The I-V Menagerie

Podcast icon Podcast: I-V Relationships

Kirchhoff's Laws provide the analytical framework for circuit analysis, but they describe only the relationships among voltages and currents at the network level. To fully determine circuit behavior, those laws must be paired with the characteristic relationship between voltage and current for each individual component. This chapter introduces that component-level perspective through the concept of the I-V curve, then examines four fundamental components: the ideal voltage source, the ideal current source, the ideal resistor, and the ideal diode.

Understanding I-V Curves

An I-V curve is a graph that plots current ($I$) on the vertical axis against voltage ($V$) on the horizontal axis for a specific component. By convention, the reference current arrow points into the positive terminal of the device, as shown in the figure below. This is called the passive sign convention and is the standard reference direction used for components whose primary role is to absorb energy.

Reading an I-V curve. Every point on an I-V curve represents one possible operating condition for the component. To interpret any such point, read off the corresponding voltage $V$ and current $I$, then compute the power $P = VI$. If $P > 0$, the component is absorbing energy at that operating point. If $P < 0$, the component is supplying energy. The sign of $P$ is determined entirely by the signs of $V$ and $I$, which is why the I-V plane divides naturally into four quadrants.

The I-V plane with the passive sign convention reference. A vertical axis labeled I and a horizontal axis labeled V divide the plane into four quadrants. A circuit element on the left shows current I flowing into the positive terminal with voltage V across it. The top-right quadrant (I greater than 0, V greater than 0) and bottom-left quadrant (I less than 0, V less than 0) are shaded red, indicating the device absorbs power where P equals VI is greater than 0. The top-left and bottom-right quadrants are shaded green, indicating the device supplies power.

Figure 1: The I-V plane with the passive sign convention reference. Red quadrants (Q1 and Q3) have $P = VI > 0$: the device absorbs power. Green quadrants (Q2 and Q4) have $P = VI < 0$: the device supplies power. A purely passive device, such as a resistor, can only ever operate in the red quadrants.

Caution: The passive sign convention reference direction shown above is used throughout this chapter for the resistor and diode. For the voltage source and current source, the current reference is defined as flowing out of the positive terminal, which is the natural direction of conventional current for a source. The quadrant coloring is therefore not applied to the source figures; instead, the operating regions are labeled directly on the I-V curve to avoid ambiguity.

The four sections that follow examine the I-V relationships of the fundamental circuit components encountered in this course.

Ideal Voltage Source

An ideal voltage source maintains a constant voltage across its terminals regardless of the current flowing through it. This behavior produces a vertical line on the I-V plane at $V = V_1$: the voltage is fixed, while the current can take any value determined by the rest of the circuit.

Conventional current flows out of the positive terminal of a source. With this reference direction, when current is positive the source is doing work on the circuit (delivering power), and when current is negative the circuit is doing work on the source (for example, charging a battery).

I-V curve of an ideal voltage source. A vertical axis labeled I and a horizontal axis labeled V are shown. A circuit element on the left shows current I flowing out of the positive terminal with voltage V equals V-sub-1 across it. A thick vertical blue line at V equals V-sub-1 represents the I-V characteristic, showing that voltage is fixed regardless of current. The upper half is annotated as delivering power and the lower half as absorbing power.

Figure 2: I-V curve of an ideal voltage source (current reference out of positive terminal). The voltage is fixed at $V_1$ for any current. The upper half corresponds to normal source operation (power delivery); the lower half corresponds to power absorption, as occurs when a battery is being charged.

The defining equation for an ideal voltage source is:

$$V = V_1 \qquad \text{(constant)}$$

Example 1: Ideal Voltage Source

An ideal 12 V voltage source is connected to a $6\,\Omega$ resistor. Calculate the current $I_0$ and the power delivered by the source.

Series circuit with a 12-volt battery on the left driving current I-sub-0 rightward along the top wire through a 6-ohm resistor on the right branch. The bottom node is grounded.

Solution

By Ohm's Law, the current through the resistor is:

$$I_0 = \frac{12}{6} = 2\,\text{A}$$

The power delivered by the voltage source is:

$$P = V \cdot I_0 = 12 \times 2 = 24\,\text{W}$$

Current flows out of the positive terminal, so $I_0 > 0$ and the source is operating in the power-delivery region of its I-V curve.

Ideal Current Source

An ideal current source maintains a constant current through it regardless of the voltage across its terminals. This behavior produces a horizontal line on the I-V plane at $I = I_1$: the current is fixed, while the voltage can take any value determined by the rest of the circuit.

As with the voltage source, the current reference is defined as flowing out of the positive terminal. The power delivered by the source therefore depends on the sign of the terminal voltage: when $V > 0$ the source delivers power to the circuit, and when $V < 0$ the circuit drives energy back into the source.

I-V curve of an ideal current source. A vertical axis labeled I and a horizontal axis labeled V are shown. A circuit element on the left shows current I equals I-sub-1 flowing into the positive terminal with voltage V across it. A thick horizontal blue line at I equals I-sub-1 represents the I-V characteristic, showing that current is fixed regardless of voltage. The right half is annotated as delivering power and the left half as absorbing power.

Figure 3: I-V curve of an ideal current source (current reference out of positive terminal). The current is fixed at $I_1$ for any terminal voltage. When $V > 0$ the source delivers power to the circuit (right half of line); when $V < 0$ it absorbs power (left half).

The defining equation for an ideal current source is:

$$I = I_1 \qquad \text{(constant)}$$

Example 2: Ideal Current Source

An ideal 3 A current source is connected to a $4\,\Omega$ resistor. Calculate the voltage across the resistor and the power delivered by the source.

Circuit with a 3-amp current source on the left and a 4-ohm resistor on the right, connected in parallel. The bottom node is grounded.

Solution

The voltage across the resistor equals the terminal voltage of the current source. By Ohm's Law:

$$V = I_1 \cdot R = 3 \times 4 = 12\,\text{V}$$

The power delivered by the current source is:

$$P = V \cdot I_1 = 12 \times 3 = 36\,\text{W}$$

The terminal voltage is positive ($V > 0$), so the current source is operating in the power-delivery region of its I-V curve.

Ideal Resistor

An ideal resistor obeys Ohm's Law: the voltage across the resistor is directly proportional to the current through it, with resistance $R$ (measured in ohms, $\Omega$) as the constant of proportionality. On the I-V plane, this relationship appears as a straight line through the origin with slope $\frac{1}{R}$.

The passive sign convention applies here: current is referenced as flowing into the positive terminal. The I-V line therefore passes through Q1 ($I > 0$, $V > 0$) and Q3 ($I < 0$, $V < 0$), both of which are red (absorbing) quadrants. The resistor never reaches Q2 or Q4, confirming that it cannot supply power.

I-V curve of an ideal resistor using the passive sign convention. A vertical axis labeled I and a horizontal axis labeled V divide the plane into four quadrants. The top-right and bottom-left quadrants are shaded red indicating power absorption. A thick blue diagonal line passes through the red quadrants from bottom-left to top-right with slope equal to 1 over R, showing that a resistor always absorbs power regardless of current direction.

Figure 4: I-V curve of an ideal resistor (passive sign convention). The I-V line passes through Q1 and Q3 only, both of which are red (power-absorbing) quadrants. A resistor always dissipates energy, regardless of current direction.

Ohm's Law takes the forms:

$$V = IR \qquad \text{or equivalently} \qquad I = \frac{V}{R}$$

Several key properties follow directly from the I-V curve. The line passes through the origin: zero voltage produces zero current. The slope equals $\frac{1}{R}$, so a larger resistance gives a shallower slope and less current for the same applied voltage. Because $I$ and $V$ always share the same sign along this line, $P = VI$ is always non-negative: a resistor converts electrical energy to heat in every operating condition and never returns energy to the circuit.

Example 3: Ideal Resistor Behavior

For a $100\,\Omega$ resistor, calculate the current and power when the applied voltage is: (a) $5\,\text{V}$, (b) $-3\,\text{V}$, and (c) $0\,\text{V}$.

Series circuit with a battery V-sub-0 on the left driving current I-sub-0 rightward along the top wire through a 100-ohm resistor on the right branch. The bottom node is grounded.

Solution

(a) $V_0 = 5\,\text{V}$ (operating in Q1):

$$I_0 = \frac{5}{100} = 0.05\,\text{A} = 50\,\text{mA} \qquad P = 5 \times 0.05 = 0.25\,\text{W}$$

(b) $V_0 = -3\,\text{V}$ (operating in Q3):

$$I_0 = \frac{-3}{100} = -0.03\,\text{A} = -30\,\text{mA} \qquad P = (-3) \times (-0.03) = 0.09\,\text{W}$$

Although both $V$ and $I$ are negative, the power is positive. Reversing the voltage simply reverses the current; the resistor still dissipates energy. This is the Q3 red quadrant in action.

(c) $V_0 = 0\,\text{V}$ (at the origin):

$$I_0 = 0\,\text{A} \qquad P = 0\,\text{W}$$

With no applied voltage there is no current and no dissipation, consistent with the I-V line passing through the origin.

Ideal Diode

Unlike the resistor, which has a linear I-V relationship, the diode has a nonlinear characteristic. An ideal diode acts as a one-way valve for current: it conducts perfectly when forward-biased (positive anode-to-cathode voltage) and blocks completely when reverse-biased (negative anode-to-cathode voltage). The schematic symbol for the diode points in the direction of conventional current flow, providing a visual reminder of this directionality.

The passive sign convention applies here, with the current reference into the positive (anode) terminal.

I-V curve of an ideal diode using the passive sign convention. The top-right quadrant is shaded red indicating power absorption. A thick blue line runs along the negative V-axis representing reverse bias where the diode blocks all current. A second thick blue line rises along the positive I-axis representing forward bias where the diode voltage is zero and current is determined by the circuit.

Figure 5: I-V curve of an ideal diode (passive sign convention). In reverse bias ($V < 0$) the diode blocks all current: the operating point sits on the negative $V$-axis. In forward bias the diode voltage is zero and the operating point rises along the $I$-axis; the circuit determines the current. Only Q1 is shaded because the ideal diode operates exclusively there during conduction.

The right-angle shape of the ideal I-V curve is the graphical expression of the diode's switching behavior: the diode is either fully off (operating on the $V$-axis) or fully on (operating on the $I$-axis), with no gradual transition. This ideal behavior is captured mathematically as:

$$V = \begin{cases} \text{any value} \leq 0, & I = 0 \quad \text{(reverse bias, blocking)} \\ 0, & I > 0 \quad \text{(forward bias, conducting)} \end{cases}$$

Real silicon diodes deviate from the ideal model in two important ways. First, a forward voltage drop of approximately 0.6–0.7 V is required before the diode conducts significantly; the turn-on is not instantaneous at $V = 0$. Second, a small reverse leakage current flows even when the diode is nominally blocking, placing the real device slightly into Q3 rather than exactly on the $V$-axis. The figure below compares both characteristics.

Comparison of ideal and real diode I-V curves. The top-right and bottom-left quadrants are shaded red. A dashed blue line represents the ideal diode: running along the negative V-axis in reverse bias then rising vertically along the positive I-axis in forward bias. A solid red line represents the real diode: nearly flat with slight reverse leakage along the negative V-axis, then curving rightward and rising steeply after the forward voltage V-sub-F of approximately 0.7 volts. The ideal curve is labeled Ideal and the real curve is labeled Real.

*Figure 6: Real (solid red) diode I-V characteristics. The ideal diode has a sharp corner at the origin. The real diode requires a forward voltage $V_F \approx 0.7\,\text{V}$ for a silicon diode before conducting significantly, and exhibits a small reverse leakage current (Q3).

Example 4: Ideal Diode in a Circuit

An ideal diode is connected in series with a $2\,\text{k}\Omega$ resistor and a 9 V battery. Calculate the current and the voltage across each component for (a) the diode connected in forward bias and (b) the diode reversed.

Series circuit with a 9-volt battery on the left driving current I rightward. On the right branch, a diode is on top in series with a 2 kilohm resistor below it, both oriented vertically.

Solution

(a) Forward bias:
The diode anode connects to the positive battery terminal. An ideal forward-biased diode acts as a short circuit: it carries any current with zero voltage across it.

$$V_\text{diode} = 0\,\text{V}$$

The full battery voltage appears across the resistor:

$$V_\text{resistor} = 9 - 0 = 9\,\text{V}$$

The circuit current is:

$$I = \frac{9}{2000} = 4.5\,\text{mA}$$

(b) Reverse bias:
With the diode reversed, it acts as an open circuit and blocks all current:

$$I = 0\,\text{A}$$

No current flows, so no voltage drop appears across the resistor. The entire battery voltage appears in reverse across the diode:

$$V_\text{resistor} = 0\,\text{V} \qquad V_\text{diode} = -9\,\text{V}$$

Note that the full supply voltage is borne by the diode in this case. For a real diode, the reverse breakdown voltage is a critical specification that must not be exceeded.

Comparing I-V Relationships

The figure below places all four ideal I-V curves on the same axes. The shape of each curve is a direct graphical expression of what the component controls.

I-V characteristic curves of four fundamental components on a common set of axes. A blue vertical line represents a voltage source with fixed voltage V. A red horizontal line represents a current source with fixed current I. A green diagonal line through the origin represents a resistor with current proportional to voltage. A magenta L-shaped curve represents an ideal diode: running along the negative V-axis in reverse bias then rising vertically along the positive I-axis in forward bias. Each curve is labeled with its component name.

Figure 7: I-V characteristics of the four fundamental components on a common set of axes. The shape of each curve encodes the component behavior: a vertical line for the voltage source (fixed $V$), a horizontal line for the current source (fixed $I$), a line through the origin for the resistor (proportional $V$ and $I$), and an L-shape for the ideal diode (unidirectional conduction).

A voltage source fixes $V$ and allows $I$ to be determined by the circuit, producing a vertical line. A current source fixes $I$ and allows $V$ to be determined by the circuit, producing a horizontal line. These two components are duals of each other: each controls one quantity and leaves the other free. A resistor imposes a proportional relationship between the two quantities, producing a line through the origin whose slope encodes the conductance $\frac{1}{R}$. A diode enforces directionality: it operates on the $V$-axis in reverse bias and on the $I$-axis in forward bias, producing the characteristic L-shape.

The table below collects the key properties of each component.

Comparing I-V Relationships
Component Fixed Free I-V shape Power role
Ideal voltage source $V = V_1$ $I$ Vertical line Source or load
Ideal current source $I = I_1$ $V$ Horizontal line Source or load
Ideal resistor $V/I = R$ — Line through origin Always absorbs
Ideal diode Direction of $I$ — L-shape Forward only

Real Components and Non-Ideal Behavior

The ideal models developed in this chapter are indispensable for building intuition and for first-order analysis, but real components deviate from ideal behavior in predictable ways that become apparent during laboratory measurements.

A real voltage source has an internal series resistance that causes its terminal voltage to drop as the load current increases; the I-V curve tilts away from vertical. This effect was examined in Chapter 3 (Secrets of DC Circuits) in the context of battery performance. A real current source has a finite parallel resistance that allows the delivered current to vary with terminal voltage; the I-V curve tilts away from horizontal. A real resistor can exhibit nonlinear behavior at extreme voltages or temperatures, and its tolerance band means the actual resistance differs slightly from its nominal value, as encountered in the Lab 2 measurements. A real diode requires a forward voltage drop of approximately 0.6–0.7 V for silicon devices before significant conduction begins, and it passes a small reverse leakage current when nominally blocking, as shown in Figure 6 above.

Recognizing these deviations is essential for connecting ideal analysis to measured results, and it is a recurring theme throughout the associated laboratory work.

Chapter Summary

This chapter introduced the I-V curve as a graphical tool for characterizing component behavior. Every point on an I-V curve represents one operating condition; the power at that point is $P = VI$. The sign of $P$ depends on the sign convention used: the passive sign convention (current into the positive terminal) applies to passive components such as resistors and diodes, while the active sign convention (current out of the positive terminal) applies to sources.

The I-V curve of an ideal voltage source is a vertical line at $V = V_1$: voltage is fixed, current is free. The ideal current source produces a horizontal line at $I = I_1$: current is fixed, voltage is free. These two components are duals. The ideal resistor produces a straight line through the origin with slope $\frac{1}{R}$; it operates exclusively in the power-absorbing quadrants and never supplies energy. The ideal diode produces an L-shaped curve: it conducts with zero voltage drop in forward bias and blocks all current in reverse bias.

Combining these component I-V relationships with Kirchhoff's Laws from Chapter 3 provides a complete toolkit for DC circuit analysis. The following chapters extend these ideas to signals that vary in time.