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Recall Last Lecture DC Analysis and Load Line Input load line is based on the equation derived from BE loop. Output load line is derived from CE loop. To complete your load line parameters: Obtained the values of I B from the BE loop Get the values of x and y intercepts from the derived I C versus V CE. Draw the curve of I B and obtained the intercept points I C and V CE (for npn) or V EC (for pnp) which is also known as the Q points

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Voltage Transfer Characteristic V O versus V i

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Voltage Transfer Characteristics - npn A plot of the transfer characteristics (output voltage versus input voltage) can also be used to visualize the operation of a circuit or the state of a transistor. Given V BEon = 0.7V, = 120, V CEsat = 0.2V, Develop the voltage transfer curve

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In this circuit, Vo = V C = V CE Initially, the transistor is in cutoff mode because Vi is too small to turn on the diodes. In cut off mode, there is no current flow. Then as Vi starts to be bigger than V BEon the transistor operates in forward-active mode. V o (V) V i (V) 5 0.7 Cutoff

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Active Mode BE Loop 100I B + V BE – V i = 0 I B = (V i – 0.7) / 100 CE Loop I C R C + V O – 5 = 0 I C = (5 – V O ) / 4 βI B = (5 – V O ) / 4 I B = (5 – V O ) / 480 Equate the 2 equations: (V i – 0.7) / 100 = (5 – V O ) / 480 β = 120 100 Vo = - 480 Vi + 836 A linear equation with negative slope

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However, as you increase Vi even further, it reaches a point where both diodes start to become forward biased – transistor is now in saturation mode. In saturation mode, V O = V CEsat = 0.2V. So, what is the starting point, x, of the input voltage, Vi when this occurs? V o (V) V i (V) 5 0.7 Cutoff Active x 0.2 Need to substitute in the linear equation V i = 1.7 V 1.7 5 Saturation and V O stays constant at 0.2V until V i = 5V To find point x, the coordinate is (x, 0.2)

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Voltage Transfer Characteristics - pnp V EC V EB Vo = V C and V E = V CC Hence, V EC = V CC – V O V O = V CC - V EC As Vi starts from 0V, both diodes are forward biased. Hence, the transistor is in saturation. So, V EC = V ECsat and Vo = V CC – V EC sat V o (V) V i (V) Vo = 4.8 saturation β = 80

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As Vi increases, V B will become more positive than V C, the junction C-B will become reverse-biased. The transistor goes to active mode. The point (point x) where the transistor start to become active is based on the equation which is derived from active mode operation

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BE Loop 200I B + 0.7 + V i – 5 = 0 I B = (4.3 – V i ) / 200 CE Loop I C R C - V O = 0 I C = V O / 8 80 I B = V O / 8 I B = V O / 640 Equate the 2 equations: (4.3 - Vi) / 200 = V O / 640 V EC V EB β = 80 200 Vo = - 640 Vi + 2752 A linear equation with negative slope

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V o (V) V i (V) Vo = 4.8 saturation Active x By increasing V i even more, the potential difference between V EB becomes less than V EBON, causing junction E-B to become reversed biased as well. The diode will be in cut off mode. V O = 0V Using the equation derived: 200 Vo = - 640 Vi + 2752 54.3 cutoff 2.8 V when Vo = 0, then, V i = 4.3 V V EC V EB β = 80 To find point x, the coordinate is (x, 4.8)

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Bipolar Transistor Biasing

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Biasing refers to the DC voltages applied to the transistor for it to turn on and operate in the forward active region, so that it can amplify the input AC signal Bipolar Transistor Biasing

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Proper Biasing Effect Ref: Neamen

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Effect of Improper Biasing on Amplified Signal Waveform Ref: Neamen

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Three types of biasing Fixed Bias Biasing Circuit Biasing using Collector to Base Feedback Resistor Voltage Divider Biasing Circuit

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Biasing Circuits – Fixed Bias Biasing Circuit The circuit is one of the simplest transistor circuits is known as fixed-bias biasing circuit. There is a single dc power supply, and the quiescent base current is established through the resistor R B. The coupling capacitor C 1 acts as an open circuit to dc, isolating the signal source from the base current. Typical values of C 1 are in the rage of 1 to 10 μF, although the actual value depends on the frequency range of interest.

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Determine the following: (a) I B and I C (b) V CE (c) V B and V C Example – Fixed Bias Biasing Circuit NOTE: Proposed to use branch current equations and node voltages

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