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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. - Physics

Sum

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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Solution

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,

\[\mu   =    r_p  \times  g_m \]

\[ =   2 . 5 \times  {10}^3  \times 2 . 78 \times  {10}^{- 3} \]

\[ =   6 . 95 \approx 7\]

Concept: Triode
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APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 19 Electric Current through Gases
Q 19 | Page 353
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