Can Dual BJT packages be reliably used as Current Mirrors
Of course you can use a dual NPN for a current mirror.  The question is only how accurate the mirror will be.  The better matched the transistors are, the more accurate the current mirror.

Using two transistors in the same package is usually better than two discrete transistors, even of the same model.  One reason for putting two transistors in a package is for better matching.  In that case, the datasheet will tell you not only the absolute range of parameters like gain and B-E voltage, but the maximum variation between the two parts.

Unless you really need the current mirror output to have the lowest possible voltage drop, you can make the mirror more accurate by giving each transistor its own emitter resistor.  Now the overall function of transistor current due to input voltage becomes more predictable and matched between the two.  The input voltage is now from the base to the other end of the emitter resistor, not just from base to emitter as with a bare transistor.  As the emitter resistor increases, the gain and B-E drops of the individual transistors become less relevant.

Here is what I'm talking about:



This is usually a good idea when making a current mirror from discrete parts.  On a IC, the transistors are well matched and resistor are hard to make.  In a discrete design, transistor matching is not nearly so good and resistors are cheap and available.

To size the resistors, you need to know what the maximum current will be and what maximum voltage drop you can tolerate.  Let's say you want a mirror to work up to 10 mA and 2 V overall drop is acceptable.  Let's say you want to keep 1 V across C-E of each transistor to keep it operating in the current sink range.  That leaves 1 V across the resistors.  By Ohms law, the resistor values are (1 V)/(10 mA) = 100 &Omega;.  Such a current mirror will be quite accurate with any two transistors of the same model out of the bin.

Note that the two resistors need not have the same value.  The Iout/Iin current gain is R1/R2 when the transistors are ideal.  In the example above, you could have Iout be 0-10 mA as before, but controlled by 0-5 mA in with R1 200 &Omega; instead of 100 &Omega;