# How to prove that $I_3=Vi/R1$ in a darlingtonpair circuit with opamp when $\beta_2\to\infty$? [closed]

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\$$?

• 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. Oct 10, 2020 at 16:02
• Set the diff. voltage at the opamp input to zero.
– LvW
Oct 10, 2020 at 16:02
• I think, only for an infinite beta value we have Ic=Ie (as the question contains the current through R3)
– LvW
Oct 10, 2020 at 16:06
• @LvW You are right, I missed that fact.
– G36
Oct 10, 2020 at 16:09
• I’m voting to close this question because it is unattempted homework Oct 10, 2020 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: -

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

• 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! Oct 10, 2020 at 17:15
• Slurp slurp slurp. Oct 10, 2020 at 17:25
• @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. Oct 17, 2020 at 12:12
• @Andy - you may want to add some Beta_2 to the slurping. Oct 17, 2020 at 12:14
• @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. Oct 17, 2020 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:

$$\text{I}_\text{b}=\text{I}_2+\text{I}_3\tag6$$

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:

$$\text{V}_+=\text{V}_-=\text{V}_\text{i}=\text{V}_5\tag8$$

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

$$\text{I}_3=\frac{\text{V}_\text{i}\beta_2}{\text{R}_1\left(1+\beta_2\right)}\tag9$$

Where I used Mathematica-code to solve for that:

In[1]:=FullSimplify[
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:

$$\lim_{\beta_2\to\infty}\text{I}_3=\lim_{\beta_2\to\infty}\frac{\text{V}_\text{i}\beta_2}{\text{R}_1\left(1+\beta_2\right)}=\frac{\text{V}_\text{i}}{\text{R}_1}\cdot\lim_{\beta_2\to\infty}\frac{1}{1+\frac{1}{\beta_2}}=\frac{\text{V}_\text{i}}{\text{R}_1}\cdot\frac{1}{1+0}=\frac{\text{V}_\text{i}}{\text{R}_1}\tag{10}$$

• 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. Oct 10, 2020 at 16:04
• @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. Oct 10, 2020 at 16:10
• @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? Oct 10, 2020 at 16:11
• @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. Oct 10, 2020 at 16:12
• @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, Oct 10, 2020 at 18:32