KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU - Control and Automation Dept. 1 4 DC BIASING BJTS (CONT D)

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1 KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 1 4 DC BIASING BJTS (CONT D) Most of the content is from the textbook: Electronic devices and circuit theory, Robert L. Boylestad, Louis Nashelsky, 11 th ed, 2013

2 KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept Voltage Divider Configuration β is temperature sensitive parameter. Its actual value is not well defined A circuit less dependent on β is desired. Voltage divider works very well on this purpose. C 2 AC input C 1 R 1 AC output R 2

3 KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 3 Thevenin Equivalent of the circuit may be needed R 1 R 1 R 2 R TH V OC V TH, R TH V OC R 2 V TH = V OC = R 2 R 1 R 2 R TH = R 1 R 2 = R 1 R 2 R 1 R 2

4 KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 4 V TH = R TH = β, = = β 1 V TH = R TH β 1 V TH = R TH β 1 R 1 V TH R TH = = R 2 The expression of = β can be used only if the tr is in the Active region!!!

5 KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 5 Example 4.6: (Voltage Divider Configuration) = 22 V R 1 = 39 k R 2 = 3.9 k = 10 k = 1.5 k C 1 = 10μF C 2 = 10μF C 3 = 50μF β=100 = 0.7 V AC input C 1 R 1 Determine,,, C 2 AC output V TH = 2 V, R TH = 3.55k = 8.38 μa = ma = V = V > 0, > 0 Our assumption that the tr. is in the active region is correct. R 2 C 3

6 KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 6 Example 4.7: Repeat the previous example using β = 50. V TH = 2 V, R TH = 3.55k = μa = 0.81 ma = V = V β [ma] [V] The circuit is relatively insensitive to the β value

7 KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 7 LoadLine Analysis The similarities with the output circuit of the emitterbiased configuration result in the same intersections for the load line of the voltagedivider configuration. = = load line equation = IC =0 = VCE =0 Two values as the intersection of the line with the axis Saturation Level The resulting equation for the saturation current (when is set to 0 V) is therefore the same as obtained for the emitterbiased configuration. B C _sat _sat 0V Regarding emmiter bias configuration: = _sat _sat _sat _sat

8 KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept Collector Feedback Configuration An improved level of stability can also be obtained by introducing a feedback path from collector to base. Although the Q point is not totally independent of beta (even under approximate conditions), the sensitivity to changes in beta or temperature variations is normally less than encountered for the fixedbias or emitterbiased configurations. The analysis will again be performed by first analyzing the base emitter loop, with the results then applied to the collector emitter loop. C 2 AC input C 1 AC output

9 KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 9 = = = (β 1) = (β 1) (β 1) = β 1 β 1 = β 1 ( ) = =

10 KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 10 = Fixedbias Configuration V = βr or R TH Voltage term ~ && or V TH && ~ or, if βr then βr βr Emitterbias Configuration = β 1 Voltage Divider Configuration V TH = R TH β 1 Collector Feedback Configuration = β 1 ( ) = β β V βr = V R The result is an equation absent of β, which would be very stable for variations in β. Because R is typically larger for the voltagefeedback configuration than for the emitterbias configuration, the sensitivity to variations in β is less. Of course, R is 0 for the fixedbias configuration and is therefore quite sensitive to variations in beta.

11 KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 11 Example 4.8: (Collector Feedback Configuration) = 10 V = 4.7 k = 250 k = 1.2 k C 1 = 10μF C 2 = 10μF C 2 Determine,,, = μa = ma = ma β=90 = 0.7 V AC input C 1 AC output = 3.69 V = 2.99 V > 0, > 0 Our assumption that the tr. is in the active region is correct.

12 KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 12 Example 4.9: Repeat previous example using a β = 135 = 8.89 μa = 1. 2 ma = ma = 2.92 V = 2.22 V β [ma] [V] Saturation Conditions Using the approximation =, we find that the equation for the saturation current is the same as obtained for the voltagedivider and emitterbias configurations. That is, LoadLine Analysis sat = Continuing with the approximation = results in the same load line defined for the voltagedivider and emitterbiased configurations.

13 KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept Emitterfollower Configuration V EE = 0 V EE = β 1 AC input C 1 C 2 AC output V EE = β 1 V EE V EE V EE = 0 = V EE

14 KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 14 Example 4.10: (Emitter Follower Configuration) V EE = 20 V = 240 k = 2 k C 1 = 10μF C 2 = 10μF β=90 = 0.7 V AC input C 1 C 2 Determine,,, AC output = μa = ma = ma = V = V > 0, > 0 Our assumption that the tr. is in the active region is correct. V EE

15 KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept Miscellaneous Bias Configurations Example 4.11: V EE = 9 V = 100 k = 1.2 k C 1 = 10μF C 2 = 10μF β=45 = 0.7 V AC input C 1 Determine,,, C 2 V EE AC output V EE V EE = 0 V EE = = V EE = x10 3 = 83μA = ma = ma V EE = 0 = V = V = V > 0, > 0 Our assumption that the tr. is in the active region is correct.

16 KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 16 Example 4.12: = 20 V V EE = 20 V R 1 = 8.2 k R 2 = 2.2 k = 2.7 k = 1.8 k Determine,,, V TH R TH C 1 = 10μF C 2 = 10μF We have to find R TH and V TH. β=120 = 0.7 V C 1 AC input R 1 R 2 AC output C 2 20 V I 8.2 k 2.2 k V TH = V OC I = V EE 40 = ma 10.4 x 103 V TH = 2. 2 x 10 3 I 20 = V 20 V R TH = R 1 R 2 = 1.73 k V EE R TH

17 KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept = 1.73 x x10 3 = 35.39μA = 4.25 ma = 4.28 ma R TH = 1.73k 20 V 40 = = V = V V 20 V

18 KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept Design Operations So far, All the elements were in place, and it was simply a matter of solving for the current and voltage levels of the configuration. The design process is one where a current and/or voltage may be specified and the elements required to establish the designated levels must be determined. Example 4.13: Given the device characteristics in the figure, determine,, and for the fixedbias configuration. = 20 V = = 8 ma R = 2.5 k C VCE =0 = = = k Standard resistor values are = 2.4 k = 470 k Using standard resistor values gives = 41.1 ma which is well within 5% of the value specified.

19 KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 19 Example 4.14: Given that Q = 2 ma and Q = 10 V, determine R 1 and for the network. = 18 V R 2 = 18 k = 1.2 k C 1 = 10μF C 2 = 10μF β=100 = 0.7 V C 2 AC input C 1 R 1 R 2 AC output V E = = 2.4 V V B = V E = 3.1 V V B = R R 1 R 2 R 1 = k 2 = V C V E V C = 12.4 V = V C = 2.8 k

20 KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 20 Example 4.15 The following emitterbias configuration has the following specifications: Q = 1 2 sat sat = 8 ma = 28 V, V C = 18 V,and β = 110. Determine, and. Q = 4mA Q = V C Q = μa = 2.5 k Regarding saturation region: = sat sat 0 V = 1 k sat = 1 β = k

21 KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept Current Mirrors The current mirror is a dc network in which the current through a load is controlled by a current at another point in the network. That is, if the controlling current is raised or lowered the current through the load will change to the same level. The discussion to follow will demonstrate that the effectiveness of the design is dependent on the fact that the two transistors employed have identical characteristics. Connected to another subcircuit Q 1 I x I Q 2 I x is the reference current, it is set by and R x and mirrored into the C of Q 2. We set I x and it is seen as I in Q 2. Q 1 and Q 2 are exactly the same transistors. = β I x = 2 β = 2 I I x =

22 KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 22 Example 4.16 Calculate the mirrored current I in the circuit = 12 V = 1.1 k β=100 = 0.7 V Q 1 I x I Q I x = = ma 1.1 x 103 I I x approximation Actual value: β I = I x = I β2 C = ma

23 KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 23 Example 4.17 Calculate the mirrored current I in the circuit = 6 V R x = 1.3 k β=100 = 0.7 V I I x = = ma 1.3 x 103 I I x approximation I x Q 2 Actual value: β I = I x = I β3 C = 3.96 ma Q 1 I Q 3

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