Input Impedance of SIMPLE Circuit

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

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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

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

IMPEDANCE is defined as deltaV/deltaI

...Jim Thompson

-- | James E.Thompson, P.E. | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona Voice:(480)460-2350 | | | E-mail Address at Website Fax:(480)460-2142 | Brass Rat | |

| 1962 | I love to cook with wine. Sometimes I even put it in the food.

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.

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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

and

output

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

...Jim Thompson

-- | James E.Thompson, P.E. | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona Voice:(480)460-2350 | | | E-mail Address at Website Fax:(480)460-2142 | Brass Rat | |

| 1962 | I love to cook with wine. Sometimes I even put it in the food.

to

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

Ratch

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.

I hope you are unconfused now. Ratch

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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).

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 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;).

--
  Keith

-----snip-------- .

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----------- 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!!"

--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer

 >
> >
> > 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)?
> > > >
> > >
> > >
> >
> >
>
>

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--------------- 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.

--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer
>
> >
> >  >
> > > >
> > > > 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)?
> > > > > >
> > > > >
> > > > >
> > > >
> > > >
> > >
> > >
> >
> >
>
>

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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?

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

volt source

terminals

Zero.

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

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.

...

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.

(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

One of my Dad's favorite Ziggy cartoons (which he had pinned to his cork board) had Ziggy looking into a mirror, asking himself rhetorically, "I wonder if I'd be modest, if I had a choice?" ;-P

Cheers! Rich

--
 "It's easy to be humble when you're successful, but it takes a lot of
 class to be arrogant when you're a total failure!" - Unknown

I read in sci.electronics.design that Ratch wrote (in ) about 'Input Impedance of SIMPLE Circuit', on Thu, 28 Apr 2005:

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.

--
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