C CRYSTAL OSCILLATOR
R
To design aYcrystal oscillator to oscillate at the specified crystal frequency.
COMPONENTS REQUIRED:
Sl. No. Components Details Specification Qty
1. Transistor SL100 1No
0.1f 2No
2. Capacitors
47µf 1No
3. Resistors 22K,4.7K,1.2K, 330
Each1No
1K Pot
4. Crystal 2MHzor1.8MHz 1No
DC Supply, CRO with Probe
CIRCUIT DIAGRAM:
Vcc = 9v
R1 Rc
1.8K Cc VO
18K
CB 0.1 f
BC109
0.1 f
Variable
1 K Pot
3.9K
470
R2 CE
RE
47f
2MHz
1.8MHz
DESIGN:
Given VCC = 9V, IC = 2mA, =50
RE: W.K.T.VRE = VCC/10 = 9/10=0.9V ------------ for biasing
IEIC=2mA
From the fig. We see that,
IERE = VRE
RE=0.9/(2x10-3)=450
Therefore RE470
RC: VCE=VCC/2=4.5V ------------for Q point to be in active region.
Applying KVL to output loop
VCC –ICRC-VCE-VRE = 0
9–2x10-3RC–45-0.9= 0
Therefore RC= 1.8k
R1&R2: From biasing circuit
VB= VBE+ VRE
=0.7+0.9
VB=1.6V
As sume10 IB flows through R1and9IB flows through R2.
W.K.T.IC=IB
2x10-3=50 IB
Therefore IB = 40 A
from the fig. we see that,
R1=VCC–VB/10 IB=9–1.6/(10x40x10-6)=18.5k
ThereforeR1 18k
R2= VB/9IB =1.6/(9x40x10-6)=4.44k
Therefore R23.9k
CE, CC, CB: Let CB=CC=0.1F
XCE= RE/10
Therefore f = 10 / (2CE RE)
Let f=100HzandW.K.TRE=470
Therefore CE=10/2f.RE=34F Therefore
CE47F.
PROCEDURE:
1. Rig up the circuits shown in the circuit diagram.
2. Before connecting the feedback network, check the circuit for
biasing conditions i.e. check VCE, and VRE.
3. After connecting the feedback network. Check the output.
4. Check for the sinusoidal waveform at output. Note down the
frequency of the output waveform and check for any deviation
from the designed value of the frequency.
5. To get a sinusoidalwaveformadjust1Kpotentiometer.
WAVEFORM:
Vo
frequency fo =1/T
RESULT:
