Energy Sources

Ideal Energy Sources:

• An ideal current source provides constant current $I_0$ independent of the voltage across it (determined by the load $R_L$).
• An ideal voltage source provides constant voltage $V_0$ independent of the current through it (determined by the load $R_L$).

Ideal sources do not exist in reality, due to such dilemmas:

• if the two ends of a voltage source is a short circuit ($R=0$), what is the output voltage?
• if the two ends of a current source is an open circuit ($R=\infty$), what is the output current?

Voltage Source:

In reality, all voltage sources (e.g., a battery or a voltage amplifier) have a nonzero internal resistance $R_0>0$ in series with the voltage source that causes the actual output voltage $V$ to be lower than $V_0$, depending on the load current $I$. The load voltage $V$ and the load current $I$ are constrained by the following two relationships imposed by the voltage source $(V_0, R_0)$ and the load $R_L$, respectively:

$$\left\{ \begin{array}{l} V=V_0-IR_0 V=R_LI \end{array} \r... ...ay}{l} I=V/(R_0+R_L) V=V R_L/(R_0+R_L) \end{array} \right.$$

The slope of the first curve is the internal resistance $R_0$ and the slope of the second curve is $R_L$. Solving these two equations we get load voltage $V$ and current $I$.

Only in the case of an ideal voltage source with $R_0=0$ will $V=V_0$. For $R_0>0$, the heavier the load, i.e., the smaller $R_L$, the larger the load current $I$, and the lower the load voltage $V$:

$$R_L \downarrow \Longrightarrow I \uparrow \Longrightarrow R_0I \uparrow \Longrightarrow V \downarrow$$

In particular, when the load is a short circuit ($R_L=0$), the output voltage of a battery is zero (instead of the specified $V_0=1.5V$), as the voltage drop across the nonzero internal resistance $R_0$ is the same as $V_0$, i.e., the electric energy of the battery is consumed internally, with no energy delivered to external circuit.

Current Source:

In reality, all current sources (e.g., a solar cell or a current amplifier) have an internal resistance $R_0<\infty$ in parallel to the current source that causes the actual output current $I$ to be lower than $I_0$, depending on the load voltage $V$. The load voltage $V$ and the load current $I$ are constrained by the following two relationships imposed by the current source $(I_0, R_0)$ and the load $R_L$, respectively:

$$\left\{ \begin{array}{l} I=I_0-V/R_0 I=V/R_L \end{array} ... ... V=I_0R_0R_L/(R_0+R_L) I=I_0R_0/(R_0+R_L)\end{array} \right.$$

The slope of the first curve is the internal resistance $R_0$ and the slope of the second curve is $R_L$. Solving these two equations we get load voltage $V$ and current $I$.

Only in the case of an ideal current source will $R_0=\infty$ and $I=I_0$. For $R_0<\infty$, the larger the load resistance $R_L$, the smaller the current $I_L$.

$$R_L \uparrow \Longrightarrow V \uparrow \Longrightarrow V/R_0 \uparrow \Longrightarrow I \downarrow$$

Equivalent Circuits and Source Transformation:

Two circuits with the same voltage-current relation (V-I characteristics) at the output port are considered equivalent, as they have the same external behavior, although they may be different internally.

A voltage source can be equivalently converted into a current source and vise versa:

• The voltage-current relation of a voltage source ($V_0, R_0$) is:
  
  $$V=V_0-R_0I=R_0(V_0/R_0-I)$$
  
  

• The voltage-current relation of a current source ($I_0, R'_0$) is:
  
  $$V=R'_0(I_0-I)=R'_0I_0-R'_0I$$
  
  

If we let $R_0=R'_0$ and $V_0=R_0I_0$ or $I_0=V_0/R_0$, then the two sources have the same V-I characteristics and they are equivalent, i.e., the voltage and current sources can be equivalently converted to each other by

$$\left\{ \begin{array}{l} R_0=R'_0 I_0=V_0/R_0 \;\;\;\mbox{or}\;\;\; V_0=I_0 R_0 \end{array} \right.$$

Finding internal resistance:

As shown above, the internal resistance $R_0$ of voltage source is the slope of the straight line $V=V_0-IR_0$, and similarly, the internal resistance $R_0$ of the current source is the slope of the straight line $I=I_0-V/R_0$. In both cases this slope can be obtained as the ratio of the open-circuit voltage (when $I=0$) and the short-circuit current (when $V=0$):

$$R_0=\frac{\mbox{open-circuit voltage}}{\mbox{short-circuit current}} =\frac{V_{oc}}{I_{sc}}$$

For a voltage source with $(V_0, R_0)$, the open-circuit voltage is $V_{oc}=V_0$ and the short-circuit current is $I_{sc}=V_0/R_0$ and their ratio is $R_0$. For a current source with $(I_0, R_0)$, the open-circuit voltage is $V_{oc}=I_0R_0$ and the short-circuit current is $I_{sc}=I_0$ and their ratio is $R_0$.

Using this method, one can determine the internal resistance $R_0$ of a given energy source theoretically. However, it may not be practical to do so experimentally, as the short circuit current is difficult to obtain without causing damage of the source (a battery, or a signal generator). In this case, we can find the voltage $V_i$ and current $I_i$ ($i=1,2$) associated with each of two load resistors $R_1$ and $R_2$. The internal resistance $R_0$ can then be found as the slope of the straight line determined by the two points $(V_1, I_1)$ and $(V_2, I_2)$.

Example 1:

A voltage source of $V_0=10V$ and $R_0=10\Omega$ can be converted to a current source of $I_0=V_0/R_0=1A$ with the same $R_0$ (and vice versa). For a load of $R_L=40\Omega$, both energy sources will provide the load current $I_L=0.2A$ and load voltage $V_L=8V$. We see that this is a good voltage source but a poor current source, as its $R_0$ is too low.

Example 2:

A current source of $I_0=1 mA$ and $R_0=8 K\Omega$ can be converted to a voltage source of $V_0=I_0 R_0=8V$ with the same $R_0$ (and vice versa). For a load of $R_L=2 K\Omega$, both energy sources will provide the load current $I_L=0.8 mA$ and load voltage $V_L=1.6V$. We see that this is a good current source but a poor voltage source, as its $R_0$ is too high.

Power Delivery/Absorption

• Voltage source
  • If the direction of the current $I$ in a circuit is such that it goes internally through a voltage source $V$ from its low potential (-) to high (+), and externally through the rest of the circuit from high to low, then the polarities of the voltage and the current are consistent (both positive or negative depending on the assumed polarity), and the source is delivering power $W=VI$.
  • If the direction of the current is reversed, then the power delivered $W=-VI$ will be negative, i.e., the voltage source is actually receiving power. A typical example is the rechargeable battery in your car that works in either of the two states.
• Current source
  • If the polarity of the voltage $V$ across a current source is such that the head of the arrow of the current source is at high potential (+) and the tail of the arrow is at low potential (-), then the polarity of the voltage and the current is consistent, and the current source is delivering power $W=VI$.
  • If the direction of the current is reversed, then the power delivered $W=-VI$ will be negative, i.e., the current source is actually receiving power.

Example 3:

In the circuit shown, the ideal current source is $I_0=1A$, and the ideal voltage source is $V_0=2V$, the resistor is $R=3\Omega$. Find the current through and voltage across each of the three components. Find the power delivered, absorbed, or dissipated by each of the three components.

The current source provides $I_0=1A$ current through the left branch (upward), while the voltage source provides $V_0=2V$ across all three components. The current through $R$ is $2V/3\Omega=2/3A$ (downward), the current through the voltage source is $I_v=1-2/3=1/3A$. Therefore

• Power dissipated by $R$ is $P_R=V_0^2/R=4/3W$,
• Power received by voltage source is $P=V_0I_V=2/3W$.
• Power delivered by current source is $P=V_0I_0=2W$ $(2/3+4/3=2)$.

(Homework) Redo the above with the polarity of $V_0$ reversed. Find:

• Power dissipated by $R$:
• Power received/delivered by voltage source:
• Power delivered/received by current source:

Verify your results so that total power delivered is equal to total power received and dissipated.

Answer

Comment: While various voltage sources such as batteries are very common in everyday life, current sources do not seem to be widely available. One type of current source is solar-cell, which generates current proportional to the intensity of the incoming light. Also, certain transistor circuits are designed to output constant current. Moreover, as discussed above, any voltage source can be converted into a current source. For example, a current source with $I_0=1$ mA and $R_0=1\;M\Omega$ can be implemented by a voltage source of $V_0=1000\; V$ in series with $R_0=1\;M\Omega$.