------------------------ VCC
        |           |
        O IC1       O IC2
        |           | 
       /             \
Vin1 -|   Q1    Q2    |- Vin2
       \             /
        |           |
        |-----------|
              |
              -
             | | Re
              -
              |
      ----------------- Vss

How would one go about calculating the value of Re above, supposing
the collector currents for both Q1 and Q2 were 1mA? (Vcc and Vss are
+5V and -5V respectively)


----- Vcc
  |
  -
 | | R
  -
  |          |
  |----\     |
   \   |    /
Q1  |------|  Q2
   /        \
  |          |
-----------------
         
In the current mirror above, if vcc equals +5V and the circuit had to
provide a 1mA current source, how would one calculate the required
value for R?

Rather than answer your question, can I recommend you purchase
?Horowitz and Hill: The Art Of Electronics'. Your question and (all?)
other future ones will then be in your hand in this superbly
accessible book.

BTW, I have no connection with H&H other than it abounds, dog eared,
in electronics labs.

Best

OK since no one else has taken pity on you :-)

First diagram. This is, as you may know, the basis of a differential
amplifier, presuming the bottom connections are the emitters. At room
temperature, for a silicon transistor to conduct between the collector
and the emitter requires in the region of 0.7 volts between the base
and the emitter. This has to be current limited though because the
base emitter (and base collector) is a diode. Nothing much happens
until the diode conduction voltage is reached but after that it
conducts quite sharply. That means that the voltage on the base will
be about 0.7 volts higher than the emitter when in conduction. So if
you require a collector current of 1mA in each transistor it would
depend upon what voltage you want the top end of the resistor to sit
at. Say you want it at zero volts (ie midway between the supplies) the
voltage on the two bases would have to be in the region of 0.7 volts
and the resistor would pass 2mA. Therefore by ohm?s law, the resistor
would need to be 2k5. Resistance = voltage divided by current. This
circuit, as it stands, would seldom be used without incorporation in
some other circuit because the two transistors will fight for
conduction. Applying the same voltage to both bases won?t give exactly
the same current in each transistor. Incidentally, the collector
current will be slightly smaller than the emitter current due to the
small base current, but in most instances this can be ignored.

Second diagram. Again assuming the bottom connections are emitters,
and the bottom rail is zero volts. To get a 1mA current sink (not
source) on the collector of the RHS transistor requires 1mA on the LHS
transistor. The voltage across the resistor will be approximately 5.0
minus 0.7, ie 4.3 volts. Therefore the resistor should be 4k3. Note
that this is a very poor current mirror, partly due to the base
currents but largely because transistor characteristics vary quite
widely. There are other effects too, such as the voltage of the
current sink supply and non-linearity of transistors. ?Horowitz and
Hill? describe much better configurations, but you could experiment.
Set the circuit up with an ammeter on the sink and heat one of the
transistors with your finger. You?ll measure the current changing.

If you want more info and I have time......

Best