Uzgodnienie co do tego obliczenia, że nie można określić, że działanie jest pointem, a nie jest, że jest to funkcja, która jest funkcjonalna i działa.

Uzgodnienie to Load Line

Thee load line presents the relationship between thee collector current (I is 1; FLT: 0 presents 3; C presents 1; FLT: 1 presents 3; Equi3; 3; and collector- emitter voltage (V present 1; Equi1; FLT: 2 presenta3; CE presentation 1; FLT: 3 presentation 3; Ecuads 3;) for a given load resistor. It is derived from the objet 's supply voltage and load resistor value.

Te equation for thee load line is:

V Xi1; Xi1; FLT: 0 Xi3; Xi3; CE XI1; Xi1; FLT: 1 XI3; = VY1; XI1; FLT: 2 XI3; XI3; CC XI1; XI1; FLT: 3 XI3; XI3; - I XI1; FLT: 4 XI3; XI3; C XI1; XI1; FLT: 5 XI3; XI3; * R XI1; FLT: 6 XI3; XI3; C XI1; FLT: 7 XIXI3; X3; X3; FLT 3;

Kalkulating thee Q- Point

Thee Q- point, or quiescent point, is the DC operating point of thee transistor where thee load line intersects thee transistor 's characteristic curves. It determinations thes e biasing conditions and linearity of thee amplifier.

Tu find thee Q- point:

  • Choose a base bias voltage or current.
  • Oblicz te kolekcje compact (I, I, I, I, I, I, I, I, I, I, FLT: 0, III, FLT: 0, III, C, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, FLT: 0, FLT: 0, III, FLT: I, C, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, FLT: 0, FLT: 0, C, I, I, C, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I, I,
  • Determine thee collector- emitter voltage (V present 1; present 1; present 1; revenge 1; revenge 3; FLT 3;) atthis present.

Te przekrojowe wartości of te load line with the transistor 's criteristic curves thee Q- point values for I contribution 1; indibu1; FLT: 0 contribution 3; C contribution 1; indibution 1; indibution 3; and V contribution 1; indibution 1; FLT: 2 contribution 3; indibution 3; CE contribution 1; indibute 1; indibute 3; indibution 3; indibusculation 3.

Praktyka Badanie

Suppose V present 1; Suppose 1; FLT: 0 presenta3; CC presenta1; Suppose 1; FLT: 0 presenta3; CC presenta1; FLT: 1 presenta3; FLT: 1 presenta3; Supporta3; Is 12V, R presenta1; FLT: 2 presenta3; Eventa3; C presenta1; FLT: 3 presentation 3; Eventa3; is 2kmbH, and the transistor 's β is 100. If thee base present is set to 20μA, then:

I BEL1; BEL1; FLT: 0 BEL3; CEL3; CEL1; CEL1; FLT: 1 BEL3; CEL3; CEL3; EL1; FLT: 2 BEL3; CEL1; CEL1; EL1; FLT: 3 BEL3; CEL3; EL3; = 100 * 20μA = 2mA.

Using thee load line equation:

V Xi1; Xi1; FLT: 0 Xi3; Xi3; CE Xi1; Xi1; FLT: 1 Xi3; Xi3; = 12V - (2mA * 2kmbH) = 12V - 4V = 8V.

Thee Q- point is approxiately at I inde1; FLT: 0 index3; FLT: 0 index3; C index1; FLT: 1 index3; FLT: 1 index3; FLT: 2 index3; FLT: 2 index3; FL3; CE index1; FLT: 3 index3; = 8V.