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

Question

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\$?

Answer 1

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: -

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

Answer 2

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:

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