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I'm learning about Wilson mirrors in the Art of Electronics Third Edition (pg. 102). It shows the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

It then explains the circuit. As part of the explanation it says this:

Transistor \$Q_3\$, by the way, does not have to be matched to \$Q_1\$ and \$Q_2\$; but if it has the same beta, then you get an exact cancellation of the (small) base current error that afflicts the simple mirror of Figure 2.55 (or the beta-enhanced mirror in Chapter 2x).

Exercise 2.17. Show that this statement is true.

Figure 2.55 is just a Widlar mirror as so:

schematic

simulate this circuit

The book doesn't explain what the "(small) base current error" is but I assume it's referring to the fact that in the Widlar mirror, \$I_P = I_{load} + I_{B_{Q1}} + I_{B_{Q2}}\$ instead of \$I_P = I_{load}\$. So I set out to solve Exercise 2.17 with that assumption.

From an analytical perspective, it makes sense to me that if \$Q_3\$'s base current is equal to \$Q_1\$'s base current, \$Q_3\$ will balance out the equation. \$I_P + I_{B_{Q1}} = I_{load} + I_{B_{Q2}}\$ (and because both base currents are equal, \$I_P = I_{load}\$). However, when I start to solve this mathematically involving \$\beta\$, it seems to me that \$\beta\$ needs to be different, not stay the same.

Here is my logic:

  1. \$ I_{B_{Q1}} = I_{B_{Q2}}, I_{C_{Q1}} = I_{C_{Q2}}, I_{E_{Q1}} = I_{E_{Q2}} \$
  2. \$ I_{B_{Q3}} = I_{B_{Q1}} \$
  3. Use \$ I_B = I_E - I_C \$ to give \$ I_{E_{Q3}} - I_{C_{Q3}} = I_{B_{Q1}}\$
  4. Use \$ I_{E_{Q3}} = I_{E_{Q1}} + I_{B_{Q1}} \$ to give \$ I_{E_{Q1}} + I_{B_{Q1}} - I_{C_{Q3}} = I_{B_{Q1}} \$
  5. Rearrange to give \$ I_{C_{Q3}} = I_{E_{Q1}} \$
  6. (Lines 2 & 5 appear to be correct according to my simulations)

Wilson mirror simulation

  1. Use \$ I_C = I_E - I_B \$ and 2. and 5. to give \$ I_{C_{Q1}} = I_{C_{Q3}} - I_{B_{Q3}} \$
  2. Use \$ I_C = \beta I_B \$ to give \$ \beta_1 I_{B_{Q1}} = \beta_3 I_{B_{Q3}} - I_{B_{Q3}} \$
  3. Substitute \$ I_{B_{Q1}} \$ with \$ I_{B_{Q3}} \$ to give \$ \beta_1 I_{B_{Q3}} = \beta_3 I_{B_{Q3}} - I_{B_{Q3}} \$
  4. Divide by \$ I_{B_{Q3}} \$ to give \$ \beta_1 = \beta_3 - 1 \$

Now we can see if the two \$ \beta \$ values were equal, the equation would not work. I've been bashing my head on this for hours. Where did I go wrong?

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I'm not sure how accurate you want to go, but there are a few things in this circuit that I believe will not make the current in both branches equal.

While \$V_{BE1} = V_{BE2}\$ means that \$I_{B1} = I_{B2}\$, in general \$V_{CE1} \neq V_{CE2}\$. This will generate a small difference in \$I_C\$ due to the Early effect.

Ignoring that difference, we can still find

$$I_p = I_{C1} + I_{B3} = \beta_1I_{B1} + I_{B3} \\ I_L = I_{C3} = \beta_3I_{B3}$$

We can also find an error in your 4th point, the KCL needs to be:

$$\begin{align} I_{E3} &= I_{C2} + I_{B1} + I_{B2} \\ &\Downarrow \\ I_{C3} + I_{B3} &= I_{C2} + 2I_{B1,2} \\ &\Downarrow \\ I_{B3} &= \frac{\beta_2 + 2}{\beta_3 + 1}I_{B1,2} \end{align}$$

This means that point 2 and 3 are incorrect as well.

So finally

$$I_p = \left(\beta_1 + \frac{\beta_2 + 2}{\beta_3 + 1}\right)I_{B1,2} \\ I_L = \beta_3\frac{\beta_2 + 2}{\beta_3 + 1}I_{B1,2}$$

These two current are equal only if

$$\beta_1(\beta_3+1) + (\beta_2 + 2) = \beta_3(\beta_2 + 2)$$

Assuming \$\beta_{1,2} = \beta_1 = \beta_2\$ we can solve that

$$\beta_3 = \beta_{1,2} + 1$$

which turns out to be the same equation anyway. Now you can continue by observing that \$\beta \gg 1\$, such that \$\beta_3 \approx \beta_{1,2}\$.

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  • \$\begingroup\$ I guess when the book says they need to have the "same" \$ \beta \$ they don't literally mean \$ \beta_1 = \beta_3 \$. :facepalm: Thanks for the clarification! \$\endgroup\$ – Aust Jul 18 '18 at 17:30
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I think there's a mistake in 4:

4. Use \$ I_{E_{Q3}} = I_{E_{Q1}} + I_{B_{Q1}} \$

I think it should be:

  1. Use \$ I_{E_{Q3}} = I_{E_{Q1}} + I_{B_{Q1}} + I_{B_{Q2}} \$

I'm too lazy to check all equations so what I did in stead is just draw the schematic and fill in the numbers. To make that easy I set the mirror's input current to 1 A and I assumed \$\beta\$ = 10. Then I get:

schematic

simulate this circuit – Schematic created using CircuitLab

Now it is easy to see how the beta compensation works, the \$ I_{E_{Q3}}\$ = 1.1 A which is 0.2 A more than the current in the other branch (0.9A). Q3 then simply diverts one base current to the other side. If Q3 did not have the same \$\beta\$ as Q1 and Q2 then this diverted current would not have the right value.

I rounded off the values of the currents, can you spot where I did that? I did this to make the values more insightful and not be bothered too much by the actual numbers. In practice this error becomes more insignificant as the value of \$\beta\$ increases. For the functionality of the circuit this rounding error does not matter, the numbers are just slightly different.

You can now check the numbers and see if they match your equations. It is possible that in the simulator \$\beta\$ is so high that you do not see the error of one base current. That's why I do this by hand with an exceptional low \$\beta\$ value.

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