In this course students learn the basic concepts of acoustics and electronics and how they can applied to understand musical sound and make music with electronic instruments. Topics include: sound waves, musical sound, basic electronics, and applications of these basic principles in amplifiers and speaker design.
Capacitors
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4.6（312 个评分）
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HM
Mar 10, 2018
Very good course, some of the math is a bit advanced, but the explanations are excellent. Thank you very much for this opportunity. I am eagerly waiting for the part 2!!
RG
Apr 18, 2017
Curso muito Bom! Completo com diversos assuntos interessantes para quem é músico e se interessa pela parte técnica. Uma boa revisão para os engenheiros eletrônicos!
从本节课中
Week 3 - Some Necessary Mathematics MB & AC Circuit Analysis
Transistors, vacuum tubes, opamps, amplification, power gain, single-stage amplifiers.
Converting sound to electrical signals –microphones and guitar pickups, converting electrical signals to sound – loudspeakers
教学方
Robert Clark
Provost and Senior Vice President for Research and Professor
Mark Bocko
Distinguished Professor and Chair
In this lecture, we're going to start working on AC circuit analysis.
And so the first thing we have to do is talk about two new circuit components,
capacitors and inductors. And then, after introducing those, we're
going to talk about decomposing signals into their sinusoidal components and the
idea of phasors. Then with all of those tools, we can
start to do some AC circuit analysis, and then using those techniques, we're going
to look at a few simple filters. Low pass and high pass filters built out
of resistors, capacitors, and inductors. We'll start by talking about capacitors
first. So, capacitors are nothing more than a
pair of metal plates that are separated by either a vacuum or maybe some kind of
a dielectric material. And there's say a gap, d, in the plates,
and on those plates, we can place some charge.
So we can put plus charge on one plate and minus charge on the other plate, and
you know that if I segregate charge in that way, I get an electric field
pointing from the plus to the minus charge, and so there's a, there's an
electric field contained inside the capacitor.
Now, the electric field in the capacitor gives rise to a voltage difference
between the terminals. So this is a device that when I put
charge on the two plates, there's a voltage difference between the two
terminals. Now, energy is stored in a capacitor in
the form of the electric field contained in the dielectric material, and you
remember from before that the electric field is the voltage divided by the
distance from the the plus charged to the minus charged plates.
Or we can rewrite that as the voltage is the electric field times that distance,
or the gap in the capacitor. Now the capacitance of this structure Is
the ability of this device to store a charge, or the capacity of the device to
store a charge. Now, we define capacitance as the number
of Coulombs per volts. So, it's Q over V.
And capacitance is measured in farads, and so, this device has a capacitance, c,
we'll cal it, and it's defined by this relationship, q is c times v.
So, if I have, if I take this device and let's say I put one volt of potential
difference on the two plates, then there will be some amount of charge stored in
the device that is equal to the capacitance times that voltage.
So a one-farad capacitor means that if I put one volt pepit, potential difference
here, I'll get one coulomb of stored charge.
Now a 1 Farad capacitor is a very big capacitor.
typical capacitor values go from just a few picofarads, 10 to the minus 12
farads, to maybe a few hundred millifarads, 10 to the minus 3.
And picofarad-type capacitors are, you'll see them in various forms in electronic
circuits. capacitors with leads maybe a few
millimeters or maybe 5 millimeters across.
the surface-mount capacitors are typically quite small, maybe 2 or 3
millimeters in size. And so these are on the small end.
Now, millifarad capacitors or hundreds of millifarad capacitors are the, the big
ones that if you open up a piece of equipment, you see some big capacitors in
there that are part of the power supply, typically, and so these may be several
inches high and a few inches in diameter. And this is just very large coil, of two
metal electrodes, and some a kind of dialectic material, often there's there's
a a liquid in the servicing as the dialectic material, and the, these are.
Actually, fairly formidable devices and you have to be careful if you charge one
up to substantial voltage and, it holds a lot of charge and you can get quit a
shock from one of these. So you have to be careful with those.
And the other thing, I'll tell you this from personal experience, is that, when
you think you've discharged it, say touching 2 terminals with a screwdriver.
you better check twice, because I it knocked myself off a chair once by
touching the one of the electrodes of a large capacitor that I thought was
discharged. I discharged it but it wasn't discharged
all the way, and there was still enough residual charge to knock me off a chair.
Okay, now I want to look at the current voltage relationship for a capacitor.
We gave you the charge voltage relationship for a capacitor which is
really just the equation that defines the capacitance of that device.
Now let start with Q equals cv that;s just remember that equation that's that's
the defining equation for a capacitor and I'm going to take and take the derivative
of this with respect to time so dQ dt is current and.
Now I'll assume that the capacitance, c, is fixed, it's not changing value.
So it's just a constant. So when I take a derivative of this
constant times the voltage, it's the constant times the derivative of the
voltage. Now, let's use this equation to look at
the transient behavior of a circuit with a resistor and a capacitor in it.
So what I want to do is find the charge, q, and the voltage across the capacitor
as a function of time. Now here's the simple circuit.
I just take a battery, connect a resister and a compassitor in series, like this
and then I have a switch and I'll say that at T equals zero, I'll throw the
switch closed and let's say before I do that I want to make the point that the
capacitor is initially uncharged. So there's no charge in the capacitor,
and obviously there's no current flowing through the circuit, because this is
open. But then at t equals 0, I throw the
switch closed, and the battery is then going to force a current through this
circuit. And the capacitor is going to charge up
and its voltage is going to increase. And we want to know exactly how that
occurs in time. I want to find this voltage, Vc, as a
function of time. So, to do that, I'll start by using
Kirchhoff's voltage law. So.
this is right after the switch is closed. in fact, this is true at all times, but
the, the voltage across the battery is, if I start down here, and I go up through
the battery, I pick up a voltage V0. And then I go through the resistor and I
drop whatever the resistor voltage is, then through the capacitor, drop whatever
that voltage is and I'm back to zero. So this then, can be rewritten as vr plus
vc equals v zero. Now, [COUGH] taking this VR the just
using Ohm's law is just i times r so this is just the voltage across the resistor.
VC is the voltage across the compasitor I can just go back here and write this v is
q over c And so that equals V0. So, I have this equation that I have to
solve for Q as a function of time. So, this is a simple differential
equation that is not any different form the ones we saw, just previously.
when we introduce the exponential function, the one difference is there's a
term here on the right that we have to deal with.
So, I want to say again that the initial condition of this one, as soon as I close
the switch is there is no charge on this capacitor at t equals 0.
And what we'll find is that the final state is that the capacitor will be
charged up and the current will stop, and so once the current stops, there won't be
any voltage drop across the resistor, and the charge, after a long time, it will
just be C times 0. Now we'll find this, this is not an
assumption, we'll find this from the solution.
This is the assumption, the initial condition equals 0.
