educationELECTRONIC QUESTIONJFET and single stage transistor amplifier

Solution to Problems on JFET and single stage transistor amplifier(EEC 224)

JFET and transistor amplifier

JFET and transistor amplifier questions on electronics III (EEC224)

In a transistor amplifier, when the signal changes by 0.02v, the base current changes by 10μA and the collector by 1mA. If collector resistance Rc = 5KΩ and RL 10KΩ. Find

(I) current gain (II) A.C load (III) voltage gain (IV) power gain

(I) current gain = ∆ic÷∆ib
Where Rc is collector resistance, RL is load resistance
And ic = collector current, ib = base current, ie = emitter current
ib = 10μA = 10 × 10–⁶A
Ic= 1mA » 10×10–³A


(I) Current gain = ∆ic÷∆ib
= [10 × 10–³] ÷[ 10×10–⁶A]=0.01÷0.00001 » 100A


(II) A.C load, Rac = RL//Rc
[RL × Rc ]÷ [RL + Rc] = [10 × 5 × 10⁶] ÷ [(10+5) ×10³][50 × 10³] ÷ [15] = 3.33 × 10³
Rac = 3.33KΩ


(III) Voltage gain, Av = current gain × [Rac÷Rin]Av = 100 × [ 3.33 ×10³] ÷ [2×10³]= 165


(IV) power gain, Ap, current gain ² × [Rac÷Rin]= 100² × [ 3.33 ×10³] ÷ [2×10³]= 16500

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Write the expression for the drain current in FET
ANSWER

ID = IDSS [ 1 – Vgs/Vp]²
ID = drain current at given Vgs
IDSS = Shorted-gate drain current
Vgs = gate-source voltage
Vgs(off) = gate-source cut off voltage.

A JFET has the following parameters IDss = 32mA, Vgs (off) = -8v, Vgs = -4.5V. Find the value of the drain current

SOLUTION

ID = ?, IDss = 32mA, Vgs (off) = -8V, Vgs = -4.5V
ID = IDSS [ 1 – Vgs/Vp]²
Pinch of voltage, Vp =negative of (Vgs(off), gate-source cut off voltage)
I.e Vp = -vgs
Therefore Vp = -(-8)
Vp = 8V
ID = 32 ×10–³ [ 1 – (-4.5V/ 8V)]²

ID = 6.12mA

Sketch the output characteristics of a FET transistor indicating the following
(I) Short-gate drain current (II) pinch off voltage, Vp (3 marks)
ANSWER


Definition of output characteristics of a FET transistor

The output characteristics of a FET transistor is the curve between drain current [ID] and drain-source voltage [Vgs] of a JFET at constant Vgs

Features of the output chara
cteristics

  1. At first, the drain current (ID), rises rapidly with drain source voltage( VDS) but then becomes constant.
  2. After pinch off voltage, the channel width becomes so narrow that depletion layers almost touch each other, the drain current (ID) passes through this layers. Increase in drain current (ID) is very small with drain source voltage (VDS) above pinch off voltage, consequently, ID becomes constant
  3. The characteristics resemble that of a pentode valve.

What is pinch off voltage,Vp:

is the point where the drain-source voltage becomes constant with the drain-source current.

What’s quiescent point:

Quiescent point is the point of intersection between A.C load and D.C load

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