written 5.8 years ago by | modified 2.8 years ago by |

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Explain Hartley oscillator. Design the same for 50KHz.

written 5.8 years ago by | modified 2.8 years ago by |

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written 2.8 years ago by |

$R_1$ and $R_2$ are used to properly bias the n-char BJT. Cc1 and cc2 are the coupling capacitors and CE is the bypass capacitor. The amplifier configuration is common emitter, hence the phase shift between p/p and o/p will be 180° the feedback network promise the additional 180° phase shift required to satisfy the condition for positive feedback.

$x_1$ = W$w_1$ $x_2$ = w$l_2$ $x_3$ = $\frac{-1}{WC3}$

For loop gain to be real, the imaginary part should be zero.

X1 + x2 + x3 = 0

W4 + wl2 - $\frac{1}{wc3}$ = 0

W (L1 + L2) = $\frac{1}{wc3}$

$w^2$ = $\frac{1}{(L1 + L2)} c3$

W = 2$\pi$ f

$\therefore$ f = $\frac{1}{2\pi}\sqrt{c3 (L1 +L2)}$

If the inductors are not wound on the same core then the mutual conductance will not be zero.

f = $\frac{1}{2\pi \sqrt c3 (L1 + L2 + 2m)}$

Application: used in local oscillator in radio and TV receiver in function generator.

Design for 50 KHz :

**Step 1 :** calculate the LC components.

Let L1 = L2 = L

f = $\frac{1}{2 \pi} \sqrt {c3 (L1 + L2)}$

= $\frac{1}{2 \pi} \sqrt {2LC3}$

Let C3 = 0.01 $\mu$ f

$[50x10^3]^2$

$\therefore$ L = 5.066 X $10^-4$ H

= 0.5066 mH

**Step 2** : baising components

Let VE = 3V, VCE = 3V, VCC =10V

Also ICQ = 1mA and $\beta$ = 50

RE = $\frac{VE}{ICQ}$ = $\frac{3v}{1mA}$ = 3k$\Omega$

IB = ICQ / $\beta$ = 1 x $10^-3$ / 50 = 20 $\mu$A

Let I2 = 10 IB = 10X20 = 200 $\mu$A

VB = VE + VBE

= 8 + 0.7 = 3.7 V

R2 = VB/I2 = 3.7/(200x$10^-6)$ = 18

R1 = $\frac{Vcc = VB}{I1}$

=$\frac{ VCC – VB}{I2 + IB}$

= 28.636.

Values,

R1 = 28.636 k $\Omega$ R2 = 18.5 k $\Omega$

RE = 3k $\Omega$

L1 = L2 = 0.5066 mH

Cc = 0.01 $\mu$ f

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