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'[OT] Impedance'
1999\03\13@123109 by Eric Oliver

flavicon
face
Ok,

This is really just an electronics question.  I hope that one of the
experts on this list will take the time to enlighten me a little.  I have
been struggling with truly understanding the concept of impedance. More
importantly, I don't understand the role it plays in the behavior of any
given circuit.  As a newbie, I sometimes try to picture the electrons
racing through a conductor, and how a component affects the behavior of
that electron.  For instance, I understand the concept of resistance ( at
least to some degree ) and ( to a lesser degree ) the concept of
capacitance.

Impedance is resistance to change ( correct ? ). To this end, I believe the
older I get the higher my impedance <g>.  Anyway, impedance is a term that
is used here on this list ( and elsewhere ) frequently in discussing all
aspects of circuit design and troubleshooting.  I thought if I read enough
that eventually I would gain a better understanding of the term by seeing
how it was used in everyday electronics problem solving.  Alas, this hasn't
happened.

Please enlighten me, oh great ones <g>.

Eric

1999\03\13@133652 by Wagner Lipnharski

picon face
Impedance is an opposition to electric charge or any other influence
that can result in this, to travel from
one place to another in a stated conductive matter.

It is exactly the same as "resistance" but it can be the result of
several physical variables.

For example, an inductor has a variable impedance based on the magnetic
field generated around it by the electric current that crosses it.  When
this magnetic field collapses it generate an inverted electric current,
that will be opposed to the one you are trying to make cross the
inductor, so that is the "impedance".

A transistor with a resistor in its emitter, has this resistor
resistance transferred to its base by a multiplication factor of the
(beta) gain of the transistor.  A transistor with a beta of 100, with a
emitter resistor of 20 Ohms, will have an impedance of 2k Ohms at its
base. So, if you apply 2Volts at the transistor base, a current of
(2-0.6)/2000 = 700uA will flow into the base and will result of an
emitter current of 100 times more, 70mA. It is not a pure resistance and
it can change based on variables, so we call it just "impedance".  (0.6
is the base-emitter junction silicon transistor voltage drop, germanium
is 0.2V).

A capacitor has some kind of "resistance" to electric current. During
the charge of its plates, there is a current flowing at its leads, it
changes according to space available at the plates for the electrons, so
it has a variable "resistance" we choose to call it "capacitive
impedance" because it is an average result in such applied alternate
frequency.

When you combine all of those elements in a circuit, each one will rule
a different impedance, but when you state a certain operation frequency,
you can determine the overall circuit impedance, calculating and
associating each component impedance.  This is the way we calculate
filters for example. The best power transfer is a method of equalize
impedance from the source to the consumer.

--------------------------------------------------------
Wagner Lipnharski - UST Research Inc. - Orlando, Florida
Forum and microcontroller web site:  http://www.ustr.net
Microcontrollers Survey:  http://www.ustr.net/tellme.htm

1999\03\13@134107 by Gerhard Fiedler

picon face
At 11:23 03/13/99 -0600, Eric Oliver wrote:
>I have
>been struggling with truly understanding the concept of impedance.

i'll give it a try. when you try to describe the behaviour of a given
circuit, you use the voltages at all the connection points (also called
"nodes") and the currents between them (in the "branches" -- do i use the
proper english terms here?)., so for any "branch," ie. for any connection
between two nodes, you have a difference in potential (a voltage
difference) between them and a current flowing between them, like this:

... (node1, v1)  -----/\/\/\----- (node2, v2)
                   R (->I)

so, if all that happens with DC only, we say

R = (v2-v1)/I = v/I

that's the "concept of resistance;" it relates voltage and current. and it
does it in a way that leaves both of them "in phase," which means that when
you look at both with a scope, they have the same form and the peaks at the
same time.

now lets put some changes with time in there and go to AC. usually, in all
"real world" situations, the peaks of current and voltage are not exactly
at the same time (although the difference may be so small that it's hard to
measure). this is caused by the effect that the resistance is not a pure
resistance, but that it is an "impedance." which means nothing more than
that it has a capacitive and/or inductive component. the capacitive
component makes the current partly dependent of the =change= of the voltage
(differential), and the inductive component makes the voltage partly
dependent on the =change= of the current (again the differential).

C = I / (dv/dt)
L = v / (dI/dt)

so, an impedance is a generalized resistance, capacitance, inductance. the
three of them are basically three special forms of impedance, the "pure"
forms. and every impedance can be described in combining these three
elements, in the appropriate serial or parallel circiut.

for example, a wire. for most purposes, you can just assume it's a mere
(ideal) conductor. sometimes, with high currents, you have to look at it as
a resistor and take the dissipated power into accout. at other times, with
high frequencies, its inductive characteristics may be important, or its
capacity (with, say, another wire). so, it often depends on what you're
looking at which parts of a "real" impedance you're using in your
calculations and which ones you find negligible. but basically, =every=
resistance, capacitance, inductance, conductor... =everything= is an
"impedance" and has all three characteristics to varying degrees.


i hope this gave you an idea -- i'm sure there are some more "enlightened"
ones out there to correct or clarify or give better examples :)

ge

1999\03\13@150528 by Sean Breheny

face picon face
Hi Eric,

At 11:23 AM 3/13/99 -0600, you wrote:
>Ok,
>
>This is really just an electronics question.  I hope that one of the
>experts on this list will take the time to enlighten me a little.  I have
>been struggling with truly understanding the concept of impedance. More
>importantly, I don't understand the role it plays in the behavior of any
>given circuit.  As a newbie, I sometimes try to picture the electrons
>racing through a conductor, and how a component affects the behavior of
>that electron.  For instance, I understand the concept of resistance ( at
>least to some degree ) and ( to a lesser degree ) the concept of
>capacitance.

Resistance is a pretty universal thing. In all cases,it is a very complex
phenomenon which manefests itself on a macroscopic scale as a linear effect
(to very good approximation). Electrical resistance comes primarily (in the
case of conductors) from interactions between the electrons in a current
flow and the molecules of the material. It is important to realize that in
a regular conductor (not a superconductor or in an electron beam) the
electrons don't stream along at the speed of light. They go MUCH slower
because they are constantly bumping into molecules and changing direction.
The net result is a drift of less than a millimeter per second! However, we
say that "electricity" moves at the speed of light because the electric
field propagates along the wire at the speed of light in that material.

To make a long story short,this "bumping" and drifting behavior in a
conductor can be and is approximated as a resistance.

As for capacitance and inductance, these are direct results of Maxwell's
equations and relativity. In the case of capacitance,there is really only
one important equation to know:

Q=CV , or Charge = Capacitance * Voltage

This means as you dump charge into a capacitor,the voltage will rise at a
rate inversely proportional to the capacitance. Sort of like a water tank
with a certain diameter. The greater the diameter(capacitance),the slower
the water level(voltage),will rise as you put in a stream of water with a
certain flow rate(Charge per second or Amps). So, if you put the equation
into differential form,you get I=C dV/dt, which states what I just
illustrated with the water tank. Another useful analogy is a spring. If you
treat current like velocity,then as you compress a spring at a cetain
velocity,it builds up force against you in the opposite direction which
increases linearly with time if you have a constant velocity while
compressing.

Inductance comes from the relationship among current,voltage,and the
magnetic field. When you have a loop or coil of wire, the current thru it
makes a magnetic field which when changed,produces a voltage around the
loop. So,if you try to change the current, the magnetic field will
change,and induce a voltage OPPOSITE to the current CHANGE you caused. The
key equation for this is:

V = -L * dI/dt

or in words, the voltage across an inductor is equal to minus the
inductance (Henries) sime the rate of current change. A good mechanical
analogy is a block of steel. <G> If you hurl the block of steel to
accelerate it from 0 to 10 meters per second,you feel a force. This is
similar to a current change from 0 to 10 amps producing a voltage in the
opposite direction.
As the block flies thru the air,there is no force on it (we are out in
space),so this is like an inductor with a constant current thru it. An
ideal inductor would produce no voltage across it. If you stop the block,
you will feel a impact or a force against you. This is just like the
"inductive kick" or voltage spike that you get when you suddenly turn off
the current thru a inductor,or the voltage which you get across it which
opposes any current change.

As stated by some of the other people who responded to your question,
resistance and reactance don't HAVE to be generated by actual physical
resistors,inductors,or capacitors,they can come from other types of
circuits,such as an amplifier. For example,using an op-amp,you can turn a
100 ohm resistor into a resistor with a different resistance by controlling
the voltage that appears across it. It is this effect that causes input and
output impedance of an amplifier o be altered by the use of feedback.

Sorry to be long winded,but these concepts are critical to understanding
impedance and its nature.

>
>Impedance is resistance to change ( correct ? ). To this end, I believe the
>older I get the higher my impedance <g>.  Anyway, impedance is a term that
>is used here on this list ( and elsewhere ) frequently in discussing all
>aspects of circuit design and troubleshooting.  I thought if I read enough
>that eventually I would gain a better understanding of the term by seeing
>how it was used in everyday electronics problem solving.  Alas, this hasn't
>happened.
>

You are mostly correct. Actually, reactance is the resistance to change in
either current(in the case of positive reactance or inductance ) or voltage
(in the case of negative reactance or capacitance). Reactance is the
imaginary part of impedance. Resistance is the real part.
So, impedance is represented like this:

Z = R + j * X

where j is the imaginary unit = sqrt ( -1 )

Most of the time when people throw around the term impedance, I find that
they are usually talking about a resistance,but they just use the term
impedance because it is more general and they are used to it. If someone
just says "the impedance is 50 ohms",they have to mean 50 ohms resistance
unless they were to specify more information such as "the impedance is 35 +
j * 35 ohms" which would mean that you have a 35 ohm resistor in series
with a 35 ohm inductive reactance.

To illustrate exactly what this complex number means,I think we should go
back to the mechanical analogy: Lets say we have a spring and you compress
it with a certain speed. At first you feel very little force. However,
after compressing it for a short period of time,you begin to feel a
force,and you continue to feel this force even if you stop compressing and
reverse and allow it to expand. Then,eventually,the force will go to zero
and you are still de-compressing it,and in fact now EXPANDING it. Again,
you are at a high rate of expansion,but don't immediately feel much force
in the opposite direction. If you applied a rate of compression which was a
sine wave,then the "bucking" force of the spring would also be a sinusoidal
wave,but with a PHASE SHIFT because of the delay in feeling the force.

The same is true for the mechanical analogy for the inductor. You push on
it hard but it doesn't initially move very fast. However,if you continue to
push,eventually it will reach a high speed,and then if you pull on it to
stop,it will take a short period of time to slow down before it actually
does stop. Same effect, a phase shift.

Resistance is different,however. The mechanical analogy to resistance is a
shock absorber. If you push hard on a shock absorber,it will immediately
reach a very slow rate of compression and that rate will not change until
you release your force,at which point it will immediately stop compressing.
There is no time delay between the application of force and its result,so
no phase shift.

Mathematically,these phase shifts come from the DERIVATIVES in the above
expressions relating V and I for the inductor and capacitor.

In the complex number Z (or impedance),Z is the ratio of Voltage to
Current. If the Voltage and Current are in phase,then their ratio is simply
a regular real number,hence the resistance part of Z. If,however, they are
out of phase by 90 deg or -90 deg (which will be the case for a pure
capacitance or a pure inductance),then there is a factor of "j" to
represent this. The actual reason for the "j" is that in the mathematics,
the general expression for a sinusoidal waveform is A*exp(j*w*t+j*phi) and
multiplying this by a "j" causes "phi" (the phase angle) to change by 90 deg.

The mathematics to see this is a bit lengthy so I won't actually go through
it,but it is easy to see if you know how to take derivatives of exponential
functions,you can just plug A*exp(...) into the expressions I = C * dV/dt
and V = -L * dI/dt and you will see how a capacitor or inductor affects a
sinusoidal wave. Z is then just the ratio of the amplitudes of V and I.
This is where those nifty little formulas Xc = 1/(2*pi*f*C) and Xl =
2*pi*f*L come from.

Horowitz and Hill explain all this in their chapter on fundamentals in "The
Art of Electronics" but it can still be a bit confusing if you forget your
calculus or never learned it.

I hope I didn't confuse things further. This topic is one of my favorites
because it is the core of something called the theory of linear
systems,which is the basic idea behind 90 % of science and especially
practical matters such as engineering. To put it very simply,the study of
linear systems is just the idea that linearization of semi-linear behavior
of a system allows you to use POWERFUL mathematical tools to describe the
behavior of the system. The ideas if Impedance, Fourier and Laplace
Transforms, s-domain analysis, etc. are all parts of this idea,and it is an
integral part of nature and the study of it,from E&M to quantum
mecahnics,the list goes on and on.

>Please enlighten me, oh great ones <g>.
>
>Eric
>

Sean

|
| Sean Breheny
| Amateur Radio Callsign: KA3YXM
| Electrical Engineering Student
\--------------=----------------
Save lives, please look at http://www.all.org
Personal page: http://www.people.cornell.edu/pages/shb7
spam_OUTshb7TakeThisOuTspamcornell.edu ICQ #: 3329174

1999\03\13@150742 by Bob Drzyzgula

flavicon
face
> Please enlighten me, oh great ones <g>.

Well, I'm only a meager one, but it has only been recently
that I've started to grasp the significance of some of these
things and I can try to give you some of the hints that
helped me along with this.

First of all, I highly recommend the book "A Practical
Introduction to Electronic Circuits", by Martin Hartley
Jones, from Cambridge U.P. (ISBN 0-521-47879-0). As
is the case with many Cambridge books, the writing is
excellent, and great care is taken to fully explain many
basic concepts that too many other books assume away, gloss
over, or explain at too deep a level. In particular, I
found Jones' explaination of impedance to be crystal-clear.

One thing that took me a while to appreciate is why
one should care quite so much about impedance, and what
the significance of "impedance matching", "high-impedance
inputs" and "low-impedance outputs" were.

In impedance matching, one is concerned about the transfer
of an electrical property across a boundry of some sort,
commonly a power supply and device, or a signal line and
a reciever.

It isn't obvious, (or it wasn't to me) what the "output
impedance" or "input impedance" of a device were.
Input impedance is pretty easy... that's just what you
would be able to measure across the input terminals of
the device; if it were a pure resistance, you could simply
take a VOM and measure the resistance through the device.
Output impedance is kind of hazier, though.  As Jones
explains, the best way to think about it is to look at
the Thevenin equivalent circuit of a device.  This goes
back to Thevenin's theorem, which basically says that
any reactive circuit (e.g. one containing resistances,
power sources, capacitance and inductance but no active
components such as transistors or diodes) can in essence be
replaced by a new circuit having a single power source and
a single impedance. If you think of the impedance being
between power source and the output of the device, and
then connect your output device to the input of another
reactive device, then what you have is simply a classic
voltage divider, albeit the AC version of such, where the
ratio will depend on the frequency of the output signal
and the complex nature of the impedances. But for a static
frequency the ratio should be fixed.

Now suppose that you had a power supply with an internal
voltage source of 5V and a 500 Ohm Thevenin-equivalent
output impedance, and suppose that you wanted to use it to
drive an 8 Ohm device that needed an input voltage of 4V in
order to power on. Would this work? Well, no, because your
resulting voltage divider would drop 500/508 of the source
voltage within the power supply itself, and 8/508 of the
source voltage across the load; the load would only ever
see a 0.078V input, and would thus never turn on. Thus, you
have a very mushy power supply [I recall, when I was just
starting to learn some of this stuff, trying to drive a 5V
microcontroller board off the +5V output of a small power
supply that was originally intended for driving an LED (OK,
y'all can snicker in unison).  What I didn't know was that
the LED header had a 680 Ohm resistor behind it to limit
the current through the LED, as is standard practice, and
of course as soon as I tried to drive a real circuit off
that header, the power supply fell right over.]

What you would really want here is for the actual output
voltage of the power supply to remain as close as possible
to the voltage of the internal voltage source. It can
pretty quickly be seen (and Jones derives) that this will
happen when the input impedance of your device is much
greater than the output impedance of the power supply. In
the case of an 8 Ohm device, you would need a power supply
with almost no output impedance. (Keep in mind that, in an
AC circuit, all these calculations will be dependant on the
frequency of the signal). This is impedance matching for
optimal transfer of voltage, and this is why you want the
inputs on your oscilliscope to have a very high impendance
across all frequencies; if this were not the case, your DUT
would just fall right over when you attached it the 'scope.

Similar analysis will show that maximum power transfer
will occur when the input and output impedances of
attached devices are equal. Jones points out that this
isn't often an interesting criterion, because it means
that as much power is dissipated in the power supply as
is dissipated in the driven device. However, it does become
useful when attaching a long transmission line to a
receiver; equating the input impedance of the reciever to
the charactaristic impedance of the cable will minimize
reflections at the end of the cable.

Jones goes on to explain that optimal current transfer will
occur when the overall impedance in the circuit is minimized;
if the output impedance of the power supply is assumed to
be fixed, then this goal is achieved by making the impedance
of the attached device as close to zero as possible. In
a sense, this is the opposite of optimal voltage transfer.

I don't write nearly as well as Jones does, and this
clearly suffers for the lack of pictures. But I hope that
it helps.

--Bob

--
============================================================
Bob Drzyzgula                             It's not a problem
.....bobKILLspamspam@spam@drzyzgula.org                until something bad happens
============================================================

1999\03\13@152023 by Sean Breheny

face picon face
Oops,

One quick IMPORTANT point I left out in what I was trying to say while on
my Impedance soapbox:

One of the biggest distinctions between resistance and reactance is that
ONLY RESISTANCE can actually dissipate power. Across a resistor, the
voltage an current are in phase and so the power absorbed is positive thru
the whole AC cycle. Across an inductor or a cap,however, the product of
voltage and current is positive for half the cycle and negative for the
other half. This means that the cap or inductor is STORING power for half
the cycle and RELEASING power for the other half, for a net gain/loss of
ZERO. Of course, real caps and inductors have resistance,too,due to the
wire,etc., so they actually DO consume power,but they are best modeled by a
resistor in series with a perfect cap or inductor.

Sean

|
| Sean Breheny
| Amateur Radio Callsign: KA3YXM
| Electrical Engineering Student
\--------------=----------------
Save lives, please look at http://www.all.org
Personal page: http://www.people.cornell.edu/pages/shb7
shb7spamKILLspamcornell.edu ICQ #: 3329174

1999\03\13@152230 by Wagner Lipnharski

picon face
Sean Breheny wrote:
[snip]
> Inductance comes from the relationship among current,voltage,and the
> magnetic field. When you have a loop or coil of wire, the current thru it
> makes a magnetic field which when changed,...

I think magnetic field and inductance effect develop
also in a plain straight conductor...

Wagner

1999\03\13@153715 by Sean Breheny

face picon face
Hi Wagner,

At 03:21 PM 3/13/99 -0500, you wrote:
>I think magnetic field and inductance effect develop
>also in a plain straight conductor...

I have often thought about this,and I'm not sure exactly what the answer
is. First of all, any time you have a current flowing, you will have  a
loop (even if the wire is straight for a long distance,something must
eventually connect it back to the supply). Yes, there will be a magnetic
field around the wire,but I don't know how you could analyze how much
voltage this will induce without taking some loop area into account,since
the voltage induced is the magnetic field (flux density) integrated over
some area, times a constant.

>
>Wagner
>

Sean

|
| Sean Breheny
| Amateur Radio Callsign: KA3YXM
| Electrical Engineering Student
\--------------=----------------
Save lives, please look at http://www.all.org
Personal page: http://www.people.cornell.edu/pages/shb7
.....shb7KILLspamspam.....cornell.edu ICQ #: 3329174

1999\03\13@154339 by Wagner Lipnharski

picon face
Bob Drzyzgula wrote:
[snip]
> In impedance matching, one is concerned about the transfer
> of an electrical property across a boundry of some sort,
> commonly a power supply and device, or a signal line and
> a reciever...
[snip]

Just using your example Bob, nothing to do with what you wrote.

Some people think that matching impedance means to connect two
circuits with equal impedance.  Years ago, I heard an electric
engineer saying that to match the two circuits we were working,
(one had lower input impedance), it was just a matter to install
a series resistor and "voila", both with same impedance.

Never in my life heard bigger stupidity.

The matching impedance means "productive" impedance, not any
other sort of miraculous solution.

Suppose you have an sound amplifier with an output impedance
of 4 Ohms, a specific waveform is delivering 8 Volts.  
Suppose you have this circuit attaching speakers of 8, 4 and 3 Ohms:

8V --- RINT 4 Ohms-----o-----REXT 8 Ohms---Gnd
Current = 8 / 4+8 = 0.66A,  Power at REXT: R×I» = 3.54W

8V --- RINT 4 Ohms-----o-----REXT 4 Ohms---Gnd
Current = 8 / 4+4 = 1A,     Power at REXT: R×I» = 4W

8V --- RINT 4 Ohms-----o-----REXT 3 Ohms---Gnd
Current = 8 / 4+3 = 1.142A, Power at REXT: R×I» = 3.91W

If the miraculous solution of that "engineer" would used
here in first example, we needed to apply a resistor of
8 Ohms in parallel with the 8 Ohms speaker, so both
circuits would be matching impedances, right?

8V --- RINT 4 Ohms-----o-----REXT 8//8 (4 Ohms)---Gnd
Current = 8 / 4+4 = 1A, Power at REXT: R×I» = 4W,
but that power would be divided by 2, between the
8 Ohms speaker and the 8 Ohms resistor. So the
productivity of that solution is at least ridiculous,
even that the amplifier would be delivering the max
possible power, the usage of it would be a disaster.

It means that the "whole" circuit impedance, what appears
at the input needs to match to the other "whole" circuit
impedance, what appears at the output, nothing different.
--------------------------------------------------------
Wagner Lipnharski - UST Research Inc. - Orlando, Florida
Forum and microcontroller web site:  http://www.ustr.net
Microcontrollers Survey:  http://www.ustr.net/tellme.htm

1999\03\13@160020 by Wagner Lipnharski

picon face
Sean Breheny wrote:
>
> Hi Wagner,
> I have often thought about this,and I'm not sure exactly what the answer
> is. First of all, any time you have a current flowing, you will have  a
> loop (even if the wire is straight for a long distance,something must
> eventually connect it back to the supply). Yes, there will be a magnetic
> field around the wire,but I don't know how you could analyze how much
> voltage this will induce without taking some loop area into account,since
> the voltage induced is the magnetic field (flux density) integrated over
> some area, times a constant.
> Sean

Yes Sean, it develops magnetic field and in the same rules and manner.
AC current meters use a coild around a plain wire to measure it.
Coiling a wire just increase the flux.

A question to think:

What happens if you build a coil with 50,000 turns of a superconductor?
Suppose the overall impedance allows a simple 1.5V battery cell to
supply 200mA of pulsing current. This arrange can generate a stupendous
magnetic flux.. right? or wrong? During the developing of the flux,
it also creates the impedance, right?
--------------------------------------------------------
Wagner Lipnharski - UST Research Inc. - Orlando, Florida
Forum and microcontroller web site:  http://www.ustr.net
Microcontrollers Survey:  http://www.ustr.net/tellme.htm

1999\03\13@162126 by Sean Breheny

face picon face
Hi again Wagner,

At 03:59 PM 3/13/99 -0500, you wrote:
>
>Yes Sean, it develops magnetic field and in the same rules and manner.
>AC current meters use a coild around a plain wire to measure it.
>Coiling a wire just increase the flux.

Yes,I was never arguing if it created a B field or not,just wondering how
to calculate the impedance if there is no loop,since there has to be some
area to integrate over to get the induced voltage.

>
>A question to think:
>
>What happens if you build a coil with 50,000 turns of a superconductor?
>Suppose the overall impedance allows a simple 1.5V battery cell to
>supply 200mA of pulsing current. This arrange can generate a stupendous
>magnetic flux.. right? or wrong? During the developing of the flux,
>it also creates the impedance, right?

Well, I'm not exactly sure what you are getting at. As far as I can see,
here's what would happen. If we assume that the 1.5v battery has an
infinite capacity but finite resistance,then the flux would build up slowly
and continue to build up until the current in the coil was 200mA (or what
ever the resistance of the battery limited it to be). If the battery didn't
have infinite capacity,then depending on the geometry of our coil, we would
reach some limiting flux where all the energy of the battery was transfered
to the coil and then the battery would act as a resistor to slowly bleed
off the energy and eventually we would get down to no flux again.

As for pulsing,we couldn't actually pulse the current,since if we tried to
suddenly shut off the current, we would get a massive inductive kick which
would probably blow up our switching circuit! We could modulate the current
in a more gentle way,though,but it would have to be done very slowly
because of the high inductance. Every time we decreased the current,we
could reclaim some of the energy because the inductor would be producing a
voltage in the same direction as the current,hence producing positive
power. Every time we increased the current,we would need to drain energy
from the battery because the current and voltage would be opposite in
sign,hence the inductor would be absorbing energy,turning it into a
magnetic field.

Sean


>--------------------------------------------------------
>Wagner Lipnharski - UST Research Inc. - Orlando, Florida
>Forum and microcontroller web site:  http://www.ustr.net
>Microcontrollers Survey:  http://www.ustr.net/tellme.htm
>
|
| Sean Breheny
| Amateur Radio Callsign: KA3YXM
| Electrical Engineering Student
\--------------=----------------
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1999\03\13@175759 by Wagner Lipnharski

picon face
Sean Breheny wrote:
> As for pulsing,we couldn't actually pulse the current,since if we tried to
> suddenly shut off the current, we would get a massive inductive kick which
> would probably blow up our switching circuit...

Well, this is the point, a simple 1.5V battery at 200mA can blow up a
switching circuit. This happens not because the power accumulated, but
because how that magnetic field is rebuild as Voltage x Current.

The collapsing magnetic field will generate an electric power almost
equivalent to what generated the field, but because the the circuit is
open, with almost none current, the voltage will increase freely very
fast, until it reaches the rupture point of the switch, or the
transistor, just frying it. This also happens because all the power
accumulated in time is applied at once over the transistor junction.  If
a single low rupture voltage component as a diode is used in parallel to
the coil, but with the polarity inverted to the applied voltage, it will
avoid the creation of such monster by shorting circuit it in a "timed"
situation. It looks like to try to blow up a kids party balloon, with a
big hole leaking pressure. The diode will dissipate that energy along
with the time the field is collapsing, so nothing happens.

Huge magnets have the problem of wire resistance, that limit the peak
current. The use of large diameter wire to reduce the resistance,
expands the coil diameter, reducing the magnetic field flux.  The
possible use of the super-conductor, reduce substantially the coil
diameter with incredible current, generates huge magnetic fields.

Imagine a 1F (yes, one Farad) 30Volts capacitor fully charged, with a
low HSR, lets say 0.01 Ohms discharging in a 0.01 Ohms inductor of super
conductive material, 10mm diameter, 5000 coils.  It will be an initial
peak of 1500 Amperes (is that right?)... during aprox 10ms? how far that
magnetic field can goes? how much damage it can create in our magnetic
media world in the vicinity?  We are talking about a momentarily
45kWatts...

It can rupture several silicon gates in the vicinity circuits, right?
One can say it is a home made poor's man EMI pulse bomb?
--------------------------------------------------------
Wagner Lipnharski - UST Research Inc. - Orlando, Florida
Forum and microcontroller web site:  http://www.ustr.net
Microcontrollers Survey:  http://www.ustr.net/tellme.htm

1999\03\13@180629 by Peter J. Bisbal

picon face
       Thank you for the explanation! The understanding finally sunk in!

====================== Original Message Follows ====================

>> Date:  13-Mar-99 13:36:25  MsgID: MC2-6DE5-69D8  ToID: 73130,2721
From:  pic microcontroller discussion list >INTERNET:PICLISTspamspam_OUTMITVMA.MIT.EDU
Subj:  Re: [OT] Impedance
Chrg:  $0.00   Imp: Norm   Sens: Std    Receipt: No    Parts: 1


Impedance is an opposition to electric charge or any other influence
that can result in this, to travel from
one place to another in a stated conductive matter.

It is exactly the same as "resistance" but it can be the result of
several physical variables.

For example, an inductor has a variable impedance based on the magnetic
field generated around it by the electric current that crosses it.  When
this magnetic field collapses it generate an inverted electric current,
that will be opposed to the one you are trying to make cross the
inductor, so that is the "impedance".

A transistor with a resistor in its emitter, has this resistor
resistance transferred to its base by a multiplication factor of the
(beta) gain of the transistor.  A transistor with a beta of 100, with a
emitter resistor of 20 Ohms, will have an impedance of 2k Ohms at its
base. So, if you apply 2Volts at the transistor base, a current of
(2-0.6)/2000 = 700uA will flow into the base and will result of an
emitter current of 100 times more, 70mA. It is not a pure resistance and
it can change based on variables, so we call it just "impedance".  (0.6
is the base-emitter junction silicon transistor voltage drop, germanium
is 0.2V).

A capacitor has some kind of "resistance" to electric current. During
the charge of its plates, there is a current flowing at its leads, it
changes according to space available at the plates for the electrons, so
it has a variable "resistance" we choose to call it "capacitive
impedance" because it is an average result in such applied alternate
frequency.

When you combine all of those elements in a circuit, each one will rule
a different impedance, but when you state a certain operation frequency,
you can determine the overall circuit impedance, calculating and
associating each component impedance.  This is the way we calculate
filters for example. The best power transfer is a method of equalize
impedance from the source to the consumer.

--------------------------------------------------------
Wagner Lipnharski - UST Research Inc. - Orlando, Florida
Forum and microcontroller web site:  http://www.ustr.net
Microcontrollers Survey:  http://www.ustr.net/tellme.htm

====================== End of Original Message =====================


Peter J. Bisbal on 13-Mar-99 at 17:43

1999\03\13@182737 by Mike Keitz

picon face
On Sat, 13 Mar 1999 15:43:03 -0500 Wagner Lipnharski
<@spam@wagnerlKILLspamspamEARTHLINK.NET> writes:

>Some people think that matching impedance means to connect two
>circuits with equal impedance.  Years ago, I heard an electric
>engineer saying that to match the two circuits we were working,
>(one had lower input impedance), it was just a matter to install
>a series resistor and "voila", both with same impedance.

Matching impedance for maximum power transfer is generally only important
for situations like microphones and antennas, where the available signal
power is small.

In most other situations a "swamping" design of low source impedance
going to high load impedance is entirely satisfactory.  This design
minimizes power loss, or in other words maximizes efficiency.  Power
supplies and power lines are designed to have as close to zero impedance
as practical.

The other major reason to match impedance is to match both ends of a
cable to its characteristic impedance, preventing signal reflections.
Signal reflections become important when as bandwidth of the signal
becomes significant against 1/ the cable length.  Resistors in series
with the signal source and in parallel with the load are often practical
for matching  to cables.  Much power will be lost in the resistors but it
is compensated for by supplying more power from the source amplifier.


>Suppose you have an sound amplifier with an output impedance=20
>of 4 Ohms, a specific waveform is delivering 8 Volts. =20

Most modern amplifiers use heavy negative feedback, which makes the
output impedance very low.  A minimum load impedance is specified to
prevent damage to the amplifier from excess output current.  Unless it
burns out, the amplifier will act as a constant voltage source.  The
power delivered to a particular speaker is just proportional to 1/Z.

Putting resistors in parallel with the speakers is pointless.  However
resistors in series can be used to reduce the power delivered to some
speakers if a combination of small and large speakers are connected in
parallel to one amplifier.


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1999\03\15@102047 by Harold Hallikainen

picon face
Listmembers might be interested in some of my articles on circuit
analysis.  Some of them are archived at
http://www.broadcast.net/~hhallika .  Go to the theory section.

Harold


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1999\03\15@154031 by John Payson

flavicon
face
|Some people think that matching impedance means to connect two
|circuits with equal impedance.  Years ago, I heard an electric
|engineer saying that to match the two circuits we were working,
|(one had lower input impedance), it was just a matter to install
|a series resistor and "voila", both with same impedance.

|Never in my life heard bigger stupidity.

If you have a modern transistor amplifier which is designed to drive
a pair of 8 ohm speakers and all you have is a pair of 4-ohm speakers,
you can connect a 4 ohm resistor in series with each speaker and then
connect them safely to the amplifier.  About half of your amp's output
power will be wasted in the resistors, but damage to the amp will be
avoided.  Imaging an amp whose full-scale output is 8 volts and whose
maximum safe current is 1 amp.

8 volt signal into 8 ohm speaker: 1 amp x 8 volts = 8 watts
8 volt signal into 4 ohm speaker: 2 amps [DANGER] x 8 volts = 16 watts
8 volt signal into 4 ohm speaker w/ resistor:
 Amplifier outputs: 1 amp x 8 volts = 8 watts
 Speaker actually gets: 1 amp x 4 volts = 4 watts

BTW, in the case of loudspeakers, things actually get a bit more comp-
licated; the impedance of a speaker varies with frequency.  One diff-
erence between "tube" amplifiers and modern transistor designs is that
tube amps often have a high impedance and will transfer more power to
a HIGHER-impedance speaker, while modern amps have a low impedance des-
ign and will transfer more power to a LOWER-impedance speaker.  Though
I've not measured the specific behaviors, I would expect many loud-
speakers would have very different frequency-response curves when driven
by tube vs transistor amps.  Wonder why I seldom see mention of that?

1999\03\15@161327 by Graeme Smith

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face
GRAEME SMITH                         email: KILLspamgrysmithKILLspamspamfreenet.edmonton.ab.ca
YMCA Edmonton

Address has changed with little warning!
(I moved across the hall! :) )

Email will remain constant... at least for now.


On Sat, 13 Mar 1999, Wagner Lipnharski wrote:

{Quote hidden}

Magnetic inductance, happens with a straight wire, but because it happens
at 90 degrees to the wire it is virtually indistinguishable from the
inductance of a straight wire... OK, just kidding, a bit...

Magnetic inductance happens in a straight wire, but it isn't as strong,
and doesn't take advantage of series inductance as well, as it would with
a coil. In a coil type inductor, the magnetic field from one wire, cuts
the next wire, with the result that it causes a slight current to try to
flow down the wire by invoking a "Back emf" or reverse voltage
(inductance creates impedence) this back emf, acts to reduce the forward
voltage, (relatively speaking), resulting in something similar to
resistance to current flow.

Since all wires are in fact antennas.... (just tuned to different
wave lengths)... Any wire that cuts the magnetic field, will experience
the inductance. When the magnetic field is growing or dying, it passes
through the wire, and creates emf/current effects in much the same manner
as a generator produces power by passing windings through a magnetic
field.

The power of the signal recieved on the wire, will of course be related to
the strength of the magnetic field, which, expands following the distance
squared law, and so, reduces at the square of the distance. Since straight
D.C. wiring does not have much of a changing magnetic field, the field
effects within a D.C. device are usually negligable.

However, an Alternating current, has rotating fields, cycling according to
the frequency of the alternation, and so, you can quite literally pick up
a 50 or 60 cycle hum, with a d.c circuit that is isolated from the A.C.
power grid, even though the only AC wires in the vicinity are straight
wires. Since some of the time, that signal is acting against the
voltage/current and some of that time the signal is acting with the
voltage/current in the D.C. device, the voltage and current will vary with
the impedence caused by the inductance... Hence the 60 cycle hum....

So... all this is in aid of linking inductance to impedance, so you
understand how magnetic impedance works, and therefore why your D.C.
circuit suddenly takes off and starts acting as if it had a variable
impedance when-ever you use the wrong combination of parts/ organization
of lines on the PCBoard.

                               GREY

1999\03\15@163020 by Gerhard Fiedler

picon face
At 14:41 03/15/99 -0600, John Payson wrote:
>BTW, in the case of loudspeakers, things actually get a bit more comp-
>licated; the impedance of a speaker varies with frequency.  One diff-
>erence between "tube" amplifiers and modern transistor designs is that
>tube amps often have a high impedance and will transfer more power to
>a HIGHER-impedance speaker, while modern amps have a low impedance des-
>ign and will transfer more power to a LOWER-impedance speaker.  Though
>I've not measured the specific behaviors, I would expect many loud-
>speakers would have very different frequency-response curves when driven
>by tube vs transistor amps.  Wonder why I seldom see mention of that?

don't the tube amps have usually output transformers, for a certain
loudspeaker impedance?

ge

1999\03\15@173423 by John Payson

flavicon
face
At 14:41 03/15/99 -0600, John Payson wrote:
>BTW, in the case of loudspeakers, things actually get a bit more comp-
>licated; the impedance of a speaker varies with frequency.  One diff-
>erence between "tube" amplifiers and modern transistor designs is that
>tube amps often have a high impedance and will transfer more power to
>a HIGHER-impedance speaker, while modern amps have a low impedance des-
>ign and will transfer more power to a LOWER-impedance speaker.  Though
>I've not measured the specific behaviors, I would expect many loud-
>speakers would have very different frequency-response curves when driven
>by tube vs transistor amps.  Wonder why I seldom see mention of that?

|don't the tube amps have usually output transformers, for a certain
|loudspeaker impedance?

They do, and the amps are definitely designed with a specific impedance
in mind.  My point was that if a transistor and tube amplifier are both
set up for an 8 ohm load, are fed a 1KHz sine wave, and have the gain
control set for 8 watts' output, the transistor amplifier will try to
output 8 volts RMS (into whatever load is actually attached) while the
tube amplifier will try to output 1 amp RMS (into whatever load is act-
ually attached).

A transistor amplifier which is designed for an 8 ohm load can drive a
16 ohm load happily (though it will only get half the power the amp
could deliver), and a tube amplifier designed for an 8 ohm load can
likewise drive a 4 ohm load.  In the extreme cases, a transistor amp-
lifier has no problem driving an open-circuited load but may be damaged
by a shorted output.  Tube amplifiers have no objection to driving a
short, but may be damaged if operated for too long open-circuit.

1999\03\15@174917 by Nigel Goodwin

flavicon
picon face
In message <4.1.19990315132440.009c8290@mail>, Gerhard Fiedler
<RemoveMElistsTakeThisOuTspamHOME.COM> writes
>At 14:41 03/15/99 -0600, John Payson wrote:
>>BTW, in the case of loudspeakers, things actually get a bit more comp-
>>licated; the impedance of a speaker varies with frequency.  One diff-
>>erence between "tube" amplifiers and modern transistor designs is that
>>tube amps often have a high impedance and will transfer more power to
>>a HIGHER-impedance speaker, while modern amps have a low impedance des-
>>ign and will transfer more power to a LOWER-impedance speaker.  Though
>>I've not measured the specific behaviors, I would expect many loud-
>>speakers would have very different frequency-response curves when driven
>>by tube vs transistor amps.  Wonder why I seldom see mention of that?
>
>don't the tube amps have usually output transformers, for a certain
>loudspeaker impedance?

Yes, valve amplifiers have an output transformer to match the (very!)
high output impedance to the speaker impedance - these are usually
tapped to provide different speaker settings, either 16, 8, or even 4
ohms. Back in the valve days 16 ohm speakers were common, now with
transistor amplifiers 8 ohms is more usual, with German systems 4 ohms
is often used.
--

Nigel.

       /--------------------------------------------------------------\
       | Nigel Goodwin   | Internet : spamBeGonenigelgspamBeGonespamlpilsley.demon.co.uk     |
       | Lower Pilsley   | Web Page : http://www.lpilsley.demon.co.uk |
       | Chesterfield    | Official site for Shin Ki Ju Jitsu         |
       | England         |                                            |
       \--------------------------------------------------------------/

1999\03\15@185624 by Bob Blick

face
flavicon
face
On Mon, 15 Mar 1999, John Payson wrote:
>
> BTW, in the case of loudspeakers, things actually get a bit more comp-
> licated; the impedance of a speaker varies with frequency.  One diff-
> erence between "tube" amplifiers and modern transistor designs is that
> tube amps often have a high impedance and will transfer more power to
> a HIGHER-impedance speaker, while modern amps have a low impedance des-
> ign and will transfer more power to a LOWER-impedance speaker.  Though
> I've not measured the specific behaviors, I would expect many loud-
> speakers would have very different frequency-response curves when driven
> by tube vs transistor amps.  Wonder why I seldom see mention of that?

Most amplifiers(tube and transistors) use negative feedback, which gives
them an effective output impedance of almost zero, so the frequency
response has little to do with the load.

Also, unless you drive your amplifier to clipping, the output power
changes at the resonant frequency will not make much different(the
amplifier would get "clean again" at the resonant frequency) since the
feedback is voltage feedback. Bozak made some amplifiers that also had
positive current feedback, but that was different.

Also, unless you drive the tube amp to clipping, operating it without a
load will not hurt it(again, I'm talking about the majority of amplifiers
having negative feedback) as long as it is not already right on the edge
of instability due to design problems.

Cheers,
Bob

1999\03\17@060952 by paulb

flavicon
face
John Payson wrote:

> My point was that if a transistor and tube amplifier are both set up
> for an 8 ohm load, are fed a 1KHz sine wave, and have the gain control
> set for 8 watts' output, the transistor amplifier will try to output 8
> volts RMS (into whatever load is actually attached) while the tube
> amplifier will try to output 1 amp RMS (into whatever load is act-
> ually attached).

 Unless, as Bob said, it uses feedback, in which case its behaviour
driving a variety of loads will approximate the transistor one.

> In the extreme cases, a transistor amplifier has no problem driving an
> open-circuited load but may be damaged by a shorted output.  Tube
> amplifiers have no objection to driving a short, but may be damaged if
> operated for too long open-circuit.

 Of course, the two exceptions here are the transistor amplifier fitted
to the Siemens 2000 16mm movie projector around 1968 which blew the
output devices *instantly* if activated without the speaker fitted, and
valve amplifiers with feedback which would tend *not* to blow the output
transformer or capacitors if operating into open circuit.
--
 Cheers,
       Paul B.

1999\03\17@093542 by Rich Graziano

flavicon
face
-----Original Message-----
From: John Payson <TakeThisOuTsupercatEraseMEspamspam_OUTCIRCAD.COM>
To: RemoveMEPICLISTspamTakeThisOuTMITVMA.MIT.EDU <PICLISTEraseMEspam.....MITVMA.MIT.EDU>
Date: Monday, March 15, 1999 3:39 PM
Subject: Re: [OT] Impedance


{Quote hidden}

John:

You are correct.  Impedance is frequency dependant.  Tube amplifiers have a
high impedance on the primary side of the outout transformer (audio
transformer).  The plate voltage, called B+, is connected through the
primary side of the transformer.  The secondary side is a low impedance
design, usually with 4, 8 and 16 ohm taps, for different speaker loads.

Regards,

Richard

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