Solving Diode Circuits | Basic Electronics

There are a couple ways of solving diode circuits and, for some of them, the diode circuit analysis is actually pretty straightforward. Josh uses a few diode circuits problems to show two possible ways to solve for diode circuits and also mentions two other methods and why you shouldn't use them. The recommended methods are either the ideal diode circuit analysis and the constant voltage drop diode analysis, and on top of discussing what they are and why they're good, we give an example of them in action. The result of these exercises yields the diode quiescent point, or diode q-point, and this is also discussed briefly. This is is the basis of the written tutorial that expands and clarifies the video on the CircuitBread website - www.circuitbread.com/tutorial...
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Table of Contents:
0:00 Introduction
0:14 What is the quiescent point, or the q-point, of a diode?
0:36 Load Line Analysis for solving circuits with diodes in them
1:15 Math model for diode circuit
2:18 Ideal diode circuit analysis with the four steps
8:39 Constant voltage drop diode example
14:15 Review of the four methods and four steps
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[@CircuitBread]

Hi all! If you want to see the written version of this tutorial, check it out on our website: www.circuitbread.com/tutorials/diode-circuit-analysis Have a great day!

[@yarenkaya7872]

Finally, a systematic solution with checkpoints! Thank you very much for the tutorial.

[@chrisoca958]

You are amazing!!! I subscribed!! Thank you very much!

[@Mehedi_buet_19]

thank you..helpful..and you speak so nicely😊

[@MERF2021]

You went to BSU?! Me too! Who was your microelectronics teacher? I had Dr Campbell she is my favorite professor! Nice to meet another alumn!

[@CircuitBread]

Awesome! I actually had Dr. Smith for microelectronics BUT! Most of the semiconductor videos we have here were run by Dr. Campbell to make sure they were accurate (any inaccuracies means that she didn't review it 😬). I had her for Circuits 1 and Semiconductor Devices and she's such an incredibly nice person. So, when we were first starting CircuitBread, I went and talked to her because I was comfortable with her and knew that she's all about trying to help students.

[@MontanaPreston]

Thanks!

[@ieatcrayons3322]

Thankyou!

[@bara8527]

keep going please you inspire me

[@CircuitBread]

Thanks Bara! We are starting our circuits analysis series very soon, which are similar to this though from the very basics of circuit analysis. I hope that it is helpful!

[@tjriz91]

I have a question. We are learning about diode circuits right now in my microelectronics class. One of our homework problems has us solving for the diode voltage and current of a diode/resistor series circuit using not only the constant voltage diode model, but also the iteration method using both the KVL equation and the exponential diode equation, along with finally a load line analysis using Python/MatLab. We are tasked with solving for these values with the applied voltages of 0.5, 0.8, 1, and 1.2 V. The values of reverse saturation current and resistance are both given. Using the iteration method for the applied voltages of 0.5 and 0.8V, the resulting values of diode voltage quickly rise above the amount of applied voltage, which obviously doesn't make sense in reality. And subsequent iterations would result in a negative current, which causes issues for the log function being used for the iteration. I know this is surely because it is before the threshold voltage that would cause current to flow. I feel comfortable in saying that no current would be flowing with the applied voltages of 0.5 and 0.8 V, but what exactly could I say about the diode voltage when the iteration breaks down?

[@ezzzyara]

Thank you so much^^^

[@CircuitBread]

You're welcome 😊

[@circuitemakes]

Interesting

[@mostofaemon8450]

Nice

[@dragontamer3334]

This is really helpful in understanding diodes and their calculations for an exam I have coming up. Just one question. How do you know if your assumptions are wrong? Are you double checking the current values you got at the end to the assumptions you made for if the diode is forward bias or reverse bias? Forward = current and Reverse = no current.

[@CircuitBread]

Yes, that's exactly it. After you've done the math, if your current is flowing the wrong way through the diode, you know that's not possible, so the assumption is incorrect. Good luck on your exam!

[@mnada72]

Thank you. Please keep up with these informative videos. Make this channel Electronics Encyclopedia 🙂 Good Luck I wish you make a similar video about transistors.

[@CircuitBread]

Hey, thanks for all your comments and kind words. Our goal is to eventually have covered all the topics but there are a LOT of them, so it's going to take some time 😃 We do have a few videos about transistors, BJTs and MOSFETs, and at the moment, we don't actually have any plans to do any more with them except as a part of general circuits topics.

[@mnada72]

@@CircuitBread Actually what I mean is how to deal with transistors in circuits as you did with diodes in this video

[@mehdicharife2335]

Why in 11:47 do you assume that there is no voltage drop across the 22k resistance ?

[@CircuitBread]

Hi Mehdi! With the open circuit, you cannot get any current flow as there's no path. No current through the resistor means that there is no voltage drop across it either, as those two are directly related. As a quick refresher, we have some Circuits 101 tutorials that we've posted this year if you want to check them out.

[@adizafran4239]

hey, thanks for the great explanations! I was wondering in the last exercise of i2=0 how does the voltage drops across D2?

[@CircuitBread]

I hope I understand your question. So, when i2 = 0 (there is no current through diode D2) why is there a voltage drop across it? It's important to note that since there is no current through the resistor, there's no voltage drop across the resistor, so the voltage potential on both sides of the resistor is the same: -9V. Since we know the voltage potential on the left side of D2 is +.7, we can see that the voltage drop goes from positive .7 to negative 9, so we get a 9.7V drop over D2. Does that help?

[@adizafran4239]

@@CircuitBread thanks for answering, from your answer I understand that even tho there is no current there is a drop in the voltage value because of the diode which is different from what happens with a resistor?

[@CircuitBread]

Sorry for the delayed response, but yes, you have to change your way of thinking between resistors and diodes. I thought ohm's law applied to everything at first and it took me a bit to realize that it doesn't work that way.

[@hannahadams3350]

Do you have any videos on how to make your forward/reverse bias assumptions? I believe this is what I struggle most with. Thank you.

[@CircuitBread]

Hey Hannah - sorry, we don't have any videos on it because it's not something you can really teach. You can get an intuitive feel as you get better at circuits on which to base your assumptions, but even then, you'll still guess wrong occasionally or even frequently. A year ago, I was chatting with the professor who currently teaches this class and, even though she's brilliant and got her PhD many years ago and has a ton of practical experience, she admitted she still makes wrong assumptions. Just don't get frustrated with yourself, make sure you have plenty of time, and it'll all be okay.

[@chrisoca958]

@@CircuitBread That means a lot, thank you very much!

[@iwantagoodnameplease]

In the second example, when D2 is considered to be reverse biased you still attribute a 0.7v drop across it. Is this still true, as it's not conducting?

[@CircuitBread]

Hey there! There shouldn't be a .7V on a reverse biased diode but I'm not sure what portion of the video you're referring to as D2 doesn't have the .7V drop across it in the second example. Can you give me a time stamp?

[@iwantagoodnameplease]

@@CircuitBread it's the part at 12:20-13:00. I'm mainly confused about how we're taking the voltage across a reverse biased diode, but I guess the voltage is still there even if it's not conducting.

[@CircuitBread]

Yeah, you're exactly right. That -9V below the 22K resistor stays there even when the diode is removed. And since there is no current flowing through the 22K resistor (since the diode is open), that same -9V is on both sides of the resistor. That's how we have that 8.3V difference across the reverse biased diode.

[@chrisoca958]

I can totally relate at 12:51 - 13:01 especially at exams hahahah

[@CircuitBread]

Haha! I had to rewatch that to know what you were talking about - yeah, that's me in any exam or, in this case, attempting to do basic math on camera 😁

[@chrisoca958]

@@CircuitBread whoa! thank you for noticing! I just wanna say that I hope your channel gain more audience in the future! your content is worth the watch!!! ^_^

[@CircuitBread]

Thanks Chris!

[@37latitude]

5:55 Why does I1 current split to make I2 and I3. Is it because voltage higher on anode for diode 2.

[@wa57s577]

Khirchoffs current law

[@0451Deus]

Feeling like my professor threw our class to the wolves, on the same day he introduced diodes he wants us solving the peak forward current through them and we have both AC and DC sources in the circuits. Any chance you have an example like that?

[@CircuitBread]

Hey Matthew, that sucks - I really hate that feeling. While we don't have anything for that explicitly, I'd recommend approaching it with superposition to simplify things: www.circuitbread.com/tutorials/solving-circuits-with-superposition-theorem Basically, find your peak forward voltage with the AC circuit and then had that to the forward voltage caused by the DC circuits. The challenge here is that you have to be sure you're biasing your diodes properly in your assumptions. Since you're planning on it being your peak forward current (and not trying to figure it out at all times), that should make it a bit easier.

[@kr4T0X]

In the last 2 examples:- You took 0.7 V drop in the forward biased diode in last example. Whereas, You didn't took 0.7 V drop in the forward biased diodes in the second last example. Why so? kindly explain please.

[@CircuitBread]

Hi Himanshu! As discussed, there are multiple ways to deal with diodes, some more accurate than others. The easiest and least accurate method is to treat a diode as if it's a short when it's forward biased, as I did in the second to last example. An easy and more accurate method is to take the .7V drop into consideration, as I did in the last example. It was just showing the difference between the two options.

[@kr4T0X]

@@CircuitBread thank you. : )

[@TKR911]

dang ! you have your own microscope for soldering ? nice. You have a nice man-cave setup!

[@CircuitBread]

Yeah, I'm definitely spoiled. It was over 10 years ago that I got that microscope but it was (relatively) inexpensive, particularly compared to a Zeiss. It's been absolutely awesome in many, many situations.

[@quincejohnsjr.9431]

how was it 409ma when you already converted ma to amps

[@CircuitBread]

What's the time stamp of the confusion?

[@derekmitt23]

I think you are using conventional flow theory and not electron flow theory. Is this correct?

[@CircuitBread]

That is correct! Other than working on things at the semiconductor level, I always use the conventional current flow.

[@95yahel]

For D2, shouldn't it be 0.7 - -9.7? ( which is 9.7 V)

[@CircuitBread]

What's the time stamp?

[@95yahel]

12:46. Awesome videos btw I told half my class to check out your videos!

[@CircuitBread]

Nice, thanks! That's a great question and can definitely be confusing. I may be off-base here, but I think the core of your confusion is you're trying to use Ohm's Law, where you calculate the current by subtracting the second voltage from the first. We're not calculating current here, we're trying to figure out the voltage drop. If that were a resistor, rather than a diode, and we were calculating current, you would get a positive number, which is correct! Because the current flow would be in the direction of our assumption. But in this case, we just know that the the voltage, using our assumption of moving from left to right, is going from .7V to -9V, so it's dropping by 9.7V. Which gives us -9.7V. Hopefully that helps!

[@95yahel]

Oh my gosh that's so basic your right! Thanks man

[@abdul-malikmustapha5394]

In the first problem, you did not consider the voltage drop across the diode

[@CircuitBread]

Hey there! At 2:22, I mention why we ignore the voltage drop across the diode. It's the simplest (though least accurate) method of solving circuits with a diode in them. Hope that clarifies things!

[@abdul-malikmustapha5394]

@@CircuitBread thanks

[@znhait]

I think drawing circuits in such a manner is confusing. Circuits are supposed to be closed and putting voltages at a node with no connection in sight throws people off.

[@CircuitBread]

Isn't that interesting how we get used to viewing circuits a certain way? But people do it different ways and it's good to get comfortable with the different common representations.

[@znhait]

@@CircuitBread I've taken a few circuit courses as an electrical engineer, so I do see the many different ways. However, the fundamentals of circuit analysis are knowing the voltage-current relationship of an element and kirchoff's voltage and current rules. It's just strange with no closed loop evident in a circuit drawing, but circuits with diodes and transistors are usually displayed as shown in your video.

[@kritisohagchakma1747]

I am confused about i1, i2 and i3 of the first math..can you please explain?

[@CircuitBread]

Hi Kriti - how strong is your background in circuits? I'd recommend checking out our new Circuits 1 series. We have a few published videos tutorials here on KZbin (here's the playlist: kzbin.info/www/bejne/sHmQZp-pZb50f5I ) but seven tutorials on our website ( www.circuitbread.com/tutorials/tags/circuit-theory ) that should make this all clear.

[@caliskanata]

complaining about your sketch ? you don't wanna see how i draw an inductor

[@CircuitBread]

😂

[@robertkeller2309]

this was absolutely terrible.