Department of Pre-University Education, Karnataka course PUC Karnataka Science Class 12
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The Plate Resistance and Amplification Factor of a Triode Are 10kω and 20. the Tube is Operated at Plate Voltage 250 V and Grid Voltage −7.5 V. the Plate Current is 10 Ma. - Physics

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Question

The plate resistance and amplification factor of a triode are 10kΩ and 20. The tube is operated at plate voltage 250 V and grid voltage −7.5 V. The plate current is 10 mA. (a) To what value should the grid voltage be changed so as to increase the plate current to 15 mA? (b) To what value should the plate voltage be changed to take the plate current back to 10 mA?

Solution

Given:-

Plate resistance,

`r_p=10KOmega`

Amplification factor,

`mu=20`

Plate voltage,

`V_p=240V`

Grid Voltage,

`V_G=7.5V`

`I_p=10"mA"`

(a) Transconductance,

\[\delta V_G  = \frac{\delta I_p}{g_m}\]

\[\delta V_G  = \frac{\left(15 \times {10}^{- 3} - 10 \times {10}^{- 3} \right)}{\mu/ r_p}\]

\[\delta V_G  = \frac{5 \times {10}^{- 3}}{20/10 \times {10}^{- 3}}\]

\[\delta V_G  = \frac{5 \times {10}^{- 3}}{2 \times {10}^{- 3}} = \frac{5}{2} = 2 . 5\]

\[ V_G  =  + 2 . 5 - 7 . 5 =  - 5 . 0  V\]

\[r_p    =    \left( \frac{\delta V_p}{\delta I_p} \right)_{V_g = \text{Constant}} \]

\[ \Rightarrow    {10}^4  = \frac{\delta V_p}{\left(10 \times {10}^{- 3} - 15 \times {10}^{- 3} \right)}\]

\[ \Rightarrow   \delta V_p    =    -  {10}^4  \times 5 \times  {10}^{- 3}  =    - 50  V\]

\[ \therefore  V_p ' -  V_p    =    - 50\]

\[ \Rightarrow    V_p ' =  - 50 +  V_p  = 200  V\]

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Solution The Plate Resistance and Amplification Factor of a Triode Are 10kω and 20. the Tube is Operated at Plate Voltage 250 V and Grid Voltage −7.5 V. the Plate Current is 10 Ma. Concept: Triode.
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