Current sources and mirrors
There are many useful circuit elements: current sources and current
mirrors are two of them. Sources and mirrors are so closely related that
it is sometimes difficult to differentiate between them. So what? - labelling
things is a function of language not of electronics!
Circuit 1
is a very standard current source. Vin causes a current to flow through
R1 into the diode. So a voltage is developed across the diode. This voltage
is also present across the base-emitter junction of the transistor, so
a current must flow in the transistor. The magnitude of the transistor
current depends on the voltage developed across the diode and this is a
function of its geometry. So the ratio of input to output current is not
easy to calculate. However it will be pretty consistent with identical
diodes and transistors because modern production causes very little spread
in device parameters. The base-emitter junction of a transistor behaves
just like a diode.
Circuit 2
Instead of a diode, you can use an LED. An LED (depending on colour, intensity
etc) drops about 2v so you can calculate the output current as shown.
Circuit 3
This is a commonly used source: Vin drives current through R1 into the
base of the second transistor, so current flows into the transistor's collector,
through it and out of its emitter. This current must flow through R2. If
the current gets too high, the first transistor turns on and robs the second
transistor of base current, so its collector current can never exceed the
value shown. This is an excellent way of either making a current source
or of limiting the available current to a defined maximum value.
Circuit 4
This is a true current mirror: feed a current (i1) into the first transistor
(usually by driving it via a resistor from a suitable voltage) and a 'mirror'
current (i2) flows in the second transistor. Provided the two transistors
are matched and of high gain, input and mirror current will match quite
closely. In fact, with modern transistors, if you use a -C gain rating
(450-900) matching of any two transistors is probably adequate for all
usual needs.
Look at the equivalent circuit of virtually any analogue IC and you
will notice this type of current source/mirror (it may be used for either)
but it works just as well with discrete transistors.
One source of errors is if the voltage across the output transistor
is high enough to cause the transistor to heat: the mirror's performance
fails because the two transistors are now no longer well matched!
This circuit can also be used with resistors in both emitters - in this
case the mirror ratio will depend on the ratio of these resistors.
An example of this.
Circuit 5
This is for the purists: circuit 4 gives an error due to the base bias
current. Here is the commonest way of reducing this error. It gives significantly
improved performance.
Circuit 6
This is another way of reducing the error but this circuit is not quite
so commonly known. If you calculate the formulae using practical transistors
(beta = 500 or so) both 5 and 6 give very close matching - but 6 is slightly
better. An example of this.
However there is a problem which sometimes occurs on current mirrors:
if the output current is flowing from a significantly high voltage (Vc),
then there may be enough power (Vc x i²) in the output transistor
of figure 5 to cause the transistor to get warm: heating destroys the balance
between the two transistors and so destroys the mirror's accuracy. In circuit
6 any heating is in the third transistor (at the top) and it is not part
of the mirror so the mirror ratio is not affected by heating in this transistor.
See 'Useful facts' below.
Circuit 7
Just to prove that I can use ICs, here is a current source (or a voltage
to current converter) using an IC. You can use a single transistor (the
PNP one) but the transistor's base current slightly alters the ratio. You
can of course use an FET (zero gate current) but these are relatively expensive.
Here is a complimentary long tailed pair biased to cancel out base current
errors. Very good - provided you have a suitable bias point for Vb!
Useful facts
There are two very useful figures that I remember about silicon diodes
are to do with the change in voltage/current. If you increase the voltage
across a diode by 60 millivolts, then the current through the diode increase
by a factor of 10 - or vice versa. Try measuring the diode's drop when
it has 100 microamps and 1 milliamp flowing: you will find the 60mV figure
quite accurate.
The second figure is to do with the change in diode drop with temperature:
increase the temperature by 1 degree K and the diode's voltage drops by
2 millivolts.
These figures hold for diodes and transistor base-emitter junctions.
