_a$? Why? Give the expression of the measured open-circuit voltage $V$ and short-circuit current $I$ in terms of the true $R_s$ and $V_s$, as well as $r_v$ and $r_a$.

Design a method to obtain the true source voltage $V_s$ and source resistance $R_s$ by a voltmeter and an ammeter with known $r_v$ and $r_a$. Give the expression of $V_s$ and $R_s$ in terms of the measured open-circuit voltage $V$, short-circuit $I$, and $r_v$ and $r_a$.

Now Assume $V_s=6V, R_s=100\Omega, r_v=10,000 \Omega, r_a=200 \Omega$ what are the measured open-circuit voltage $V$, and the short-circuit current $I$? Given $V$, $I$, and the known $r_v$ and $r_a$, how do your get the true $V_s$ and $R_s$ using your method above? Show your numerical computations.

Example 1 (Homework)

Usually the internal resistances of the voltmeter and ammeter are not readily known (and the values may change depending on the scale used). As another method to find $R_s$ and $V_s$ of a voltage source, we can measure the voltage $V_L$ across two different load resistors $R_L$ connected to the voltage source. If the values of $R_L$ are significantly smaller than that of the internal resistance $r_v$ of the voltmeter, the voltmeter can be considered to be ideal with $r_v\rightarrow \infty$.

Assume when the load resistor is $R_L=1 k\Omega$, the voltage across it is found to be $V_L=9.1V$, then a different load $R_L=2 k\Omega$ is used and the voltage across it is $V_L=9.52V$. Find $V_s$ and $R_s$ of the voltage source.

Answer