Editing Current mirror (section)

==== Output resistance ====
[[File:Mirror output resistance.PNG|thumb|300px|Figure 5: Small-signal circuit to determine output resistance of mirror; transistor Q<sub>2</sub> is replaced with its [[hybrid-pi model]]; a test current ''I''<sub>X</sub> at the output generates a voltage ''V''<sub>X</sub>, and the output resistance is ''R''<sub>out</sub> = ''V''<sub>X</sub> / ''I''<sub>X</sub>.]]
An idealized treatment of output resistance is given in the footnote.<ref group="nb">An idealized version of the argument in the text, valid for infinite op amp gain, is as follows. If the op amp is replaced by a [[nullor]], voltage ''V''<sub>2</sub> = ''V''<sub>1</sub>, so the currents in the leg resistors are held at the same value. That means the emitter currents of the transistors are the same. If the ''V''<sub>CB</sub> of Q<sub>2</sub> increases, so does the output transistor β because of the [[Early effect]]: ''β'' = ''β''<sub>0</sub>(1 + ''V''<sub>CB</sub> / ''V''<sub>A</sub>). Consequently the base current to Q<sub>2</sub> given by ''I''<sub>B</sub> = ''I''<sub>E</sub> / (''β'' + 1) decreases and the output current ''I''<sub>out</sub> = ''I''<sub>E</sub> / (1 + 1 / ''β'') increases slightly because ''β'' increases slightly. Doing the math,
: <math>\begin{align}
 \frac{1}{R_\text{out}} &= \frac{\partial I_\text{out}}{\partial V_\text{CB}}
 = I_\text{E} \cdot \frac{\partial}{\partial V_\text{CB}} \left(\frac{\beta}{\beta + 1}\right)
 = I_\text{E} \cdot \frac{1}{(\beta + 1)^2} \cdot \frac{\partial\beta}{\partial V_\text{CB}} \\
 &= \frac{\beta I_\text{E}}{\beta + 1} \cdot \frac{1}{\beta} \cdot \frac{\beta_0}{V_\text{A}} \cdot \frac{1}{\beta + 1} \\
 &= I_\text{out} \cdot \frac{1}{1 + \frac{V_\text{CB}}{V_\text{A}}} \cdot \frac{1}{V_\text{A}} \cdot \frac{1}{\beta + 1} \\
 &= \frac{1}{(\beta + 1) r_0},
\end{align}</math>
where the transistor output resistance is given by r<sub>O</sub> = (''V''<sub>A</sub> + ''V''<sub>CB</sub>) / ''I''<sub>out</sub>. That is, the ideal mirror resistance for the circuit using an ideal op amp [[nullor]] is ''R''<sub>out</sub> = (''β'' + 1''c'')''r''<sub>O</sub>, in agreement with the value given later in the text when the gain → ∞.</ref> A small-signal analysis for an op amp with finite gain ''A''<sub>v</sub> but otherwise ideal is based upon Figure 5 (''β'', ''r''<sub>O</sub> and ''r''<sub>π</sub> refer to Q<sub>2</sub>). To arrive at Figure 5, notice that the positive input of the op amp in Figure 3 is at AC ground, so the voltage input to the op amp is simply the AC emitter voltage ''V''<sub>e</sub> applied to its negative input, resulting in a voltage output of −''A''<sub>v</sub> ''V''<sub>e</sub>. Using [[Ohm's law]] across the input resistance ''r''<sub>π</sub> determines the small-signal base current ''I''<sub>b</sub> as:
: <math> I_\text{b} = \frac{V_\text{e}}\frac{r_\pi}{A_\text{v} + 1} \ .</math>

Combining this result with Ohm's law for <math>R_\text{E}</math>, <math>V_\text{e}</math> can be eliminated, to find:<ref group="nb">As ''A''<sub>v</sub> → ∞, ''V''<sub>e</sub> → 0 and ''I''<sub>b</sub> → ''I''<sub>X</sub>.</ref>
:<math> I_\text{b} = I_\text{X} \frac{R_\text{E}}{ R_\text{E} + \frac{r_\pi}{A_\text{v} + 1} }.</math>

[[Kirchhoff's voltage law]] from the test source ''I''<sub>X</sub> to the ground of ''R''<sub>E</sub> provides:
: <math> V_\text{X} = (I_\text{X} + \beta I_\text{b)} r_\text{O} + (I_\text{X} - I_\text{b} )R_\text{E}.</math>

Substituting for ''I''<sub>b</sub> and collecting terms the output resistance ''R''<sub>out</sub> is found to be:
: <math>R_\text{out} = \frac{V_\text{X}}{I_\text{X}} = r_\text{O} \left( 1 + \beta \frac{R_\text{E}}{R_\text{E} + \frac{r_\pi}{A_\text{v} + 1}} \right) + R_\text{E} \|\frac{r_\pi}{A_\text{v} + 1}.</math>

For a large gain ''A''<sub>v</sub> ≫ ''r''<sub>π</sub> / ''R''<sub>E</sub> the maximum output resistance obtained with this circuit is
: <math>R_\text{out} = (\beta + 1)r_O,</math>
a substantial improvement over the basic mirror where ''R''<sub>out</sub> = ''r''<sub>O</sub>.

The small-signal analysis of the MOSFET circuit of Figure 4 is obtained from the bipolar analysis by setting ''β'' = ''g''<sub>m</sub> ''r''<sub>π</sub> in the formula for ''R''<sub>out</sub> and then letting ''r''<sub>π</sub> → ∞. The result is
: <math>R_\text{out} = r_\text{O} \left[1 + g_\text{m} R_\text{E}(A_\text{v} + 1)\right] + R_\text{E}.</math>

This time, ''R''<sub>E</sub> is the resistance of the source-leg MOSFETs M<sub>3</sub>, M<sub>4</sub>. Unlike Figure 3, however, as ''A''<sub>v</sub> is increased (holding ''R''<sub>E</sub> fixed in value), ''R''<sub>out</sub> continues to increase, and does not approach a limiting value at large ''A''<sub>v</sub>.