Diode Circuit Analysis
So we've learned about diodes in previous tutorials. But today we're going to be solving circuits with diodes and then we're going to be finding what the quiescent or Q-point is for those diodes and we're going to be talking about the benefits of each of these methods.
The Q-point is basically that point around which you don't expect a whole lot of variation in the current or the voltage. And with diodes, there's a couple of different ways to approach it. And we're going to go over them really quick, we're going to talk about two of them, talk about why we're not going to be doing them. And then we're going to do two examples with two of the other methods. So again, we're looking for the voltage across it and the current through it.
Load Line Analysis
So the first thing that you can do, and one of the ones that we don't recommend is called the load line analysis. And that's where you take the diode that you're going to be using in the circuit and you physically take it into a lab, put a voltage across it and measure the current through it. And as you increase the voltage, you create your graph to show exactly what the current is when you're at a certain voltage across that diode and then you use that in your actual circuit and you solve for things and you move things around. And it's 1) huge pain to get the data in the first place and 2) it's still a huge pain to use later. You know, depending on what you're doing, it may be necessary, probably not.
The second thing you can do is actually a mathematical model. With this math model, you can see that you have the voltage drop across the diode on the left hand of the equation, and then you have a bunch of other stuff. So you have the VS which is the source voltage across the diode, R is the resistance of the resistor that is in series with the diode. And then you have your IS, which is the current through the diode and resistor. But you also notice on the right hand of the equation, there's another VD, another voltage drops.
So this is an iterative thing, where you have to go through and not only do you have to have all of those other things figured out, you then have to iterate and try and find exactly what that voltage drop is going to be over that.
So it's a pain and again, you probably don't need it. It doesn't make a whole lot of sense, but it is one method of figuring out what that voltage drop across the diode will be if you have all of the other information anyway. So kind of dumb, but I just wanted to introduce those because there are multiple ways of approaching this. And just because there's a lot of ways doesn't mean that you should use all of them. Again, if you're going to be doing your PhD thesis, then do what you got to do.
Ideal Diode Model
But the third way that we're going to talk about this is the simplest way and in another tutorial we talked about this. And this is where we just assume that if a diode is forward biased, there is no voltage drop across it. And if it's reverse bias, then the voltage is whatever, it just happens to be in the circuit. So as we do this, there's going to be four steps that we follow one when solving a circuit with the ideal diode model.
So here, I'm going to create a circuit. I got this from taking a microelectronics class at Boise State, just pulling off there and if I were wise, I would have practiced this beforehand and we will see if it goes to complete heck in a hand basket again. And I am creating this circuit, so that we can try to go through those four steps that I just mentioned.
So the first thing just created the circuit. Now I have to look at the circuit. And wow, that is a terrible resistor that, that may be one of the worst resistors I've ever made in my entire life. So we're looking at the circuit, and we're deciding which diode is going to be forward biased and which diode is going to be reverse biased. And so I have to just kind of guess, looking at it and think, okay, I've got nine volts here. And so if these were both forward biased, I would have zero volts here, which would mean that I'd have current flowing this way that makes sense, and then I'd have a negative six volts there. But then is that being a much smaller resistor than that? Is that going to make it so I'm actually getting flow this way? Or is the flow going to be going back that way? And you just have to look at it and make that assumption and that is the very first step.
So on this one, I'm going to look at this and say, I believe that this is going to be forward so it's going to be treated as a short circuit in this ideal diode model, and then this will also be treated as a short circuit in the ideal model. So let's recreate this with those assumptions.
So we're going to go down, and that actually didn't get any better. And then we'll put a note there, show that as a short circuit to ground. And then we'll go there and show that as a short circuit to 43k and -6 volts, +9 volts, 22k, and 43k. So that's our second step is now that we've made our assumptions we've recreated the circuit.
The third step is now doing some basic circuit analysis to see if our assumptions actually work.
So in this case, we are going to have basically, two... let me do my ohm's law in the corner like I always do, just keep things straight in my head, we're going to have our two currents right here. I call this I1, and I2. And I1 is simply nine volts minus zero volts over 22k and then I2 is going to be zero volts minus -6 volts over 43k. And so let me throw that into a calculator really quick because there's no way I'm gonna be able to do that in my head. And that comes out to be about 409 microamps. And this comes out to be about 140 microamps. Okay, so that looks good.
Now one of the interesting things here is that I1 is not the current that goes through this diode. It's the current that goes through the diode and also through this diode. So we're going to mark that as I3 and we are simply going to find I3 as I3 equals I1 minus I2. So that is going to be 409 microamps minus 140 microamps and that will give us a number that, again, I can't do in my head, 269 microamps, okay.
So now we have actually solved and we have an assumption that we are going to have zero volts across both of these diodes and we have assumed that we are going to have 140 microamps through that and 269 microamps through that way. So our quiescent point or Q-point for, let's do some retcon, for D1 is our voltage is going to be zero volts and our current is going to be 269 microamps and D2… Wow, I've just I'm having one of those days, D2, subs of two is also going to be zero volts and 140 microamps.
Now, frankly, the the way I did this, I often see it done another way where you actually have the current first I'm used to just doing volts and amps. But you might see it where people say, hey actually do current first so 140 microamps and then zero volts in the that will be your Q-point for D1 and your Q-point for D2.
Now this is where we get to our final step in the process. And this is where we look at it and say, Does this make sense? Now for there to be zero volts across these diodes, we need to have a positive current in these directions that we've established. So as we look at this and we say all right, D1 has a positive 269 microamps in that direction and D2 has a positive 140 microamps in that direction. And we are assuming that those are both zero volts. All of these numbers make sense with the way we've set up. So we can actually look at this and say, Yes, my assumption that both of these are forward biased is correct.
And so this is how you do it with an ideal model, when you're trying to solve for the diode, and it's pretty straightforward in terms of making it simple, just shorting that stuff. And the only difference between this in the fourth and final method that we're going to go over is that in the constant voltage drop model, we have to make things more interesting in there. You can't just put shorts or opens, well, I guess you can still put opens, but let's get into that right now.
Constant Voltage Drop Model
So let's do another circuit. So this time, we're going to start with +6 volts. So have our node right there, have our 43k right there, another node. And if we want to zoom through this, that's totally fine with me. You know, I've had people tell me that I should have been a lot of things, but an artist was never one of them. I don't even know. Was I trying to put -9? -9 volts. Okay, is that legible? I think that's legible. Okay, then we have our 22k. So now we are set up for our constant voltage drop (CVD) style.
So same exact four steps that we did in the previous one of first looking at it, making some assumptions. Second, drawing out the model with our assumptions. Third, solving for that model. And then fourth, going back, seeing if our assumptions work. And if they do great, we move forward. If not, we go back and change our assumptions and do the entire thing over again.
But the difference is with this should have pointed this out. This is very, very crucial difference between this model and the ideal model is in this one, we're going to assume that it's a 0.7 volt drop. So if we are forward biased, it's not going to just be alright, forward bias. Let's put in a short we're actually going to make model it with a voltage drop, a voltage source, of 0.7 volts. So six volts, I'm still going to make an approximation that this is going to be about zero volts. If that is forward, in reality, it'd be 0.7. And that would put 0.7, -9, I think that this is going to be forward biased and this is going to be reverse biased.
So now on step two - create that model, six volts down... why is that so bad. And then I put in my voltage source of 0.7 volts and then I put this as an open to -9 volts and 22k, step two is done.
Step three, let's actually solve for this. Now with that being gone that makes this pretty simple. So let's see if I can, in this case, I1 will be the same thing as going through that and then we have I2 which is going to be pretty straightforward because that's going to be zero because it's open but we model it we have I1. Let's see what should I do this, I1 equals six minus 0.7 over 43k divided by 43,000. It's gonna give me 123 microamps and it is going to put this voltage right here it's going to be 0.7, and this voltage at this node is going to be -9 volts and we assume I2 is just going to be zero because nothing can flow through that.
Okay, so now let's take this and establish our Q-points. So we are going to again retcon this because I don't do things in order like I should, D1 one equals… Let's put this as a current first 123 microamps and 0.7 volts. And then D2's Q-point is going to be zero amps and -8.3… Nope, nope, I totally got that backwards because I cannot do math in my head. So that is 0.7 minus and I went the wrong direction. So this should be -9.7. Now the beautiful thing is that it doesn't change the way that it's biased. It just changes the fact that I was totally screwed up.
So now again, did everything taking our time breathing, making sure we're not making mistakes. Come on, Josh. Don't make mistakes. Now we verify that this is actually set up. So we assumed that D1 was going to be forward biased. And we assumed D2 was going to be reverse biased. So as we look at this, we have a positive current through it, and then that 0.7 across it just because that's what we have. And then D2, we have zero current through in a big -9.7 volts across it.
So both of these are correct in that we assumed this was going to be forward biased, and it is, and we assume this is going to be reversed biased, and it also is. Now again, either of these have been wrong, we'd have to go back, you'd have to say, all right, I was wrong here. Let's assume that is forward biased. And we'd have to recreate this, redo all that, which would of course, increase the chance of me being unable to do simple arithmetic again. But hopefully, that is more straightforward.
So again, the only difference between the constant voltage drop and the ideal model is the fact that you put in a voltage source to say, okay, we're losing 0.7, or whatever your assumption is, 0.7 volts across this diode. And in most cases, it won't make a difference, but on occasion it will, it definitely will make things more complicated for you.
So four different types of ways of solving four diodes in an equation, four steps in the way we do the ideal and the constant voltage drop. And despite this, despite the iteration, and despite the fact that I can't do math, this isn't that bad. Doing the constant voltage drop and doing the ideal model are pretty straightforward. So, just take your time, go through them. Make sure that at the end, you look at it and say, does it actually make sense? So hopefully that helped. If it didn't confuse you, that is fantastic. Hope you liked this, subscribe to our channel. We love people to subscribe and ask questions and learn more about electronics. We'll catch you on the next one.
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