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)?

 

[mike]

gaya.patel@gmail.com wrote:

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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with links. Delete this sig when replying.
..
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ht<removethis>tp://www.geocities.com/SiliconValley/Monitor/4710/

 

[G Patel]

gaya.pa...@gmail.com wrote:

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?)

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)?

 

Jim Thompson wrote:

On 25 Apr 2005 17:05:59 -0700, "G Patel" <gaya.patel@gmail.com
wrote:
gaya.pa...@gmail.com wrote:

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?)



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 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

 

[Tim Wescott]

Jim Thompson wrote:

On 25 Apr 2005 17:05:59 -0700, "G Patel" <gaya.patel@gmail.com> wrote:


gaya.pa...@gmail.com wrote:

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?)



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 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.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

 

[G Patel]

John Woodgate wrote:

I read in sci.electronics.design that Ratch <Watchit@comcast.net
wrote
(in <7sGdne-JJLJ7XPDfRVn-2A@comcast.com&gt about 'Input Impedance of
SIMPLE Circuit', on Mon, 25 Apr 2005:

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?

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).

--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
'What is a Moebius strip?'
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

 

[G Patel]

John Woodgate wrote:

I read in sci.electronics.design that G Patel <gaya.patel@gmail.com
wrote (in <1114523530.168984.18040@o13g2000cwo.googlegroups.com&gt
about
'Input Impedance of SIMPLE Circuit', on Tue, 26 Apr 2005:

Although I'm a Computer Science student, I do understand Thevenin
and
Laplace from my EE course.

NO, YOU DON'T.

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).

Thévenin applies to ANY two terminals. That's why I say you DON'T
understand. It is the substance of your enquiry.

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).

No, we DON'T always use incremental values. It depends on what we
want
to determine.

But Thévenin works for ANY two terminals. It doesn't 'know' whether
the
terminals are an input or an output.


From all the resources I've seen, the THEVENIN IMPEDANCE = OUTPUT
IMPEDANCE (the impedance related to the THEVENIN VOLTAGE SOURCE). Can

you point me to a reference that related THEVENING IMPEDANCE to INPUT
IMPEDANCE?

--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
'What is a Moebius strip?'
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

 

[John Woodgate]

I read in sci.electronics.design that Jim Thompson
<thegreatone@example.com> wrote (in
<4ljt61h8itnpgi45oakbkgrn5v8q8mhgl3@4ax.com&gt about 'Input Impedance of
SIMPLE Circuit', on Tue, 26 Apr 2005:

Shall I go on ?

Ratch redefined 'input impedance' to mean 'input current/input voltage',
so it varies in the example from near 100 ohms to infinity and back
again as you change what is connected to the terminals. This is not a
helpful concept.
--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
'What is a Moebius strip?'
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

 

[Ratch]

"Don Kelly" <dhky@peeshaw.ca> wrote in message
news:73Gbe.1139246$Xk.955550@pd7tw3no...

"Ratch" <Watchit@comcast.net> wrote in message
news:boSdnRPub4n1VPDfRVn-sg@comcast.com...

"G Patel" <gaya.patel@gmail.com> wrote in message
news:1114473959.547108.117520@o13g2000cwo.googlegroups.com...
gaya.pa...@gmail.com wrote:
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?)


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
---------

No, it makes the input "impedance" 100 ohms. Essentially you have a
Thevenin
model of 100 ohms in series with a 20V source.

The fact that it is a Thevenin "source" is not changed by the external
source (or load) applied. (i.e hook a 100 ohm, 20V Thevenin source to a
load
which has an active source of 20V. the current will be 0. Does that make
the impedance infinite? No- it just means that the internal and external
voltages add up to 0. )
If you take your definition of impedance, then the accepted term
"Thevenin
impedance" of a source is incorrect.

You can use the Z(s)=V(s)/I(s) quite happily if there is no source other
than the external driving source-works like a hot damn with passive,
linear
elements. Using this Z(s) in series with an internal V(s) is also happy
but
don't call it an impedance as it doesn't behave like one (particularly a
linear one where Laplace is meaningful), but behaves like a source behind
an
impedance- and that's how you model it, using Laplace or otherwise.

I read and understand what you are saying. However, what about op
amps? We all have seen countless derivations of the input impedance of
inverting and non-inverting configurations, which assume a input voltage,
calculate the expected current, divide the two and proclaim such and so
input impedance. Now we are dealing with internal dependent voltage
sources. Can you square that method with what I was trying to do? Ratch

If sufficiently masochistic, you may treat it as a non-linear "impedance"
,
throw away linear circuit analysis (including Laplace) and solve KVL and
KCL
non-linear, often differential, equations. You can also cut your lawn
with
manicure scissors.
--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer




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)?

 

[Ratch]

"John Woodgate" <jmw@jmwa.demon.contraspam.yuk> wrote in message
newsf8kvbET7ybCFwqd@jmwa.demon.co.uk...

I read in sci.electronics.design that Jim Thompson
thegreatone@example.com> wrote (in
4ljt61h8itnpgi45oakbkgrn5v8q8mhgl3@4ax.com&gt about 'Input Impedance of
SIMPLE Circuit', on Tue, 26 Apr 2005:
Shall I go on ?

Ratch redefined 'input impedance' to mean 'input current/input voltage',
so it varies in the example from near 100 ohms to infinity and back
again as you change what is connected to the terminals. This is not a
helpful concept.

Isn't that what is done to calculate the input impedance of op amps?
See my post to Don Kelly above. Ratch

--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
'What is a Moebius strip?'
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

 

[G Patel]

Fred Bloggs wrote:

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).

Your problem is that you are thinking too much and not paying
attention
to the obvious. You have been short changed by the university if they

allowed you to walk out of a networks course with your level
confusion.
The Thevenin/Norton reduction theorems say no such thing about an
input/output orientation, the theorems statements are about
*equivalent*
two terminal networks. When the equivalent network contains an active

source then it obviously makes no sense to talk about the terminals
being an input or output because it can dissipate as well deliver
energy, isn't that right. The equivalent impedance is an input
impedance
when you zero the internal source and it's an output impedance when
you
zero the external sources, then let the superposition theorem take it

from there. So you seem to be missing the fact that a single network
port can be an input and output simultaneously. Vin/Iin reduces to a
constant only when Vin is the only source.

I guess my prof's mere Vin/Iin isn't a broad enough definition of input
impedance (as I suspected). Thank you for clearing things up (I'll
ignore the insults in your post).

 

[Dr. Polemic]

On Thu, 28 Apr 2005 07:25:40 -0700, Jim Thompson <thegreatone@example.com> wrote:

On Thu, 28 Apr 2005 05:30:43 -0700, Dr. Polemic <nospam@aol.com
wrote:

On Thu, 28 Apr 2005 04:47:06 -0500, "Ratch" <Watchit@comcast.net> wrote:


"Dr. Polemic" <spamnot@aol.com> wrote in message
news:a1s071h1v8hc5acn18ivupu1v91eaiun61@4ax.com...
On Wed, 27 Apr 2005 20:42:29 -0500, "Ratch" <Watchit@comcast.net> wrote:


"Dr. Polemic" <spamnot@aol.com> wrote in message
news:bld0719uds2tifhkb3tm8epspr4ndtlopl@4ax.com...
On Mon, 25 Apr 2005 23:53:54 -0500, "Ratch" <Watchit@comcast.net
wrote:


"G Patel" <gaya.patel@gmail.com> wrote in message
news:1114473959.547108.117520@o13g2000cwo.googlegroups.com...
gaya.pa...@gmail.com wrote:
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?)


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

Ok, so then if we have a 1 volt internal source, and apply 1 volt
externally, the impedance is infinite. Likewise, if we have a 1
picovolt internal source and apply 1 picovolt externally, the
impedance is still infinite. And finally, continuity considerations
would argue that if we have a 0 volt internal source and apply 0 volts
externally, the impedance is infinite by the same reasoning, contrary
to what you said above. Thus, we have proved that the impedance of a
100 ohm resistor is infinite.

When there is zero internal voltage and zero applied voltage, there
is
zero input current. Now 0/0 is indefinite or indeterminate, so one
cannot
determine the impedance with those values.


Or, does the impedance suddenly become 100 ohms when the internal
voltage is zero and we apply zero volts externally?

Impedance can only be measured when either voltage or current is not
zero?

But we can't
measure any current to divide into the applied voltage in this case,
so how can we measure the impedance?

If you cannot measure the current, it must be zero. If the voltage
in
not zero, then you can calculated a impedance value. If the voltage is
also
zero, the impedance is indeterminate.

You're making this up as you go along, aren't you? Can you quote one
authoritative source
which suggests that impedance can be indeterminate? I mean real impedance
and not a
mathematical construct. Maybe when zero volts is applied to a resistor,
its impedance
becomes evanescent. Sort of like, if a tree falls in the forest when no
one is there to
hear it, does it make a sound? Maybe if no one is measuring an impedance,
it isn't there.
I showed how, using your reasoning, if the internal and external voltages
in the example
under discussion decrease together, from 20 volts, to 1 volt, to 1
picovolt, to 1
femtovolt, etc., the impedance remains infinite. Now, 0/0 may be
indeterminate, but as V
-> 0, V/0 approaches a limit of "infinity" in the same sense as you used
the word. So,
clearly, your reasoning leads us to the conclusion that a 100 ohm resistor
has infinite
impedance with zero applied voltage. And if the impedance isn't infinite
with zero
applied and internal voltages sources, but it is with 1 picovolt sources,
how does it
suddenly jump to 100 ohms as the voltage goes to zero?

I and others in this NG are trying to reach an agreement on what input
impedance is. If impedance is simply input voltage divided by input
current, then in the case where the internal voltage is zero, V/I will
always be 100 ohms and no indeterminancy will occur. If an internal voltage
is present, no matter how slight, then the applied voltage can be adjusted
to cancel the input voltage and the input impedance will be calculated as a
infinite value. The crux of the matter is whether the internal voltage of
the circuit affects the input impedance. In other words, is impedance a
dynamic quantity when internal voltages are present?

Well, what have you and others decided about this question? I have seen you agree in some
measure to explanations given by John Woodgate and Jim Thompson, but what is the answer to
the question in the last line of the paragraph just above? Is there a theory without the
inconstencies found in your line of reasoning?

As for indeterminate
impedance, I probably should have said that impedance cannot be calculated
when both voltage and current are of zero value. Certainly it exists. It
can be likened to attempting to measure a resister with an ohmmeter equipped
with a dead battery. Ratch

Suppose then that the 100 ohm resistor is removed from the "box", but the 20 volt source
remains and is connected to the terminals. What is the impedance at the input terminals
of the "box" now?

Zero.

Not if you try to measure the impedance by applying 20 volts externally according to
Ratch. Pasting in his earlier explanation, we see how the reasoning would go:

"...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."

So it would seem that a voltage source can have infinite impedance under certain
circumstances, even though the texts tell us to replace voltage sources with a short and
current sources with an open.

The general rule for determining impedance requires all DC voltage and
current sources to be set to ZERO value.

...Jim Thompson