Source and Load

It is often needed to concatenate two circuits in series (cascade), by connecting the output port of the first circuit, considered as the source, to the input port of the second circuit, considered as the load. Here are some examples:

• The source is a voltage or current source, and the load is a resistor.
• The source is a sensor and the load is an amplifier.
• The source is signal (waveform) generator that produces various waveforms (sinusoids, saw tooth, square wave, etc.), and the load is an oscilloscope to display such waveforms.
• Two voltage amplifiers can be cascaded to achieve greater amplification gain to amplify some weak signals.
• One or more filters can be cascaded to filter signals of different frequencies.

In general, an active circuit containing active components (e.g., transistors and operational amplifiers) as well as passive components (e.g., $R$, $C$ and $L$) can be modeled by an input resistance $R_{in}$ and a voltage source $v_{out}$ in series with an output resistance $R_{out}$, where $v_{out}=Av_{in}$ is proportional to the input voltage $v_{in}$ and $A$ is the voltage gain. When two such circuits are cascaded, $R_{out}$ of the first circuit is the internal resistance of the source, and $R_{in}$ of the second circuit is the resistance of the load.

Consider the following three examples:

• Maximize voltage delivery

  The output resistance $R_{out}$ of the course and the input resistance $R_{in}$ of the load form a voltage divider:

  

  For the load to receive maximize voltage, we want

  • the output (internal) resistance $R_0=R_{out}$ of the source circuit to be as small as possible, so that little voltage drop across it will be caused when the load circuit draws a large curren from the source even if its input (load) resistance $R_L=R_{in}$ is small.

  • the input (load) resistance $R_L=R_{in}$ of the load circuit to be as large as possible, so that it draws little current from the source, causing small internal voltage drop across its output resistance $R_0=R_{out}$, even if it is large.

• Maximize power delivery

  Some times we need to maximize the power (instead of the voltage) delivered from the source to the load. For example, the power amplifier of a stereo system needs to deliver high power to the speakers as the load. Given the internal resistance $R_0=R_{out}$ of the source and the load resistance $R_L=R_{in}$, the power received by the load is:

  $$P_L=I^2 R_L=\left(\frac{V_0}{R_0+R_L}\right)^2 R_L =V_0^2\frac{R_L}{(R_0+R_L)^2}$$ (118)

  • Given $R_L$, we can minimize $R_0$ to maximize the delivered power $P_L$:

    $$P_L=V_0^2\frac{R_L}{(R_0+R_L)^2}\stackrel{R_0\rightarrow 0}{\Longrightarrow} \frac{V_0^2}{R_L}$$ (119)

  • On the other hand, given $R_0$, we need to find the optimal value for $R_L$ to maximize $P_L$, by solving the following equation:

    $$\frac{d}{dR_L} P_L(R_L) =V_0^2\frac{(R_0+R_L)^2-2R_L(R_0+R_L)}{(R_0+R_L)^4} =V_0^2\frac{R_0-R_L}{(R_0+R_L)^3}=0$$ (120)

    We see that only when $R_L$ is equal to the internal resistance $R_0$, will it receive maximum power.

  The maximum power received by $R_L=R_0$ is:

  $$P_L=\frac{V_0^2}{(R_0+R_L)^2} R_L\bigg\vert _{R_L=R_0}=\frac{V_0^2}{4R_0}$$ (121)

  the load current is

  $$I=\frac{V_0}{R_0+R_L}=\frac{V_0}{2R_0}$$ (122)

  and the total power delivered by the voltage source $V_0$ is

  $$P_0=V_0 I=\frac{V_0^2}{2R_0}=2P_L$$ (123)

  The internal resistance $R_0=R_L$ consumes the same amount of power as the load $R_L=R_0$. The figure below shows the power $P_L$ consumed by the load resistor $R_L=x$ when $R_0=2$. We see that if $R_L$ is either smaller than or greater than $R_0$, $P_L$ is reduced.

  

  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 $P_0$:

  $$\eta=\frac{P_L}{P_0}=\frac{I^2 R_L}{I^2 (R_0+R_L)}=\frac{R_L}{R_0+R_L}$$ (124)

  When $R_L=R_0$ and the load receives maximal power, but the efficiency is only $50\%$. We can improve the efficiency by increasing $R_L$ so that $\eta$ approaches 1 when $R_L \gg R_0$. But in this case the power received by the load is reduced.

  For example, if $R_L=2R_0$, the efficiency becomes:

  $$\eta=\frac{R_L}{R_0+R_L}\bigg\vert _{R_L=2R_0}=\frac{2}{3}>\frac{1}{2}$$ (125)

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

  $$P_L=I^2R_L=\frac{V_0^2}{(R_0+R_L)^2}R_L=V_0^2\frac{2R_0}{(R_0+2R_0)^2} =\frac{2V_0^2}{9R_0} < \frac{V_0^2}{4R_0}$$ (126)

  In some applications with small power, efficiency can be sacrificed to maximize the load power.

• Minimize loss in power transmission line:

  

  A power transmission line delivers power from power generation (a power plant) to power consumption (e.g., a city). The concern is no longer delivering maximum power (as long as needed power is delivered), but to achieve maximum efficiency in the sense that the power loss along the power transmission line is minimized.

  We denote the resistance of the transmission line by $R_T$ and the total load resistance of the power consumption by $R_L$. Also denote the voltage on the consumer's side by $V_L$. We have

  • Power consumption by the load $R_L$:

    $$P_L=V_L I\;\;\;\;\;\; \;\;\;\;\;\;\; I=\frac{P_L}{V_L}$$ (127)

  • Power dissipation along the transmission line $R_T$:

    $$P_T=R_T I^2=R_T \;\left(\frac{P_L}{V_L}\right)^2=\frac{R_T\,P_L^2}{V_L^2}$$ (128)

  As the transmission resistance $R_T$ is fixed (already 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. In practice, the voltage $V_L$ can be as high as a few hundred $kV$ (69 kV, 115 kV, 230 kV, 500 kV, 765 kV).

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:

• Maximize voltage delivery (or minimize loading 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 $V_L$.