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Lesson 1: Diode

# Diode graphical solution

We solve a diode circuit graphically by plotting a diode i-v curve and resistor to find the intersection. Graphical solution methods are a common way to work with non-linear components like diodes. Created by Willy McAllister.

## Want to join the conversation?

• I did not properly understand the meaning of the load line. What exactly is its importance?
• When we make a current-voltage plot (i-v curve) for an isolated resistor using Ohm's Law, the plot is straight line through the origin with a positive slope of 1/R. That's the i-v plot we get for a resistor all by itself.

For the circuit in this video, the resistor is one of three components, connected between the diode and the voltage source. The resistor is connected to the positive voltage supply. We did some circuit analysis and algebra, and came up with an equation of a line involving the resistor. When that line is plotted, it turns out it doesn't go through the origin, and it ends up with a negative slope. This is just what comes out of the circuit analysis algebra. It happens when a resistor has its top end connected to a positive supply voltage. That resistor connection turns out to be quite common in electronics. A resistor in this position is sometimes referred to as a load resistor and the resulting straight-line negative-slope plot has the nickname load line. [The term "load* isn't restricted to just this position for a resistor. It comes up in other contexts, too.]

You might want to check out the nearby article for another form of the same explanation: https://www.khanacademy.org/science/electrical-engineering/ee-semiconductor-devices/ee-diode/a/ee-diode-circuit-element
• My question could be silly. How is 3/330 is 9 mA?
• 3/330 = 0.0090909... If we round this it becomes 0.009.
"m" is "milli-" the numerical prefix that stands for 10^-3.
0.009 in scientific notation is 9 x 10^-3, which can be written as 3mA. (The A stands for ampere.)
• Sorry I am a bit confused about the direction of the current. I thought that current could only flow in the forward direction, but in the video why is it that we can calculate it in the reverse direction of the diode?
• The orange arrow actually not referring to the current direction, it refers to the voltage across 'x'. X in the video is Vd & Vr, or voltage across the diode & voltage across the resistor.

You may solve the circuit regardless of where we start (to calculate), as long as it covers the whole (circuit) loop.

Hope that helps.
• How can we have voltage but have no current? I'm confused.
(1 vote)
• IHMO,

My (personal) analogy :
Voltage = existence of "pressure",
Current = existence of "flow"

A flow will need a path, while pressure doesn't need one. Thus, It is ok to have voltage without current.

I hope this helps. :)

p/s : In reality there is no such thing as absolute zero current or absolute zero voltage at any point. If there is a charge, there will be potential difference(or voltage or Electric field). If the charges moves, then it is a current flows. take V = IR, or R = V/I .
Since the resistance, R rarely is 0 Ω [ V = 0 V ] or ∞ Ω [ I = 0 A ], so does V or I are rarely 0. It depends on which scale are we looking at. If in high power transmission, 6.8mA can be negligible in the calculations, but not in the nanoworld, where 6.8 mA means a swarm of 4.24 × 10^16 electrons per second. which is very significant in any chip design. :)
• Is it true that whenever you have the current going in the proper direction through the diode you can usually safely assume that V_D always equals 0.7 V? And then you solve the rest of the circuit after putting that in for the V_D drop? (As long as it's known that there is more than 0.7 V supplied to it?)
• Why the forward drop of a diode is not the quiescent voltage ( The operating voltage across a diode) but a constant 0.7 Volts?
(1 vote)
• The diode operates according to its i-v equation. That equation is plotted at . The curve has an exponential shape, so the voltage isn't actually constant as the current goes up.

However... that exponential curve rises pretty fast. Engineers take a simplified view of the curve. If you hold that plot out at arms length and look at it, the steep upward part of the curve is around 0.7V for silicon diodes with "normal' magnitude currents (say in the 1-20mA range). The voltage is somewhere between 0.6 and 0.8V, so 0.7 is a good all-around guess at what the voltage will be. It is rare to need more precision than that.

For LED diodes the forward voltage isn't 0.7, it is some other value. (LED's are not made of silicon, they are made of different elements depending on the color). But all LEDs have that sharp rising behavior, so it is convenient to summarize that whole curve with a constant voltage.
• At of the video, the current increases rapidly when it is around 0.7 volts and above in the graph. Is there a reason why the current suddenly increases around the point? Thanks.
• That's a tough question to answer in detail without going into the quantum mechanics of semiconductor junctions.

One thing you have to be careful of with exponential curves is how they appear to have a 'knee' at some point in the curve. There really isn't a knee, the exponential is increasing everywhere. The 'knee' is kind of an optical illusion determined by the scale of the plot you are looking at, and the fact that we use linear scales on the i-v plot. If you plot the exact same function on a log-log plot it comes out a straight line with no knee.

Visit this url and notice how one plot is smooth and the other has a knee, just from changing the scale. https://www.wolframalpha.com/input/?i=e%5Ex
(1 vote)
• I have this one question regarding BJT .In case of transistor why is it in active mode when the first junction is in forward mode and another is in reverse?
(1 vote)
• The brilliant trick of a bipolar transistor is that the base region (the middle layer of the 3-layer sandwich) is very thin, on the order of a millionth of a meter or less. Let's track the electron current for an NPN bipolar transistor: When the Base-Emitter junction is forward biased, electrons flow into the base from the emitter. When the electrons reach the P material of the base, they all of a sudden become "minority carriers", the -electrons are in the +p material. The base is so thin, this big swarm of minority carriers gets swept up by the reverse biased Base-Collector junction and most of them flow over to the collector. A small amount of electron current flows out of the base connection. For every electron that flows out the base, 50-200 electrons head towards the collector.

The BJT is constructed so it's able to dump a whole lot of minority carriers into a reversed biased depletion zone. They get caught up in the strong electric field that's trying to evacuate minority carriers. That's what forms the collector current.

The fine details of how this works is part of the field of "solid state physics". It involves a lot of quantum mechanics and a really cool phenomena called "quantum tunneling".
• Suppose you were to flip the diode in the reverse direction. How would you find the voltage across the diode then?
(1 vote)
• To solve the circuit graphically with a reversed diode, you draw the diode curve flipped around the current axis (draw the rising part of the diode curve is to the left of the current axis). In this case you will notice the diode line and the resistor load line intersect at v = 3v and i = 0. That's the answer.

You can also solve it using the diode formula. Instead of v_D in the exponent you put in -v_D because the diode is turned around. You have to solve two equations simultaneously, this diode equation and an Ohm's law equation for the resistor. Up in the exponent the value of q/kT is 26mV (at room temp). Any reasonable value you put in for -v_D will result in a large negative exponent, so the exponential term pretty much goes away and you are left with just -Is for the diode current (which is pretty close to 0).
(1 vote)
• how can i found the efficiency of half wave rectification?
(1 vote)
• What is your measure for "efficiency"? A half-wave rectifier keeps half of the sine wave (all the positive humps) and throws away all the negative humps. In that regard it uses 50% of the available energy. If you want to account for the small diode voltage you need to subtract 0.7V from the peak voltage.
(1 vote)