# CALCULATING THE BIAS POINT OF THE STABILIZED CIRCUIT Electronics Help

If the stabilized bias design in Figure 6-1 truly makes the bias point independent of parameter values, then it should be possible to determine the quiescent current and voltage without using any parameter values in the computations. The analysis procedure that follows will demonstrate that this is indeed the case. Refer to Figure
6-4. Note in Figure 6-4 that the base-to-ground voltage VB is determined by the voltage-divider network composed of resistors R. and Rz connected between Vcc and ground. If we neglect the loading effect of Rill looking into the base, then ( n, ) V Vcc RI + R2 (6-10) Assuming a silicon transistor, it is apparent from the figure that VI{ = VB – 0.7. Then the emitter current is the emitter-to-ground voltage VI{ divided by RIO: Calculating the Bias Point of the Stabilized Circus 205 This computation gives us the quiescent very anyone me same as the quiescent collector current: Ir =IE (6-12) By constructing a Evening equivalent circuit of the voltage-divider network an riling Kirchhoff’s voltage law through it and the base-emitter junction, it can be shown that the e ct value of the quiescent collector current is ((RI ~2 RJ Vee – VBE) Ie = {JIB = {J ,Lordly + ({J + l)RE (6-12a) T find the quiescent value of VCE, we write Kirchhoff’s voltage law from through Re and R/i to obtain (6-13) Thus, (6- 14) or, since Ves V<.t: – (Re + RE) Note that the equation of the  load line is = -( 1 ) Vet.’ + c- , Re + RE . Re + R/i Assuming that the bias circuit shown in Figure 6-5 is well stabilized against variations in {J, find and VCE at the Q-point. Solution. From equation 6-10, ( VB…. 1 X 1()3 ) 1 X 1()3+ 3.8 X 1()3, 12 V = 2.5 V Then V£ = VII – 0.7 V = 1.8 V and l e = (1.8 V)/(470 0 ) = 3.83 mA Thus. the quiescent collector current is about 3.83 mA. From equation 6-15. 12 – (3.83 X 10-.1 A)(1.5 kO + 4700) := 4.45 V. Notice that no transistor parameter values were used in the preceding computations. Recall that we neglected the loading effect of R(Figure 6-4) on the computation of VII (equation 6-10). Following a procedure similar to In::e used to derive equation 5-53, it is easy to show that R”, = ({3 + l)Rt. + Vllf/lll {3Rf.” «(1-17) If Rin is not neglected, the value of VB is (approximately) V-(~)V /I – R, + ;” cc Since R”, depends on {3, equation 6-18 shows that V/I and hence the Q-point do depend to some extent on {3. However. in a well-stabilized circuit, R, is large and RII = R, II R2 is small, so R2  The whole point here is that R, and Rz should be small enough in comparison to {3Rt. that the voltage V II is unaffected by
changes in {3. .In effect. the voltage divider should look as much like an ideal voltage source as possible (i.e., hold V«constant), regardless of how the load {3 changes. VB stays essentially constant if R2 is small and R£ is large. (Draw a equivalent circuit looking out from the base and convince yourself of this fact.) This perspective is simply another way of viewing the already-established result  that R£ should be large and RII small to achieve good (3-stability. The next example shows how the stability deteriorates when the voltage-divider resistors have the same ratio as in the previous example, but are each increased tenfold in value.Suppose R, and R~ in Figure’ 6-5 arc increased to 38 kO and 10 kf], respectively. Find Ie when (I) {3 = 50 and (2) {3= 150. Solution
1. When {3 = 50. R”,(3(Rf.) = (SO)(470 H) = 23.5 kO. From equation 6-18. [ (10 X l(n II (23.5 X 10′) ] 12 V VB”‘” 38 X 101 + (to X tO~)II (23.5 X 103) = (7 X 1O~) 12 V = 1.R7 V 45 X l(l’ VI. = (1.87 – 0.7)V = 1.17 Y v. 1.17 V Ie- ” Ie = -e 2.49 mA 470 n 2. When {3 = 150. R”, (3R£ = (150)(4700) = 70.5 kil. Then [ (10 X 10.1) II (70.5 X 10-‘) J? VB ~ 38 X 1()3 + (10 X 1(}1) II (70.5 X t01) L V = (8.76X 1) 12=2.2″V 46.76 X 103 ) Small-Signal Performance of the Bias-Stabilized Circuit 207 VE = (2.25 – O.7)V = 1.55 V Ir = Y.f = 1.55 V = 3.30 mA . RE 470 n With the larger values of RI and R2, the increase in {3 causes  to increase  about 33%. Note that the ratio RBI R, in this example is (R, II R2)/ R£ = 16.84, which is somewhat large for good stability.

Posted on November 18, 2015 in Bias Design in Discrete and Integrated Circuits