Input Impedance of SIMPLE Circuit

Yes, but the Thévenin/Norton impedance is not necessary the input/output impedance when active voltage/current sources are present. This was proved in the example he submitted. Ratch

Apply 19V to the input terminals... I=-10mA (flowing toward you)

Apply 20V to the input terminals... I=0A

Zin = deltaV/deltaI = 1/10mA = 100 ohms

Apply 25V to the input terminals... I=50mA

Apply 30V to the input terminals... I=100mA

Zin = deltaV/deltaI = 5/50mA = 100 ohms

Shall I go on ?:-)

...Jim Thompson

Look around the beginning of this thread. The OP submitted a example circuit.

Yes, they do. If you read the example, you will see that the input varies with respect to the the input voltage because of the internal voltage source. The Thévenin/Norton impedances are not affected by the active sources, but they are not necessarily the input/output impedances.

The "incremental impedance" will vary in the example circuit depending on the input voltage. Ratch

Then please be so kind to point out, using the previous example, where I erred. Ratch

depending

Show me! Show me! Use the example. Ratch

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

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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Careful there. AC analysis (Laplace) considers a DC value as an AC ZERO. So looking into this network with Laplace aforethought gives the same result. (As it should.)

...Jim Thompson

I read in sci.electronics.design that Ratch wrote (in ) about 'Input Impedance of SIMPLE Circuit', on Tue, 26 Apr 2005:

Yes. I also know that he doesn't know the implications of his naive question.

I didn't say it was incorrect. To the OP, it is most likely incomprehensible and thus confusing, which means that it's of negative help in furthering his understanding.

I read in sci.electronics.design that G Patel wrote (in ) about 'Input Impedance of SIMPLE Circuit', on Tue, 26 Apr 2005:

NO, YOU DON'T.

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

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.

I call that a bifocal.

Al

Jim, I really appreciate you proving your point instead of resorting to hateful invective like some of the posters in this NG do. Yes, I realize that I misspoke about the incremental impedance being dependent on the input voltage. I do know better and should have seen that. The Laplace method gives the total impedance, and shows that it does vary widely according to the input voltage. Again, I should have checked the incremental impedance more carefully. Ratch

I read in sci.electronics.design that G Patel wrote (in ) about 'Input Impedance of SIMPLE Circuit', on Tue, 26 Apr 2005:

You don't have to believe me. Just continue in ignorance. And arrogance.

I read in sci.electronics.design that Ratch wrote (in ) about 'Input Impedance of SIMPLE Circuit', on Tue, 26 Apr 2005:

Which example was that? Since the theorems require sources to be disabled when determining the impedance 'seen' from the terminals, they don't influence that impedance.

I think you may be referring to the difference between incremental and 'DC' values of voltage and current, but that isn't relevant to T/N, because they include an assumption of linearity.

I always thought that this was one of the eay-to-understand concepts of 'tronics. Methinks young 'Ratch' has missed the stick all together.

You need to be registered to view that image. That sucks.

Put it elsewhere where it will be globally viewable without the need to register.

Ratch

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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.

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 snipped-for-privacy@peeshaw.ca remove the urine to answer

I read in sci.electronics.design that Ratch wrote (in ) about 'Input Impedance of SIMPLE Circuit', on Tue, 26 Apr 2005:

current?

I don't see any point in continuing. You have re-defined 'input [load] impedance' and 'output [source] impedance' to make your argument.

The **theorems** say that you can replace any 2-terminal network by either a voltage generator in series with an impedance or a current generator in parallel with that impedance. The generator voltage is equal to the open-circuit voltage at the terminals; the generator current is equal to the short-circuit current at the terminals and the impedance is equal to the impedance measured at the terminals with all generators inside the network disabled (voltage generators shorted, current generators open-circuited.).

They don't make ANY additional claims about what happens when you connect some other network to those terminals. You have to work that out for yourself, using the equivalent circuit defined by one or other of the theorems, Ohm's and Kirchoff's Laws, superposition etc.

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 read in sci.electronics.design that Jim Thompson wrote (in ) about 'Input Impedance of SIMPLE Circuit', on Tue, 26 Apr 2005:

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.