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Explain Foster seeley discriminationtor with near diagram.
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Foster seeley is called as phase discriminator. Both winding's are tuned to FC. The o/p voltage changes according to phase difference between 2 winding's.

Working:

• The voltage applied to each diode is vector sum of primary & corresponding half secondary v/g.

• The o/p v/g depends on the vector sum of voltages applied to each diode.

• $R_3$ & $R_4$ are adjusted such that $C_3$ & $C_4$ become negligible.

• The ckt therefore has C, Ls & $C_4$ across primary winding.

• $L_3$ is choke and therefore has large reactance, thereby making capacitors C and G redundant.

$\therefore$ $VL_3 = \frac{j \times L_3}{j \times L_3 - j (xc + xc_4)} .V_{12}$

$= \frac{j \times L_3}{j \times L_3} . \ V_{12}$

$XL_3 \ \gt \gt \ (XC + XG)$

$\therefore$ $VL_3 \ = V_{12}$

• Primary current, IP can be given as,

$IP = \frac{V_{12}}{jwL_1}$

• This induces voltage at secondary as,

Vs = I jwm Ip

$= I jwm \frac{V_{12}}{jwL_1}$

$V_s = I \frac{m}{L_1} V_{12}$

• The secondary ckt is

$\therefore$ $Vab = \frac{-j \times c_2}{R + j (xl_2 - xc_2)} \ V_s$

$= \frac{-j \times c_2 (- \frac{m}{L_1} V_{12})}{R + jx}$ . . . . . $( Vs = \frac{-m}{4} V_{12})$

$\therefore$ $V_{ab} = \frac{j \times c_2 \ m }{(R+ j X) L_1}$ . . . . $X = XL_2 - X c_2$

• The v/g applied to $D_1$ and $D_2$ are given by

$V_{ao} = V_{ac} + V_{12} = 1/2 V_{ab} + V_{12}$

$V_{bo} = V_{bc} + V_{12} = 1/2 V_{ab} + V_{12}$

$V_o = V_{ab} = V_{ao} - V_{bo}$

$V_o \ \alpha \ V_{ao} - V_{bo}$

1] Case : fin = fc

• when fin = fc & secondary is tuned to fc, due to parallel resonance,

$XL_2 = XC_2$ $\therefore$ X = 0

$\therefore$ $V_{ab} = \frac{X c_2 \ m}{RL_1} V_{12} \lt 90˚$

$V_{ab}$ leads $V_{12}$ by 90˚

$\therefore$ $V_{ab}$ leads $V_{12}$ by 90˚

1/2 $V_{ab}$ lags $V_{12}$ by 90˚

$\therefore$ $V_{ao} = V_{bo}$

$\therefore$ $V_o = 0$

2] Case : fin > fc

$XL_2 \gt XC_2$

$\therefore$ X becomes inductive.

$\therefore$ $V_{ab} = \frac{XC_2 \ m \ V_{12} \lt 90}{ L_1 | Z | \theta}$ . . . z = R + gx

$\therefore$ $V_{ab} = \frac{XC_3 \ m}{L_1 | z|} V_{12} \lt 90 - \theta$

$\therefore$ $V_{ab}$ leads $V_{12}$ by 90 - $\theta$

1/2 $V_{ab}$ leads $V_{12}$ by 90 - $\theta$

• 1/2 $V_{ab}$ lags $V_{12}$ by 90 + $\theta$

$\therefore$ $V_{ao} \gt V_{bo}$

$\therefore$ $Vo \ os$ Positive.

3] Case: fin < fc

$CL_2 \lt XC_2$

X is capacitive.

$\therefore$ $V_{ab}$ leads $V_{12}$ by 90 + $\theta$

1/2 $V_{ab} leads$V_{12}$by 90 +$\theta$-1/2$V_{ab} lags $V_{12} by 90 -$\theta\thereforeV_{ao} \lt V_{bo}V_o\$ goes negative with decrease in fc.