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I have a hard time understanding what an operating point and load line represent and why do we need them at all.

After reading some wikibedia,

  • What I got is this load line represent response of the linear circuit attached to the diode for changing in voltage across it.

But, this doesn't seem to answer my question about

  • why do we even need this and what does it represent exactly in a little more simple detail.
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    \$\begingroup\$ Suppose you have a bathtub you want to fill up with water. The rate at which the faucet runs water into the tub is 4 gallons per minute. The tub is full when it holds 35 gallons. You can take out a sheet of paper, make the y-axis the number of gallons and the x-axis is time. You can draw a sloped line starting at 0,0 that represents the 4 gallons per minute for the water rate. You can draw a horizontal line across the chart representing the (same for all time) 35 gallon mark that represents "full." Where these two lines intersect, if you drop down to the x-axis, is the time it takes to fill. \$\endgroup\$
    – jonk
    Apr 23 '20 at 5:48
  • \$\begingroup\$ Do some research about diodes and their use other than wikipedia, you might find more info. \$\endgroup\$
    – Solar Mike
    Apr 23 '20 at 5:48
  • \$\begingroup\$ Now, you might ask "what's the point?" Well, it's one way to get a quick, approximate answer without having to know how to do much algebra. All you need is a ruler and the ability to put a point at 4 gallons at one minute. Then draw the sloped line. No algebra. Just using a ruler. The 35-gallon horizontal line is equally easy to draw. No algebra. Where they cross gives you the answer. Without algebra. Sometimes, people prefer that approach. And it is "good enough.' Of course, we have computers today. \$\endgroup\$
    – jonk
    Apr 23 '20 at 5:50
  • \$\begingroup\$ Diodes and BJTs and MOSFETs and JFETs are non-linear devices and their equations are non-trivial to mess with. Sometimes, it's "good enough" to just grab some easy curves from a chart that is handed to you on a datasheet and, since resistors act very much like "straight lines", just draw a resistor line through the mess and see where it all hits. I don't think I can say it much simpler. I could spend a few hours drawing up examples based upon problems you present until you get it. But.. then that would not be "simple detail." Just do a few problems until it sinks in. It will. \$\endgroup\$
    – jonk
    Apr 23 '20 at 5:54
  • \$\begingroup\$ That said, some folks never use load lines. Some use them frequently. I used to use them (on vacuum tube charts.) But computers have made it far, far simpler. So maybe that's why you are asking? Computers make things a lot easier now? \$\endgroup\$
    – jonk
    Apr 23 '20 at 6:00
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Load lines are useful in understanding and calculating the operating point of a circuit which would otherwise require complex calculations. Diodes have a non-linear relationship between I and V and we often use an approximation such as "2 V drop across the LED" when making calculations. We can do better if we have a load line chart.

enter image description here

Figure 1. A simple LED circuit.

enter image description here

Figure 2. Loadlines for various resistors on 5 V supply. Image source: LEDnique.

Here we can quickly calculate the operating point of an LED.

  • If we used a 180 Ω resistor on a RED LED we would expect to get 18 mA through the LED.
  • If we wanted 10 mA through a blue LED we can see that the 270 Ω resistor would do the job.

No calculations involved (once the graph has been created). To create the graph draw the lines from the supply voltage point on the V-axis and the \$ \frac V R \$ point on the I-axis.

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I have a hard time understanding what an operating point and load line represent and why do we need them at all.

Do not worry because it is not just you who have this problem. For example, I first started thinking of this concept as a student in the 70's... and I continue thinking of it even now when explaining it to my students... and even when writing this answer. Below are my insights (some of them shared in the cited Wikipedia page). Тhey can help you not only to formally know... but to truly understand what these formal terms mean...


1. Ideal source and load. The most elementary electric circuit consists only of two 2-terminal elements connected to each other - a source and a load. A typical example is a voltage source connected to a resistor (Ohm's experiment) - Fig. 1a. This humble circuit is described by a simple equation, which in this case is Ohm's law - V = I.R.

Ideal voltage source - resistor

Fig. 1. An "ideal" voltage source supplying a resistor (a picture from 90's); E is the voltage V. The triangle with catheti VA and IA, and angle tan R, is an excellent geometrical interpretation of Ohm's law (more meaningfull than the so-called "Ohm's triangle").


Another (less common) example is a current source connected to a resistor (another version of Ohm's experiment) - Fig. 2a.

Ideal current source - resistor

Fig. 2. An "ideal" current source supplying a resistor (a picture from 90's); the circle on the left is the current source. The triangle with catheti VA and IA, and angle tan R geometrically represents Ohm's law.

The idea. Each of these elements can be graphically represented in the coordinate system by its IV curve (the set of all possible pairs of voltage and current). In the first example above (Fig. 1b), the IV curve of the voltage source (the source line) is a vertical line shifted to right in the case of positive voltage; the IV curve of the resistor (the load line) is an inclined (to right) line passing through the origin of the coordinate system. Since the voltage across both elements and the current through them are the same, we can impose their IV curves on each other in the same coordinate system. The intersection point A of the two curves represents the graphical solution of the circuit equation and is named operating point. Its coordinates determine the instant values of the voltage and current of the circuit.

In the dual example (Fig. 2b), the IV curve of the current source (the source line) is a horizontal line shifted up in the case of positive current; the IV curve of the resistor (the load line) is the same as above.


Varying voltage, constant resistance (Fig. 3). This graphical representation can be used to visualize the circuit operation when we vary the voltage, resistance or even both. For example, when exploring the passive voltage-to-current converter, the voltage source IV curve moves in the horizontal direction (translates) while the resistor IV curve stays immovable:

V-to-i-graph

Fig. 3. Varying the voltage as an input quantity in the Ohm's circuit (voltage-to-current converter)

Besides linear (ohmic) resistors, all sorts of non-linear elements and even sources can serve as a load (diodes, transistors, etc.). For example, this setup is used to measure the (horizontal part of) transistor output characteristic.


Constant voltage, varying resistance (Fig. 4). When exploring the passive resistance-to-current converter, the resistor IV curve rotates while the voltage source IV curve stays immovable:

R-to-i-graph

Fig. 4. Varying the resistance as an input quantity in the Ohm's circuit (resistance-to-current converter)


Varying current, constant resistance (Fig. 5). In the dual configuration (a current source connected to a resistor), when exploring the passive current-to-voltage converter, the current source IV curve moves in the vertical direction (translates) while the resistor IV curve stays immovable:

I-to-v-graph

Fig. 5. Varying the current as an input quantity in the dual Ohm's circuit (current-to-voltage converter)


Constant current, varying resistance (Fig. 6). When exploring the passive resistance-to-voltage converter, the resistor IV curve rotates while the current source IV curve stays immovable:

R-to-v-graph

Fig. 6. Varying the resistance as an input quantity in the dual Ohm's circuit (resistance-to-voltage converter)

As above, all sorts of non-linear elements and even sources can serve as a load. For example, this setup is used to measure the (vertical part of) diode IV curve.


2. Real source and load. The graphical solution is a result of intersection between two curves in the operating point; so this technique can be applied to a circuit of two elements connected to each other. In circuits with more elements, we can reduce the complex circuit by means of equivalent transformations to an equivalent circuit consisting of two parts - a real voltage source and a load. A typical example of this technique is the voltage divider configuration (Fig. 7):

Voltage divider load line

Fig. 7. The voltage divider operation geometrically represented by two intersected IV curves (a picture from 90's). E1 is the voltage V of the voltage source; UR1 and UR2 are the voltage drops VR1 and VR2 across the resistors R1 and R2.

Here, two elements in series - the "ideal" voltage source E1 and the upper resistor R1, are combined into a new "composed" element E1R1. It can be thought as a real voltage source with voltage E1 and internal resistance R1. Its IV curve is a line shifted to right by E1 and inclined to left by an angle R1. It is obtained by subtracting R1 IV curve (an inclined to right line passing through the origin of the coordinate system) from E1 IV curve (a vertical line shifted to right with E1) - Fig. 8:

Composed IV curve

Fig. 8. Revealing what the "load line" actually is...


This configuration is widely used to illustrate and roughly calculate transistor amplifying stages where the name of "load line" came from. For example, in the circuit of the common-emitter stage - Fig. 9, we can combine the collector resistor Rc and the voltage source E (the power supply VCC) into a real voltage source ERc with voltage E and internal resistance Rc.

Transistor load line

Fig. 9. In the common-emitter stage, the network of power supply and collector resistor in series can be thought as a real voltage source with voltage E and internal resistance Rc. The transistor acts as a varying current-stable resistor.

The other element - the transistor, will act as a current-stable non-linear load. So, in this arrangement, a real voltage source with linear internal resistance drives a load with non-linear resistance. Only, the conventional viewpoint is that the transistor drives the load Rc; hence the name "load line".


3. Two real sources. If there is another voltage source, it can be combined with the load. For example, in the case of R1-R2 voltage divider, if we insert another voltage source between R2 and ground, we obtain the useful circuit of a resistor voltage summer. It is used in op-amp inverting circuits - Fig. 10.

IV inverting amplifier

Fig. 10. Op-amp inverting amplifier illustrated graphically (a picture from 90's). Ein (VIN) and R1 form the one element; EA (Vout) and R2 form the other element. The question is, "What is source and what load line here?"


Finally, what was the "load line"?

Simply speaking, "load line" is the IV curve of a resistor and voltage source in series.

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    \$\begingroup\$ Don't forget to provide the name of the original author and a link to the original articles for any images used that aren't your own. See How to reference material written by others. \$\endgroup\$
    – Transistor
    Apr 25 '20 at 20:24
  • \$\begingroup\$ Thanks for the response. I made this illustrations in the late 90's when I decided to begin sharing my circuit philosophy with my colleagues, students and readers. \$\endgroup\$ Apr 25 '20 at 21:05
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    \$\begingroup\$ Credit yourself then! \$\endgroup\$
    – Transistor
    Apr 25 '20 at 21:09
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In amplifiers, we expect even the tiniest input voltage to be linearly amplified.

The random electron-movement noise floor will be in the low to moderate microVolts, and we expect/require the amplifier to respond even to the noise, because later circuits may introduce methods (such as GPS uses) to rescue signals weaker than the random noise.

Thus the amplifier MUST BE ON AT ALL TIMES. The value of the voltage across our transistors, and the current thru our transistors, is the OPERATING POINT.

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