How to Bias Bjts for Fun and Profit

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.

V CC

R1 RC IC

IB + + V + CE V – BE – I ≈ I + E C VB R2 VE 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.

β Since II= (1) CEβ + 1

β >> ≈ = and 1, we have IICEas shown in the figure above. We will assume IICE 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 + β vi RB rπ 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, RRRB 12|| . 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 rRoC|| . However, ≈Ω>> rkRoC10 , 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 RIC C34 V CC V CE . 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 Q- point, the minimum specified DC current gain is βmin = 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

V 10 V ===CC 33.V . (2) E 3 3

– 4 –

== But VIRIREEECE, so

===VE 33. Ω RE 066.k . (3) IC 5

We select the closest standard value, or = Ω RE 680 . (4)

To find RC, we use Kirchhoff's voltage law (KVL) to get

=+ + VVIRRCC CE C() C E . (5)

The only unknown in eq. (5) is RC, so we solve for RC and get

− = VVCC CE − RC RE IC 10− 4 = −=−=068....k 12 068 052 Ω . (6) 5

We use the closest standard value, RC = 510 Ω.

To achieve VE = 3.3 V [eq. (2)], we must have

=+ VVBBEE V =+=07.. 33 40 .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

I I = C , (8) B β

– 5 –

I I = C . (9) B max β min

We must have I1 much larger than this, so we set

I II≥=20 20 C . (10) 1 B max β min

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, V I I = CC≥ 20 C , (11) 1 + β RR12 min

β +≤minVCC or RR12 . (12) 20IC

But, from the voltage divider, we must also have

R 2 VV==40.V (13) + CC B RR12 from eq. (7). We use the limiting equality in eq. (12) to write eq. (13) as

20IR C 2 = V . (14) β B min

We solve for the unknown R2: β ()()100 4.0 ==minVB =Ω R2 4.0 k . (15) 20IC ()() 20 5.0

Finally, again using the limiting equality in eq. (12), we solve eq. (12) for the only remaining unknown – 6 –

resistor value, R1:

β ()()100 10 =−=−=−=ΩminVCC RR124.0 10 4.0 6.0 k . (16) 20IC ()() 20 5.0

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

R1 RC IC + IB + V VCC + CE + – – VBE – + IE ≈ IC V B R2 V E 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 –

V CC

RC IC

IB + V + CE + – RTH VBE – ≈ + IE IC VTH VE RE – –

Figure 4. The standard BJT biasing configuration with the base-bias circuit replaced by its Thevenin equivalent.

RR ()()5.6 3.9 In Figure 4, RRR==||12 = =Ω 2.3 k (17) TH 12 ++ RR125.6 3.9

R 3.9 and VV==2 ()10 = 4.1V . (18) TH++ CC RR12 5.6 3.9

We apply Kirchhoff’s voltage law around the base-emitter loop to get

−−−= VIRVIRTH B TH BE E E 0 . (19)

We use eq. (8) to eliminate IB from eq. (19) and

β + 1 II= (20) ECβ

to eliminate IE. We solve the result for IC and get – 8 –

VV− I = β TH BE . (21) C ++β RTH()1 R E

ββ== If we assume min 100 for the 2N3904, eqs. (4), (17), and (18) substituted into eq. (21) give us

4.1− 0.7 I ==()100 4.8 mA , (22) C 2.3++()() 100 1 0.68

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

=− +=− + =−= VVIRRCE CC C() C E 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

=+ VVIRCC CE C E . (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 –

− − = VVCC CE = 10 4. 0 = Ω RE 12.k , (25) IC 50. which coincidentally is a standard value. We now have

== = VIRECE()()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:

V CC

RC ICC I1

IC R1 I B +

VCE + + VBE – V R – B 2 –

Figure 5. Biasing without an emitter resistor. – 10 –

Note from Fig. 5 that = VVBBE0.7 V (27)

=+ and ICCII C 1 . (28)

Once again, we design the circuit such that IBmax << I1 so that IB can be neglected. This requires the use of eq. (11) again: I I ≥ 20 C . (11) 1 β min

With IB negligible, we have = VVBBE IR12

or, once we have chosen I1 with the aid of eq. (11),

= VBE R2 (29) I1

With IB negligible, we see from Fig. 5 that

=+ VIRRCE 11() 2 , (30)

=−VCE or R12R . (31) I1

Finally, to calculate RC we use eq. (28) and Ohm’s Law to derive

− =+=VVCC CE IIICC C 1 , RC

– 11 –

VV− or R = CC CE . (32) C + IC I1

Unfortunately, this biasing scheme is sensitive to temperature variations. To show this, we rewrite eq. (29) as

= VBE I1 . (33) R2

Substituting this into eq. (30) gives

+  ==+RR12 R 1 VVCE BE1 V BE . (34) RR22

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: VCC I1

ICC Rx IC2 R1 IB2 +

Q2 + RC I R2 C1 Vy I Vx B1 + – RB VCE – Q1 –

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 − = VVBE2 T between base-emitter voltage and collector current in Q2 ( IIeCS2 ). To design this bias circuit, note that

=+ =+ ICCII C12 C II C 11 B (35)

since IC2 = IB1. But IB1 << IC1, so ≅ ICCI C1 . (36)

=−+ Therefore VVCE CC() RRI C x C1 . (37)

The value of Rx can be determined from

+=− VVyEBCCxC21 V RI . (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.