How to Bias BJTs for Fun and Profit
Standard BJT biasing configuration:
The standard biasing configuration for bipolar junction transistors, sometimes called the “H-bias
configuration” because the resistors form the outline of the letter H, is shown in Figure 1 below. This
configuration can be used for all BJT amplifiers (common emitter, common base, and common
collector), although in the common-collector configuration we usually set RC = 0.
VCC
I1
R1
RC
IB
+
VB
R2
+
VBE
–
+
VE
IC
+
VCE
–
IE ≈ IC
RE
–
–
Figure 1. Standard BJT biasing configuration.
Biasing procedure:
Biasing a BJT means establishing the desired values of VCE and IC so that the amplifier will have the
proper gain, input impedance, undistorted output voltage swing, etc. These values of VCE and IC are
known as the quiescent operating point or Q-point. The values of VCE and IC required are determined
– 2 –
from inspection of the BJT's data sheet and load line analysis.
IC =
Since
β
IE
β +1
(1)
and β >> 1 , we have I C ≈ I E as shown in the figure above. We will assume I C = I E to bias the
transistor. The bias configuration shown above actually sets the value of IE .
Each resistor in the configuration shown above plays a role in biasing the transistor. In addition, RC
also sets the output resistance of common-emitter amplifiers, as can be seen in Figure 2 below, which is
the small-signal equivalent circuit (valid at frequencies where the parasitic capacitances of the BJT are
negligibly small and if RE is bypassed by a suitable capacitor) of the amplifier shown in Figure 1 above.
ib
+
vi
RB
rπ
+
β ib
ro
RC
–
vo
–
Figure 2. The AC small-signal equivalent circuit of the amplifier shown in Figure 1.
This AC equivalent circuit is not used for DC biasing!!
In Figure 2, RB = R1|| R2 . If vi = 0 (the condition for calculating output resistance), then ib = 0, the
controlled current source is off (i.e., it is an open circuit), and the output resistance is ro || RC . However,
ro ≈ 10 k Ω >> RC , so ro can be neglected and the output resistance is approximately RC. If the output
impedance of a common-emitter amplifier is a critical design specification, set RC first. However, make
sure that RC I C ≤ 3VCC 4 − VCE . If this cannot be achieved at the desired Q-point, a new Q-point must
be selected.
The only purpose of RE is to provide DC feedback to stabilize the Q-point against variations in the β
– 3 –
(also known as hFE) of the BJT. Since β can vary by as much as five to one from one BJT to another
(even if the BJTs have the same 2N number), large unit-to-unit variations in Q-point could result unless
this DC feedback is present. When the resistor values in the above figure are selected properly, β
variations will have a negligibly small effect on the Q-point. Instead, unit-to-unit variations in Q-point
will be due almost entirely to resistor tolerance. The voltage VE across RE sets IE, which is
approximately equal to IC. To provide sufficient feedback, VE is usually chosen to be one-quarter to
one-third of VCC. Because VBE ≈ 0.7 V (silicon devices), VE is set by the voltage divider formed by R1
and R2. However, the alert reader will note from Figure 1 above that R1 and R2 are not in series, and
hence do not form a voltage divider! Only if the maximum value of IB is much less than I1, so that IB can
be neglected, will R1 and R2 act like a voltage divider. The circuit designer must establish this condition
by the appropriate choice of values for R1 and R2.
BJT biasing example:
Consider a 2N3904 BJT with VCC = 10 V and a desired value of VCE = 4.0 V. Assume that the
output resistance of the amplifier is not critical.
From the data sheet for the 2N3904 we find that maximum DC current gain occurs when IC is in the
range of 3 to 10 mA. With no specification on amplifier bias current to concern us, we select IC = 5 mA.
Our Q-point, therefore, is VCE = 4.0 V and IC = 5 mA. Also from the data sheet, we see that at this Qpoint, the minimum specified DC current gain is β min = 100.
Since there is no specification on output resistance, we will determine the value of RE first. We
apply the engineering rule of thumb given above and select
VE =
VCC
10
=
= 3.3 V .
3
3
(2)
– 4 –
But VE = I E RE = I C RE , so
RE =
VE
3.3
=
= 0.66 kΩ .
IC
5
(3)
We select the closest standard value, or
RE = 680 Ω .
(4)
To find RC, we use Kirchhoff's voltage law (KVL) to get
VCC = VCE + I C ( RC + RE ) .
(5)
The only unknown in eq. (5) is RC, so we solve for RC and get
RC =
VCC − VCE
− RE
IC
10 − 4
. − 0.68 = 0.52 kΩ .
− 0.68 = 12
5
=
(6)
We use the closest standard value, RC = 510 Ω.
To achieve VE = 3.3 V [eq. (2)], we must have
VB = VBE + VE
= 0.7 + 3.3 = 4.0 V .
(7)
Now, to make the combination of R1 and R2 act like a voltage divider, we must make the largest possible
value of IB, i.e., IBmax , negligibly small compared to I1. Since
IB =
IC
,
β
(8)
– 5 –
I B max =
IC
.
β min
(9)
We must have I1 much larger than this, so we set
I1 ≥ 20 I B max = 20
IC
.
β min
(10)
Now that we have designed our circuit such that IB is negligibly small even under the worst-case
condition of minimum β , we can pretend that there is no connection between R1 and R2 and the base of
the BJT. In that case,
I1 =
VCC
I
≥ 20 C ,
R1 + R2
β min
R1 + R2 ≤
or
β minVCC
.
20 I C
(11)
(12)
But, from the voltage divider, we must also have
R2
VCC = VB = 4.0 V
R1 + R2
(13)
from eq. (7). We use the limiting equality in eq. (12) to write eq. (13) as
20 I C R2
= VB .
β min
(14)
We solve for the unknown R2:
R2 =
(100 )( 4.0 ) = 4.0 kΩ .
β minVB
=
20 I C
( 20 )( 5.0 )
(15)
Finally, again using the limiting equality in eq. (12), we solve eq. (12) for the only remaining unknown
– 6 –
resistor value, R1:
R1 =
β minVCC
(100 )(10 ) − 4.0 = 10 − 4.0 = 6.0 kΩ .
− R2 =
20 IC
( 20 )( 5.0 )
(16)
Since neither 4.0 kΩ nor 6.0 kΩ is a standard value, we select the next smaller [so that the inequality in
eq. (12) will be satisfied] standard values, R1 = 5.6 kΩ and R2 = 3.9 kΩ.
To confirm that the BJT is biased at the chosen Q-point, we redraw Figure 1 in the form shown in
Figure 3.
VCC
RC
R1
+
VCC
IB
+
VBE
+
–
VB
R2
–
+
VE
–
IC
+
VCE
–
IE ≈ IC
RE
–
Figure 3. An equivalent way of drawing the circuit of Figure 1.
We take the Thevenin equivalent circuit looking from the base of the BJT toward R1 and R2 to convert
this figure to the equivalent form:
– 7 –
VCC
RC
IB
RTH
+
VTH
+
+
VBE –
+
VE
–
–
IC
VCE
–
IE ≈ IC
RE
Figure 4. The standard BJT biasing configuration with the base-bias circuit replaced by its Thevenin
equivalent.
In Figure 4,
and
RTH = R1 || R2 =
VTH =
( 5.6 )( 3.9 )
R1R2
=
= 2.3 kΩ
R1 + R2
5.6 + 3.9
R2
3.9
VCC =
(10 ) = 4.1 V .
R1 + R2
5.6 + 3.9
(17)
(18)
We apply Kirchhoff’s voltage law around the base-emitter loop to get
VTH − I B RTH − VBE − I E RE = 0 .
(19)
We use eq. (8) to eliminate IB from eq. (19) and
IE =
to eliminate IE. We solve the result for IC and get
β +1
I
β C
(20)
– 8 –
IC = β
VTH − VBE
.
RTH + ( β + 1) RE
(21)
If we assume β = β min = 100 for the 2N3904, eqs. (4), (17), and (18) substituted into eq. (21) give us
IC = (100 )
4.1 − 0.7
= 4.8 mA ,
2.3 + (100 + 1)( 0.68 )
(22)
only 4 percent less than the desired value of 5 mA. We substitute this value of IC, the value of RE from
eq. (4), and the value of RC (510 Ω) into eq. (5) and solve for VCE to get
VCE = VCC − IC ( RC + RE ) = 10 − ( 4.8 )( 0.51 + 0.68 ) = 10 − 5.7 = 4.3 V ,
(23)
7.5 percent more than the desired value.
Modified bias configuration:
In high-frequency amplifiers RC might be replaced by an inductor. In common-collector amplifiers
(also known as emitter followers) RC is frequently replaced by a short circuit. In either case, RC = 0 in
DC biasing calculations.
We will repeat the above example for the case when RC = 0. In this case, we do not get to choose VE.
Instead, with RC = 0, eq. (5) reduces to
VCC = VCE + I C RE .
(24)
Since we determined VCE and IC when we selected the Q-point, the only unknown in this equation is RE.
We solve eq. (24) for RE and get
– 9 –
RE =
VCC − VCE
10 − 4.0
=
= 1.2 kΩ ,
IC
5.0
(25)
which coincidentally is a standard value. We now have
VE = I C RE = ( 5.0 )(1.2 ) = 6.0 V ,
(26)
and the rest of the design proceeds as above.
Biasing without an emitter resistor:
Sometimes amplifier stability considerations or the lack of space force the designer to bias a BJT
without an emitter resistor. The only purpose of the emitter resistor is to provide negative DC feedback
to stabilize the circuit against β variations. If the circuit has no emitter resistor, the required negative
feedback must be provided another way, such as shown in the following circuit:
VCC
RC
I1
R1
+
VB
R2
ICC
IC
IB
+
VBE
–
+
VCE
–
–
Figure 5. Biasing without an emitter resistor.
– 10 –
Note from Fig. 5 that
VB = VBE 0.7 V
(27)
ICC = I C + I1 .
(28)
and
Once again, we design the circuit such that IBmax << I1 so that IB can be neglected. This requires the
use of eq. (11) again:
I1 ≥ 20
IC
.
β min
(11)
With IB negligible, we have
VB = VBE I1R2
or, once we have chosen I1 with the aid of eq. (11),
R2 =
VBE
I1
(29)
With IB negligible, we see from Fig. 5 that
VCE = I1 ( R1 + R2 ) ,
or
R1 =
VCE
− R2 .
I1
Finally, to calculate RC we use eq. (28) and Ohm’s Law to derive
ICC = IC + I1 =
VCC − VCE
,
RC
(30)
(31)
– 11 –
RC =
or
VCC − VCE
.
IC + I1
(32)
Unfortunately, this biasing scheme is sensitive to temperature variations. To show this, we rewrite
eq. (29) as
I1 =
VBE
.
R2
(33)
Substituting this into eq. (30) gives
VCE =
R

R1 + R2
VBE =  1 + 1VBE .
R2
 R2

(34)
Since VBE varies –2.2 mV/ºC, VCE will vary by the factor 1 + R1 /R2 greater than this. This could be an
unacceptable variation in a system that must function from –30ºC to +70ºC.
Usually, the emitter resistor must be eliminated to insure AC stability only in UHF, microwave, and
millimeter-wave circuits. In such circuits the BJT being biased is often relatively expensive (> $0.50
each). Therefore it can be cost-effective to use a low-cost, low-frequency BJT to bias the high-frequency
BJT. Such a scheme is shown in the following figure:
I1
R1
ICC
IB2
+
Q2
+
R2
Rx
IC2
VCC
RC
Vy
–
RB
IB1
Q1
IC1
+
Vx
VCE
–
–
Figure 6. Active biasing of a BJT without an emitter resistor.
– 12 –
In this circuit, the BJT being biased is Q1. The only function of the low-cost, low-frequency pnp BJT Q2
is to provide the base bias current IB1 to Q1.
The negative feedback in this circuit occurs as follows: If something causes IC1, the collector current
in Q1, to increase, the voltage Vx decreases (because the voltage drop across Rx increases due to the
increased collector current). However, the voltage Vy is fixed by the voltage divider formed by R1 and
R2. Therefore the base-emitter voltage of Q2 decreases, reducing Q2’s collector current IC2, which is
also the base current IB1 to Q1. The reduction of IB1 returns IC1 to its original value. A very small
change in IC1 can produce a relatively large change in IC2 = IB1 due to the exponential relationship
between base-emitter voltage and collector current in Q2 ( IC 2 = I S e −VBE 2 VT ).
To design this bias circuit, note that
ICC = IC1 + IC 2 = IC1 + I B1
(35)
ICC ≅ I C1 .
(36)
VCE = VCC − ( RC + Rx ) I C1 .
(37)
since IC2 = IB1. But IB1 << IC1, so
Therefore
The value of Rx can be determined from
V y + VEB 2 = VCC − Rx I C1 .
(38)
Do not, however, make Rx so large that the desired VCE cannot be achieved. In many cases RB can be
eliminated since the resistance looking into the collector of Q2 is large.