0
59views
The JFET shown in figure 3b has parameter IDSS=8mA and Vp=4V. Determine VG, IDSQ, VGSQ and VDSQ
1 Answer
0
1views
  • The Drain Source Voltage of a JFET is given by the equation

$V_{{DS}_Q}=V_{DD}-I_D(R_D+R_S)=5-(5\times10^{-3})(3+0.5)\times10^3=-12.5V$

  • For a Voltage Divider JFET the Voltage between Source and Gate is given as

$V_{{GS}}=-I_DR_S =-8V$

  • By Voltage Division we have the Voltage between Gate and Ground as

$V_{G_Q}=\dfrac{R_2V_{DD}}{R_1+R_2}=\dfrac{1200}{200 \times10^3}=6mV$

  • The Drain Current is given by

$I{D_Q}=\dfrac {I_{DSS}}{2}=4mA$

 

 

Please log in to add an answer.

Continue reading...

The best way to discover useful content is by searching it.

Search