Input Impedance of SIMPLE Circuit

[Fred Bloggs]

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.

 

[Ratch]

Z(s)=(V/s)/I(s)=100V/V-20
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

 

[Ratch]

I read in sci.electronics.design that Jim Thompson


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

 

[Don Kelly]

-----snip--------
.. behind

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

-----------
The general case is the Thevenin model. The op amp is a particular case -
the op amp "source" is a dependent source which can be eventually related to
the input source such that the effective input loop voltage is Vg(1-BF)
where BF is the feedback bugger factor. Then Zin =Vg(1-BF)/Iin.becomes
independent of the magnitude of Vg and Iin- i.e.it behaves like an
impedance. (sure you can use a BF on the current or the physical input
resistance- the effect is the same).

Note also that the op-amp "small signal model involves a number of
simplifications and practical approximations. The derivations make these
approximations. "If the gain is very large"..., etc. ". You are trying to
model a device on the basis of approximating the actual terminal behaviour -
i.e. postulating the innards of a black box which looks the same from the
outside. Whether the postulation bears any relationship, internally, to the
device is unimportant. This is also true with Thevenin and Norton models.

In the particular case (op-amp) the Thevenin model effectively behaves as an
impedance because there is only one actual independent source involved- the
voltage applied to the input terminals). It is quite reasonable to go from
the general to the particular but going the other way is following a path
which should be marked with signs "Here there be dragons!!"

 

[Dr. Polemic]

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.

Or, does the impedance suddenly become 100 ohms when the internal
voltage is zero and we apply zero volts externally? But we can't
measure any current to divide into the applied voltage in this case,
so how can we measure the impedance? Maybe we need an electronic
L'Hospital's rule since otherwise we would have to evaluate 0 volts/0
amps. Or maybe there is no impedance when the applied voltage is
zero. You've confused me now.

 

[Ratch]

Ratch
-----------
The general case is the Thevenin model. The op amp is a particular case -
the op amp "source" is a dependent source which can be eventually related to
the input source such that the effective input loop voltage is Vg(1-BF)
where BF is the feedback bugger factor. Then Zin =Vg(1-BF)/Iin.becomes
independent of the magnitude of Vg and Iin- i.e.it behaves like an
impedance. (sure you can use a BF on the current or the physical input
resistance- the effect is the same).

Note also that the op-amp "small signal model involves a number of
simplifications and practical approximations. The derivations make these
approximations. "If the gain is very large"..., etc. ". You are trying to
model a device on the basis of approximating the actual terminal behaviour -
i.e. postulating the innards of a black box which looks the same from the
outside. Whether the postulation bears any relationship, internally, to the
device is unimportant. This is also true with Thevenin and Norton models.

In the particular case (op-amp) the Thevenin model effectively behaves as an
impedance because there is only one actual independent source involved- the
voltage applied to the input terminals). It is quite reasonable to go from
the general to the particular but going the other way is following a path
which should be marked with signs "Here there be dragons!!"

Nevertheless, the internal source voltage of the op amp does have a
profound effect on the input/output of the op amp. And we don't have to
stick to op amps either. Tubes and transistors change their input/output
impedances depending on their amplification factors too. But how would one
determine the impedance of the example circuit presented by the OP which was
a 100 ohm resistor in series with a 20 volt opposing voltage. If that
circuit was black box so you did not know what it contained, would you
determine the impedance incrementally by measuring the change in voltage
divided by the change in current. Or make a voltage vs. current curve from
several measured points? The curve will not be a straight line, but could
one not be able to say that the derivative of the curve at a particular
point is the incremental impedance at that point? Ratch

 

[Ratch]

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.

Maybe we need an electronic
L'Hospital's rule since otherwise we would have to evaluate 0 volts/0
amps. Or maybe there is no impedance when the applied voltage is
zero. You've confused me now.

I hope you are unconfused now. Ratch

 

[Jim Thompson]

IMPEDANCE is defined as deltaV/deltaI

...Jim Thompson

 

[keith]

IMPEDANCE is defined as deltaV/deltaI

I agree that imput/output impedance is a small-signal concept, but then
impedance does change based on bias conditions (f the driver and the
drivee.

 

[Jim Thompson]

I agree that imput/output impedance is a small-signal concept, but then
impedance does change based on bias conditions (f the driver and the
drivee.

NOT in the example that everyone is all-confused over.

...Jim Thompson

 

[Don Kelly]

that to (particularly related as

Nevertheless, the internal source voltage of the op amp does have a
profound effect on the input/output of the op amp. And we don't have to
stick to op amps either. Tubes and transistors change their input/output
impedances depending on their amplification factors too. But how would one
determine the impedance of the example circuit presented by the OP which was
a 100 ohm resistor in series with a 20 volt opposing voltage. If that
circuit was black box so you did not know what it contained, would you
determine the impedance incrementally by measuring the change in voltage
divided by the change in current. Or make a voltage vs. current curve from
several measured points? The curve will not be a straight line, but could
one not be able to say that the derivative of the curve at a particular
point is the incremental impedance at that point? Ratch

---------------
I agree that the internal voltage does have a profound effect. However, this
internal voltage is directly related to the applied source voltage through
such things as the amplifier gain and the feedback network. I implied as
much.
Certainly this is also true for tubes and transistors. The models we use are
simplifications based on small deviations about an operating point.The small
signal parameters are "linearised" to make solution easier. Useful only
because there is a fairly good degree of linearity about any normal
operating point in such devices. Change this operating point, DC supply
voltage or signal strength and lots of things can change.

However, in a given situation- which model works best?

In the circuit shown there is an impedance of 100 ohms in series with 20V.
Suppose that we apply an external voltage Vin
Then Vin-20=100I
Now apply Vin +dV
Vin +dV-20 =100(I+dI) which results in dV/dI =100 or dI/dV =0.01
Incremental impedance is constant at 100 ohms for any value of Vin

Note that, in this example, current IS a linear function of the input
voltage. I=0.01Vin - 0.2
This IS a straight line passing through I=0 at Vin =20v. I suggest that you
plot a few points.

The V/I ratio is not constant because of the bias voltage. Now we have two
choices:

a) Use a Thevenin model of 100 ohms in series with a 20V source. This is a
linear model, which is very easy to incorporate into a wider circuit model
and make use of any and all network theorems that are in our tool kit. Piece
of cake. Laplace methods are valid. Do the job in 5 minutes and break off
for a beer.

b)Use the Z =Vin/I which is a very badly behaved nonlinear resistance in
this case. Now try to solve in conjunction with a source circuit. You have
Kirchoff's Laws to work with but forget all the useful tools in the kit bag,
including Laplace. In simple cases a graphical approach works well-
otherwise, iterate and hope that your iterative algorithm doesn't blow up.
Forget the beer, you haven't any spare time.

 

[G Patel]

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]

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.


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


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?

 

[Ratch]

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? 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

 

[John Woodgate]

I and others in this NG are trying to reach an agreement on what
input impedance is.

I already explained that you are trying to *re-define* input [load]
impedance in a different way from what is almost universally used in the
discipline. Most people here don't like excessive formality, but
sometimes it's necessary.

The normal definition of 'input impedance' is [formally 'is equal to']
the Thévenin/Norton equivalent impedance. It is NOT the ratio of input
current to input voltage if the input circuit (unusually) includes an
*independent* voltage or current source.

Furthermore, there isn't necessarily a 'one size fits all' explanation
for what goes on at input terminals. Consider an amplifier for an
electret microphone which has just two connections, requiring an
external drain resistor for the FET head amp. Use Courier font:

+---------+-------+----------+--- +5 V
| | | |
R1 R2 R5 |
| | +----- out |
| | |/ |
+----o---C1----+-----| C2
mic |O | |\e |
+----o | | |
| R3 R4 |
| | | |
| | | |
| | | |
+---------+-------+----------+--- 0 V

For DC, we just see 5 V in series with R1, but for signals, we see the
effect of every component. R5 has very little effect if its value is not
very incorrect, and the +5 V rail is ground/earth for signals, due to
C2. The impedance presented by the transistor base is beta*R4, closely
enough. The presence of, say, 1.6 V DC on the base is irrelevant.

 

[Ratch]

I and others in this NG are trying to reach an agreement on what
input impedance is.

I already explained that you are trying to *re-define* input [load]
impedance in a different way from what is almost universally used in the
discipline. Most people here don't like excessive formality, but
sometimes it's necessary.

The normal definition of 'input impedance' is [formally 'is equal to']
the Thévenin/Norton equivalent impedance. It is NOT the ratio of input
current to input voltage if the input circuit (unusually) includes an
*independent* voltage or current source.

Furthermore, there isn't necessarily a 'one size fits all' explanation
for what goes on at input terminals. Consider an amplifier for an
electret microphone which has just two connections, requiring an
external drain resistor for the FET head amp. Use Courier font:

+---------+-------+----------+--- +5 V
| | | |
R1 R2 R5 |
| | +----- out |
| | |/ |
+----o---C1----+-----| C2
mic |O | |\e |
+----o | | |
| R3 R4 |
| | | |
| | | |
| | | |
+---------+-------+----------+--- 0 V

For DC, we just see 5 V in series with R1, but for signals, we see the
effect of every component. R5 has very little effect if its value is not

very incorrect, and the +5 V rail is ground/earth for signals, due to
C2. The impedance presented by the transistor base is beta*R4, closely
enough. The presence of, say, 1.6 V DC on the base is irrelevant.
--

Thank you. Your definition of input impedance and comment about
independant voltage/current was succinct and ...instructive. Ratch

 

[Dr. Polemic]

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? 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?

 

[Jim Thompson]

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.

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

...Jim Thompson

 

[Rich Grise]

Zero.

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

Boy, you PhDs sure do go the long way around to get an answer. ;-P
I'm "only" a tech, and I determined by inspection in one glance that the
input impedance of that circuit is 100 ohms. This is corroborated by, as
far as I know, the "Thevenin/Norton Resistance", in which the 20 VDC is
replaced by a short.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/thevenin.html#c3

(I've held my figurative tongue, because when I first saw the question,
"What's the input impedance of this circuit?", I answered to myself,
"100 ohms," but I waited, because I wasn't absolutely sure and didn't
want to be a dork under my techie name. Thanks!)

Cheers!
Rich

 

[John Woodgate]

I read in sci.electronics.design that Rich Grise <[email protected]>

(I've held my figurative tongue, because when I first saw the question,
"What's the input impedance of this circuit?", I answered to myself,
"100 ohms," but I waited, because I wasn't absolutely sure and didn't
want to be a dork under my techie name. Thanks!)

You don't need to be too modest. It's a 'tech' question, which is why I
moaned about the introduction of Laplace and other exotica. The thread
was lengthened by my assumption that 'Ratch' was a perverse tech rather
than the (newly enlightened) student he now appears to be.