SATHEE: Semiconducting Devices Question 63

Semiconducting Devices Question 63

Question: The plate current in a triode is given by $I_{p} = 0.004 {(V_{p} + 10 V_{g})}^{3 / 2} m A$ where Ip, Vp and Vg are the values of plate current, plate voltage and grid voltage, respectively. What are the triode parametersm, rp and gm for the operating point at $V_{p} = 120 v o l t$ and $V_{g} = - 2 v o l t$ ?

Options:

A) 10, 16.7 kW, 0.6 m mho

B) 15, 16.7 kW, 0.06 m mho

C) 20, 6 kW, 16.7 m mho

D) None of these

Show Answer

Answer:

Correct Answer: A

Solution:

$I_{p} = 0.004 {(V_{p} + 10 V_{g})}^{3 / 2}$

Therefore $\frac{Δ I_{p}}{Δ V_{g}} = 0.004 [\frac{3}{2} {(V_{p} + 10 V_{g})}^{1 / 2} \times 10]$

Therefore $g_{m} = 0.004 \times \frac{3}{2} {(120 + 10 \times - 2)}^{1 / 2} \times 10$

Therefore $g_{m} = 6 \times 10^{- 4} m h o = 0.6 m m h o$

Comparing the given equation of Ip with standard equation $I_{p} = K {(V_{p} + μ V_{g})}^{3 / 2}$

we get m = 10 Also from m = rp x gm

Therefore $r_{p} = \frac{μ}{g_{m}} = \frac{10}{0.6 \times 10^{- 3}}$

Therefore $r_{p} = 16.67 \times 10^{3} Ω = 16.67 k Ω .$

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Semiconducting Devices Question 63

Options:

Answer:

Solution:

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Semiconducting Devices Question 63

Question: The plate current in a triode is given by Ip=0.004 (Vp+10Vg)3/2mA where Ip, Vp and Vg are the values of plate current, plate voltage and grid voltage, respectively. What are the triode parametersm, rp and gm for the operating point at Vp=120volt and Vg=−2volt ?

Options:

Answer:

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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