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.

Answer 1

EXPLANATION:

In the question, it is given:

Plate voltage, VP = 250 V

Grid voltage, VG = -20 V

• (a) The plate current varies as shown in the question,

ip = 41(250-140)1.41ip = 41 × (110)1.41ip = 30.98 mA

• (b) ip = 41(Vp+7VG)1.41

When this equation is differentiated, we obtain:

diP = 41 × (Vp+7Vg)0.41 x 1.41× (dVp+7dVg)                          (1)

Plate resistance: rp = Constant = dVpdip Vg  

From equation (1), dVpdip =106 × 2.51 × 10-3dVpdip = 2.5 KΩ

• (c) From above:

As dIP = 41 × (250+7 × (-20)) 0.41 ×1.41 × 7dVg,

gm = (dIpdVg)VP = Constant

From equation (1), 17dIpdVgVP = Constant = 41×1.41×(250+7×(-20))0.41dIpdVgVP  

= 41 × 1.41 × (110)0.41 × 7 mhodIpdVgVP  

= Constant = 2.78 milli mho

• (d) Amplification factor,

μ = rp × gm = 6.95, which is 7.