Recall Last LectureDC 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 IB from the BE loopGet the values of x and y intercepts from the derived IC versus VCE. Draw the curve of IB and obtained the intercept points IC and VCE (for npn) or VEC (for pnp) which is also known as the Q points

Voltage Transfer CharacteristicVO versus Vi

Voltage Transfer Characteristics - npnA 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 VBEon = 0.7V, = 120, VCEsat = 0.2V, Develop the voltage transfer curve

In this circuit, Vo = VC = VCEInitially, 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 VBEon the transistor operates in forward-active mode. Vo (V)Vi (V)5

Active ModeBE Loop100IB + VBE Vi = 0IB = (Vi 0.7) / 100

CE LoopICRC + VO 5 = 0IC = (5 VO) / 4IB = (5 VO) / 4IB = (5 VO) / 480

Equate the 2 equations: (Vi 0.7) / 100 = (5 VO) / 480

= 120 A linear equation with negative slope

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, VO = VCEsat = 0.2V. So, what is the starting point, x, of the input voltage, Vi when this occurs?

Vo (V)Vi (V)5Need to substitute in the linear equation Vi = 1.7 V1.7and VO stays constant at 0.2V until Vi = 5V

Voltage Transfer Characteristics - pnpVo = VC and VE = VCCHence, VEC = VCC VO VO = VCC - VEC As Vi starts from 0V, both diodes are forward biased. Hence, the transistor is in saturation. So, VEC = VECsat and Vo = VCC VEC satVo (V)Vi (V)Vo = 4.8 = 80

As Vi increases, VB will become more positive than VC, 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

BE Loop200IB + 0.7 + Vi 5 = 0IB = (4.3 Vi ) / 200

CE LoopICRC - VO = 0IC = VO / 880 IB = VO / 8IB = VO / 640

Equate the 2 equations: (4.3 - Vi) / 200 = VO / 640

= 80 A linear equation with negative slope

Vo (V)Vi (V)Vo = 4.8xBy increasing Vi even more, the potential difference between VEB becomes less than VEBON, causing junction E-B to become reversed biased as well. The diode will be in cut off mode. VO = 0V

Using the equation derived:

2.8 Vwhen Vo = 0, then, Vi = 4.3 V = 80

Bipolar Transistor Biasing

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

Proper Biasing EffectRef: Neamen

Effect of Improper Biasing on Amplified Signal WaveformRef: Neamen

Three types of biasingFixed Bias Biasing CircuitBiasing using Collector to Base Feedback ResistorVoltage Divider Biasing Circuit

Biasing Circuits Fixed Bias Biasing CircuitThe 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 RB.The coupling capacitor C1 acts as an open circuit to dc, isolating the signal source from the base current.

Typical values of C1 are in the rage of 1 to 10 F, although the actual value depends on the frequency range of interest.

Determine the following: (a) IB and IC (b) VCE (c) VB and VC Example Fixed Bias Biasing CircuitNOTE: Proposed to use branch current equations and node voltages

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