The plate resistance and the amplification factor of a triode are $10 K Ω$ 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 votage be changed to take the plate current back to 10 mA?

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Solution

$r_{p} = 10 K Ω = 10^{3} Ω μ = 20, V_{p} = 250 V, V_{0} = 7.5 V, I_{p} = 10 m A (a) g_{m} = (\frac{δ I_{p}}{δ V_{g}}) δ V_{p} = \frac{δ I_{p}}{δ V_{g}} = \frac{(15 \times 10^{- 3} - 10 \times 10^{- 3})}{\frac{μ}{r_{p}}} = \frac{5 \times 10^{- 3}}{\frac{20}{10} \times 10^{- 3}} = \frac{5 \times 10^{- 3}}{2 \times 10^{- 3}} = \frac{5}{2} = 2.5 r_{0}^{'} = + 2.5 - 7.5 = - 5 V .$
$r_{p} = (\frac{δ V_{p}}{δ I_{p}}) \Rightarrow 10^{4} = \frac{δ V_{p}}{(15 \times 10^{- 3} - 10 \times 10^{- 3})} \Rightarrow δ V_{p} = 10^{4} \times 5 \times 10^{- 3} = 50 V V_{p}^{'} - V_{p}^{p} = 50 \Rightarrow V - P = - 50 + V_{p} = 200 V$

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rp=10KΩ=103Ωμ=20,Vp=250V,V0=7.5V,Ip=10mA(a)gm=(δIpδVg)δVp=δIpδVg=(15×10−3−10×10−3)μrp=5×10−32010×10−3=5×10−32×10−3=52=2.5r′0=+2.5−7.5=−5V. rp=(δVpδIp)⇒104=δVp(15×10−3−10×10−3)⇒δVp=104×5×10−3=50VV′p−Vpp=50⇒V−P=−50+Vp=200V