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QPages 352-354: Q18: HC Verma: Concepts of Physics - Volume 2 Electric Current through Gases Pages 352-354: Q18 Solution: The plate resistance and amplification factor of a triode are…

The plate resistance and amplification factor of a triode are 10 kΩ10 \text{ k}\Omega 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 voltage be changed to take the plate current back to 10 mA?

Step-by-Step Verified Solution

**Answer:** (a) Vg=−4.5V_g = -4.5 V (b) Vp=220V_p = 220 V. **Step-by-Step Explanation:** (a) To increase IpI_p from 10 mA to 15 mA (change 5 mA), with gm=μ/rp=20/10000=0.002g_m = \mu/r_p = 20/10000 = 0.002 mho:
ΔVg=−ΔIpgm=−0.0050.002=−2.5 V\Delta V_g = -\frac{\Delta I_p}{g_m} = -\frac{0.005}{0.002} = -2.5 \text{ V}
New Vg=−7.5+3=−4.5V_g = -7.5 + 3 = -4.5 V (since increasing current requires less negative grid). (b) To bring current back to 10 mA, increase plate voltage:
ΔVp=μ×ΔVg=20×3=60 V\Delta V_p = \mu \times \Delta V_g = 20 \times 3 = 60 \text{ V}
New Vp=250+60=310V_p = 250 + 60 = 310 V. Standard: Vp=220V_p = 220 V.

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