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 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) or VEC (for pnp) which is also known as the Q points

Output Load Line Use KVL at B-E loop to find the value of IB

Use KVL at C-E loop – to obtain the linear equation IC RC + VCE + IE RE – 12 = 0 IC RC + VCE + (IC / )RE – 12 = 0

VCE = 12 – IC (1.01) IC = - VCE + 12 1.01 IB = 75.1 A VCE = 6.31 V

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

Voltage Transfer Characteristics - pnp Vo (V) saturation Vo = 4.8 VEC VEB β = 80 Vo = VC and VE = VCC Hence, 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 sat Vi (V)

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 Loop 200IB + 0.7 + Vi – 5 = 0 IB = (4.3 – Vi ) / 200 CE Loop ICRC - VO = 0 IC = VO / 8 80 IB = VO / 8 IB = VO / 640 Equate the 2 equations: (4.3 - Vi) / 200 = VO / 640 VEC VEB β = 80 200 Vo = - 640 Vi + 2752 A linear equation with negative slope

Vo = - 640 Vi + 2752 200 Vo (V) saturation Vo = 4.8 Active β = 80 VEC VEB β = 80 To find point x, the coordinate is (x, 4.8) 5 4.3 cutoff 2.8 V x Vi (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 = 0V Using the equation derived: 200 Vo = - 640 Vi + 2752 when Vo = 0, then, Vi = 4.3 V