I have the following circuit:


simulate this circuit – Schematic created using CircuitLab

How can I prove that \$I_3=Vi/R1\$ when \$\beta_2\to\infty\$?

  • \$\begingroup\$ Hopefully you can see that the current through R3 is basically that through R1. Mentally erase the transistors so that the op-amp is driving R1 directly, and ask yourself what an op-amp is supposed to do. Then recognize that the transistors are there just to give the op-amp greater power drive capability than it has on its own. \$\endgroup\$ – Chris Stratton Oct 10 '20 at 16:02
  • \$\begingroup\$ Set the diff. voltage at the opamp input to zero. \$\endgroup\$ – LvW Oct 10 '20 at 16:02
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    \$\begingroup\$ I think, only for an infinite beta value we have Ic=Ie (as the question contains the current through R3) \$\endgroup\$ – LvW Oct 10 '20 at 16:06
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    \$\begingroup\$ @LvW You are right, I missed that fact. \$\endgroup\$ – G36 Oct 10 '20 at 16:09
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    \$\begingroup\$ I’m voting to close this question because it is unattempted homework \$\endgroup\$ – Chris Stratton Oct 10 '20 at 16:14

You don't need to prove it if you accept that an op-amp (even quite modestly priced op-amps) that use negative feedback, seek to ensure that the two input voltages (+Vin and -Vin) are identical.

In effect, what ever voltage you apply to +Vin (Vi in your example) results in the same Vi appearing on -Vin. This inevitably means that: -

$$I_{R_3} = \dfrac{V_i}{R_1}$$

  • \$\begingroup\$ This is the answer, at least the answer of an engineer, or carpenter. Jan's answer is the answer of the King of Kings. Choose wisely! \$\endgroup\$ – Neil_UK Oct 10 '20 at 17:15
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    \$\begingroup\$ Slurp slurp slurp. \$\endgroup\$ – Andy aka Oct 10 '20 at 17:25
  • \$\begingroup\$ @Neil_UK 's marginally rude response notwithstanding, Andy's answer, while not actually wrong, fails to note the role of Beta2 in the problem. Andy correctly comments about the opamp's gain BUT they are asking about I_R3 and not about I_R1. I2 exceeds I3 by Ib_Q2. For infinite Beta2 these are the same. For some transistors beta can be low enough to matter in some cases. | Andy knows this - he just didn't mention it. \$\endgroup\$ – Russell McMahon Oct 17 '20 at 12:12
  • \$\begingroup\$ @Andy - you may want to add some Beta_2 to the slurping. \$\endgroup\$ – Russell McMahon Oct 17 '20 at 12:14
  • \$\begingroup\$ @RussellMcMahon well, I stayed clear of it because although the OP mentioned a "darlington" (and that overcomes the beta thing by another factor of 100 or so), the OP's circuit isn't strictly speaking a darlington configuration. \$\endgroup\$ – Andy aka Oct 17 '20 at 13:42

First, I will use Mathematica to solve your problem.

Well, we are trying to analyze the following circuit:


simulate this circuit – Schematic created using CircuitLab

When analyzing a transistor we need to use the following relations:

  • $$\text{I}_\text{T}=\text{I}_0+\text{I}_2\tag1$$
  • $$\text{I}_1=\text{I}_\text{T}+\text{I}_3\tag2$$
  • Transistor gain \$\beta_1\$: $$\beta_1=\frac{\text{I}_2}{\text{I}_0}\tag3$$
  • Transistor gain \$\beta_2\$: $$\beta_2=\frac{\text{I}_3}{\text{I}_\text{T}}\tag4$$
  • Emitter voltage \$\text{T}_1\$: $$\text{V}_{\text{BE}_1}=\text{V}_1-\text{V}_2\tag5$$
  • Emitter voltage \$\text{T}_2\$: $$\text{V}_{\text{BE}_2}=\text{V}_2-\text{V}_5\tag5$$

Using KCL, we can write:


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

$$ \begin{cases} \text{I}_1=\frac{\text{V}_5}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_\text{b}-\text{V}_3}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_\text{b}-\text{V}_4}{\text{R}_3} \end{cases}\tag7 $$

Now, using an ideal opamp, we know that:


Now, it is not hard to solve for \$\text{I}_3\$:


Where I used Mathematica-code to solve for that:

 Solve[{IT == I0 + I2, 
   I1 == IT + I3, \[Beta]1 == I2/I0, \[Beta]2 == I3/IT, 
   VBE1 == V1 - V2, VBE2 == V2 - Vi, Ib == I2 + I3, I1 == Vi/R1, 
   I2 == (Vb - V3)/R2, I3 == (Vb - V4)/R3}, {Ib, IT, I0, I1, I2, I3, 
   V1, V2, V3, V4}]]

Out[1]={{Ib -> (Vi (\[Beta]1 + \[Beta]2 + \[Beta]1 \[Beta]2))/(
   R1 (1 + \[Beta]1) (1 + \[Beta]2)), IT -> Vi/(R1 + R1 \[Beta]2), 
  I0 -> Vi/(R1 + R1 \[Beta]1 + R1 \[Beta]2 + R1 \[Beta]1 \[Beta]2), 
  I1 -> Vi/R1, 
  I2 -> (Vi \[Beta]1)/(
   R1 + R1 \[Beta]1 + R1 \[Beta]2 + R1 \[Beta]1 \[Beta]2), 
  I3 -> (Vi \[Beta]2)/(R1 + R1 \[Beta]2), V1 -> VBE1 + VBE2 + Vi, 
  V2 -> VBE2 + Vi, 
  V3 -> Vb - (R2 Vi \[Beta]1)/(
    R1 + R1 \[Beta]1 + R1 \[Beta]2 + R1 \[Beta]1 \[Beta]2), 
  V4 -> Vb - (R3 Vi \[Beta]2)/(R1 + R1 \[Beta]2)}}

Now, when \$\beta_2\to\infty\$, we get:


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    \$\begingroup\$ The point of an academic assignment is to work things out for onself with paper and pencil and an understanding of the concepts. Not to use software for what the student is supposed to be learning to do manually, and certainly not to have someone on an internet site simply hand you the answer. \$\endgroup\$ – Chris Stratton Oct 10 '20 at 16:04
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    \$\begingroup\$ @ChrisStratton This is not true! I am allowed to use Matlab for example and Mathematica is kind of the same, so you're comment is not usefull. \$\endgroup\$ – Jan Oct 10 '20 at 16:10
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    \$\begingroup\$ @Derik what did you actually learn from this posting which simply handed you the answer? Was it what your professor hoped you would, when they assigned the problem? \$\endgroup\$ – Chris Stratton Oct 10 '20 at 16:11
  • \$\begingroup\$ @Derik When I was studying for my degree in electrical engineering I was also allowed to use software in practical problems. For test problems of course not, but if you understand the concepts it is good to learn how to program. \$\endgroup\$ – Jan Oct 10 '20 at 16:12
  • \$\begingroup\$ @Derik How are you parsing this sentence that I write. Letter by letter, or word by word? Compare the answer with high upvotes to the one you've accepted. The former deals with electronic design 'words', Jan's deals with letters, the hard way. Think h o w y o u w a n t t o l e a r n i n t h e f u t u r e, and whether words are worth t h e e f f o r t t o m a s t e r, \$\endgroup\$ – Neil_UK Oct 10 '20 at 18:32

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