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प्रश्न
The plate current, plate voltage and grid voltage of a 6F6 triode tube are related as ip = 41 (Vp + 7 Vg)1.41
Here, Vp and Vg are in volts and ip in microamperes.
The tube is operated at Vp = 250 V, Vg = −20 V. Calculate (a) the tube current, (b) the plate resistance, (c) the mutual conductance and (d) the amplification factor.
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उत्तर
Given:-
Plate voltage, VP = 250 V
Grid voltage, VG = -20V
(a) As given in the question, plate current varies as,
\[i_p = 41( V_p + 7 V_g )^{1 . 41} \]
\[ i_p = 41(250 - 140 )^{1 . 41} \]
\[ i_p = 41 \times (110 )^{1 . 41} \]
\[ i_p = 30984 . 71 \mu A = 30 . 98 \text{ mA}\]
(b) \[i_p = 41( V_p + 7 V_G )^{1 . 41}\]
Differentiating this equation, we get:-
\[d i_P = 41 \times 1 . 41 \times ( V_p + 7 V_g )^{0 . 41} \times (d V_p + 7d V_g ) ................(1)\]
Plate resistance is defined as:-
\[ r_p = \left( \frac{d V_p}{d i_p} \right)_{V_g = \text{Constant}} \]
From equation (1),
\[\frac{d V_p}{d i_p} = \frac{1 \times {10}^6}{41 \times 1 . 41 \times {110}^{0 . 41}}\]
\[\frac{d V_p}{d i_p} = {10}^6 \times 2 . 51 \times {10}^{- 3} \]
\[\frac{d V_p}{d i_p} = 2 . 5 \times {10}^3 \Omega = 2 . 5 K\Omega\]
(c) From above:-
As \[d I_{{}_P} = 41 \times 1 . 41 \times (250 + 7 \times ( - 20) )^{0 . 41} \times 7d V_{g,} \]
\[ g_m = (\frac{d I_p}{d V_g} )_{V_P =\text{ Constant}} \]
From equation (1),
\[ \frac{1}{7} \left( \frac{d I_p}{d V_g} \right)_{V_P =\text{ Constant}} = 41 \times 1 . 41 \times (250 + 7 \times ( - 20) )^{0 . 41} \]
\[ \left( \frac{d I_p}{d V_g} \right)_{V_P =\text{ Constant}} = 41 \times 1 . 41 \times (110 )^{0 . 41} \times 7\text{ mho}\]
\[ \left( \frac{d I_p}{d V_g} \right)_{V_P =\text{ Constant}} = 41 \times 1 . 41 \times 6 . 87 \times 7\text{ mho}\]
\[ \left( \frac{d I_p}{d V_g} \right)_{V_P =\text{ Constant}} = 2 . 78\text{ milli mho}\]
(d) Amplification factor,
\[\mu = r_p \times g_m \]
\[ = 2 . 5 \times {10}^3 \times 2 . 78 \times {10}^{- 3} \]
\[ = 6 . 95 \approx 7\]
