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Input Impedance of SIMPLE Circuit

Can anyone figure out the input impedance of the circuit attached here:
http://physicsforums.com/showthread.php?t=73070 ?

Does anyone know a THOROUGH definition of input impedance? output
impedance?

Output impedance is equivalent to the thevenin impedance, which mean we
turn off all voltage/current sources. But for input impedance, what if
the circuit has voltage/current sources, what do we do (like the
circuit I have attached)?
 
J

Jim Thompson

Jan 1, 1970
0
Can anyone figure out the input impedance of the circuit attached here:
http://physicsforums.com/showthread.php?t=73070 ?

Does anyone know a THOROUGH definition of input impedance? output
impedance?

Output impedance is equivalent to the thevenin impedance, which mean we
turn off all voltage/current sources. But for input impedance, what if
the circuit has voltage/current sources, what do we do (like the
circuit I have attached)?

I don't respond to problem definitions posted on sites requiring
registration. Sorry.

...Jim Thompson
 
M

mike

Jan 1, 1970
0
Can anyone figure out the input impedance of the circuit attached here:
http://physicsforums.com/showthread.php?t=73070 ?

Does anyone know a THOROUGH definition of input impedance? output
impedance?

Output impedance is equivalent to the thevenin impedance, which mean we
turn off all voltage/current sources. But for input impedance, what if
the circuit has voltage/current sources, what do we do (like the
circuit I have attached)?

Impedance is impedance. It defines the vector relationship between
voltage and current at the port. Doesn't matter whether it's "input" or
"output" or whether you've "turned off" sources. Turning on the sources
won't affect the impedance, but it may require you to use different
measurement techniques.
mike

--
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T

Tim Wescott

Jan 1, 1970
0
Jim said:
I don't respond to problem definitions posted on sites requiring
registration. Sorry.

...Jim Thompson

I second that. Jim's language is more diplomatic than what I had in
mind, so I'll let it stand for me, too.
 
B

Bob Monsen

Jan 1, 1970
0
Can anyone figure out the input impedance of the circuit attached here:
http://physicsforums.com/showthread.php?t=73070 ?

Does anyone know a THOROUGH definition of input impedance? output
impedance?

Output impedance is equivalent to the thevenin impedance, which mean we
turn off all voltage/current sources. But for input impedance, what if
the circuit has voltage/current sources, what do we do (like the
circuit I have attached)?

Post the gif/jpg on the newsgroup alt.binaries.schematics.electronic
 
J

Jim Thompson

Jan 1, 1970
0
Jim said:
On 25 Apr 2005 16:26:05 -0700, [email protected] wrote:
[snip]

I don't respond to problem definitions posted on sites requiring
registration. Sorry.

...Jim Thompson

I second that. Jim's language is more diplomatic than what I had in
mind, so I'll let it stand for me, too.

WOW! It's been a long time since someone referred to me as
"diplomatic" ;-)

...Jim Thompson
 
G

G Patel

Jan 1, 1970
0
Can anyone figure out the input impedance of the circuit attached
here:

Correction: Let me draw you a text version of the circuit:

100 ohms
o---------/\/\/\----|
|
|
-----
| + | 20Vdc
| - |
Zin -> -----
|
|
|
o-------------------|




That's a voltage source in the circuit. Of course this could be a more
complex circuit with more voltage sources in different places, but this
two element circuit serves to explain my confusion. I don't know what
the input impedance is for circuits like this (for output impedance of
complex circuits with sources I can simply short voltage sources or
open current sources before calculating, but what would I do for input
impedance?)
 
J

Jim Thompson

Jan 1, 1970
0
here:

Correction: Let me draw you a text version of the circuit:

100 ohms
o---------/\/\/\----|
|
|
-----
| + | 20Vdc
| - |
Zin -> -----
|
|
|
o-------------------|




That's a voltage source in the circuit. Of course this could be a more
complex circuit with more voltage sources in different places, but this
two element circuit serves to explain my confusion. I don't know what
the input impedance is for circuits like this (for output impedance of
complex circuits with sources I can simply short voltage sources or
open current sources before calculating, but what would I do for input
impedance?)

Impedance is generally defined incrementally, so a 1V CHANGE at the
input will produce a 1/20 Amp change in current... thus the input
impedance is 20 ohms, irrespective of the voltage source behind it ;-)

...Jim Thompson
 
Jim said:
Impedance is generally defined incrementally, so a 1V CHANGE at the
input will produce a 1/20 Amp change in current... thus the input
impedance is 20 ohms, irrespective of the voltage source behind it ;-)

...Jim Thompson

Yes, ie 100 ohms, so the OP is clear.

NT
 
W

Winfield Hill

Jan 1, 1970
0
Jim Thompson wrote...
Might help if I could read ;-)

Oops, time for another trip to the drugstore for stronger reading
glasses. Or perhaps a trip to the kitchen so the good wife can
give them a good cleaning. <sigh>
 
T

Tim Wescott

Jan 1, 1970
0
Jim said:
Impedance is generally defined incrementally, so a 1V CHANGE at the
input will produce a 1/20 Amp change in current... thus the input
impedance is 20 ohms, irrespective of the voltage source behind it ;-)

...Jim Thompson

-- and this is why you short a voltage source and open a current source,
on an input _or_ output. Change the voltage by 1V into a voltage source
and the incremental current change is infinite. Change the voltage by
1V into a current source and the incremental current change is zero.
 
R

Ratch

Jan 1, 1970
0
Can anyone figure out the input impedance of the circuit attached here:
http://physicsforums.com/showthread.php?t=73070 ?

Does anyone know a THOROUGH definition of input impedance? output
impedance?

Yes, the general definition of impedance is Z(s)=V(s)/I(s). Where Z(s),
V(s), and I(s) are the Laplace transform of voltage, current and Impedance,
If you can calculate the voltage and current Laplace transforms, then you
can find the impedance provided the circuit is linear, which it usually is.
This will work for alternating or DC sources. Ratch
 
R

Ratch

Jan 1, 1970
0
G Patel said:
here:

Correction: Let me draw you a text version of the circuit:

100 ohms
o---------/\/\/\----|
|
|
-----
| + | 20Vdc
| - |
Zin -> -----
|
|
|
o-------------------|




That's a voltage source in the circuit. Of course this could be a more
complex circuit with more voltage sources in different places, but this
two element circuit serves to explain my confusion. I don't know what
the input impedance is for circuits like this (for output impedance of
complex circuits with sources I can simply short voltage sources or
open current sources before calculating, but what would I do for input
impedance?)


OK, let's do the problem. Assume a constant DC voltage is applied to
the terminals. V(s)=V/s. Writing the loop equation we get
100*I(s)=V/s-20/s=(V-20)/s==>I(s)=(V-20)/100s . Z(s)=(V/s)/I(s)=100V/V-20 .
Notice from the last equation, that if the 20 volt source is removed, the
impedance value becomes 100 ohms resistive regardless of the input source
voltage. Also from the last equation, it can be seen that if the input
voltage is 20 volts, the impedance is infinite. This is because the 20
volts of input voltage balances the 20 volt source voltage, so that no
current exists in the circuit, thereby making the impedance infinite. Ratch
 
J

John Woodgate

Jan 1, 1970
0
Yes, the general definition of impedance is Z(s)=V(s)/I(s). Where
Z(s), V(s), and I(s) are the Laplace transform of voltage, current and
Impedance, If you can calculate the voltage and current Laplace
transforms, then you can find the impedance provided the circuit is
linear, which it usually is. This will work for alternating or DC
sources. Ratch

The OP doesn't understand Thévenin and you want to sell him Laplace? Is
this just an ego trip or do you really not understand the KISS
Principle?
 
R

Ratch

Jan 1, 1970
0
John Woodgate said:
The OP doesn't understand Thévenin and you want to sell him Laplace? Is
this just an ego trip or do you really not understand the KISS
Principle?

Did you read the OP's request? "Does anyone know a THOROUGH definition
of input impedance? output impedance?" He did not ask for a KISS, and
therefore incomplete definition. So I gave a complete, general, thorough,
and correct definiton. In another post, I worked his example and explained
the result to show him how it is done. Ratch
 
G

G Patel

Jan 1, 1970
0
John said:
The OP doesn't understand Thévenin and you want to sell him Laplace? Is
this just an ego trip or do you really not understand the KISS
Principle?

No not KISS. Didn't you read my OP? I wanted to go passed the simple
cases that everyone talks about and delve deaper. I was taught the
calculation method for input impedances and have used it before but
then I started questioning how voltage and current sources would be
dealt with. Then I realized that I could make circuits (like the
simple one I posted) that would not have a contast Vin/Iin ratio. This
is why I wanted a more STRICT definition of input impedance.

Although I'm a Computer Science student, I do understand Thevenin and
Laplace from my EE course. Thevenin applies to output impedances, and
the method has specific treatments for voltage/current sources when
looking BACK into a circuit. But I always wondered the method required
to treat voltage and current sources for looking INTO a circuit (INPUT
IMPEDANCE). V/I for this circuit is not constant, but other posters
have noted that in such a case we only worry about the incremental
effect on I for each Volt (which is the same result you get if you
short voltage sources and open current sources - much like when
calculating Thevenin output impedances).
 
R

Ratch

Jan 1, 1970
0
John said:
The OP doesn't understand Thévenin and you want to sell him Laplace? Is
this just an ego trip or do you really not understand the KISS
Principle?

No not KISS. Didn't you read my OP? I wanted to go passed the simple
cases that everyone talks about and delve deaper. I was taught the
calculation method for input impedances and have used it before but
then I started questioning how voltage and current sources would be
dealt with. Then I realized that I could make circuits (like the
simple one I posted) that would not have a contast Vin/Iin ratio. This
is why I wanted a more STRICT definition of input impedance.

Although I'm a Computer Science student, I do understand Thevenin and
Laplace from my EE course. Thevenin applies to output impedances, and
the method has specific treatments for voltage/current sources when
looking BACK into a circuit. But I always wondered the method required
to treat voltage and current sources for looking INTO a circuit (INPUT
IMPEDANCE). V/I for this circuit is not constant, but other posters
have noted that in such a case we only worry about the incremental
effect on I for each Volt (which is the same result you get if you
short voltage sources and open current sources - much like when
calculating Thevenin output impedances).

I hope the Laplace definition of impedance and previous example have
helped you. It is correct for input impedances, output impedances, and
anything in between. It also works for active voltage/current sources, as
the example illustrates. Ratch
 
J

Jim Thompson

Jan 1, 1970
0
Jim Thompson wrote...

Oops, time for another trip to the drugstore for stronger reading
glasses. Or perhaps a trip to the kitchen so the good wife can
give them a good cleaning. <sigh>

I have to buy a new pair... dropped them on a concrete floor, and have
a scratch right across the line of sight :-(

...Jim Thompson
 
B

Ben Bradley

Jan 1, 1970
0
I don't respond to problem definitions posted on sites requiring
registration. Sorry.

Ahem. Did you see the bugmenot site in the earlier thread? And I
didn't add this login, it was already there:

http://bugmenot.com/view.php?url=http://physicsforums.com

It was hardly worth going there. The first pic is two terminals
that go to a 20V battery in series with a 100 ohm resistor. Just a
SWAG, but I'd say the input impedance of that circuit is 100 ohms.
Second pic, "Zin = V / I Formula works with Phasor Algebra also."
But I'll have to ask Science Officer Spock how Phasor Algebra works. I
know it has to do with the current through a Flux Capacitor.
 
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