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?(5 votes)
- 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(8 votes)
- My question could be silly. How is 3/330 is 9 mA?(2 votes)
- 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.)(8 votes)
- 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?(3 votes)
- 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.(3 votes)
- How can we have voltage but have no current? I'm confused.(1 vote)
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. :)(4 votes)
- 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?)(2 votes)
- 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 at2:00. 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.(3 votes)
- At1:25of 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.(2 votes)
- 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".(2 votes)
- 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)
- [Voiceover] Now I wanna use a diode in a circuit, and we'll see how we solve circuits that include these nonlinear diodes in them. So I have a circuit here with a battery and a resistor and a diode here, and it's gonna be a special kind, it's gonna be an LED diode, so it's gonna give off light. This is a kind of diode that's manufactured to generate photons of light when it has a current flowing through it in the forward direction. They're pretty cool and you see them all the time in electronic components. So we're gonna figure out how to use an LED in a real simple little circuit. And in our circuit here, we're gonna have a resistor of 330 ohms, and we'll make this battery three volts, so it's like two double-A cells. And what we wanna find out is how much current's gonna flow around this circuit. And then we label this here, let me label the voltages across our components. We'll have VD, which is this voltage here, and we'll have VR, which is the resistor voltage, and that's that voltage right there. So now I'm gonna show you a plot of the iv curve of our diode. You can see right here, around 0.7 volts, the current rises rapidly when there's 0.7 volts across the diode. And let's start out by analyzing this, let's start with the same tools that we always have, which is, let's try to write some current laws for these two things. So for the diode, we write a current law that looks like this. The current is equal to IS times e to the qv on kT minus one. So that's the iv characteristic for the diode, where this is V diode right there, and the corresponding equation for the resistor is i equals V resistor over 330 ohms. That's just Ohm's Law for the resistor, and i in both cases is this i right here. Now if I wanted to, I could set these two expressions equal to each other and solve somehow for VD and VR, but what we're gonna do instead is we're gonna solve this by graphing, by a graphical method. Here's a graph for the diode, and this is the VD scale, this is V diode, and this is i up here. And so what I wanna do is I wanna plot the resistor curve around here as well. I wanna plot the resistor iv curve on this same plot. Now, in this expression, I have VR instead of VD, so let me see if I can work on VR here. Let me try to figure out VR in terms of VD. So I can derive VR as three volts minus VD. And let's put this into this i expression here. The Ohm's Law expression now becomes i equals, VR is replaced with three minus VD, all over 330 ohms. And let's work on this a little more. i equals three over 330 minus VD over 330, and this is starting to look like the equation of a line. Let me write this to recognize it as a equation of a line. Minus VD over 330 plus three over 330 is nine milliamps. So this is the equation of a line, and the slope is right here, is minus one over 330, and the i-intercept is nine milliamps. So let's see if we can plot this line. This is actually called a Load Line. That's just a nickname for this kind of expression that you get when you have a resistor connected to a fixed power supply above and the resistor is hanging down from it. You get this characteristic equation of a line that has a negative slope, which is really distinctive. Let's see if we can plot this line. Now, it's a line, so all we have to do is find two points that solve the line and then we'll be able to draw the line. So if I set VD equal to zero, then i equals nine milliamps. So here's VD equal to zero, and it'll go through nine milliamps. So that's one point on the line. But what else can I set to zero? I can set i equal to zero, which means I'm on the voltage axis. And what I'm actually gonna do is I'm just gonna look at my circuit and figure this out in my head. So I'm setting i equal to zero, and i equal zero means there's no current in this resistor, which means there's no voltage drop across that resistor. And that means that this voltage here is the same as this voltage here. And I know the voltage here, the voltage here is three volts, so that means the voltage here is three volts, and that's because I know the current is zero. So let's go over and put that point on a line. So when i is zero, v is three volts. So there's another point on the line. And now we have two points and we can draw a line between 'em, like that. And what we've drawn is the Load Line for this 330-ohm resistor. You remember, back over here we said we could solve these two equations by setting the two is equal to each other, and that's basically where do these two lines intersect? And they intersect right here. That's the solution to our problem. So this intersection point is the solution. It's where the resistor current and the diode current are the same, and that's that point there. Now I can just read off my answer. Right there is about 0.7 volts, and the current over here, if I read off the current, just straight across there, it's about 6.8 milliamps of current. So now we actually just solved our circuit using a graphical technique. And what that says is, let me erase this a bit to clean it up. Let me take out these two things here. That was just the resistor Load Line that we were talking about. And now for our solution, we have i equals 6.8 milliamps, and V, V diode equals about 0.7 volts. So that's how you do a graphical solution with a diode.