written 4.7 years ago by | • modified 4.7 years ago |

**Mumbai University > Electronics Engineering > Sem 4 > Discrete Electronic Circuits**

**Marks:** 10M

**Year:** May 2015

**1 Answer**

0

5.6kviews

For the given circuit find IDq and VGSq. Also state in which operating region circuit works.

written 4.7 years ago by | • modified 4.7 years ago |

**Mumbai University > Electronics Engineering > Sem 4 > Discrete Electronic Circuits**

**Marks:** 10M

**Year:** May 2015

ADD COMMENT
EDIT

0

228views

written 4.7 years ago by | • modified 4.7 years ago |

**Given:**

$IDSS = 10mA$

$V_P = - 4V$

**Solution:**

We know that,

$IDS = IDSS(1- \frac{VGS}{V_P})^2$

$IDS = 10m(1 + \frac{VGS}{4})^2$

Put different values of VGS and obtain IDS.

VGS(V) | IDS (mA) |
---|---|

0 | 10 |

-0.5 | 7.65 |

-1 | 5.625 |

-1.5 | 3.90 |

-2 | 2.5 |

-2.5 | 1.4 |

-3 | 0.625 |

-3.5 | 0.156 |

-4 | 0 |

Apply KVL from $R_2$ to ground through gate and source,

$VR_2 - VGS - IDS \times R_S = 0$...........(1)

Put $IDS = 0$ in equation (1) we get,

$VR_2 = VGS = \frac{VDD \times R_2}{R_1 + R_2}$

$VR_2 = VGS = \frac{16 \times 110K\Omega}{910K\Omega + 110K\Omega}$

$VR_2 = VGS = 1.72V$.......will point on X-axis.

Put $VGS = 0$ in equation (1) we get,

$IDS = \frac{VR_2}{R_S}$

$IDS = \frac{1.72V}{1.1K\Omega}$

$IDS = 1.56mA$.......will point on Y-axis.

From graph $IDSq = 3.2mA, VGSq = -1.7V$

Apply KVL from VDD to ground through drain and source,

$VDD - IDSq(R_S + R_D) - VDSq = 0$

$16 - 3.2m(2.2K\Omega + 1.1K\Omega) = VDSq$

$VDSq = 5.44V$

**$VDSq \gt |V_P| \gt 4 V$, FET is operating in pinch off region.**

**Answer:**

- $IDSq = 3.2mA$
- $VGSq = -1.7V$
**Operating in Pinch off region**

ADD COMMENT
EDIT

Please log in to add an answer.