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The plate current, plate voltage and grid voltage of a 6F6 triode tube are related as
ip=41(Vp+7Vg)1.41
Where Vp and Vg are in volts and ip in microamperes. The tube is operated at Vp=250V,Vg=20V. Calculate (a) The tube current (b) the plate resistance, (c) the mutual conductance and (d) the amplification factor.

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Solution

Given: Vp=250V;V2=20VIp=41(Vp+7Vg)1.41=41(250140)1.41=41×(110)1.41=30984μA=30mA(b)Ip=41(Vp+7V0)1.41
Differentiating now, we get
=dip=41×1.41×(Vp+7V4)0.41×(dVp+7dVg)Now,rp=dVpdipordVpdip=1×10641×1.41×1100.41=106×2.51×103=2.5×108Ω=2.5KΩ.
(c) From Above,
dI2=41×1.41×6.87×7dvggm=dIpdVg=41×41×6.87×7μmho=2780μmho=2.78milli mho
(d) Amplification factor
μ=rp×gm=2.5×102×2.78×103=6.957


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