KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU - Control and Automation Dept. 1 4 DC BIASING BJTS (CONT D II ) Most of the content is from the textbook: Electronic devices and circuit theory, Robert L. Boylestad, Louis Nashelsky, 11 th ed, 2013
KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU - Control and Automation Dept. 2 4.11 Current Sources The concept of a power supply provides the starting point in our consideration of current source circuits. A practical voltage source is a voltage supply in series with a resistance. An ideal voltage source has R = 0, whereas a practical source includes some small resistance. A practical current source is a current supply in parallel with a resistance. An ideal current source has R =, whereas a practical current source includes some very large resistance.
KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU - Control and Automation Dept. 3 An ideal current source provides a constant current regardless of the load connected to it. There are many uses in electronics for a circuit providing a constant current at a very high impedance. R 1 I 1 I B I C I 2 V - BE I 1 = I 2 I B I 2 - I E V EE R 2 V EE value is > 0 R E V CE V EE I 1 I 2 = R 1 R 2 I 2 R 2 = V BE I E R E I E = I 2R 2 V BE I C = I E R E β β 1 I E
KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU - Control and Automation Dept. 4 4.12 Switching of the Transistors The application of transistors is not limited solely to the amplification of signals. Through proper design, transistors can be used as switches for computer and control applications. The network of next example can be employed as an inverter in computer logic circuitry Example 4.18 V C? V CC = 5V = 68 k = 0.82 k β=125 V BE = 0.7 V - V CB I B V BE I C - - I E V CC V CE V C Proper design for the inversion process requires that the operating point switch from cutoff to saturation along the load line. For our purposes we will assume that I C = I CEO 0 ma when I B = 0 μa. In addition, we will assume that V CE = V CEsat 0 V rather than the typical 0.1 V to 0.3 V level. Transistor switches between I C = 0 ma, V CE = 5 V and I C = I Csat, V CEsat = 0 V
KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU - Control and Automation Dept. 5 When = 5 V : 5 I Csat = = 6.098 ma 0.82 x 103 If I Csat < βi B or I B > I Csat β then the transistor is in the saturation region When = 5 V, the transistor will be on and the design must ensure that the network is heavily saturated by a level of I B greater than that associated with the I B curve appearing near the saturation level. I B = 0.7 6.098 x 103 = 63 μa > 68 x 103 125 V CE 0 V V C 0 V 48.78 μa When = 0 V : I B = 0 μa I C = 0 ma V CE 5 V V C 5 V
KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU - Control and Automation Dept. 6 5 BJT AC ANALYSIS Most of the content is from the textbook: Electronic devices and circuit theory, Robert L. Boylestad, Louis Nashelsky, 11 th ed, 2013
KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU - Control and Automation Dept. 7 5.1 Introduction We will examine the ac response of the BJT amplifier by reviewing the models most frequently used to represent the transistor in the sinusoidal AC domain. One of our first concerns in the sinusoidal ac analysis of transistor networks is the magnitude of the input signal. It will determine whether small-signal or large-signal techniques should be applied. There is no set dividing line between the two, but the application and the magnitude of the variables of interest relative to the scales of the device characteristics will usually make it quite clear which method is appropriate. The small-signal technique is considered right now. There are three models commonly used in the small-signal ac analysis of transistor networks: 1. the model, 2. the hybrid π model, 3. and the hybrid equivalent model. We will emphasize the model.
KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU - Control and Automation Dept. 8 5.2 Amplification in the AC Domain i ac pp i c pp due to current amplification The peak value of the oscillation in the output circuit is controlled by the established DC level. In general, therefore, proper amplification design requires that the DC and AC components be sensitive to each other s requirements and limitations. However, it is extremely helpful to realize that: The superposition theorem is applicable for the analysis and design of the DC and AC components of a BJT network, permitting the separation of the analysis of the DC and AC responses of the system. We can make a complete dc analysis of a system before considering the AC response. Once the DC analysis is complete, the AC response can be determined using a completely AC analysis. It happens, however, that one of the components appearing in the AC analysis of BJT networks will be determined by the DC conditions, so there is still an important link between the two types of analysis.
KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU - Control and Automation Dept. 9 5.3 BJT Transistor Modeling A model is a combination of circuit elements, properly chosen, that best approximates the actual behavior of a semiconductor device under specific operating conditions. Once the ac equivalent circuit is determined, the schematic symbol for the device can be replaced by this equivalent circuit and the basic methods of circuit analysis applied to determine the desired quantities of the network. Let us assume for the moment that the small-signal ac equivalent circuit for the transistor has already been determined. Because we are interested only in the AC response of the circuit, all the DC supplies can be replaced by a zero-potential equivalent (short circuit) because they determine only the DC (quiescent level) of the output voltage and not the magnitude of the swing of the AC output. V CC R 1 C 2 R s C 1 V s R 2 R E C 3
KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU - Control and Automation Dept. 10 V CC zero-potential equivalent of the network C 2 R 1 R 1 R s C 1 R s V s R 2 R E C 3 V s R 2 R E All DC supplies can be replaced by a zero potential equivalent Use short circuits for DC voltage sources Use open circuits for DC current sources Coupling capacitors C 1 and C 2, by-pass capacitor C 3 chosen to have very small reactance at the frequency of application. Therefore they can be replaced by short circuits in many applications.
KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU - Control and Automation Dept. 11 R i R o : output current : input current : input impedance (impedance when looking into the system.) : output impedance (impedance when looking back to the system.) Small signal modeling: very common analysis technique which is used to approximate behavior of non-linear devices with lineaquations. A V = : Voltage gain of the system A i = : Current gain of the system
KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU - Control and Automation Dept. 12 zero-potential equivalent of the network zero-pot. eq. of the network (re-drawn) R 1 R s R s * R 2 R 1 R 2 V s R E V s * Now, we can insert small signal AC model of BJT AC equivalent of the network can be found after: 1. All DC supplies can be replaced by a zero potential equivalent. short circuits for DC voltage sources open circuits for DC current sources 2. Replace all capacitors with short circuits. 3. Remove bypassed components. 4. Redraw the network in a more convenient form.
KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU - Control and Automation Dept. 13 5.4 model BJT equivalent circuit. I c βi b Diode can be replaced with resistor = = V be I b I b V be = I e = I c I b = β 1 I b = β 1 = 26mV/I E, I b V be I e V ce b I b I E : emitter current we obtained from DC analysis r o = V CE I C (given in the data sheet) I c c βi b r o model of CE e
KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU - Control and Automation Dept. 14 The equivalent circuit of will be used throughout the analysis to follow for the common-emitter configuration. Typical values of β run from 50 to 200, with values of typically running from a few hundred ohms to a maximum of 6 k to 7 k. The output resistance is typically in the range of 40 k to 50 k. = 26mV/I E = I E : emitter current we obtained from DC analysi z o I e I c e I e I c c r o model of CB αi e b This is a pnp transistor! The direction of the collector current in the output circuit is now opposite to that of the defined output current. Because the output current is opposite to the defined direction, you will find in the analysis to follow that there is no phase shift between the input and output voltages. For the common-emitter configuration there is a 180 phase shift.
KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU - Control and Automation Dept. 15 5.5 Fixed Bias Configuration (CE) Note that the input signal is applied to the base of the transistor, whereas the output is off the collector. In addition, recognize that the input current is not the base current, but the source current, and the output current is the collector current. The small-signal AC analysis begins by removing the DC effects of V CC and replacing the DC blocking capacitors C 1 and C 2 by short-circuit equivalents V CC zero-potential equivalent of the network C 2 C 1 I b I c r o small-signal AC equivalent of the network using model - βi b
KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU - Control and Automation Dept. 16 = ( ) = r o is determined when = 0 Mostly r o is greater than more than a factor of 10. = βi b (r o ) r o - I b = = β (r o = (r o A v = = (r o ) If r o 10 V CC A v is negative there is a 180 phase shift between and Example 5.1 C 2 a) Determine b) Calculate c) Calculate and A v for r o = d) Calculate and A v for r o = 50 k C 1 V CC = 12 V = 470 k = 3 k C 1 = 10 μf C 2 = 10 μf β=100 V BE = 0.7 V
KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU - Control and Automation Dept. 17 a) Regarding DC conditions ; I B = 24.04 μa I E = 2.428 ma 26 mv = = 10.71 I E b) = 100 10.71 = 1.071 k c) = (r o ) ( 3 k ) = 3k A v = = (r o ) = 3x103 10.71 = 280.11 d) = r o (50k 3 k ) = 2.83k A v = = (r o ) = 2.83x103 10.71 = 264.24
KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU - Control and Automation Dept. 18 5.6 Voltage Divider Configuration (CE) V CC R 1 C 2 small-signal AC equivalent of the network using model I b I c I C 1 i R 1 R 2 βi b r o R 2 R E C 3 - = R 1 R 2 ( ) = r o Mostly r o is greater than more than a factor of 10. = βi b (r o ) I b = = β (r βr o ) = (r e r o ) e A v = = (r o ) A v is negative there is a 180 phase shift between and
KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU - Control and Automation Dept. 19 Example 5.2 V CC R 1 I C 1 i R 2 R E C 2 C 3 a) Determine b) Calculate c) Calculate and A v for r o = d) Calculate and A v for r o = 50 k V CC = 22 V R 1 = 56 k R 2 = 8.2 k = 6.8 k R E = 1.5 k C 1 = 10 μf C 2 = 10 μf C 3 = 20 μf β=90 V BE = 0.7 V a) Regarding DC conditions ; I E = 1.41 ma 26 mv = = 18.44 I E b) = 90 18.44 = 1.66 k = (R 1 R 2 ) = 1.35 k c) = r o ( 6.8 k ) = 6.8 k A v = = (r o ) = 6.8x103 18.44 = 368.76 d) = r o (50k 6.8 k ) = 5.98 k A v = = (r o ) = 5.98x103 18.44 = 324.3
KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU - Control and Automation Dept. 20 5.7 Emitter Bias Configuration (CE) V CC small-signal AC equivalent of the network using model I b I c C 2 C 1 βi b R E R E - Z b Emitter resistor is not by-passed With no r o effect. In most application for the sake of simplification its effect can be ignored. r o 10 R E if r o r 1 o can be ignored
KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU - Control and Automation Dept. 21 = I b I b βi b R E Z b = = βr I e 1 β R E b = Z b β is much greater than 1, R E is greater than Z b βr E = βr E When is zero, I b = 0 βi b = 0, = R c = βi b = I b I b βi b R E A v = βi b = I b I b βi b R E R E A v is negative there is a 180 phase shift between and
KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU - Control and Automation Dept. 22 Example 5.3 V CC I C 1 i R E If R 3 is by-passed C 2 a) Determine b) Calculate c) Calculate and A v a) = 26 mv I E V CC = 20V = 470 k = 2.2 k R E = 0.56 k C 1 = 10 μf C 2 = 10 μf β=120 V BE = 0.7 V r o = 40 k = 5.99 b) = ( ) = 717.70 c) = = 2.2 k a) Regarding DC conditions ; I B = 35.89 μa, I E = 4.35 ma 26 mv = = 5.99 I E r o 10 R E.? r o 1.? 40x 10 3 10 2.2 0.56 x 10 3 True 40x 10 3 5.99 1 True We can ignore r o b) Z b = = βr I e 1 β R E = 67.92 k b = Z b = 470 k 67.92 k = 59.34 k c) = = 2.2 k A v = = 3.69 R E A v = = 367.28
KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU - Control and Automation Dept. 23 5.8 Collector Feedback Configuration (CE) V CC small-signal AC equivalent of the network using model C 2 I R i B I 1 C 1 I b I c βi b - With no r o effect. In most application for the sake of simplification its effect can be ignored. r o 10 if r o can be ignored
KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU - Control and Automation Dept. 24 I R i B I 1 If I b I c βi b - = = = re 1 1 β When is zero, I b = 0,βI b = 0, I e = 0 = = I 1 βi b βi b I 1 βi b I b = We have to relate with to find I 1 = I 1 = = βi b = β = 1 = 1 1 = I b = I 1 = 1 1 1 1 = = = 1 1 re = 0 A V = =
KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU - Control and Automation Dept. 25 Example 5.4 C 1 V CC a) Determine b) Calculate c) Calculate and A v C 2 V CC = 9V = 180 k = 2.7 k C 1 = 10 μf C 2 = 10 μf β=200 V BE = 0.7 V r o a) Regarding DC conditions ; I B = 11.53 μa, I E = 2.32 ma 26 mv = = 11.21 I E r o 10 True True We can ignore r o b) = 1 βr = 560.5 C c) = = 2.6 k A v = = = 240.86