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Understanding Current-Voltage Curves

This article discusses I-V curves for passive components, voltage sources, and current sources.

This article discusses I-V curves for passive components, voltage sources, and current sources.

The current (I) - voltage (V) relationship of electrical components can often provide insight into how electronic devices are used. More specifically, many non-linear devices such as diodes and transistors are used inoperating regionsin which they behave like ideal components—such as current sources, voltage regulators, and resistors.

An understanding of I-V curves often provides insight into knowing how the device operates and helps us know how to operate a device in a way that enables the required functionality.

We'll begin by looking at how to obtain an I-V curve for any component.

获得I-V曲线

Method 1: Voltage Sweeps

The current-voltage (i-v) relationship for a device is a current measured for a given voltage. For devices that do not supply power, I-V curves are obtained by using linear voltage sweeps. Voltage sweeps involve the linear variation of the voltage, to obtain the corresponding measured output current. Because it is impossible to physically sweep through all the voltages in an instant, it is important to understand that these measurements are made with respect to time as well.

Figure 1.1 illustrates the translation of voltage sweep with respect to time (v vs t) onto the X-axis of the current-voltage graph (我vs v). It is important to understand that thev vs t信息隐含在我vs vcurve. The notion of time is relevant for components that respond to a change in voltage (such as a capacitor) rather than the instantaneous voltage (as with a resistor).

Figure 1.1(a):线性扫描电压(V)对time (t);(b):电流(i) - 电压(v)曲线中的相应电压扫描。

If you have a device that supplies voltage or current, such as a battery or a solar panel or a regular power supply, you cannot change the voltage across the device, because there is a specific voltage or current being generated by the device. For these devices, I-V curves are obtained by load switching.

Method 2: Load Switching

Load switchingisa method that involves measuring the current supplied by the power source for varying load resistance. The load is the device that electrical power is being delivered to, where power is defined as $$P = V \times I$$.

通常,将电阻用作测量电流或电压源传递的功率的负载,因为它们是线性设备,它们没有表现出磁滞性能,即电阻器的操作不取决于其先前的状态。由于设备可以以较小的电阻(1-10Ω)以及较大的电阻值(10-1000kΩ)的速度运行,因此电阻在对数上变化,即从10到100到1000到1000,等等。

The current that is supplied by the power supply is measured by an ammeter for each value of load resistance—shown in Figure 1.2(b)—and the voltage across the load is measured using a voltmeter—shown in Figure 1.2(c).

Figure 1.2(a)Schematic circuit for load switching I-V curve measurements; shown here is an example of an ideal voltage source. The value of RLis varied over a large range and for each value of resistance the voltage(b)and current(c)are measured.

不te that in Figure 1.2(c), the (base 10) logarithmic value of current decreases linearly. This is because a resistor is governed by Ohm's law, and this is an ideal voltage source at a fixed voltage VS; as the resistance value increases logarithmically, the current value decreases logarithmically. Load switching techniques are used to measure the I-V characteristics of devices and circuits that supply power, such as voltage regulator circuits, solar cells, and batteries.

理想组件的I-V曲线

使用线性电压扫描和负载开关,我们will now look at the I-V curves of ideal components. In general, if the device requires power to operate, the voltage sweep method is used.

On the other hand, if the device acts as a source of power, the load switching method is used. Based on their basic definitions, we can derive the I-V curves of ideal passive components (resistors, capacitors, and inductors) using the concept of linear voltage sweeps. We will use the concept of load switching for the I-V curves of anidealvoltage source and anideal当前来源。

Ideal Resistor

Let's start with one of the more familiar ideal components: the resistor.

The resistor is a component that represents a linearrelationship between voltage and currentas dictated by Ohm's law, i.e., $$V=I \times R$$. The graphical representation on the I-V curve of the Ohm's law equation is a straight line passing through the origin, as shown in Figure 2.

Figure 2.The I-V curve of an ideal resistor is a straight line that passes through the origin.

Ideal Voltage Source

An ideal voltage source is a component that can provide a fixed voltage regardless of the current delivered to the load.

For example, let's say a voltage source supplying 10V is connected across a resistor. If the value of the resistor is $$10k\Omega$$, then the current drawn by the resistor from the voltage supply will be dictated by Ohm's law, which is $$ I = \frac{V}{R} = \frac{10V}{10k\Omega} = 1mA$$. If the value of the resistor is 1Ω, then the current drawn will be 10A!

A real voltage supply is limited with respect to the amount of current it can supply for a given voltage, but an ideal voltage source is not. Therefore, the I-V curve for an ideal voltage supply will be a straight line parallel to the Y-axis (see Figure 3). An empirical I-V curve for a real voltage source would be obtained using the load-switching method.

Figure 3.理想电压源的I-V曲线是平行于电流(i)轴的直线,即无论通过设备的电流如何,电压都不会改变。

Zener二极管是一种非线性的被动装置,当以反向偏置操作时,它用作电压调节器。反向偏置的齐纳二极管的(理想化的)I-V曲线表明,它保持在特定电压(由制造过程确定),而不管电流通过它如何。我们将在未来文章中查看Zener二极管的I-V曲线。

理想的当前来源

An ideal current source is a component that can provide a fixed current regardless of the voltage across the component, itself. In other words, a 5A ideal current source would deliver exactly 5A to a 1Ω load resistor or to a 1 kΩ resistor, even though the second resistor would generate a voltage drop of 5000V! This is highly impractical but, nonetheless, ideal current sources are useful tools in circuit analysis.

The I-V curve for an ideal current source is a straight line parallel to the X-axis (see Figure 4). An empirical I-V curve for a real current source would be obtained using the load-switching method.

图4。The I-V curve of an ideal current source is a straight line parallel to the voltage axis; i.e., the current flowing from the source is the same regardless of the voltage across it.

Though "current supplies" are not nearly as common as voltage supplies, many analog transistor circuits are biased using aconstant-current source. Also, a MOSFET operating in the saturation region exhibits behavior similar to that of a (voltage-controlled) current source.

Ideal Capacitor

In a resistor, the voltage is determined by the resistance and the current flowing through the resistor. Capacitors and inductors are fundamentally different in that their current-voltage relationships involve the rate of change. In the case of a capacitor, the current through the capacitor at any given moment is the product of capacitance and therate of change(i.e., the derivative with respect to time) of the voltage across the capacitor.

$$ i = c \ cdot \ frac {\ text {d} v} {\ text {d} t} $$

Because we are using a linear voltage sweep, the current through the capacitor is constant when the voltage is increasing or decreasing. When the voltage changes from a positive slope (shown in blue in Figure 5) to a negative slope (orange), the direction of the current reverses; this is represented in the current vs. time plot as a change from the positive-current section of the graph to the negative-current section of the graph.

Figure 5 (a)Linear voltage sweep and(b)the corresponding capacitor current vs. time.

The I-V relationship of an ideal capacitor is shown in Figure 6. The magnitude of the current is constant, but two horizontal lines are needed because the direction of the current changes depending on whether the voltage is moving from V1 to V2 or V2 to V1. When the voltage has a positive rate of change, the current is positive (indicated by the blue arrowhead); when the voltage has a negative rate of change, the current is negative (indicated by the orange arrowhead).

Figure 6.i-vcurve for an ideal capacitor based on the voltage sweep shown in Figure 5.

Ideal Inductor

The voltage across an inductor is the product of inductance and the rate of change of the current flowing through the inductor:

$$ v = l \ cdot \ frac {\ text {d} i} {\ text {d} t} $$

This means that the current is proportional to the integral of voltage, and this is what we see in the following plots. The current increases in magnitude as the (negative) area under the voltage curve increases. But when the voltage crosses the time axis, the positive area under the curve starts to balance out the negative area under the curve, and this causes the current magnitude to decrease toward zero.

Figure 7 (a)Linear voltage sweep and(b)the corresponding inductor current vs. time.

不te the difference between the capacitor and the inductor: With a capacitor, current is proportional to the derivative of voltage, and thus a linear voltage sweep translates to constant current. With an inductor, current is proportional to the integral of voltage, and thus a linear voltage sweep translates to a quadratic shape in the current vs. time plot.

The I-V relationship of an ideal inductor is shown in Figure 8. The magnitude of the current gradually increases and then decreases as the voltage moves from V2 to V1 or from V1 to V2. The direction of the current is negative as the voltage moves from V1 to V2 and positive as the voltage moves from V2 to V1.

Figure 8.基于图7所示的电压扫描的理想电感器的I-V曲线。

概括

The table below summarizes some of the insights we have obtained by looking at the I-V curves of several ideal devices. In a future article, we will look at I-V curves of non-linear devices.

设备 Requires Power? i-vMethod Description of the Graph Passes Through Origin?
IdealResistor Yes Voltage Sweep Straight line passing through origin Yes
IdealVoltage Source Load Switching Straight line parallel to current axis
Ideal当前来源 Load Switching 直线与电压轴平行
Ideal电容器 Yes Voltage Sweep Closed rectangle around origin
IdealInductor Yes Voltage Sweep Closed parabolic loop around origin
5 Comments
  • S
    swn12 January 19, 2017

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  • imre February 25, 2017

    感谢您的文章。如果我在图8中正确理解。I1应该为负,I2呈阳性。另外,根据我在图7中的计算。Imin应该为(1/(2l*(v1-v2)/∆t))*v1^2和imax应为(1/(2l*(v2-v1)/∆t))*v2^2。您能澄清一下吗?

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