ELEC 351L
Electronics II Laboratory
Spring 2004
Lab #6: BJT Current Mirrors
Introduction
Resistor networks are usually used to bias BJT and FET amplifiers made with discrete devices
(i.e., those not fabricated on integrated circuit chips). One classic example is the emitterdegeneration technique used in BJT amplifiers. Resistor biasing is impractical in integratedcircuit (IC) amplifiers, because large-value resistors require a large amount of space compared to
that required for transistors. Instead, biasing is often accomplished on IC chips using an
assembly of transistors and small-value 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 investigate the performance characteristics of one kind of
current mirror.
Theoretical Background
Pure voltage and current sources do not exist in the physical world. It is 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) sources can be modeled by Thévenin and Norton
equivalent circuits (TECs and NECs).
RTH
vTH
+
vL
+
−
RL
iN
RN
iL
RL
−
(a)
(b)
Figure 1. a) Thévenin and b) Norton equivalent circuits. A “good” voltage
source is usually represented by a TEC with a small RTH value, and a “good”
current source is usually represented by an NEC with a large RN value.
In the case of a good voltage source, the Thévenin equivalent resistance RTH is small enough that
the voltage vL 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
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is one in which the Norton equivalent resistance RN is large enough that the load current iL is
almost equal to the Norton equivalent current iN. A good current source is characterized by
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. One such circuit that
can be made with discrete devices is shown in Figure 2. This is essentially a common-emitter
amplifier with the resistor normally connected between the collector and VCC replaced by a load
(although the load could simply be a resistor). Since there is no signal applied to the base, the
BJT draws a constant load bias 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 nearly-constant base voltage VB regardless of the value of
the base current IB. The Norton equivalent resistance RN of this current source circuit is derived
in the course textbook (c.f. Sec. 7.4.5) and is given by
RN  RE r  R1 R2   rce 
 F RE rce
r  R1 R2  RE
,
where r is the incremental base-emitter resistance, and rce is the incremental collector-emitter
resistance. This expression applies to the biasing case; the small-signal Norton equivalent
resistance rn is given by the same expression with F changed to o. Note that, if rce → ∞, then
RN → ∞, as expected. The last term in the expression for RN is the dominant one as long as r is
a few kilohms or less. The value of RN (and 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 requires large-value resistors (R1 and R2 are typically large), it can bias only
one amplifier stage, and it requires a relatively large voltage drop (at least 1-2 V) between the
collector and VEE. The latter characteristic limits the voltage swing that can be developed across
the load.
VCC
load
R1
IL
VB
IB
R2
RE
VEE
Figure 2. Current source based on a common-emitter amplifier circuit.
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The circuit of Figure 3 overcomes some of these disadvantages and is widely used in integrated
circuits, especially in operational amplifiers. The transistors Q1 and Q2 are fabricated as a
“matched pair,” which means that their F and Vf values (as well as other parameters) are nearly
equal. In an IC 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 = VBE1, which sets
VCE1 = 0.7 V, a value larger than the typical saturation threshold of 0.2-0.3 V. The emitter
resistors RE1 and RE2 are also matched. Therefore, because the transistors are matched and RE1 =
RE2, the base-emitter voltages are equal; that is, VBE1 = VBE2. This implies that the base currents
are equal as well, 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 each almost equal. The values of VT for both
transistors are also almost equal, because both transistors are at essentially the same temperature.
Finally, the relationship IB1 = IB2 implies that IC1 = IC2, because the F values for 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,” or “mirrors,” the
value of IREF. The value of IREF is easy to control, since, by Ohm’s law and KVL,
I REF 
VCC  VEE  VCE1 VCC  VEE  0.7
.

R1  RE1
R1  RE1
In an integrated circuit, the emitter resistors are typically unnecessary. However, if they are
necessary, they are small enough in value that they do not take up much area on the chip.
VCC
IREF
R1
+
Q1
VCE1
−
load
VB
Q2
+
−
IL
VBE1
RE1
RE2
VEE
Figure 3. A simple current mirror circuit. In an IC, the emitter resistors RE1 and
RE2 often can be omitted.
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The Norton equivalent resistance of the current mirror shown in Figure 3 is given by
R N  RE  rce 
 F RE rce
r  R1

r
F
 RE

,
where RE = RE1 = RE2, r = r = r, rce = rce1 = rce2., and F = F1 = F2. Again, the last term
dominates if r is not too large. The small-signal value rn is given by the same expression if F is
changed to o. If emitter resistors are not used (which is often the case in IC current mirrors),
then RN = rce. In any event, the minimum value of RN is rce. Although this circuit uses three
resistors, the emitter resistors can usually be made relatively small (a few kilohms at most), and
even R1 is typically 10-20 k or less. Also, if emitter resistors are not used, the voltage across
the collector-emitter terminals of Q2 can drop almost to the saturation limit of 0.2-0.3 V without
serious degradation in the performance of the circuit as a current source. Finally, as shown in
Figure 4, additional transistors can be added to the circuit with their bases connected to the base
of Q1 (in the same way Q2 is connected) to provide additional current sources for other amplifier
circuits on the chip.
VCC
IREF
R1
Load A
ILA
Load B
ILB
Load C
ILC
VB
Q1
RE1
QA
QB
QC
REA
REB
REC
VEE
Figure 4. Biasing more than one amplifier using a single current mirror. All four
transistors have their bases tied together, and all four emitter resistors have the
same value. The reference and load currents are related by IREF = ILA = ILB = ILC.
Although only three loads are controlled in this circuit, in theory any number of
loads could be controlled.
Experimental Procedure
As you make entries in your notebook, explain why you are performing the tasks described
below. Also, explain the implications of your results as you obtain them. In other words,
explain to the reader why the steps are being carried out and what the results mean.
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
Construct the current mirror shown in Figure 3 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. (The procedure outlined below can be used if you have to find your own.)
Design the circuit so that IREF = 2 mA. Make the circuit less dependent on device parameters
by designing for a drop of 1 V across each emitter resistor.

Verify that the circuit acts as a good current source for a wide range of load conditions by
measuring and recording the current through a variety of resistors RL connected as loads.
The resistor values should span a range that is a few orders of magnitude wide. (Why?)
However, RL should not be too large. (How large is too large, and why?) In order to
improve the accuracy of your data, you should measure the actual value of each resistor you
use as a load. Summarize your IL vs. RL measurements (and important related or intermediate
results) in a well-organized data table. For each measurement, you should allow enough time
for voltages and/or currents to stabilize. (Why? Why are they changing?)

Using the IL vs. RL data you have collected, calculate the Norton equivalent current IN and
resistance RN of the current mirror circuit. (You should explain the method you use to make
your calculations, and you should discuss the implications of your results). To guide you in
your analysis, you may refer to the equivalent circuit of the current mirror, power supplies,
and load shown in Figure 5a.
VCC
VCC
+
RL
IL
R1
VRC
RC
−
IN =
IREF
RN
VEE
(b)
(a)
Figure 5. (a) Equivalent circuit of current mirror with load. (b) Suggested test
configuration for matching BJTs.
Procedure for Matching Transistors
The circuit shown in Figure 5b can be used to help match BJTs. Select values of VCC, R1, and RC
so that, for an average value of F, a collector current of around 2 mA flows through the BJT and
so that VRC is small enough to keep the BJT well out of saturation. (VCE could be measured
instead.) Matched transistors have VRC values that lie close together. The current mirror circuit
is very robust; BJTs that have values within ±10% or so of each other should be close enough.
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