written 2.6 years ago by |
There are two major reasons for which we require biasing in BJT
1] To maintain the position of Q-point.
2] Variation of IC d.r.t. $\beta$ VBE & temperature.
1] Q pt position:
Ideally Q pt shall be maintained at the middle of DC load line. If Q pt shifts np or down there is a possibility of clipping off the o/p waveform
$\therefore$ Biasing is necessary.
2] Variation in IC:
Q pt is a function of IC & VCE. Q (VCE, IC)
For Q pt to become stable VCE and IC must have stable values.
But IC = Function (temperature, $\beta$ and VBE) As temp/$\beta$/ VBE changes results change in IC so Q pt also changed.
1] IC against temperature.
pi (oc) = i/p DC power
$Vcc \times Ic$ - - - - (1)
PQ $\rightarrow$ o/p across transistor $VCE \times Ic$
$\because$ Q pt is @ middle of load line $VCE = \frac{Vcc}{2}$
$PQ = \frac{Vcc}{2} \times Ic$ = 50% of Pi dc - - -- (2)
$\therefore$ 50% of power is dropped across $T_r$. of taken power.
$\therefore$ Temp $\uparrow$
$\rightarrow$
Ic majority $\uparrow$
Ic = Ic minority + Ic majority
$\downarrow$
$Ic \uparrow$
$\downarrow$
$PQ \alpha Ic$
$\therefore$ $PQ \uparrow$
It may result into thermal runway.
2] Ic variation wrt r $\beta$
$Ic = \beta IB$
$\beta \rightarrow$ device parameter.
Change in device may change $’\beta’$ so Ic changes.
3] Ic variation against cut is vtg (VBE)
VBE $\uparrow$ if Temp $\uparrow$
By 2.5mv/˚c $\downarrow$
So IB $\downarrow$
$\downarrow$
$\because$ Ic = $\beta$ IB
$\downarrow$
Ic changes Q pt may change.