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I'm trying to find the resistance \$Z_{CE}\$ looking into C, E terminals of the circuit below when \$V_{IN} = 0\$. I introduced a current source between two terminals C, E and then calculate the voltage across the terminals to get the equivalent resistance. It can also be solved by a number of other circuit analysis techniques like nodal analysis, mesh analysis.
However, the result is quite messy in a sense that it's written a form that doesn't give us much insight to approximate.
(for example, if R1 and R2 are in parallel, the result above is written as R1*R2/(R1 + R2) not R1 || R2.
It's clear that writing R1 || R2 is more meaningful as it allows us to make necessary approximation based on the relative size of R1 and R2. )
Question:
Is there a way to get the resistance looking into CE in a form of series or parallel resistances?

enter image description here

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15
  • \$\begingroup\$ Open circuit the current source and short the voltage source then add up the resistances left in parallel and series combinations. \$\endgroup\$
    – Andy aka
    Jun 19 at 16:25
  • \$\begingroup\$ @Andyaka I belived it's not that simple due to existing of the controlled current source. You cannot ignore it. \$\endgroup\$
    – emnha
    Jun 19 at 16:34
  • \$\begingroup\$ It does not contribute anything to the impedance, believe it or not. Additionally, if Vin is zero then Ib is zero and Ic is therefore also zero and rE is infinity. Zce = r0 || (Re + Rc) \$\endgroup\$
    – Andy aka
    Jun 19 at 16:36
  • \$\begingroup\$ @Andyaka Re || (re + Rb/(1+beta)) +Rc and this sum is again in parallel with ro? \$\endgroup\$
    – emnha
    Jun 19 at 16:41
  • \$\begingroup\$ @Andyaka I think it is not correct. I'll check it again but I remember the exact impedance is much more complex. \$\endgroup\$
    – emnha
    Jun 19 at 16:43
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Here a solution with the Extra Element Theorem (EET). with the dependant current source as extra element (See Vorperian's book).

To get the resistance a voltage vx is injected and the current ix is determined.

The other way to inject a current and get the voltage is possible too. But with them I get a complicate sub circuit.

Ro is removed first. Rx is the resistance without Ro.

Alltogether we will get:

$$ R = Ro || Rx = Ro || \frac{1}{Gx} $$

$$ Gx = \frac{ix}{vx} = H0 * \frac{(1 + \beta * An )}{(1 + \beta * Ad)} $$

Determine H0:

With dependant current source removed, ie. $$ \beta=0 $$.

$$ H0 = ix/vx = \frac{1}{(RC + RE || (Rb + Re))} $$

Determine Ad:

The dependant current source is replaced with a indepenent current source with changed polarity. vx is removed, that is shorted.

$$ Ad = \frac{Re}{( Re + Rb + (RE || RC) )} $$

Determine An:

The dependant current source is replaced with a indepenent current source with changed polarity. vx, ix is nulled. That means ix is zero, and vx is any value. -> vx is opened.

$$ An = \frac{( Re + RE )}{( Rb + Re + RE )} $$

Result:

$$ R = Ro || (RC + RE || (Rb + Re)) * \frac{(1 + \beta * \frac{Re}{(Re + Rb + (RE || RC))} )}{(1 + \beta * \frac{( Re + RE )}{( Rb + Re + RE)})} $$

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  • \$\begingroup\$ Nice. I'm thinking about applying 2EET for Rc and Re. I think it may work. \$\endgroup\$
    – emnha
    Jun 20 at 16:48
  • \$\begingroup\$ This is a good answer and the obtained formula is correct. However, I am not sure if using the EET here is not an overkill as Lemmy would yell. Here, by temporarily disconnecting \$r_o\$ and determining the resistance with the \$I_T\$ current source is three simple equations away. Then the paralleling of \$r_o\$ with this intermediate result brings the correct expression. The EET is truly efficient when the identified extra element is bothering the analysis and cannot be simplified via an intermediate sketch. This is a good exercise though. \$\endgroup\$ Jun 24 at 12:52
  • \$\begingroup\$ @VerbalKint it's easy to write equations and solve for the equivalent resistance but the result is just only good if you want a numerical value. It's hard to get any insight to design. \$\endgroup\$
    – emnha
    Jun 24 at 19:02
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First, I will present a method that uses Mathematica to solve this problem. When I was studying this stuff I used the method all the time (without using Mathematica of course).

Well, we are trying to analyze the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

$$ \begin{cases} \text{I}_2=\text{I}_\text{a}+\text{I}_1\\ \\ \text{I}_3=\text{I}_2+\text{I}_8\\ \\ 0=\text{I}_\text{a}+\text{I}_4+\text{I}_7\\ \\ \text{I}_8=\text{I}_4+\text{I}_6\\ \\ \text{I}_7=\text{I}_5+\text{I}_6\\ \\ \text{I}_1=\text{I}_3+\text{I}_5 \end{cases}\tag1 $$

When we use and apply Ohm's law, we can write the following set of equations:

$$ \begin{cases} \text{I}_1=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_1-\text{V}_2}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_2}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_3-\text{V}_2}{\text{R}_4}\\ \\ \text{I}_5=\frac{\text{V}_3}{\text{R}_5}\\ \\ \text{I}_6=\frac{\text{V}_3-\text{V}_2}{\text{R}_6} \end{cases}\tag2 $$

Now, we can set up a Mathematica-code to solve for all the voltages and currents:

In[1]:=Clear["Global`*"];
FullSimplify[
 Solve[{I2 == Ia + I1, I3 == I2 + I8, 0 == Ia + I4 + I7, 
   I8 == I4 + I6, I7 == I5 + I6, I1 == I3 + I5, I1 == (Vi - V1)/R1, 
   I2 == (V1 - V2)/R2, I3 == V2/R3, I4 == (V3 - V2)/R4, I5 == V3/R5, 
   I6 == (V3 - V2)/R6}, {I1, I2, I3, I4, I5, I6, I7, I8, V1, V2, V3}]]

Out[1]={{I1 -> (-Ia R2 R4 (R3 + R5) - 
      Ia (R3 R4 + 
         R2 (R3 + R4 + R5)) R6 + (R4 (R3 + R5) + (R3 + R4 + 
            R5) R6) Vi)/((R1 + R2) R3 R4 + (R1 + R2 + 
         R3) R4 R5 + (R3 (R4 + R5) + R1 (R3 + R4 + R5) + 
         R2 (R3 + R4 + R5)) R6), 
  I2 -> (Ia R3 R4 R5 + Ia R1 R4 (R3 + R5) + Ia R3 R5 R6 + 
      Ia R1 (R3 + R4 + R5) R6 + 
      R4 (R3 + R5) Vi + (R3 + R4 + R5) R6 Vi)/((R1 + R2) R3 R4 + (R1 +
          R2 + R3) R4 R5 + (R3 (R4 + R5) + R1 (R3 + R4 + R5) + 
         R2 (R3 + R4 + R5)) R6), 
  I3 -> (Ia R1 R4 R6 - Ia R2 R5 (R4 + R6) + R5 R6 Vi + 
      R4 (R5 + R6) Vi)/((R1 + R2) R3 R4 + (R1 + R2 + 
         R3) R4 R5 + (R3 (R4 + R5) + R1 (R3 + R4 + R5) + 
         R2 (R3 + R4 + R5)) R6), 
  I4 -> -((R6 (Ia R1 R3 + Ia (R1 + R2 + R3) R5 + 
          R3 Vi))/((R1 + R2) R3 R4 + (R1 + R2 + 
           R3) R4 R5 + (R3 (R4 + R5) + R1 (R3 + R4 + R5) + 
           R2 (R3 + R4 + R5)) R6)), 
  I5 -> (-Ia (R2 R3 R4 + R2 R3 R6 + (R1 + R2 + R3) R4 R6) + 
      R3 (R4 + R6) Vi)/((R1 + R2) R3 R4 + (R1 + R2 + 
         R3) R4 R5 + (R3 (R4 + R5) + R1 (R3 + R4 + R5) + 
         R2 (R3 + R4 + R5)) R6), 
  I6 -> -((R4 (Ia R1 R3 + Ia (R1 + R2 + R3) R5 + 
          R3 Vi))/((R1 + R2) R3 R4 + (R1 + R2 + 
           R3) R4 R5 + (R3 (R4 + R5) + R1 (R3 + R4 + R5) + 
           R2 (R3 + R4 + R5)) R6)), 
  I7 -> (Ia R4 (-(R1 + R2) R3 - (R1 + R2 + R3) R5) - 
      Ia (R2 R3 + (R1 + R2 + R3) R4) R6 + 
      R3 R6 Vi)/((R1 + R2) R3 R4 + (R1 + R2 + 
         R3) R4 R5 + (R3 (R4 + R5) + R1 (R3 + R4 + R5) + 
         R2 (R3 + R4 + R5)) R6), 
  I8 -> -(((R4 + R6) (Ia R1 R3 + Ia (R1 + R2 + R3) R5 + 
          R3 Vi))/((R1 + R2) R3 R4 + (R1 + R2 + 
           R3) R4 R5 + (R3 (R4 + R5) + R1 (R3 + R4 + R5) + 
           R2 (R3 + R4 + R5)) R6)), 
  V1 -> (Ia R1 (R2 R4 (R3 + R5) + R3 R4 R6 + 
         R2 (R3 + R4 + R5) R6) + (R3 R4 R5 + R2 R4 (R3 + R5) + 
         R3 (R4 + R5) R6 + R2 (R3 + R4 + R5) R6) Vi)/((R1 + 
         R2) R3 R4 + (R1 + R2 + R3) R4 R5 + (R3 (R4 + R5) + 
         R1 (R3 + R4 + R5) + R2 (R3 + R4 + R5)) R6), 
  V2 -> (R3 (Ia R1 R4 R6 - Ia R2 R5 (R4 + R6) + R5 R6 Vi + 
        R4 (R5 + R6) Vi))/((R1 + R2) R3 R4 + (R1 + R2 + 
         R3) R4 R5 + (R3 (R4 + R5) + R1 (R3 + R4 + R5) + 
         R2 (R3 + R4 + R5)) R6), 
  V3 -> (-Ia R5 (R2 R3 R4 + R2 R3 R6 + (R1 + R2 + R3) R4 R6) + 
      R3 R5 (R4 + R6) Vi)/((R1 + R2) R3 R4 + (R1 + R2 + 
         R3) R4 R5 + (R3 (R4 + R5) + R1 (R3 + R4 + R5) + 
         R2 (R3 + R4 + R5)) R6)}}

Now, we can find:

  • \$\text{V}_\text{th}\$ we get by finding \$\text{V}_3-\text{V}_2\$ and letting \$\text{R}_6\to\infty\$: $$\text{V}_\text{th}=-\frac{\text{R}_4\left(\text{R}_3\left(\text{V}_\text{i}+\text{I}_\text{a}\text{R}_1\right)+\text{I}_\text{a}\text{R}_5\left(\text{R}_1+\text{R}_2+\text{R}_3\right)\right)}{\left(\text{R}_1+\text{R}_2\right)\left(\text{R}_3+\text{R}_4+\text{R}_5\right)+\text{R}_3\left(\text{R}_4+\text{R}_5\right)}\tag3$$
  • \$\text{I}_\text{th}\$ we get by finding \$\text{I}_6\$ and letting \$\text{R}_6\to0\$: $$\text{I}_\text{th}=-\frac{\text{R}_4\left(\text{R}_3\left(\text{V}_\text{i}+\text{I}_\text{a}\text{R}_1\right)+\text{I}_\text{a}\text{R}_5\left(\text{R}_1+\text{R}_2+\text{R}_3\right)\right)}{\text{R}_4\left(\text{R}_3\left(\text{R}_1+\text{R}_2\right)+\text{R}_5\left(\text{R}_1+\text{R}_2+\text{R}_3\right)\right)}\tag4$$
  • \$\text{R}_\text{th}\$ we get by finding: $$\text{R}_\text{th}=\frac{\text{V}_\text{th}}{\text{I}_\text{th}}=\frac{\text{R}_4\left(\text{R}_3\left(\text{R}_1+\text{R}_2\right)+\text{R}_5\left(\text{R}_1+\text{R}_2+\text{R}_3\right)\right)}{\left(\text{R}_1+\text{R}_2\right)\left(\text{R}_3+\text{R}_4+\text{R}_5\right)+\text{R}_3\left(\text{R}_4+\text{R}_5\right)}\tag5$$

Where I used the following Mathematica-codes:

In[2]:=FullSimplify[
 Limit[((-Ia R5 (R2 R3 R4 + R2 R3 R6 + (R1 + R2 + R3) R4 R6) + 
       R3 R5 (R4 + R6) Vi)/((R1 + R2) R3 R4 + (R1 + R2 + 
          R3) R4 R5 + (R3 (R4 + R5) + R1 (R3 + R4 + R5) + 
          R2 (R3 + R4 + R5)) R6)) - ((R3 (Ia R1 R4 R6 - 
         Ia R2 R5 (R4 + R6) + R5 R6 Vi + 
         R4 (R5 + R6) Vi))/((R1 + R2) R3 R4 + (R1 + R2 + 
          R3) R4 R5 + (R3 (R4 + R5) + R1 (R3 + R4 + R5) + 
          R2 (R3 + R4 + R5)) R6)), R6 -> Infinity]]

Out[2]=-((R4 (Ia R1 R3 + Ia (R1 + R2 + R3) R5 + R3 Vi))/(
 R3 (R4 + R5) + R1 (R3 + R4 + R5) + R2 (R3 + R4 + R5)))

In[3]:=FullSimplify[
 Limit[-((R4 (Ia R1 R3 + Ia (R1 + R2 + R3) R5 + 
         R3 Vi))/((R1 + R2) R3 R4 + (R1 + R2 + 
          R3) R4 R5 + (R3 (R4 + R5) + R1 (R3 + R4 + R5) + 
          R2 (R3 + R4 + R5)) R6)), R6 -> 0]]

Out[3]=-((R4 (Ia R1 R3 + Ia (R1 + R2 + R3) R5 + 
    R3 Vi))/((R1 + R2) R3 R4 + (R1 + R2 + R3) R4 R5))

In[4]:=FullSimplify[%2/%3]

Out[4]=((R1 + R2) R3 R4 + (R1 + R2 + R3) R4 R5)/(
R3 (R4 + R5) + R1 (R3 + R4 + R5) + R2 (R3 + R4 + R5))
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I don't think the method of "looking into the terminals" to find the equivalent resistance works in this case, due to the dependent current source. Look at this excerpt from Electrical Engineering Principles and Applications by Allan Hambley:-

enter image description here

Although I am starting to doubt this conclusion when I look at the comments of the question.

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  • \$\begingroup\$ Anyone care to explain the downvote? \$\endgroup\$
    – Carl
    Jun 20 at 11:00
  • 1
    \$\begingroup\$ I'm not the downvoter, but probably because this post doesn't try to answer the question. It is more of a comment about the approach of the OP. It may be relevant to the discussion, but it is not an answer (although it may be a part af an answer). Moreover it is almost just a citation from a book. It is an answer to some comments from AndyAka to the OP, but not to the OP. \$\endgroup\$ Jun 21 at 7:27
  • \$\begingroup\$ @LorenzoDonati--Codidact.com I can see where you are coming from and I am not trying to start a debate, but I think I'm the only one who actually answers OP's question. OP is specifically asking if he can find the equivalent resistance by looking into the terminals to which the answer is no. All the other answer's on here involve node equations or inserting test sources which is not what the OP initially asked. \$\endgroup\$
    – Carl
    Jun 21 at 8:07
  • \$\begingroup\$ @Antonio51 Why are you asking me what it means? Just use Google or ask OP if you don't know what it means. \$\endgroup\$
    – Carl
    Jun 21 at 11:23
  • \$\begingroup\$ Sorry. Question was not addressed to you ... :( .... Calculating is writing also equations. I did not understand "one post". \$\endgroup\$
    – user288518
    Jun 21 at 11:27
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MappleSoft Licensee . see https://www.maplesoft.com/products/Maple/students/ Owner Maple sheet software

Ok. Complicated to see the result as shown. More understandable by following the "classic" procedure ... in the theoretical framework :) But hey, it's only the result that counts. On the other hand, when the diagram is more complicated, it suffices to write "all" the equations. But verification requires at least a "simulation".

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  • \$\begingroup\$ There should be an equation involving Uc. \$\endgroup\$
    – ErikR
    Jun 20 at 12:48
  • \$\begingroup\$ Depend how you write yours equations. It is not mandatory. What you see is part of the sheet. You see that I used Uc in calculating Z. It is obtain through ic and Rc, simply. \$\endgroup\$
    – user288518
    Jun 20 at 13:34
  • \$\begingroup\$ For writing the equations, never include a loop through the beta source ... Here there is 2 loops only and three nodes , for 5 variables. \$\endgroup\$
    – user288518
    Jun 20 at 13:36
  • \$\begingroup\$ Then proceed by elimination & substitution of variables as usual. Here, I have just to call "solve( )" function ... :) \$\endgroup\$
    – user288518
    Jun 20 at 13:42
  • \$\begingroup\$ It would just be nice to know what your complete set of equations are - I'd like to plug them into python's sympy package and play around with them. \$\endgroup\$
    – ErikR
    Jun 20 at 13:44
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To answer your explicit question: you start from the false assumption that any equivalent resistance can be expressed as a combination of parallel or series resistances. There are even simple passive resistive circuits that cannot.

For example, try to calculate the equivalent resistance of this one:

schematic

simulate this circuit – Schematic created using CircuitLab

Speaking about the circuit you posted: \$r_o\$ is in parallel to whatever equivalent resistance you find between nodes C-E with \$r_o\$ removed, so this simplifies a little bit the calculations. This can be seen by simple inspection.

Then you could also try to place a voltage source, instead of a current source, between C-E as a test source. Then you would need to calculate the current by the circuit from that test source, which is \$ \beta I_b - I_c \$ (\$r_o\$ removed).

See if this make your formulas less threatening. YMMV.

In general, what you want is to obtain formulas in which repeated terms and quantities are not uselessly repeated. Of course this is not always possible, so there is no standard method behind it. It's more art than science.

Moreover the same formula can be rearranged to show better some dependency from a parameter or a ratio of parameters (e.g. a ratio of resistances). Again, there is no standard way to do it, besides being fluent with algebraic manipulation.

More often, in that kind of "exercises", it is more important to being able to compare one's result with the "result of the book" and see that they are the same formula (or, more mathematically, they are equivalent expressions/equations, because they can be converted into one another by algebraic manipulations).

And, yes, one of the skill needed to manipulate those formulas is being able to recognize the expansion of a "parallel" operation, and this sometimes requires some tricks (like multiplying numerator and denominator by the same term).

From a technical perspective, it is more important that you learn how to simplify those formulas in special conditions so that in practice you can use approximated formulas.

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  • 1
    \$\begingroup\$ For that circuit, use the triangle-star equivalence theorem. \$\endgroup\$
    – user288518
    Jun 20 at 11:02
  • \$\begingroup\$ What is YMMV ? :) \$\endgroup\$
    – user288518
    Jun 20 at 13:57
  • \$\begingroup\$ @Antonio51 GIYF (Google Is Your Friend). :-) \$\endgroup\$ Jun 20 at 14:52
  • \$\begingroup\$ @Antonio51 Triangle-Star is an alternative, but I can't remember those damned formulas by heart. :-) So I prefer to apply a test voltage source, split it in two, separating the upper node into two and proceed from there, e.g. with superimposition. Longer, but simpler formulas and I don't have to look for them in a book/internet or whatever. \$\endgroup\$ Jun 20 at 14:56
  • \$\begingroup\$ @Antonio51, BTW, with a bit of experience you don't have to rewrite the circuit, but you can do that (apply superimposition) in your head and come up with the formula for the current right away. \$\endgroup\$ Jun 20 at 14:59

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