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Load/Source Impedance

U

Uriah

Jan 1, 1970
0
I am still having a hard time understanding Load and Source impedance.
When someone says you need a low ouput Impedance and high imput
impedance to transfer the most signal? Are they saying a low output
Impedance is wired to the input impedance in parallel so that low side
is not changed much by a high impedance R in parrallel?

Does low Impedance mean high current? And High I mean low current?
Does low I mean high V? And high I mean low V? But if you have a
voltage with a resistor in series and you raise the value of R to 10XR
(Isn't R really I) the current drops so you have lower impedance but
doesn't the voltage drop on the resistor increase so you have high V?

Another example from a book I am reading talks about sending data over
a data bus from a CPU to ROM and says "you need Low impedance from the
CPU to drive the byte firmly on to the bus" And " High impedance from
the ROM to avoid loading the bus while receiving the byte" How do you
get low I? Does this also mean Low I means high V? How does high
Impedance not load a bus?

Can someone explain this? I am getting confused. I have gone over this
and I guess
some wires are getting crossed:)

Thankss
Uriah
 
P

Phil Allison

Jan 1, 1970
0
** Warning Will, danger, danger ....

I am still having a hard time understanding Load and Source impedance.
When someone says you need a low ouput Impedance and high imput
impedance to transfer the most signal? Are they saying a low output
Impedance is wired to the input impedance in parallel so that low side
is not changed much by a high impedance R in parrallel?


** Basically - yes

Does low Impedance


** Do not isolate words from their context - this is where YOU are going
wrong all the time.

mean high current? And High I mean low current?


** NEVER use the letter " I " to mean other than current.

This is where YOU are also going wrong all the time.


Saying a source is "low impedance" indicates that it has the *capacity * to
deliver significant current into a load, a current level that is significant
*in the circumstances*.

With high frequencies, the main worry with loading is not the input
impedance of the next device - but stray capacitance in connecting cables
and PCB tracks.


This is a formula you need to know:

I = C dv/dt

If a signal voltage jumps by 5 volts in 10 nanoseconds,

then dv/dt = 500 exp6 V/s

If the stray load capacitance is say, 50 picofarads, then

I = 0.025 amps or 25 milliamps.



....... Phil
 
F

flank

Jan 1, 1970
0
Uriah,
It sounds to me like you don't understand what impedance means.
Impedance can be loosely defined as a resistance to current flow.
Resistance and impedance are very similar in that they are both used to
describe something which limits current flow. I imagine you know what
resistance is. In DC circuits, you are usually dealing with
resistances.

Example: If you take a 9V battery and connect a 100 k-ohm
resistor across the terminals, the battery will supply 9V/100000 ohms
of current. Given the current capacity of the battery, this is a high
resistance (impedance), meaning it allows a small amount of current
relative to the capacity of the battery. On the other hand, if you
place a 10 ohm resistor across the same terminals, you will then be
drawing 0.9A from the battery. This is a lot of current for the
battery, and it is considered to be a large load (low
resistance/impedance).

Low resistance or impedance >> large load
High resistance or impedance >> small load

Now the difference between impedance and resistance is in a sense, just
semantics. Impedance sort of means a resistance which varies with
frequency. Resistance doesn't vary with frequency. A capacitor
becomes highly resistive at low frequencies, and it is minimally
resistive at high frequencies. An inductor does the opposite. It
ideally has zero resistance in DC circuits (where the frequency is 0),
and it becomes highly resistive at high frequencies.

Now the idea behind a low output impedance and a high input impedance
is based on the idea that your input signal is going into an amplifier,
not a speaker. High input impedance will prevent your output signal
from having to produce more current than it is capable of. In other
words, a high input impedance will retain the integrity of your signal.


I have no idea if this is going to help. I hope so.

flank
 
C

Chris

Jan 1, 1970
0
Uriah said:
I am still having a hard time understanding Load and Source impedance.
When someone says you need a low ouput Impedance and high imput
impedance to transfer the most signal? Are they saying a low output
Impedance is wired to the input impedance in parallel so that low side
is not changed much by a high impedance R in parrallel?

Does low Impedance mean high current? And High I mean low current?
Does low I mean high V? And high I mean low V? But if you have a
voltage with a resistor in series and you raise the value of R to 10XR
(Isn't R really I) the current drops so you have lower impedance but
doesn't the voltage drop on the resistor increase so you have high V?

Another example from a book I am reading talks about sending data over
a data bus from a CPU to ROM and says "you need Low impedance from the
CPU to drive the byte firmly on to the bus" And " High impedance from
the ROM to avoid loading the bus while receiving the byte" How do you
get low I? Does this also mean Low I means high V? How does high
Impedance not load a bus?

Can someone explain this? I am getting confused. I have gone over this
and I guess
some wires are getting crossed:)

Thankss
Uriah

Hi, Uriah. Look at this circuit (view in fixed font or M$ Notepad):

|
| Signal
| Source Load
| .----------------. .------------.
| | | | |
| |VCC-o Vo ___ | | |
| | __--o-|___|-o----------o---o----o |
| |GND-o Zo | | | + |
| | | | | .-. |
| | | | |Zi| | |
| | | | | | | Vi |
| | | | | '-' |
| | | | | | |
| | | | | | |
| | | | | | - |
| | '-----------o----------o---o----o |
| | | | |
| '----------------' '------------'
|
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

There's a signal source and a load. In this diagram, for the sake of
discussion, the signal source Vo (Vout, the output signal itself)
switches between Vcc and GND like a digital signal. This works for
analog signals, too. All of your questions, in one form or another,
all have to do with transmitting the signal Vo, and the question of
what's happening at Vi (Vin, the input or load voltage). Zo (Zout, the
output impedance) is a characteristic of the signal source, and Zi
(Zin, the input or load impedance) is a characteristic of the load.

There are a couple of points worth mentioning about this circuit:

* If you want maximum Vi for a given Vo, you want Zo (output
Impedance) to be as low as possible, and the load impedance Zi to be as
high as possible. It's just a voltage divider question. It's that
simple.

* The current flowing from the signal source to the load is just the
voltage divided by the sum of the impedances. Ohm's law works for AC,
too. It's that simple.

* If your signal source is a microprocessor pin, and your load is a bus
with many CMOS inputs, you'll usually find the load impedance is
primarily capacitive. This means that Vi will rise as the capacitance
is charged up by the signal source, using Mr. Allison's equation. The
speed with which Vi rises to Vcc or falls to GND (very important for
many signals) is dependent on the source impedance Zo. The higher the
impedance, the slower the cap will get charged up. This could mess up
a clock signal, or another signal with specified maximum rise/fall
time.

But the load impedance Zi is also important in determining rise and
fall times. The more you increase the load capacitance Zi for a given
Zo and signal, the slower Vi will rise and fall. So you want to
endeavor to keep the bus impedance (load impedance Zi) as high as
possible (in other words, try to keep the bus capacitance to a
minimum).

I hope this has answered your immediate questions. There's a lot more
really basic stuff to learn here. Your question also indicates you
might not be entirely clear on what impedance is. It actually sounds
like you need to back up the truck. I'd suggest you get back to the
textbooks (assuming you bought them here <8-o ), turn off the iPod, and
work at really understanding what's written there. Take your time, and
take pencil and paper notes while you're reading. Work out sample
problems on your own. Use any supplemental materials. Ask the
professor questions and make yourself a pest. If your access to the
instructor is limited or you're too far behind, get some help from an
upperclassman or tutor. Do what you can to assimilate the knowledge
that's there.

And after the emergency is over, read the covered material before
class, and ask questions as soon as you get lost.

Good luck
Chris
 
C

Chris

Jan 1, 1970
0
Chris wrote:
many signals) is dependent on the source impedance Zo. The higher the
impedance, the slower the cap will get charged up. This could mess up
a clock signal, or another signal with specified maximum rise/fall
time.
Good luck
Chris

The *lower* the impedance (more load capacitance), the slower the cap
will get charged up by the source.

Oops. Sorry.

Chris
 
M

mc

Jan 1, 1970
0
Uriah said:
I am still having a hard time understanding Load and Source impedance.
When someone says you need a low ouput Impedance and high imput
impedance to transfer the most signal? Are they saying a low output
Impedance is wired to the input impedance in parallel so that low side
is not changed much by a high impedance R in parrallel?

Simple example.

Suppose you are driving a 9-volt, 0.1-amp light bulb. That is a load with a
load impedance of 90 ohms.

And you're doing it with a 9-volt battery with a 1-ohm resistor in series
with it. That is a source with a source impedance of 1 ohm.

The 90-ohm resistance of the light bulb and the 1-ohm resistor attached to
the battery (or inside it) form a voltage divider. The bulb gets 90/91 of
the voltage.

If that 1-ohm resistor were 90 ohms, the bulb would only get 90/180 of the
voltage.
 
R

Robin

Jan 1, 1970
0
Uriah said:
I am still having a hard time understanding Load and Source impedance.
When someone says you need a low ouput Impedance and high imput
impedance to transfer the most signal? Are they saying a low output
Impedance is wired to the input impedance in parallel so that low side
is not changed much by a high impedance R in parrallel?

Does low Impedance mean high current? And High I mean low current?
Does low I mean high V? And high I mean low V? But if you have a
voltage with a resistor in series and you raise the value of R to 10XR
(Isn't R really I) the current drops so you have lower impedance but
doesn't the voltage drop on the resistor increase so you have high V?

Another example from a book I am reading talks about sending data over
a data bus from a CPU to ROM and says "you need Low impedance from the
CPU to drive the byte firmly on to the bus" And " High impedance from
the ROM to avoid loading the bus while receiving the byte" How do you
get low I? Does this also mean Low I means high V? How does high
Impedance not load a bus?

Can someone explain this? I am getting confused. I have gone over this
and I guess
some wires are getting crossed:)

Thankss
Uriah


A "low impedance" source is "powerful" and has plenty of "drive". A
"high-impendance" source is weaker with less "drive".

e.g.
Case A
A battery lights a bulb: then you can say "The battery has enough drive
to light the bulb."

Case B
But some time later the bulb gets dimmer. What has happened? Evidently
the battery is getting flat but what does that mean? It means that the
battery has *less* drive than it before or "less power".

The new battery has "low output impedance" while the flat battery has
"higher output impedance". Let's measure the that output impedance in
both cases.

1) Take the new battery and measure its voltage (with nothing except
your volt meter connected to it). Say it is 1.50 Volts.
2) Repeat 1) while the bulb is connected. Say it is 1.40 Volts.
3) And measure the current. Say it is 100 mA.

Now we can calculate the output "impedance" (it is resistance really).
You can imagine the battery when nothing is connected, it is having an
easy time, doing no work and so it can hold its output at maximum of
1.5 volts. But once the bulb is connected, then it struggles to force
current through the hot filament and its output is loaded down to say
1.4 volts.

We say "this drop of 0.1 Volts inside the battery is happening across
the batteries output impedance" and we can say that it equals 1 Ohm
because Ohms law says V/I = R and so 0.1 Volts divided by 100 mA is
equal to 1 Ohm.

Evidently this thing called "output impedance" is not a resistor as
such but is a sort of build of waste matter while the battery is doing
work and this build up causes the loss of drive and the fall in output
voltage. If you loaded the battery more with say two bulbs in parallel
then the above calculation might produce an output impedance of 1.1
Ohms i.e. if all depends on the load e.g. if you short the battery out,
then evidently the output voltage will be 0 Volts and if the current
was 1 Amps then the output impedance would be

1.5 / 1 = 1.5 Ohms

In case B the voltage might be 0.4 volts (under load) and the current
might be 50 mA. If the unloaded voltage was 1.4 volts then

1 / 0.05 = 20 Ohms

is the output impedance.

**********************************************************************
The load impedance (resistance really)

1) In Case A and B the load impedance is (mainly) the restance of the
bulb's filament (when hot) which is lowish.

2) In Case A and B without the bulb connected, the load was resistance
of your voltmeter only and this will be very high maybe 10,000,000 Ohms
if you are using a digital multimeter or 200,000 Ohms if you are using
an old fashioned AVO. I.e. both very high.

Voltmeters are designed to have high resistances... so that they don't
load the thing they are measuring... and give a falsely low reading :)

Cheers
Robin
 
M

Michael A. Terrell

Jan 1, 1970
0
mc said:
Simple example.

Suppose you are driving a 9-volt, 0.1-amp light bulb. That is a load with a
load impedance of 90 ohms.

And you're doing it with a 9-volt battery with a 1-ohm resistor in series
with it. That is a source with a source impedance of 1 ohm.

The 90-ohm resistance of the light bulb and the 1-ohm resistor attached to
the battery (or inside it) form a voltage divider. The bulb gets 90/91 of
the voltage.

If that 1-ohm resistor were 90 ohms, the bulb would only get 90/180 of the
voltage.


You say you teach electronics, then you give someone this lousy
example? A light bulb is a non linear device, so you will not get what
you stated. It will confuse the hell out of anyone who tries it as a
hands on experiment. Now, if you said, We can use a 90 ohm resistor to
simulate...


--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida
 
M

mc

Jan 1, 1970
0
You say you teach electronics, then you give someone this lousy
example? A light bulb is a non linear device, so you will not get what
you stated. It will confuse the hell out of anyone who tries it as a
hands on experiment. Now, if you said, We can use a 90 ohm resistor to
simulate...

Admittedly, I oversimplified. But I wanted to keep things simple.
Supplying power to a light bulb is a common, ordinary thing to do; supplying
power to a resistor is not. To teach effectively, I like to avoid
abstruseness.
 
M

Michael A. Terrell

Jan 1, 1970
0
mc said:
Admittedly, I oversimplified. But I wanted to keep things simple.
Supplying power to a light bulb is a common, ordinary thing to do; supplying
power to a resistor is not. To teach effectively, I like to avoid
abstruseness.


Any experiment that can't be done on the bench should be avoided. At
the very least you could have pointed out that the device is non-linear
rather than post pure BS.

I was on a committee a few years ago to choose new textbooks for an
electronics course and they were all full of this crap. We ended up
choosing the one with the least mistakes and sending a list of errors to
the publisher. It wasn't the best overall book, but it doesn't instill
confidence when a teacher has to hand out thick errata sheets with the
textbooks.


--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida
 
M

mc

Jan 1, 1970
0
Michael A. Terrell said:
Any experiment that can't be done on the bench should be avoided. At
the very least you could have pointed out that the device is non-linear
rather than post pure BS.

In general, I agree with you, and it was an omission on my part -- I should
have added "in practice, the resistance of a light bulb changes with
temperature, so this calculation doesn't describe an actual one."

I agree that experiments should be able to be done on the bench, and in fact
I'm drafting some text materials in which every circuit, without exception,
has all the component values indicated and can be breadboarded. It used to
drive me crazy that all the circuits in the books lacked most of the
component values.
I was on a committee a few years ago to choose new textbooks for an
electronics course and they were all full of this crap. We ended up
choosing the one with the least mistakes and sending a list of errors to
the publisher. It wasn't the best overall book, but it doesn't instill
confidence when a teacher has to hand out thick errata sheets with the
textbooks.

Actually, I'd like to see some of your critiques. Did you find any in The
Art of Electronics? (I have no connection with that book other than that I
like it... but it's probably too hard for beginners.) Under "Bad circuits"
in one place they list a circuit that is not one of their students'
mistakes, but is taken from an earlier textbook by Diefenderfer.
 
U

Uriah

Jan 1, 1970
0
I want to thank everyone for taking the time to reply to my post. I
have been reading them over and will need to go over them a few times.
There is allot there. It is starting to clear up. I don't know
calculus so when I see the I=C dv/dt I am clueless but that is for
another time. I think I am on the right track and once I go through
this a few more times I think it will be clear. Once again thanks to
everyone for taking the time to explain. I really appreciate it.
Uriah
 
M

Michael A. Terrell

Jan 1, 1970
0
mc said:
In general, I agree with you, and it was an omission on my part -- I should
have added "in practice, the resistance of a light bulb changes with
temperature, so this calculation doesn't describe an actual one."

I agree that experiments should be able to be done on the bench, and in fact
I'm drafting some text materials in which every circuit, without exception,
has all the component values indicated and can be breadboarded. It used to
drive me crazy that all the circuits in the books lacked most of the
component values.


Actually, I'd like to see some of your critiques. Did you find any in The
Art of Electronics? (I have no connection with that book other than that I
like it... but it's probably too hard for beginners.) Under "Bad circuits"
in one place they list a circuit that is not one of their students'
mistakes, but is taken from an earlier textbook by Diefenderfer.


These books were all for a two year vocational electronics course,
and the state didn't have AOE on the list. I may be ale to contact the
man who was the teacher and see what he remembers. I turned all of the
materials back in to the school board's book committee and I don't even
have a list of the titles. They replaced the electronics course with a
computer course the following semester and tossed all of the books, test
equipment and supplies into a construction dumpster. They also left a
dozen, new 4 channel analog scopes sitting out in the rain. The
bastards.


--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida
 
R

Robin

Jan 1, 1970
0
Michael said:
I was on a committee a few years ago to choose new textbooks for an
electronics course and they were all full of this crap. We ended up
choosing the one with the least mistakes and sending a list of errors to
the publisher. It wasn't the best overall book, but it doesn't instill
confidence when a teacher has to hand out thick errata sheets with the
textbooks.

Things have not changed much then, since Feynmann's chapter (probably
titled) "judge a book by its cover" in his (definitely titled) "Surely
you're joking Mr Feynmann"

Cheers
Robin

(One of the choosen books was blank. All the pages were blank. The only
print was on the front and back covers. No one had noticed because no
one had read it, except Feynmann.)
 
M

mc

Jan 1, 1970
0
These books were all for a two year vocational electronics course,
and the state didn't have AOE on the list. I may be ale to contact the
man who was the teacher and see what he remembers. I turned all of the
materials back in to the school board's book committee and I don't even
have a list of the titles. They replaced the electronics course with a
computer course the following semester and tossed all of the books, test
equipment and supplies into a construction dumpster. They also left a
dozen, new 4 channel analog scopes sitting out in the rain. The
bastards.

Was it a state-funded institution? Most states have an office that solicits
suggestions of ways to save money and reports of egregious waste. That
would be worth reporting.
 
P

Pooh Bear

Jan 1, 1970
0
mc said:
Was it a state-funded institution? Most states have an office that solicits
suggestions of ways to save money and reports of egregious waste. That
would be worth reporting.

It's an absolutely shocking waste of money.

Graham
 
F

flank

Jan 1, 1970
0
Michael said:
Any experiment that can't be done on the bench should be avoided. At
the very least you could have pointed out that the device is non-linear
rather than post pure BS.

I don't think the explanation was difficult to understand. As far as a
benchtop experiment is concerned, the light bulb would heat up fast
enough (and reach its steady state characteristics) that the experiment
would appear to have a linear load. If the experiment was done with an
oscilloscope, designed to capture the first second or two, then the
experiment would appear to be non-linear.
I was on a committee a few years ago to choose new textbooks for an
electronics course and they were all full of this crap. We ended up
choosing the one with the least mistakes and sending a list of errors to
the publisher. It wasn't the best overall book, but it doesn't instill
confidence when a teacher has to hand out thick errata sheets with the
textbooks.

You sound quite knowledgeable. You should consider writing a better
textbook. I also teach electronics, and I use the Hayt, Kemmerly,
Durbin book, which is by far the best I have found, and it has many
errors and truly horrible homework problems. Good luck with your book.
 
M

Michael A. Terrell

Jan 1, 1970
0
flank said:
I don't think the explanation was difficult to understand. As far as a
benchtop experiment is concerned, the light bulb would heat up fast
enough (and reach its steady state characteristics) that the experiment
would appear to have a linear load. If the experiment was done with an
oscilloscope, designed to capture the first second or two, then the
experiment would appear to be non-linear.


Sorry, but when you have the 90 ohm resistor in series with the lamp
in the second part, there isn't enough current flow to allow the
filament to come to full operating temperature, so it won't be 90 ohms,
unless you double the supply voltage.

from the problem that was posted:

* If that 1-ohm resistor were 90 ohms, the bulb would only get 90/180 of
the
* voltage.

You sound quite knowledgeable. You should consider writing a better
textbook. I also teach electronics, and I use the Hayt, Kemmerly,
Durbin book, which is by far the best I have found, and it has many
errors and truly horrible homework problems. Good luck with your book.


I've also taught vocational electronics, back in the '60s. I ws
offered a full time job teaching at another vocational school in the mid
'70s, but they wanted me to move into their district, and they money
just wasn't there. in the late '70s I was offered a job teaching at one
of those TV and industrial tech "mills" where they tried to hammer the
basics into their heads in a very short course. Their equipment and
materials were 30 years out of date and I was not impressed by any of
their graduates that I had to show how things really worked.

As far as writing a textbook, the teacher and I were were negotiating
with a publisher during the book selection process, but the deal fell
through. I was working about 60 hours a week, and the teacher had
outside interests, so they deadline they wanted along with an
insultingly small price it just didn't happen. It would have cost me
money to write the book by their deadline because I would have had to
turn down all the overtime to be able to complete it on time. Now, I am
on a 100% disability. The VA tells me that I am allowed some small
income, but I still have now answer as to what I can earn and not lose
my pension and medical care.

I think I would rather see it done as an interactive CDROM book with
a large section of experiments and test questions to encourage
additional self study. I learned the basics by reading every
electronics book I could find, starting when I was 10 years old. The
ARRL handbooks, used college textbooks and the then common monthly
electronics magazines and assorted tools took up every cent I earned.
The authors had a way of teaching you electronics while keeping it
interesting, yet leaving you hungry for more knowledge. I will see if I
can find a copy of the book you are talking about. There is still one
vocational electronics course in the area that might have some newer
books. I lost my collection of electronics textbooks when I was laid
off five years ago, and the company closed before I could get my
personal property out. The company hired a number of wet behind the
ears techs every year, and some needed refreshing fairly often.


--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida
 
M

Michael A. Terrell

Jan 1, 1970
0
mc said:
Was it a state-funded institution? Most states have an office that solicits
suggestions of ways to save money and reports of egregious waste. That
would be worth reporting.


They claimed that it was the contractors fault, the contractor
claimed that they were told everything they wanted had been moved. A
lot of finger pointing and no results, as usual. At least the moron
that closed the course is gone. They claimed that the older items had
been offered to other schools in the state, but no one wanted any of
it. They also claimed that the water damaged equipment was an accident,
then they had the nerve to auction it off without telling anyone about
it sitting in the rain for days. You have no idea about the amount of
money wasted by school systems while they cry for more money. I've seen
complete school intercom systems at auction that were replaced when they
were only a few years old. When I serviced that type of equipment
through my commercial sound and industrial electronics business some
systems were 50 years old and still in daily use.


--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida
 
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