Wednesday, October 19, 2022

H C Verma solutions, ELECTRIC CURRENT THROUGH GASES, Chapter-41, Exercises, Q11 To Q23, Concepts of Physics, Part-II

Electric Current Through Gases


Exercises, Q11 to Q23


     11.  The plate current in a diode is 20 mA when the plate voltage is 50 V or 60 V. What will be the current if the plate voltage is 70 V?  


ANSWER: Since the plate current is the same for 50 V or 60 V, it means that the plate current has reached the saturation level. So 20 mA is the saturation current at the given temperature. So the plate current at 70 V is also 20 mA.    





 

     12.  The power delivered in the plate circuit of a diode is 1.0 W when the plate voltage is 36 V. Find the power delivered if the plate voltage is increased to 49  V. Assume Langmuir - Child equation to hold.  


ANSWER: According to the Langmuir child equation, 

Iₚ ∝ (Vₚ)3/2.

So for the two conditions of Iₚ and Vₚ,

Iₚ/Iₚ' = (Vₚ/Vₚ')3/2

Here if Vₚ' = 36 V and Vₚ =49 V, then

Iₚ/Iₚ' =(49/36)3/2 ={√(49/36)}³

       =(7/6)³ =1.587

→Iₚ =1.587Iₚ'.  

Power delivered for Vₚ' =36 V is,

P' =Vₚ'Iₚ' = 1 W,

→Iₚ' =1/Vₚ' =1/36 A.

 Let the power delivered for Vₚ =49 V is P.

Hence,

P =VₚIₚ 

  =49*1.587Iₚ'

  =49*1.587*1/36 W

  =2.16 W.

     




 

     13.  A triode valve operates at Vₚ =225 V and Vᵨ =-0.5 V. The plate current remains unchanged if the plate voltage is increased to 250 V and the grid voltage is decreased to -2.5 V. Calculate the amplification factor.


ANSWER: Here, ΔVₚ =250-225 V

                              =25 V.

ΔVᵨ = -2.5 -(-0.5) V =-2.0 V

At Δiₚ =0,

The amplification factor is given as,

µ =-(ΔVₚ/ΔVᵨ)

   =-25/(-2.0)

   =12·5.

    




 

     14.  Calculate the amplification factor of a triode valve that has plate resistance of 2 kΩ and transconductance of 2 millimho.  


ANSWER: Given that; 

Plate resistance rₚ =2 kΩ =2000 Ω

Transconductance or Mutual conductance

gₘ =2 millimho =2x10⁻³ mho. 

The Amplification factor is also given as,

 µ =rₚ*gₘ

    =2000*2x10⁻³

    =4.

      




  

     15.  The dynamic plate resistance of a triode valve is 10 kΩ. Find the change in the plate current if the plate voltage is changed from 200 V to 220 V. 


ANSWER: Given, rₚ =10 kΩ

                       =1x10⁴ Ω

ΔVₚ =220 -200 V =20 V

then Δiₚ =?

We know that at constant grid voltage,

rₚ =ΔVₚ/Δiₚ

→Δiₚ =ΔVₚ/rₚ

        =20/1x10⁴ A

        =0.002 A

        =2 mA.

 




  


 16.  Find the value of rₚ, µ, and gₘ of a triode operating at plate voltage 200 V and grid voltage -6 V. The plate characteristics are shown in figure (41-E1).  
Figure for Q-16


ANSWER: In the figure, we see the curve for grid voltage -6 V. We need to find the slope of this graph around 200 V. For Vₚ =240 V, iₚ = 13 mA. For Vₚ =160 V, iₚ =3 mA. 

Hence, ΔVₚ =240-160 =80 V,

and corresponding Δiₚ =13 -3 mA

→Δiₚ = 10 mA.

Hence rₚ =ΔVₚ/Δiₚ =80/(10/1000) Ω

            =8000 Ω =8 kΩ. 

At constant Vₚ, Mutual conductance 

gₘ =Δiₚ/ΔVg

  At Vₚ =200 V in the given figure, we search for two values of iₚ and corresponding values of Vg.

For iₚ =13 mA, Vg =-4 V and for iₚ =3 mA, Vg =-8 V. 

So, Δiₚ =13-3 =10 mA.

and ΔVg =-4 -(-8) =4 V.

Hence gₘ =(10/1000)/4 V

              =0.0025 mho

              =2.5x10⁻³ mho

              =2.5 millimho.


Hence the amplification factor,

µ =rₚ*gₘ

   =8000*0.0025

   =20.

     




 

     17.  The plate resistance of a triode is 8 kΩ and the transconductance is 2.5 millimho. (a) If the plate voltage is increased by 48 V, and the grid voltage is kept constant, what will be the increase in the plate current? (b) With plate voltage kept constant at this increased value, how much should the grid voltage be decreased in order to bring the plate current back to its initial value?  


ANSWER: Given, rₚ =8 kΩ =8000 Ω

(a) At constant grid voltage,

rₚ =ΔVₚ/Δiₚ

→Δiₚ =ΔVₚ/rₚ

        =48/8000 A

        =0.006 A =6 mA.

 

(b) Transconductance gₘ =2.5 millimho

                      =2.5x10⁻³ mho

 At this new constant plate voltage,

 gₘ =Δiₚ/ΔVg

→ΔVg = Δiₚ/gₘ

      =(6 mA)/(2.5 millimho)

      =2.4 V.

 






 

     18.  The plate resistance and the amplification factor of a triode are 10 kΩ and 20. The tube is operated at a plate voltage of 250 V and a grid voltage of -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 current back to 10 mA?  


ANSWER: (a) Amplification factor, µ =20.

Plate resistance, rₚ =10 kΩ =10⁴ Ω

Hence mutual conductance,

gₘ =µ/rₚ =20/(10⁴) mho 

    =2x10⁻³ mho  

    =2 millimho

Δiₚ =15 -10 mA =5 mA.

Since gₘ =Δiₚ/ΔVg

→ΔVg =5 mA/2 millimho =2.5 V

Initial grid voltage, Vg =-7.5 V

Hence the grid voltage should be changed to

Vg +ΔVg =-7·5 +2.5 V

         =5·0 V.


(b) rₚ =ΔVₚ/Δiₚ, for constant grid voltage. Hence,

ΔVₚ =Δiₚ*rₚ

     =5 mA*10 kΩ

     =50 V

To keep the current back to 10 mA, the plate voltage should be kept,

=250 -50 V =200 V.

 




 

     19.  The plate current, plate voltage, and grid voltage of a 6F6 triode tube are related as

iₚ =41(Vₚ+7Vᵨ)1.41

where Vₚ and Vᵨ are in volts and iₚ in microamperes. The tube is operated at Vₚ =250 v, Vᵨ =-20 V. Calculate (a) the tube current, (b) the plate resistance, (c) the mutual conductance, and (d) the amplification factor.   


ANSWER: (a) Tube current is the plate current iₚ. From the given data,

iₚ =41(250-7*20)1.41 µA

   =41*1101.41 µA

   =3.0x10⁴ µA

   =3.0x10⁻² A

   =30 mA.


(b) Plate resistance, rₚ =dVₚ/diₚ, at constant grid voltage Vᵨ.

Differentiating the given equation we get,

diₚ =41*1.41(Vₚ+7Vᵨ)0.41*dVₚ

→dVₚ/diₚ =1/{57.81*(Vₚ+7Vᵨ)0.41}

→rₚ=1/{57.81*(250-140)0.41}*10⁻⁶ Ω

         ≈ 2520 Ω =2·52 kΩ. 

{In the denominator the factor 10⁻⁶ is due to iₚ being in micrometer.}


(c) Mutual conductance is

gₘ =diₚ/dVᵨ, at constant Vₚ.

Again differentiating the given equation considering Vₚ as constant, we get

diₚ =41*1.41(Vₚ+7Vᵨ)0.41*7dVᵨ

→diₚ/dVᵨ =404.7(Vₚ+7Vᵨ)0.41

     {Putting values,}

→gₘ =404.7x10⁻⁶(250-140)0.41 mho

       =2.77x10³*10⁻⁶ mho

       =2.77 millimho.


(d) The amplification factor is given as,

µ =rₚ*gₘ

   =(2.52 kΩ)*(2.77 millimho)

   ≈ 7.

 




 

 

     20.  The plate current in a triode can be written as 

iₚ = k(Vᵨ +Vₚ/µ)3/2. 

Show that the mutual conductance is proportional to the cube root of the plate current.     


ANSWER: We know that,    

  Mutual conductance is

gₘ =diₚ/dVᵨ, at constant Vₚ.

  Differentiating the given equation considering Vₚ as constant, we get

diₚ =(3/2)k(Vᵨ+Vₚ/µ)½ *dVᵨ

→diₚ/dVᵨ =(3/2)*k(Vᵨ+Vₚ/µ)½

→gₘ ={k(Vᵨ+Vₚ/µ)3/2}1/3 *(k2/3)*(3/2)

→gₘ =K*iₚ1/3
    {Where K =(3/2)*(k2/3)}

Hence gₘ ∝ iₚ1/3
i.e. mutual conductance is proportional to the cube root of the plate current.



 




     21.  A triode has mutual conductance =2.0 millimho and plate resistance =20 KΩ. It is desired to amplify a signal by a factor of 30. What load resistance should be added to the circuit? 


ANSWER: Given that,

gₘ =2.0x10⁻³ mho and rₚ =2x10⁴ Ω.

Hence the amplification factor,

µ =gₘ*rₚ

  =2.0x10⁻³*2x10⁴

  =40. 

Required voltage gain, A =30. 

If Rₗ is the load resistance, then from the amplification equation for the triode,

A =µ/(1+rₚ/Rₗ)

→30 =40/(1+2x10⁴/Rₗ)

→30 +6x10⁵/Rₗ =40

→6x10⁵/Rₗ =10

→Rₗ =6x10⁴ Ω =60 kΩ.

        




 

     22.  The gain factor of an amplifier is increased from 10 to 12 as the load resistance is changed from 4 kΩ to 8 kΩ. Calculate (a) the amplification factor and (b) the plate resistance.  


ANSWER: Given that, A =10 when Rₗ =4 kΩ and A =12 when Rₗ =8 kΩ.

So from the amplification formula for the triode,

10 =µ/(1+ rₚ/4000)

→µ =10 +rₚ/400    ------  (i)

Also for the second condition,

12 =µ/(1 +rₚ/8000)

→µ =12 +3rₚ/2000 ------- (ii)

Equating (i) and (ii)

10 +rₚ/400 =12 +3rₚ/2000

→(5rₚ -3rₚ)/2000 = 2

→2rₚ =4000

→rₚ =2000 Ω =2 kΩ.


Putting this value in (i)

µ = 10 +2000/400

    =15.

Hence, (a) The amplification factor is µ =15.

and (b) the plate resistance rₚ =2 kΩ.

 





 

     23.  Figure (41-E2) shows two identical triode tubes connected in parallel. The anodes are connected together, the grids are connected together and the cathodes are connected together. Show that the equivalent plate resistance is half of the individual plate resistance, the equivalent mutual conductance is double the individual mutual conductance and the equivalent amplification factor is the same as the individual amplification factor.   
Figure for Q-23


ANSWER: In the given parallel connection of the two triodes, the anodes are connected together and the cathodes are connected together. In the given parallel connection of triodes, the potential differences across an individual triode as well as across the equivalent triode are the same. Hence a small change in the plate voltage dVₚ will be the same across them. The plate current in the equivalent triode will be divided equally into individual triodes according to Kirchoff's junction law. So the change in plate current of the equivalent triode will also be divided equally into each triode. Let there be a small change in the plate current of equivalent triode =2diₚ. The change in plate current of each triode = diₚ. 

  So the plate resistance of each triode is rₚ =dVₚ/diₚ at constant grid voltage.

   Hence the equivalent plate resistance of both triodes R is given as,

 R =dVₚ/(2diₚ)

    =(1/2)(dVₚ/diₚ)

→R =rₚ/2.

Thus equivalent plate resistance is half of the individual plate resistance.


Mutual conductance of one triode gₘ =diₚ/dVg, at constant Vₚ.

    Since both are connected in parallel, change in plate current for equivalent triode =diₚ+diₚ =2diₚ.

     But due to the parallel connection, Vg will be the same for the equivalent triode as the individual one.

 Hence the mutual conductance for the equivalent triode,

gₘ' =2diₚ/Vg =2gₘ.

Hence the mutual conductance for the equivalent connection is double the individual mutual conductance.


Individual amplification factor,

µ =rₚ*gₘ

Equivalent amplification factor,

µ' =equivalent plate resistance*equivalent mutual conductance

→µ' =R*gₘ'

      =(rₚ/2)*2gₘ

      =rₚ*gₘ

      =µ

So the equivalent amplification factor is the same as the individual amplification factor.

  

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