1
\$\begingroup\$

This is an example calculating output resistance that drives the inductor.

Source:Linear Circuit Transfer Functions: An Introduction to Fast Analytical Techniques by Christophe P. Basso

What I am confused is why the input and output are grounded references so they are connected together here as indicated in the red circles?

For calculating output resistance seen by the inductor, should we only short the input 1 so the output resistance will be infinite?

enter image description here

enter image description here

Update:

Why not just calculate seen by the inductor as normal (short input voltage source and leave everything else remain the same as below? So the resistance seen by inductor is infinity.

enter image description here PS: Google book link - page 4:

https://books.google.co.kr/books?id=WGBFjgEACAAJ&printsec=frontcover#v=onepage&q&f=false

Or this online version:

http://dl.4mohandes.com/book/ba/Linear_Circuit_Transfer_Functions.pdf

\$\endgroup\$
1
  • \$\begingroup\$ I can see the download link given by Anhnha for an "online version" leads to the illegal download of a copyrighted document. Does this contribution comply with the ethic we expect from posters on this site? \$\endgroup\$ Commented Mar 9, 2017 at 12:28

2 Answers 2

1
+50
\$\begingroup\$

He's finding the Thevenin resistance across the inductor. It's a two port network and the circuit shown in the figure is equivalent circuit of it. It's equivalent ckt will have lower port of input and output grounded. See here for all type of equivalent circuits. Since capacitor is getting open circuited and i/p voltage source short circuited we are getting Thevenin resistance as R1 + R2.

\$\endgroup\$
8
  • \$\begingroup\$ Thank you. However, what I am confused is why lower port of input and output arevgrounded here? Because if it is grounded in the fig 1.3 then C and R2 are useless here. I think the input resistance seen by inductor should be infinite. \$\endgroup\$
    – emnha
    Commented Oct 15, 2016 at 9:41
  • \$\begingroup\$ I/p voltage is applied with reference to lower terminal of i/p port, similarly o/p voltage is taken with reference to lower o/p terminal which are grounded in equivalent circuit. \$\endgroup\$
    – Vedanshu
    Commented Oct 15, 2016 at 9:53
  • \$\begingroup\$ If you still not getting it take Op-Amp as an example. Even though we apply input voltage to the two terminals but when solving questions we take the negative terminal as reference and similarly we take a ground at the output for reference to the output voltage. \$\endgroup\$
    – Vedanshu
    Commented Oct 15, 2016 at 10:12
  • \$\begingroup\$ Thanks. Could you see my update and question in my post? Normally to calculate resistance seen by a component, what we do is set the input source to zero (voltage sources are shorted and current sources are open) and leave everything else remain the same as the picture above? \$\endgroup\$
    – emnha
    Commented Oct 15, 2016 at 10:26
  • \$\begingroup\$ If that was an exact circuit then you were right. But this is an equivalent of a two port network. For solving such type of questions we take two references one at the input and other at the output. Then we short the voltage source and open the current source. In some texts they are shown as a single reference also but there we have some dependent source and no other linkage between input and output other than dependent source. \$\endgroup\$
    – Vedanshu
    Commented Oct 15, 2016 at 10:36
1
\$\begingroup\$

It was confusing to me at first , when your suggestion to solve this trivial network seems more obvious. Butthis method just uses equivalent circuit methods to demonstrate how it works.

The advantage comes with more complex internal networks and many loops by "loop conversion" methods into a simple single loops, so that computing transfer functions from input/output and reactive part removed permits the impedance or admittance to result in simpler transfer functions to define the DC response, infinite f and mid-range AC Poles and Zero's without complex alegbra.

He calls it a ** low entropy method** of simplifying by Thevenin equivalent circuits then finding the equivalent impedance of each reactive part to identify more easily the other elements they resonate with in a current loop. From this he writes simple LaPlace transforms using Z(s) for impedance from the reduction of network elements into simpler loops.

I can't say I am proficient in this method, but I understand the concept of finding port admittance while grounding the other port sides (one side at a time, not together) and admit this is more of a comment than an answer where more space is available.

I use a low entropy methods all the time, converting nonlinear semiconductors into linear models with ESR in order to estimate load regulation, ripple , LED currents, thermal runaway thresholds with physical thermal resistance and electrical ESR, ESR/load ratios, and many other linearization methods of nonlinear circuits.

Here he uses Driving Point voltages with 0 impedance, artificial ground reference points for partial loop admittance and other methods to eliminate complex algebraic terms that have neglible effects in order to find pole/zeros that dominate a transfer function and is adopted with a passion by his reviewer at ON Semi.

\$\endgroup\$
1
  • \$\begingroup\$ Thank you. For this, "I can't say I am proficient in this method, but I understand the concept of finding port admittance while grounding the other port sides (one side at a time, not together) and admit this is more of a comment than an answer where more space is available" how do you explain the case two port sides are grounded here? \$\endgroup\$
    – emnha
    Commented Oct 21, 2016 at 16:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.