# Current mirror

Given, a complicated darlington circuitry:

We now replace the darlington pairs with single NPN transistors by defining: $I_C=I_B \cdot a$ with $a=(\beta+1)\cdot\beta$ and $I_E=I_B \cdot b$ with $b=(\beta+1)^2$

Now we set:
$I_{KD} = I_{B1} + I_{C3} = I_{B1} + I_{B3} \cdot a$ and $I_{RL}=I_{C4}=I_{B4}\cdot a$ and $I_{B2}=I_{B5}$ and $I_{KD} + I_{C1} + I_{RL} = I_{E5} + I_{E2}$
$I_{KD} + I_{B1} \cdot a + I_{RL} = I_{B5} \cdot b + I_{B2} \cdot b = 2\cdot I_{B5}\cdot b = 2\cdot I_{B2}\cdot b$
$\Rightarrow I_{RL}=- a \cdot I_{KD} - I_{KD} + a^2 \cdot I_{B3} + 2 \cdot b \cdot I_{B2}$

And resolve the current equation:
$I_{B3} \cdot b = I_{E3} = I_{C5}+I_{B2}+I_{B5}$
$=I_{C5}+2 \cdot I_{B2}$
$=I_{C5}+2 \cdot I_{B5}$
$=a \cdot I_{B5}+2 \cdot I_{B5}=(a+2)\cdot I_{B5}=(a+2)\cdot I_{B2}$

$\Rightarrow I_{B2} = I_{B5} = \frac{1}{a+2} \cdot I_{E3} =\frac{b}{a+2} \cdot I_{B3}$
$I_{E1} = I_{B1} \cdot b = I_{B3}+I_{B4}$

$I_{C2} = I_{B2} \cdot a$
$= I_{E4}=I_{B4} \cdot b$
$\Rightarrow I_{B4} = \frac{a}{b}\cdot I_{B2}=\frac{a}{b}\cdot\frac{b}{a+2} \cdot I_{B3}$

$I_{RL}=\frac{a^2 b}{a^2 b + 2 a b +2}$