Input Impedance of SIMPLE Circuit

Apr 25, 2005 64 Replies

agree in some

the answer to

without the

volt source

input terminals

according to

balances the

making the

short and

I think the "infinite" crowd have themselves thoroughly confused by the differences in meaning between _impedance_ and _Thevenin_equivalence_.

Heaven help us when they discover Norton ;-)

...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.

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

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.

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

Reggie's Theorem -

To calculate the input impedance of any network simply replace internal current and voltage sources with their own internal impedances and carry on as normal.

Can't imagine what all the reams of fuss from the old wives is all about.

It's not Vin/Iin (I suspect you've misunderstood your prof). It's (delta Vin)/(delta Iin). Put a 1Mohm load on the input node and measure the

*change* in voltage divided by the *change* in current. Now make that resistor infinite (limit).
Keith

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