Presentation on theme: "Recall Last Lecture DC Analysis and Load Line"— Presentation transcript:
1Recall 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 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
3Voltage 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 VBEon = 0.7V, = 120, VCEsat = 0.2V, Develop the voltage transfer curve
4In this circuit, Vo = VC = VCE Cutoff50.7Vi (V)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.
5Active Mode BE Loop 100IB + VBE – Vi = 0 IB = (Vi – 0.7) / 100 CE Loop ICRC + VO – 5 = 0IC = (5 – VO) / 4βIB = (5 – VO) / 4IB = (5 – VO) / 480Equate the 2 equations:(Vi – 0.7) / 100 = (5 – VO) / 480β = 120100Vo = Vi + 836A linear equation with negative slope
6Vo (V)Cutoff50.7Active5SaturationTo find point x, the coordinate is (x, 0.2)0.2Vi (V)x1.7However, 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?Need to substitute in the linear equation Vi = 1.7 Vand VO stays constant at 0.2V until Vi = 5V
7Voltage Transfer Characteristics - pnp Vo (V)saturationVo = 4.8VECVEBβ = 80Vo = VC and VE = VCCHence, VEC = VCC – VO VO = VCC - VECAs Vi starts from 0V, both diodes are forward biased. Hence, the transistor is in saturation. So, VEC = VECsat and Vo = VCC – VEC satVi (V)
8As 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
9BE Loop 200IB + 0.7 + Vi – 5 = 0 IB = (4.3 – Vi ) / 200 CE Loop ICRC - VO = 0IC = VO / 880 IB = VO / 8IB = VO / 640Equate the 2 equations:(4.3 - Vi) / 200 = VO / 640VECVEBβ = 80200Vo = ViA linear equation with negative slope
10Vo = - 640 Vi + 2752 200 Vo (V) saturation Vo = 4.8 Active β = 80 VECVEBβ = 80To find point x, the coordinate is (x, 4.8)54.3cutoff2.8 VxVi (V)By 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 = 0VUsing the equation derived:200Vo = Viwhen Vo = 0, then, Vi = 4.3 V
14Effect of Improper Biasing on Amplified Signal Waveform Ref: Neamen
15Three types of biasingFixed Bias Biasing CircuitBiasing using Collector to Base Feedback ResistorVoltage Divider Biasing Circuit
16Biasing 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 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 μF, although the actual value depends on the frequency range of interest.
17Example – Fixed Bias Biasing Circuit Determine the following: (a) IB and IC (b) VCE (c) VB and VCNOTE: Proposed to use branch current equations and node voltages