Saturday, July 24, 2010

Dj Record Live On Adobe Audition

Modelos para señales fuertes

continuous flows in the emitter and collector are in normal operation by: The internal power base is mainly by diffusion and
  • Where: I
  • E is the current emitter.
    I C is the collector current. Α
    T is the direct current gain in common base configuration. (From 0.98 to 0.998)
    ES
    is the reverse saturation current of base-emitter diode (in the order of 10
    -15 -12
    I_\mathrm{E} = I_\mathrm{ES} \left(e^{\frac{V_\mathrm{BE}}{V_\mathrm{T}}} - 1\right) to 10 amps)
    I_\mathrm{C} = \alpha_T I_\mathrm{ES} \left(e^{\frac{V_\mathrm{BE}}{V_\mathrm{T}}} - 1\right)
    V
    T is the thermal voltage k
    J_p(Base) = \frac{q D_p p_{bo}}{W} \left[e^{\frac{V_{EB}}{V_T}}\right]
    T / q
    • (approximately 26 mV at room temperature ≈ 300 K). V BE
    • is base-emitter voltage. W is the width of the base.
    • The collector current is slightly less than the emitter current, because the value of αT is very close to 1.0. In the bipolar junction transistor is a small variation of the base-emitter current generates a large change in collector-emitter current. The relationship between the current collector-emitter to the base-emitter is called gain, β or hFE. A β value of 100 is typical for bipolar transistors smaller. In a typical configuration, a very weak signal current flowing through the base-emitter to control the current emitter-collector. α β is related through the following relationships:
    • issuer Efficiency: Other equations are used to describe the three streams in any region of the transistor are expressed below . These equations are based on the transport model of a bipolar junction transistor.
    • Where: i
    • C is the collector current.
    • i B
    • is the base current.
    • i
    E is the emitter current.
    \alpha_T = \frac{I_\mathrm{C}}{I_\mathrm{E}}
    F \beta_F = \frac{I_\mathrm{C}}{I_\mathrm{B}}
    β is the active gain common-emitter (20 to 500)
    \beta_F = \frac{\alpha_T}{1 - \alpha_T}\iff \alpha_T = \frac{\beta_F}{\beta_F+1}
    R
    \eta = \frac{J_p(Base)}{J_E} β is the inverse common-emitter gain (0 to 20)
    I  i_\mathrm{C} = I_\mathrm{S}\left(e^{\frac{V_\mathrm{BE}}{V_\mathrm{T}}} - e^{\frac{V_\mathrm{BC}}{V_\mathrm{T}}}\right) - \frac{I_\mathrm{S}}{\beta_\mathrm{R}}\left(e^{\frac{V_\mathrm{BC}}{V_\mathrm{T}}} - 1\right)
    S
     i_\mathrm{B} = \frac{I_\mathrm{S}}{\beta_\mathrm{F}}\left(e^{\frac{V_\mathrm{BE}}{V_\mathrm{T}}} - 1\right) + \frac{I_\mathrm{S}}{\beta_\mathrm{R}}\left(e^{\frac{V_\mathrm{BC}}{V_\mathrm{T}}} - 1\right) is the reverse saturation current (in the order of 10
    -15 -12
     i_\mathrm{E} = I_\mathrm{S}\left(e^{\frac{V_\mathrm{BE}}{V_\mathrm{T}}} - e^{\frac{V_\mathrm{BC}}{V_\mathrm{T}}}\right) + \frac{I_\mathrm{S}}{\beta_\mathrm{F}}\left(e^{\frac{V_\mathrm{BE}}{V_\mathrm{T}}} - 1\right)
    to 10 amps)
    • V T is the thermal voltage k
    • T / q (approximately 26 mV at room temperature ≈ 300 K).
    • V BE is the base-emitter voltage.
    • V BC is the base-collector voltage.
    • Posted by: Emmanuel Angel

    0 comments:

    Post a Comment