Okay, here we are, here's a slide where I promise you we do a calculation. So here's my formula fr equals one over two pi LC, I just plug in my numbers. The only thing here I want to mention is if you look at this 10 to the minus six here, when I find the square root of that, I just divide the minus six by two. And then when that gets placed up here into the numerator, the sign goes from a minus three to plus three, and eight times 20 is 160. Right there, and now I just do my math 10159 times 10 to the minus three odd time to the plus three I'm sorry, is 159. The square root 160 is 12 dot six five kilohertz, not kilohertz hertz.

And there's my answer. All right, so I've done one. The other thing here that I'm showing you is we can transpose these, this formula here, and get C equals one over one over four pi frequency of residence squared times L. And basically what I've done is I've broken this down for you. So one over two pi squared is actually 0254. So all you really need to do is is get the frequency squared times the inductor and that'll give you see Same thing over here but instead of looking, trying to find c over here, we're looking for L Same deal and what this is good for Right here, what value of c? And this is a little bit of a problem, what value of c resonates with 239?

Micro Henry's l at 1000 kilohertz? Well, I got to find C. So I, so basically I go, I do my math. And then I find out that it's 106, Pico farads. And if you remember, Pico is 10 to the minus 12. Right there. So that's what I get.

All right, so take a look at it. There's some problems that I've that I've placed in this course. And I think there's one where I asked you to look at the resonant frequency, and I do give you the answer. So take a look at that. And again, we have done math problems. Pre prior to this.

This is somewhat of an advanced topic, you should have honed your math skills before you got here, either taking some of my previous courses or, or felt as though you've had a basic understanding of math and a basic understanding of electronic principles. All right? If this is your very, very, very first course on electronics, you picked it off the web and he says, I'm going to learn it, you're probably lost, because you don't have a background for this. And I don't mean to come down on anyone. But some of the courses I put on they, they tell me I'm too basic, or I didn't understand. You need to go on building blocks, okay.

You can expect to go 1000 miles an hour, the minute you, you hop on a bicycle for the first time. Alright, so you got to build you got to build the foundation. All right. Okay, enough said let's clear the slide and go on to the next one. All right, we're talking about the cue of the circuit. And we've already we already mentioned this previously.

In the last section where we we we showed you how you can Find the total Z of a parallel resonant circuit by using Q. Alright. And q also will give you the sharpness of the curve meeting up here. Alright. So the higher the queue, the sharper the curve is and and you'll see that it's one or two slides coming up. But what I want you to know and we've already covered is q equals X sub l over Rs. And we know from the previous section, that RS is the actual coil resistance of the coil.

All right, or quite honestly, if I wanted to manipulate the Q, all right. What I could do is put a resistor in series with the coil windings and that would increase my resistance. And what would that do that would lower the queue and I would flatten out the frequency and you'll see what I mean in the next couple of slides here. Alright, so right here on this slide here, okay, the higher the queue the sharp of the curve, even though I didn't put it down there, the lower the queue queue Q, the more flatter the curve is, all right, q equals X sub l divided by Rs. Rs is the series resistance of the coil. All right, in this slide, we're talking about the cue of this circuit, okay, in the queue of the circuit defines the sharpness of the peak at my resonant frequency fr right here.

All right, notice how sharp that is. It comes to a nice point. All right. So If I have a high Q, this gets much sharper. If I have a lower q over here, this gets much wider, it's not as sharp at the peak. And again, I've got a slide that will show you that.

It see the next one or the one after, and you'll see what I mean. Again, q equals X sub l over Rs. Rs is the coil resistance. It's the resistance in the windings of the coil. If I want to play with Rs, isn't it possible that if I wanted to increase the value of Rs, I could put a resistor in series, right with the coil. And now I've increased Rs.

What would happen to my q? Because if if this if this gets higher, this here goes lower. So My curve won't be a sharp and maybe there's a reason why I want that maybe I want a large, I'm going to use a new term here, bandwidth. What was showing you here for whatever reason, maybe I want to increase the bandwidth. So the higher the queue, the shop of the curve, the lower the queue, the more broader the curve. All right, and we'll leave it at that.

And we're going to go on to the next slide and look at it. Alright, on this slide, we Well, we've taken the previous slide, we've expanded it, and we're talking about we added bandwidth, all right, and bandwidth is the resonant frequency divided by q. All right, but before I go on to that, here is my bandwidth right there. Notice we have f1 and f2. All right. That is said 70.7% of the peak.

So let's say that my current here was like was showing you 100 milliamp hours. So f1 which is the first frequency in my bandwidth because my bandwidth goes from here to here. So F one will be the first frequency of that bandwidth, and that that value will be 70.7% of my maximum current. Or if I'm talking about voltage, it would be voltage, all right. So the largest or the the f1 is is the first frequency in my bandwidth. f two is the second frequency in my bandwidth over here.

That also will be 70.7% sent right there. Okay. All right. And from here to here is my complete bandwidth. Alright, so for instance, on this example, since my resonant frequency is 50 kilohertz, my cue is 2.5. I do the math.

And my bandwidth from f1 to f2 is 20 kilohertz. Now I can find the frequency of f1 and find the frequency f2. So when I show you here, f1 would be my resonant frequency minus the bandwidth divided by two so it'd be 50. All right, minus 20 divided by two, I do the math that's 10 kilohertz. So I come up with 40. And over here, I add it and that comes 60.

So my, my bandwidth I have a I have a 20 kilohertz band. With that's the width from F one to F co I, but the frequency of F one is 40 kilohertz. And the frequency of f Q is 60 kilohertz if I take the difference between them 60 minus 40. That's 20. And these are both kilohertz. So it's 20 kilohertz.

That's my bandwidth. That's my bandwidth. Alright, Nuff said. Let's go on. Okay, on this one here, the only thing I wanted to do is remember at the beginning, I told you we'd have a slide about the with my bandwidth. Well, well, that's what I'm showing you here.

All right. He was at I have a 10 kilohertz bandwidth. if q is 40, I have a 20 kilohertz bandwidth. And if q was 10, I get an 80 kilohertz bandwidth. All right, let me clear the slide off, I want to just emphasize one other thing. So if this is f one, f two over there, notice my bandwidth is tighter.

My bandwidth is larger. My bandwidth is larger. Alright. So that's what I meant. Were I if I play with the cue of my coil, I can change the bandwidth. And how do I change the cue of the coil?

I change it by adding a resistor. All right. That's how I hide change that. All right, that set up. We're going to stop here. We've done presidents.

All right, we've done both series in parallel resonance. The last section in this This module will be filters. All right. So see over there, hope you enjoyed it so far.