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ELECTRONIC CIRCUITS- I

ELECTRONIC CIRCUITS- I. DC Biasing Circuits. The ac operation of an amplifier depends on the initial dc values of I B , I C , and V CE . By varying I B around an initial dc value, I C and V CE are made to vary around their initial dc values.

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ELECTRONIC CIRCUITS- I

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  1. ELECTRONIC CIRCUITS- I

  2. DC Biasing Circuits The ac operation of an amplifier depends on the initial dc values of IB, IC, and VCE. By varying IB around an initial dc value, IC and VCE are made to vary around their initial dc values. DC biasing is a static operation since it deals with setting a fixed (steady) level of current (through the device) with a desired fixed voltage drop across the device.

  3. Purpose of the DC biasing circuit • To turn the device“ON” • To place it in operation in the region of its characteristic where the device operates most linearly, i.e. to set up the initial dc values of IB, IC, and VCE

  4. Voltage-Divider Bias • The voltage – divider (or potentiometer) bias circuit is by far the most commonly used. • RB1, RB2  voltage-divider to set the value of VB , IB , C3  to short circuitac signals to ground, while not effect the DC operating (or biasing) of a circuit (RE stabilizes the ac signals)  Bypass Capacitor

  5. Graphical DC Bias Analysis

  6. DC Load Line The straight line is know as the DC load line Its significance is that regardless of the behavior of the transistor, the collector current IC and the collector-emitter voltage VCE must always lie on the load line, depends ONLY on the VCC, RC and RE (i.e. The dc load line is a graph that represents all the possible combinations of IC and VCE for a given amplifier. For every possible value of IC, and amplifier will have a corresponding value of VCE.) It must be true at the same time as the transistor characteristic. Solve two condition using simultaneous equation  graphically Q-point !! What is IC(sat) and VCE(off) ?

  7. Q-Point (Static Operation Point) • When a transistor does not have an ac input, it will have specific dc values of IC and VCE. • These values correspond to a specific point on the dc load line. This point is called the Q-point. • The letter Q corresponds to the word (Latent) quiescent, meaning at rest. • A quiescent amplifier is one that has no ac signal applied and therefore has constant dc values of IC and VCE.

  8. Q-Point (Static Operation Point) • The intersection of the dc bias value of IB with the dc load line determines the Q-point. • It is desirable to have the Q-point centered on the load line. Why? • When a circuit is designed to have a centered Q-point, the amplifier is said to be midpoint biased. • Midpoint biasing allows optimum ac operation of the amplifier.

  9. DC Biasing + AC signal • When an ac signal is applied to the base of the transistor, IC and VCE will both vary around their Q-point values. • When the Q-point is centered, IC and VCE can both make the maximum possible transitions above and below their initial dc values. • When the Q-point is above the center on the load line, the input signal may cause the transistor to saturate. When this happens, a part of the output signal will be clipped off. • When the Q-point is below midpoint on the load line, the input signal may cause the transistor to cutoff. This can also cause a portion of the output signal to be clipped.

  10. DC Biasing + AC signal

  11. DC and AC Equivalent Circuits Bias Circuit DC equivalent circuit AC equivalent circuit

  12. AC Load Line • The ac load line of a given amplifier will not follow the plot of the dc load line. • This is due to the dc load of an amplifier is different from the ac load.

  13. AC Load Line What does the ac load line tell you? • The ac load line is used to tell you the maximum possible output voltage swing for a given common-emitter amplifier. • In other words, the ac load line will tell you the maximum possible peak-to-peak output voltage (Vpp) from a given amplifier. • This maximum Vpp is referred to as the compliance of the amplifier. (AC Saturation Current Ic(sat) , AC Cutoff Voltage VCE(off) )

  14. AC Saturation Current and AC Cutoff Voltage

  15. Amplifier Compliance • The ac load line is used to tell the maximum possible output voltage swing for a given common-emitter amplifier. In another words, the ac load line will tell the maximum possible peak-to-peak output voltage (VPP) from a given amplifier. This maximum VPP is referred to as the compliance of the amplifier. • The compliance of an amplifier is found by determine the maximum possible of IC and VCE from their respective values of ICQ and VCEQ.

  16. Maximum Possible Compliance

  17. VPP = 2ICQrC Compliance The maximum possible transition for VCE is equal to the difference between VCE(off) and VCEQ. Since this transition is equal to ICQrC, the maximum peak output voltage from the amplifier is equal to ICQ rC. Two times this value will give the maximum peak-to-peak transition of the output voltage: VPP = the output compliance, in peak-to-peak voltage ICQ = the quiescent value of IC rC= the ac load resistance in the circuit (A)

  18. VPP = 2VCEQ (B) Compliance When IC = IC(sat)­, VCE is ideally equal to 0V. When I­C = ICQ, VCE is at VCEQ. Note that when IC makes its maximum possible transition (from ICQ to IC(sat)), the output voltage changes by an amount equal to VCEQ. Thus the maximum peak-to-peak transition would be equal to twice this value: • Equation (A) sets the limit in terms of VCE(off). If the value obtained by this equation is exceed, the output voltage will try to exceed VCE(off), which is not possible. This is called cutoff clipping, because the output voltage is clipped off at the value ofVCE(off). • Equation (B) sets of the limit in terms of IC(sat). If the value obtained by this equation is exceed, the output will experience saturation clipping.

  19. Cutoff and Saturation Clipping When determining the output compliance for a given amplifier, solve both equation (A) and (B). The lower of the two results is the compliance of the amplifier.

  20. Example • For the voltage-divider bias amplifier shown in the figure, what is the ac and dc load line. Determine the maximum output compliance.

  21. Transistor Bias Circuits

  22. Objectives • Discuss the concept of dc biasing of a transistor for linear operation • Analyze voltage-divider bias, base bias, and collector-feedback bias circuits. • Basic troubleshooting for transistor bias circuits

  23. Introduction For the transistor to properly operate it must be biased. There are several methods to establish the DC operating point. We will discuss some of the methods used for biasing transistors as well as troubleshooting methods used for transistor bias circuits.

  24. The DC Operating Point The goal of amplification in most cases is to increase the amplitude of an ac signal without altering it.

  25. The DC Operating Point For a transistor circuit to amplify it must be properly biased with dc voltages. The dc operating point between saturation and cutoff is called the Q-point. The goal is to set the Q-point such that that it does not go into saturation or cutoff when an a ac signal is applied.

  26. The DC Operating Point Recall that the collector characteristic curves graphically show the relationship of collector current and VCE for different base currents. With the dc load line superimposed across the collector curves for this particular transistor we see that 30 mA of collector current is best for maximum amplification, giving equal amount above and below the Q-point. Note that this is three different scenarios of collector current being viewed simultaneously. Slope of the dc load line?

  27. The DC Operating Point With a good Q-point established, let’s look at the effect a superimposed ac voltage has on the circuit. Note the collector current swings do not exceed the limits of operation(saturation and cutoff). However, as you might already know, applying too much ac voltage to the base would result in driving the collector current into saturation or cutoff resulting in a distorted or clipped waveform. (Example 5-1)

  28. Voltage-Divider Bias Voltage-divider bias is the most widely used type of bias circuit. Only one power supply is needed and voltage-divider bias is more stable( independent) than other bias types. For this reason it will be the primary focus for study.

  29. Voltage-Divider Bias Apply your knowledge of voltage-dividers to understand how R1 and R2 are used to provide the needed voltage to point A(base). The resistance to ground from the base is not significant enough to consider in most cases. Remember, the basic operation of the transistor has not changed.

  30. Voltage-Divider Bias In the case where base to ground resistance(input resistance) is low enough to consider, we can determine it by the simplified equation RIN(base) = DCRE We can view the voltage at point A of the circuit in two ways, with or without the input resistance(point A to ground) considered.

  31. Voltage-Divider Bias For this circuit we will not take the input resistance into consideration. Essentially we are determining the voltage across R2(VB) by the proportional method. VB= (R2/R1 + R2)VCC

  32. Voltage-Divider Bias We now take the known base voltage and subtract VBE to find out what is dropped across RE. Knowing the voltage across RE we can apply Ohm’s law to determine the current in the collector-emitter side of the circuit. Remember the current in the base-emitter circuit is much smaller, so much in fact we can for all practical purposes we say that IE approximately equals IC. IE≈ IC

  33. Voltage-Divider Bias Although we have used npn transistors for most of this discussion, there is basically no difference in its operation with exception to biasing polarities. Analysis for each part of the circuit is no different than npn transistors.

  34. Base Bias This type of circuit is very unstable since its  changes with temperature and collector current. Base biasing circuits are mainly limited to switching applications.

  35. Emitter Bias This type of circuit is independent of  making it as stable as the voltage-divider type. The drawback is that it requires two power supplies. Two key equations for analysis of this type of bias circuit are shown below. With these two currents known we can apply Ohm’s law and Kirchhoff's law to solve for the voltages. IB≈ IE/ IC ≈ IE ≈( -VEE-VBE)/(RE + RB/DC)

  36. Collector-Feedback Bias Collector-feedback bias is kept stable with negative feedback, although it is not as stable as voltage-divider or emitter.With increases of IC, less voltage is applied to the base. With less IB ,IC comes down as well. The two key formulas are shown below. IB = (VC - VBE)/RB IC = (VCC - VBE)/(RC + RB/DC)

  37. Summary • The purpose of biasing is to establish a stable operating point (Q-point). • The Q-point is the best point for operation of a transistor for a given collector current. • The dc load line helps to establish the Q-point for a given collector current. • The linear region of a transistor is the region of operation within saturation and cutoff.

  38. Stability Factor

  39. Operating Regions

  40. Saturation Region Active Region Ib = 60μA Ic = 10mA Ib = 50μA Ic = 8mA Ib = 40μA Ic = 6mA Ib = 30μA Ic = 4mA Ib = 20μA Ic = 2mA Cut-off Region Ic 0 V 24 V Vce

  41. Typical junction voltages In the saturation region Ic >  Ib For active region Vce > Vce(sat)

  42. Vcc = 10 V Rc 2 K Rb 300 K Ic • Problem • Rb = 300 K • Calculate Ib, Ic & Vce if  = 100 for the Silicon transistor. Find the region of operation • Hint • Vbe = 0.7 V • Answer • Ib = 31 A Ic=3.1mA Vce = 3.8 V Active

  43. Vcc 10 V Vbb -5 V • Problem • Leakage current Io = 2 A at 250 C • Calculate Rb, if the Ge transistor remains in cut-off at 750 C • Hints • Leakage current doubles for every 100 C • I’o = Io . 2i/10 • i = t2 – t1 • Vbe(cut-off) = -0.1V Rc Rb 5.6 K 270 K Io • Answer • Rb = 76.6 K

  44. Vcc 10 V Vbb -1 V • Problem • If Vbb = 1 V, Rb = 50 K, upto what temperature, the transistor will remain in cut-off ? (Room temp. = 250 C • Hints • Find Io’ • I’o = Io . 2i/10 • i = t2 – t1 • Find t2 Rc Io Rb 5.6 K 50 K • Answer • t2 = 56.70 C

  45. +Vcc 10 V • Problem • Show that the transistor is in saturation region • Hints • In saturation Ic is not equal to  Ib • Vbe(sat) = 0.8 V • Ie = Ib + Ic • Find Ib & Ic Ib Ic 100K 2K  100 1K Ie • Answer Ib = 58.9 A Ic = 3.24mA

  46. Ie Ic - - - C E - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - B _ _ + + Ib Vbe Vcb Common Base Configuration Input Output Here the input is applied at the Emitter & the output taken from the Collector In this arrangement Base is common to the input & output This is called Common Base configuration

  47. Vcc Output Input Rb1 Rc output input Re Rb2 Ie Ic Common Base Configuration Re Rc Ib _ _ + + Vee Vcc The circuit can be re-drawn as shown, with input at Emitter & output at Collector Vb is obtained using Rb1 & Rb2 This is called potential divider arrangement

  48. Ie Ic Output Vcc Ib _ _ + + Rb1 Vee Vcc Input Output Rc Input Re Rb2 Common Emitter Configuration E C B The circuit has been re-configured with input at Base & output at Collector The Emitter is common to input & output This is called Common Emitter configuration

  49. Ico Vee _ _ + + Reverse Saturation current Ico Vcc When Emitter is open, the base & collector act as a reverse biased diode Since CB junction is reverse biased there will not be any Ic However, there will be a current due to the minority charge carriers This is called Reverse Saturation Current Ico

  50. Ie Icbo _ _ + + Vee Vcc Reverse Collector Saturation current Icbo Icbo is the leakage current that flows at the collector due to the minority charge carriers, in the common base mode  Is the current gain in the CB mode

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