ELEC 351L
Electronics II Laboratory
Spring 2002
Lab 6: BJT Current Mirror
Introduction
Resistor networks are usually used to bias BJT and FET amplifiers made with discrete devices
(that is, individual devices as opposed to groups of devices on integrated circuit chips). The
classic example is the emitter degeneration technique used in BJT amplifiers that requires at least
two resistors connected to the base and one resistor connected to the emitter. Resistor biasing is
impractical in amplifiers fabricated on integrated circuit chips, because an integrated resistor
requires a large amount of space compared to that for an integrated transistor. Instead, biasing is
often accomplished on IC chips using an assembly of transistors and resistors that collectively
act as a current source. Only one or two resistors are typically required. This type of circuit is
known as a current mirror, and just one of them with minor extensions can often provide the
biasing for all of the amplifiers on a chip. In this lab experiment you will compare the
performance of a resistor-based biasing circuit to a current mirror.
Theoretical Background
Pure voltage and current sources do not exist in the physical world. It is currently impossible to
design a circuit or create a device that can maintain a given voltage across itself or current
through itself regardless of what is connected to it. However, there are many devices or circuits
(batteries and power supplies, for example) that come close to acting as pure voltage and current
sources. As shown in Figure 1, these “good” (not perfect) voltage and current sources can be
modeled by Thévenin and Norton equivalent circuits, respectively. In the case of a good voltage
R th
+
Vth
RL
_
In
(a)
Rn
RL
(b)
Figure 1. a) Thévenin and b) Norton equivalent circuits. “Good” voltage sources are
represented by Thévenin equivalent circuits with small Rth, and “good” current sources are
represented by Norton equivalent circuits with large Rn.
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source, the Thévenin equivalent resistance Rth is small enough that the voltage across the load is
almost equal to the Thévenin equivalent voltage Vth. Thus, the main criterion for a good voltage
source is to have Rth << RL. Conversely, a good current source is one in which the Norton
equivalent resistance Rn is large enough that the current through the load is almost equal to the
Norton equivalent current In. A good current source has Rn >> RL.
Although there are no individual devices that act as good current sources, there are several
simple circuits consisting of a handful of resistors and transistors that do. (These simple circuits
are offered in single packages by a few manufacturers.) One such circuit that is often made with
discrete devices is shown in Figure 2. This is essentially a common-emitter amplifier circuit
with the resistor normally connected between the collector and VCC replaced by a load. Since
there is no signal applied to the base, the BJT draws a constant load current IL, the value of which
is determined by the bias resistors R1, R2, and RE. As in the common-emitter amplifier, the
values of R1 and R2 are chosen to create a stiff bias network that maintains a constant base
voltage regardless of the value of the base current. The Norton equivalent resistance of this
circuit is derived in the course textbook (c.f. Sec. 7.4.5) and is given by
Rn  RE r  R1 R2   ro 
 o RE ro
r  R1 R2  RE
.
The last term in this equation is the dominant one. The value of Rn is typically a few megohms,
so the circuit of Figure 2 acts as a very good current source. However, it is not practical for use
in integrated circuits, because it uses several resistors, it can bias only one amplifier stage, and it
requires a relatively large voltage drop (roughly 1-2 V minimum) between the collector of the
BJT and VEE. The latter characteristic limits the voltage swing available across the load.
VCC
+
load
vL
_
R1
iL
R2
RE
VEE
Figure 2. Current source based upon common-emitter amplifier circuit.
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The circuit of Figure 3 overcomes these disadvantages and is widely used in integrated circuits,
especially in operational amplifiers. The transistors Q1 and Q2 should be a “matched pair,”
meaning that their o values and other parameters are nearly equal. Since this circuit is intended
for use in integrated circuits, the matching of parameters is not difficult. Note that the collector
of Q1 is shorted to its base, which in effect turns Q1 into a diode. The short also provides a path
for the two base currents. Q1 operates in the constant current region because VCE1 = VBE2, which
sets VCE1 = 0.7 V, a value larger than the typical saturation threshold of 0.2 V. The base-emitter
terminals of both transistors are in parallel, so VBE1 = VBE2. This implies that the base currents of
both transistors must be equal, because the base-emitter voltage controls the value of the base
current according to the diode equation

I B  I EO eVBE
VT

1 ,
where IEO and  are the reverse saturation current and emission coefficient, respectively, of the
base-emitter pn-junction, and VT is the thermal voltage. Since the two transistors are matched,
the parameters IEO and  for both transistors are very close in value. The values of VT for both
transistors are practically equal as well, because the temperatures of both transistors are
essentially the same. Finally, the relationship IB1 = IB2 implies that IC1 = IC2, because the o
values of the matched transistors should be almost equal. Thus, IL = Iref. Since the collector
current of Q1 matches that of Q2, this circuit is called a current mirror. The value of IL “reflects”
the value of Iref. The value of Iref is easy to control and is given by
I ref 
VCC  VEE  0.7
.
R1
VCC
+
load
I ref
R1
vL
_
IL
Q1
Q2
VEE
Figure 3. A simple current mirror circuit.
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The Norton equivalent resistance of the current mirror shown in Figure 3 is simply ro2, the
collector-emitter resistance of Q2. Note that this circuit uses only one resistor, and the voltage
across the collector-emitter terminals of Q2 can drop almost to the saturation limit of 0.2 V
without serious degradation in the performance of the circuit as a current source. Also,
additional transistors can be added in parallel with Q2 and with their bases connected to the base
of Q1 to provide additional current sources for other amplifier circuits on the chip.
Experimental Procedure
 Construct the simple current source shown in Figure 2 using power supply voltages of
VCC = +10 V and VEE = 10 V. Use a 2N3904 BJT, which has the pin-out shown in Figure 4.
Design the circuit to act as a current source with a value of 2 mA. Make the circuit independent
of device parameters by designing for a drop of 1 V across the emitter resistor. Design a stiff
base bias network under the assumption that the parameter o never falls below a value of 50.
Lead identification for 2N3904 in TO-92 package
(top view)
E
B
C
Figure 4. Identification of terminals for 2N3904 npn BJT.
 Verify that the circuit acts as a good current source for a wide range of load conditions by
measuring and recording the voltage across and current through a variety of resistors RL
connected as loads. The resistor values should span a wide enough range to obtain a measurable
change in the load current. Sketch or print out a plot of IL vs. RL. From the data you collect,
calculate the Norton equivalent resistance Rn of the circuit.
 Construct the current mirror shown in Figure 3 again using power supply voltages of
VCC = +10 V and VEE = 10 V. Use a closely-matched pair of 2N3904 BJTs. The pairs will
either be available from the instructor, or you will have to find a matched pair on your own from
the supply of BJTs. As with the first current source, design this circuit to supply a load current IL
of 2 mA.
 Repeat the verification procedure you used with the first current source by sketching or
printing out a plot of IL vs. RL for a wide range of load resistor values. As before, calculate the
Norton equivalent resistance Rn of the current mirror.
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