Recall Lecture 11 DC Analysis and Load Line Input load line is based on the equation derived from KVL at BE loop. Output load line is derived from KVL at CE loop. To complete your load line parameters: Obtained the values of IB from the BE loop Get 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) which is also known as the Q points

Voltage Transfer Characteristic VO versus Vi

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 VBEon = 0.7V,  = 120, VCEsat = 0.2V, Develop the voltage transfer curve

In this circuit, Vo = VC = VCE Cutoff 5 0.7 Vi (V) In this circuit, Vo = VC = VCE 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 VBEon the transistor operates in forward-active mode.

Active Mode BE Loop 100IB + VBE – Vi = 0 IB = (Vi – 0.7) / 100 CE Loop ICRC + VO – 5 = 0 IC = (5 – VO) / 4 βIB = (5 – VO) / 4 IB = (5 – VO) / 480 Equate the 2 equations: (Vi – 0.7) / 100 = (5 – VO) / 480 β = 120 100 Vo = - 480 Vi + 836 A linear equation with negative slope

Vo (V) Cutoff 5 0.7 Active 5 Saturation To find point x, the coordinate is (x, 0.2) 0.2 Vi (V) x 1.7 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? Need to substitute in the linear equation  Vi = 1.7 V and VO stays constant at 0.2V until Vi = 5V

Bipolar Transistor Biasing

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

Proper Biasing Effect Ref: Neamen

Effect of Improper Biasing on Amplified Signal Waveform Ref: Neamen

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

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 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.

Example – Fixed Bias Biasing Circuit Determine the following: (a) IB and IC (b) VCE Assume VBE(on) = 0.7 V, and  = 50 (a) VB – VE = 0.7 VB = 0.7V IB = (12 – 0.7) / 240k = 0.0471 mA IC = 50 (0.0471) = 2.355 mA VB = ?? VE = 0V (b) IC = 2.355 mA = (12 – VC) / 2.2k VC = 6.819 V VCE = 6.819 – 0 = 6.819 V NOTE: Proposed to use branch current equations and node voltages

Biasing using Collector to Base Feedback Resistor IC + IB = IE IB IC IE Find RB and RC such that IE = 1mA , VCE = 2.3 V, VCC = 10 V and  =100. NOTE: Proposed to use branch current equations and node voltages

Biasing using Collector to Base Feedback Resistor IE = 1mA , VCE = 2.3 V, VCC = 10 V and  =100. IB = (IE / (+1) = 0.0099 mA (VC – VB ) / RB= IB but VC = VCE and VB = VBE = 0.7 V (2.3 – 0.7) / RB = 0.0099 mA RB = 161.6 k (VCC – VC ) / RC = IE RC = 7.7 k VC VB VE = 0V