The Widlar source was designed by and named for Bob
(Robert) Widlar, who worked for National Semiconductor and produced
a number of very profitable designs. The Widlar source is an important
analog building block. When combined with other blocks such as the
differential pair and the current mirror, it can be used in the design
of operational amplifiers (op amps), operational transconductance
amplifiers, and other such circuits.
A Widlar current source is used in analog integrated circuit
design to produce a small driving current without the use of large
resistors. This is particularly important because it is difficult
to produced large resistors on chip that have good tolerances, meaning
that it is very difficult to get predictable, reliable currents using
large resistors.
The Widlar source uses two transistors fabricated with their bases
connected together, and two small resistors. One of the transistors
has its base connected to its collector, forming a diode connected
transistor.
The input current is fed into the collector of the diode connected
transistor, and a small current results at the collector of the other
transistor. A crude ASCII circuit diagram may help:
+Vcc
----
|
|-------------
| |
| | |
| | Ie1 |
\ \|/ |
/ |
\ R1 Load |
/ | | Ie2
| | \|/
|------ |
C1\ | /C2
\ | /
\| | |/
Q1 |------| Q2
/|B1 B2|\
/ \
E1/ \E2
| |
| \
| /
| \ R2
| /
| |
| |
-------------|
|
---
- GND
I've labeled the base, emitter, and collector terminals of Q1 and Q2
because I'm stuck on how to do the diode triangle thing that appear on
the emitter terminal in ASCII...
Okay, if we assume that the transistors are operating in their
active regions (which is a good assumption as long as the voltage at
the collector of Q2 is greater that about 0.7 V for bipolar devices), then the input current Ie1 through R1 is
given by:
Ie1 = (Vcc - 0.7) / R1
The voltage at the base and collector of Q1 is assumed to be 0.7
when Q1 is in its active region and when the operating temperature is
about 20 degrees Celsius.
Having the input current, we can find the output current iteratively
using the Widlar design equation:
I2 = (Vt/R2) ln (I1/I2)
Vt can be assumed to be 26mV at room temperature, and we know I1
from the previous calculation. If we are designing toward a target
output current, we would plug that current in and solve for
R2. Otherwise, an iterative solution (done by plugging in values for
I2 and seeing if the result matches) is required.
The derivation of the Widlar design equation is quite straight
forward; it comes from writing a loop equation for the bottom loop
in the diagram, using the Ebers-Moll equations to model the
transistors (you can't assume a 0.7 V drop across the emitter-base
junction in this case). Interested parties can look it up in a
textbook on analog IC's, such as Microelectronic Circuits by Sedra & Smith.
References:
Memory, and my ENEL 465 class notes.