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Question

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 = -20 V

(a) As given in the question, plate current varies as,
ip= 41(Vp+7 Vg)1.41ip= 41(250-140)1.41ip=41×(110)1.41ip=30984.71 μA=30.98 mA

(b) ip=41(Vp+7VG)1.41
Differentiating this equation, we get:
diP = 41×1.41×(Vp+7Vg)0.41×(dVp+7dVg) (1)Plate resistance is defined as: rp= dVpdip Vg = ConstantFrom equation (1), dVpdip=1×10641×1.41×1100.41dVpdip=106×2.51×10-3dVpdip=2.5×103 Ω=2.5

(c) From above:
As dIP=41×1.41×(250+7×(-20))0.41×7dVg,gm = (dIpdVg)VP = ConstantFrom equation (1), 17dIpdVgVP = Constant =41×1.41×(250+7×(-20))0.41dIpdVgVP = Constant= 41×1.41×(110)0.41×7 mhodIpdVgVP = Constant= 41×1.41×6.87×7 mhodIpdVgVP = Constant=2.78 milli mho
(d) Amplification factor,
μ = rp×gm= 2.5×103×2.78×10-3= 6.957

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