Electric Current Through Gases
Questions for Short Answer
1. Why is conduction easier in gases if the pressure is low? Will the conduction continue to improve if the pressure is made as low as nearly zero?
ANSWER: When the pressure is low, the density of a gas is also low. So the free-conducting electrons move longer before colliding with another molecule. Thus they face less resistance to moving. So the conduction is easier at low pressure.
The conduction does not improve if the pressure is made as low as zero because there are very few molecules to get ionized. Also, the electrons emitted from the cathode have a mean free path greater than the length of the tube and they strike the walls of the tube, and the conduction stops.
2. An AC source is connected to a diode and a resistor in series. Is the current through the resistor AC or DC?
ANSWER: Since a diode allows current to flow only in one direction through it, the current through the resistor will be DC. But for each AC cycle, the DC current in the resistor will be only half the cycle, and the other half cycle will have zero current (when the current in AC is in the reverse direction).
3. How will the thermionic current vary if the filament current is increased?
ANSWER: If the filament current is increased, the temperature of the cathode will increase resulting in more emission of electrons. This will result in an increase in thermionic current.
4. Would you prefer a material having a high melting point or a low melting point to be used as a cathode in a diode?
ANSWER: Since the thermionic emission from a diode occurs at high temperatures, a high melting point material with good conductance will be preferred.
5. Would you prefer a material having a high work function or a low work function to be used as a cathode in a diode?
ANSWER: Since work function is the amount of energy used to detach an electron from an atom, a material with a low work function will be preferred so that the electrons needed for the current in a diode are available at the low expense of energy.
6. An isolated metal sphere is heated to a high temperature. Will it become positively charged due to thermionic emission?
ANSWER: Since the metal sphere is isolated, due to the thermionic emission of electrons the metal sphere will have a net positive charge remaining. So it will become positively charged.
7. A diode valve is connected to a battery and a load resistance. The filament is heated so that a constant current is obtained in the circuit. As the cathode continuously emits electrons, does it get more and more positively charged?
ANSWER: No. Since a constant current flows in the circuit, there is a continuous supply of electrons from the negative terminal of the battery/source to the cathode. So it does not get more and more positively charged.
8. Why does thermionic emission not take place in nonconductors?
ANSWER: For thermionic emission, a material should have a low work function so that free electrons are easily available. Nonconductors have very high work functions and electrons are not easily detached from the atoms. So the thermionic emission does not take place.
9. The cathode of a diode valve is replaced by another cathode of double the surface area. Keeping the voltage and temperature conditions the same, will the plate current decrease, increase, or remain the same?
ANSWER: According to the Richardson-Dushman equation, the number of thermionic electrons ejected by a metal is proportional to its surface area. Since the cathode of a diode valve emits thermions, doubling its surface area keeping the temperature and other conditions the same, double the thermions will be ejected. So the plate current will increase.
10. Why is the linear portion of the triode characteristic chosen to operate the triode as an amplifier?
ANSWER: In the linear portion of the triode characteristic, the change in the voltage across the load resistance is proportional to the input signal but with a much higher amplitude. This quite amplifies the input signal. That is why the linear portion of the triode characteristic is chosen for the operation of the amplifier.
OBJECTIVE-I
1. Cathode rays constitute a stream of
(a) electrons
(b) protons
(c) positive ions
(d) negative ions.
ANSWER: (a).
EXPLANATION: The cathode is connected to the negative terminal of the source. When a high potential difference is applied across the discharge tube, free electrons increase on the cathode, and at very low pressure in the tube, a beam of electrons rushes from the cathode toward the anode. This beam is called a cathode ray.
2. Cathode rays are passing through a discharge tube. In the tube, there is
(a) an electric field but no magnetic field
(b) a magnetic field but no electric field
(c) an electric as well as a magnetic field
(d) neither an electric field nor a magnetic field.
ANSWER: (c).
EXPLANATION: Cathode rays are itself a result of a high electric field present in the discharge tube. Also, cathode rays are beams of fast-moving electrons that also ionize gas molecules present in the tube. These charged particles have also their electric fields around them. Since a moving charged particle also produces a magnetic field around it, there is also a magnetic field present in the tube.
3. Let iₒ be the thermionic current from a metal surface when the absolute temperature of the surface is Tₒ. The temperature is slowly increased and the thermionic current is measured as a function of temperature. Which of the following plots may represent the variation in (i/iₒ) against (T/Tₒ). 
The figure for Q-3
ANSWER: (d).
EXPLANATION: From the Richardson-Dushman equation, the thermionic current is directly proportional to the square of the absolute temperature and directly proportional to e-φ/kT. Hence the current will increase exponentially with an increase in temperature and only plot (d) relates to it.
4. When the diode shows the saturated current, dynamic plate resistance is
(a) zero
(b) infinity
(c) indeterminate
(d) different for different diodes.
ANSWER: (b).
EXPLANATION: Dynamic plate resistance is the ratio of the change in plate voltage to the corresponding change in the current. Since at saturation, there is no increase in current with an increase in plate voltage, the dynamic plate resistance becomes infinity.
5. The anode of a thermionic diode is connected to the negative terminal of a battery and the cathode to its positive terminal.
(a) No appreciable current will pass through the diode.
(b) A large current will pass through the diode from the anode to the cathode.
(c) A large current will pass through the diode from the cathode to the anode.
(d) The diode will be damaged.
ANSWER: (a).
EXPLANATION: In this case, the electrons are pushed back towards the cathode, and no current flows in the external circuit. Hence option (a).
6. A diode, a resistor, and a 50 Hz AC source are connected in series. The number of current pulses per second through the resistor is
(a) 25
(b) 50
(c) 100
(d) 200.
ANSWER: (b).
EXPLANATION: Since the diode will allow current only in one direction, only half cycle of a cycle of AC source will show current as a current pulse. Hence the number of current pulses per second will be equal to the frequency of the AC source. Option (b) is correct.
7. A triode is operated in the linear region of its characteristics. If the plate voltage is slightly increased, the dynamic plate resistance will
(a) increase
(b) decrease
(c) remain almost the same
(d) become zero.
ANSWER: (c).
EXPLANATION: The dynamic plate resistance
rₚ =ΔVₚ/Δiₚ
In the linear region of the triode characteristic, ΔVₚ ≈ Δiₚ, hence rₚ will remain almost the same.
8. The plate current in a triode valve is maximum when the potential of the grid is
(a) positive
(b) zero
(c) negative
(d) nonpositive.
ANSWER: (a).
EXPLANATION: When the potential of the grid is made positive, it helps electrons emitted from the cathode to accelerate towards the plate. Thus increasing the current. Hence option (a) is correct.
9. The amplification factor of a triode operating in the linear region depends strongly on
(a) the temperature of the cathode
(b) the plate potential
(c) the grid potential
(d) the separations of the grid from the cathode and the anode.
ANSWER: (d).
EXPLANATION: When the triode is operating in the linear region of its characteristics, the plate voltage, grid voltage, and temperature of the cathode are already specified. It is the separation of the grid from the cathode or the anode that affects the amplification factor. If the grid is nearer to the cathode, the electrons are accelerated strongly. Option (d) is correct.
OBJECTIVE-II
1. Electric conduction takes place in a discharge tube due to the movement of
(a) positive ions
(b) negative ions
(c) electrons
(d) protons.
ANSWER: (a), (b), (c).
EXPLANATION: Due to cosmic rays, some ions are always present in a gas. When a potential difference is applied across a discharge tube, the ions are accelerated and collide with other molecules and strike the outer loose electrons. Thus the molecules become positive ions and the released electrons after going through some distance may attach to some neutral molecule making it a negative ion. Thus positive ions, negative ions, and electrons are present in the discharge tube that moves under the effect of the electric field and conduction takes place. No free proton is present in the discharge tube. Options (a), (b) and (c) are correct.
2. Which of the following is true for cathode rays?
(a) It travels along straight lines.
(b) It emits an X-ray when strikes a metal
(c) It is an electromagnetic wave.
(d) It is not deflected by a magnetic field.
ANSWER: (a), (b).
EXPLANATION: Cathode rays are actually a stream of fast-moving electrons. They travel in a straight line and when they strike a metal X-ray is emitted. Options (a) and (b) are true. It is not an electromagnetic wave and due to the moving charge, it is deflected by a magnetic field. The other two options are not correct.
3. Because of the space charge in a diode valve,
(a) the plate current decreases
(b) the plate voltage increases
(c) the rate of emission of thermions increases
(d) the saturation current increases.
ANSWER: (a).
EXPLANATION: The space between the cathode and anode contains electrons and hence, is negatively charged. It is called the space charge. New electrons coming out of the cathode are repelled back by this space charge which reduces the plate current. Option (a) is correct.
It does not affect the plate voltage, rate of emission of thermions, and saturation current.
4. The saturation current in a triode valve can be changed by changing
(a) the grid voltage
(b) the plate voltage
(c) the separation between the grid and the cathode
(d) the temperature of the cathode.
ANSWER: (d).
EXPLANATION: When all the electrons emitted by the cathode are collected by the anode, the saturation current occurs. At this point, whether we increase the plate voltage or grid voltage, the number of electrons is not going to increase, and hence the saturation current is constant. The only way to increase the number of electrons emitted by the cathode is to increase the temperature of the cathode so that the number of thermions increases. Hence only option (d) is correct.
5. Mark the correct options.
(a) a diode valve can be used as a rectifier.
(b) a triode valve can be used as a rectifier.
(c) A diode valve can be used as an amplifier.
(d) A triode valve can be used as an amplifier.
ANSWER: (a), (b), (d).
EXPLANATION: Since a diode valve and a triode valve allows the current to flow in one direction only, these two can be used as a rectifier. Options (a) and (b) are correct.
Due to the presence of a grid in a triode valve, the output signal is controlled in proportion to the input signal. Hence it can be used as an amplifier. In a diode valve, there is no grid and hence the output signal cannot be controlled. Option (d) is correct, and option (c) is not correct.
6. The plate current in a diode is zero. It is possible that
(a) the plate voltage is zero
(b) the plate voltage is slightly negative
(c) the plate voltage is slightly positive
(d) the temperature of the filament is low.
ANSWER: All.
EXPLANATION: If the plate voltage is zero, then there will be no current in the diode. Hence option (a) is possible.
If the plate voltage is negative, then there will be no plate current because the diode does not allow current in the reverse direction. Option (b) is possible.
When the plate voltage is slightly positive, it may not be able to overcome the effects of space charge. In this case, also, the plate current will be zero. Option (c) is possible.
When the temperature of the filament is low, there will not be sufficient thermions emission and the plate current will be zero. Option (d) is also possible.
So all options are correct.
7. The plate current in a triode valve is zero. the temperature of the filament is high. It is possible that
(a) Vᵨ>0, Vₚ>0
(b) Vᵨ>0, Vₚ<0
(c) Vᵨ<0, Vₚ>0
(d) Vᵨ<0, Vₚ<0.
ANSWER: (b), (c), (d).
EXPLANATION: When the temperature of the filament is high and both the plate voltage and the grid voltage are positive, the electrons will travel from cathode to anode and the current will not be zero. Option (a) is not correct.
If the plate voltage is negative but the grid voltage is positive or vice versa, then the electrons will not be able to reach the anode and the plate current will be zero. Options (b) and (c) are correct.
If both the plate voltage and the grid voltage are negative, then the thermions from the filament will be repelled towards the cathode itself and no plate current will appear. Option (d) is correct.
Exercises
1. A discharge tube contains helium at low pressure. A large potential difference is applied across the tube. Consider a helium atom that has just been ionized due to the detachment of an atomic electron. Find the ratio of the distance traveled by the free electron to that by the positive ion in a short time dt after the ionization.
ANSWER: Let the electric field near the ionized Helium atom be E. Due to the detachment of an atomic electron, there will be a positive charge equal to 'e' on the ion and a negative charge on the detached electron equal to '-e'. So both will experience an equal and opposite electric force,
F =eE
Let the mass of the electron =mₑ and mass of the ion =mₘ, then
Acceleration of the electron,
aₑ =Force/mass =eE/mₑ
Distance traveled by this electron in time dt,
S =½aₑ(dt)²
And the acceleration of the ion,
aₘ =eE/mₘ
So the distance traveled by the ion,
S' =½aₘ(dt)²
So the required ratio is
S/S' =aₑ/aₘ
=(eE/mₑ)/(eE/mₘ)
=mₘ/mₑ
Now, mₘ =mass of two protons and two neutrons, neglecting the mass of the remaining electron.
≈4*1.67x10⁻²⁷ kg
=6.68x10⁻²⁷ kg
and mₑ =9.1x10⁻³¹ kg
Hence the required ratio
S/S' =mₘ/mₑ
=6.68x10⁻²⁷/9.1x10⁻³¹
=66800/9.1
=7340.
2. A molecule of gas, filled in a discharge tube, gets ionized when an electron is detached from it. An electric field of 5.0 kV/m exists in the vicinity of the event. (a) Find the distance traveled by the free electron in 1 µs assuming no collision. (b) If the mean free path of the electron is 1.0 mm, estimate the time of transit of the free electron between successive collisions.
ANSWER: (a) Force on the electron,
F =eE
Acceleration of the electron if its mass =m,
a =F/m =eE/m
The distance traveled by the electron in time t,
S = ut+½at²
=0 +½(eE/m)t²
=½*(1.6x10⁻¹⁹*5000/9.1x10⁻³¹)*(10⁻⁶)² m
=440 m.
(b) Let the time of transit of free electrons between successive collisions =t. From,
S =½at²
Here, S =1.0 mm =0.001 m
a =eE/m
=1.6x10⁻¹⁹*5000/9.1x10⁻³¹ m/s²
=8.79x10¹⁴ m/s²
Hence,
0.001 =½*8.79x10¹⁴*t²
→t² =0.002/(8.79x10¹⁴)
=2.28x10⁻¹⁸
→t =1.50x10⁻⁹ s
=1.50 ns.
3. The mean free path of electrons in the gas in a discharge tube is inversely proportional to the pressure inside it. The Crookes dark space occupies half the length of the discharge tube when the pressure is 0.02 mm of the mercury. Estimate the pressure at which the dark space will fill the whole tube.
ANSWER: In the Crookes dark space the electrons emitted from the cathode travel a distance equal to the mean free path before they collide and cause the cathode to glow. So according to the question, the length of the Crookes dark space is inversely proportional to the pressure inside the given discharge tube. Let the length of the Crookes dark space = L and the pressure inside the discharge tube =p, then
L ∝ 1/p
So for the two cases when Crookes's dark spaces are L₁ and L₂ and the corresponding pressures are p₁ and p₂,
L₁ ∝ 1/p₁ and
L₂ ∝ 1/p₂. Hence
L₁/L₂ =p₂/p₁.
Let the length of the discharge tube =l. Given, L₁ = l/2 and p₁ =0.02 mm of mercury, L₂ =l then p₂ =?
So,
(l/2)/l =p₂/0.02
→p₂ =0.02/2 =0.01 mm of mercury.
4. Two discharge tubes have identical material structures and the same gas is filled in them. The length of one tube is 10 cm and that of the other tube is 20 cm. Sparking starts in both tubes when the potential difference between the anode and cathode is 100 V. If the pressure in the shorter tube is 1.0 mm of mercury, what is the pressure in the longer tube?
ANSWER: According to Paschen's law, the sparking potential in discharge tubes is a function of the product of the pressure of the gas and the length of the tube. So,
Sparking potential V=f(pd)
Since the sparking potential in the given case is the same for both tubes, hence products of pressure and length in both tubes must be equal. Thus,
p₁d₁ =p₂d₂
→p₂ =p₁d₁/d₂
=(1.0 mm)*(10 cm)/(20 cm)
=0.50 mm of mercury.
5. Calculate n(T)/n(1000 K) for tungsten emitter at T = 300 K, 2000 K, and 3000 K where n(T) represents the number of thermions emitted per second by the surface at temperature T. Work function of tungsten is 4.52 eV.
ANSWER: From the Richardson-Dushman equation, the thermionic current is given as
i =n(T)*e' =AST²e-φ/kT
where e' is the charge on an electron and 'e' is the exponential e. S is the surface area, A is a constant related to the nature of the metal, T is the absolute temperature of the surface, φ is the work function of the metal and k is the Boltzmann constant. Considering two temperatures of the metal surface T and T', the ratio of the number of thermions emitted,
n(T)/n(T') =(T/T')²eφ/kT'-φ/kT
=(T/T')²e(1/T'-1/T)φ/k
Here T' =1000 K,
φ =4.52 eV =4.52*1.6x10⁻¹⁹ J
k =1.38x10⁻²³ J/K
Hence φ/k =5.24x10⁴ K
So,
n(T)/n(1000 K) =(T/1000)²e5.24x10⁴(1/1000 -1/T)
Thus,
n(300 K)/n(1000 K)=(0.3)²e5.24x10⁴*(-7/3000)
=0.09*e^-122.3
=6.91x10⁻⁵⁵
n(2000 K)/n(1000 K) =2²*e5.24x10⁴*0.0005
=4*e26.2
=9.56x10¹¹.
And for T =3000 K,
n(3000 K)/n(1000 K) =3²*e5.24x10⁴(2/3000)
=9*e34.93
=1.33x10¹⁶
6. The saturation current from a thoriated-tungsten cathode at 2000 K is 100 mA. What will be the saturation current for a pure tungsten cathode of the same surface area operating at the same temperature? The constant A in the Richardson-Dushman equation is 60x10⁴ A/m²-K² for pure tungsten and 3.0x10⁴ A/m²-K² for thoriated-tungsten. The work function of pure tungsten is 4.5 eV and that of the thoriated tungsten is 2.6 eV.
ANSWER: From the Richardson-Dushman equation, saturation current,
i =AST²e-φ/kT
In the case of Thoriated Tungsten Cathode, i =100 mA =0.1 A, T =2000 K, A =3.0x10⁴ A/m²-K². Hence,
0.1 =3.0x10⁴S*(2000)²e-2.6x1.6x10⁻¹⁹/kT --(i)
For the pure Tungsten Cathode,
A=60x10⁴ A/m²-K², φ =4.5 eV. So the saturation current,
i =60x10⁴S*(2000)²e-4.5x1.6x10⁻¹⁹/kT -- (ii)
Dividing (ii) by (i),
i/0.1 =20*e(-1.9x1.6x10⁻¹⁹)/kT
{Here, k =1.38x10⁻²³ J/K, T =2000 K}
→i =2*e-11.01 A =3.3x10⁻⁵ A
=33x10⁻⁶ A =33 µA.
7. A tungsten cathode and a thoriated-tungsten cathode have the same geometrical dimensions and are operated at the same temperature. The thoriated-tungsten cathode gives 5000 times more current than the other one. Find the operating temperature. Take relevant data from the previous problem.
ANSWER: For the Tungsten Cathode, current
i =AST²e-φ/kT
For the Thoriated Tungsten cathode, current
i' =A'ST²e-φ'/kT
But given that, i' =5000i. So by taking the ratio,
i'/i =(A'/A)e(φ-φ')/kT
→5000 =(3/60)e(4.5-2.6)*1.6x10⁻¹⁹/kT
→e1.9*1.6x10⁻¹⁹/kT =20*5000 =10⁵
Taking the log of both sides, we get
1.9*1.6x10⁻¹⁹/kT =5*ln (10) =11.513
→kT =3.04x10⁻¹⁹/11.513
→T =3.04x10⁻¹⁹/(11.513*1.38x10⁻²³)
=1913.4 K.
8. If the temperature of the tungsten filament is raised from 2000 K to 2010 K, by what factor does the emission current change? The work function of tungsten is 4.5 eV.
ANSWER: T =2000 K, T' =2010 K.
Emission current for T',
i' =AST'²e-φ/kT',
and for T,
i =AST²e-φ/kT.
Hence the factor by which the emission current change is,
i'/i =(T'/T)²e(1/T -1/T')φ/k
=(2010/2000)²e(1/2000-1/2010)4.5*1.6x10⁻¹⁹/1.38x10⁻²³
=1.01*e0.13
=1.15.
9. The constant A in the Richardson-Dushman equation for tungsten is 60x10⁴ A/m²-K². The work function of tungsten is 4.5 eV. A tungsten cathode having a surface area of 2.0x10⁻⁵ m² is heated by a 24 W electric heater. In a steady state, the heat radiated by the cathode equals the energy input by the heater and the temperature becomes constant. Assuming that the cathode radiates like a black body, calculate the saturation current due to thermions. Take Stefen constant =6x10⁻⁸ W/m²-K⁴. Assume that the thermions take only a small fraction of the heat supplied.
ANSWER: The energy given to the cathode by the heater is radiated fully in a steady state. Since the cathode radiates like a black body, its emissivity is 1. From the Stefens law of black body radiation, power radiated by a black body is,
P =𝞂AT⁴
→T⁴ =P/𝞂A
=24/(6x10⁻⁸*2.0x10⁻⁵)
=2x10¹³ =20x10¹²
→T =√(20 *10³
=2114.7 K
Saturation current due to thermions,
i =AST²e-φ/kT
=60x10⁴*2x10⁻⁵*(2114.7)²*e-4.5*1.6x10⁻¹⁹/(1.38x10⁻²³*2114.7) A
=12*4.472x10⁶*e-24.672 A
=0.001 A
=1.0 mA.
10. A plate current of 10 mA is obtained when 60 volts are applied across the diode tube. Assuming the Langmuir-Child equation iₚ ∝ Vₚ3/2 to hold, find the dynamic resistance rₚ in this operating condition.
ANSWER: The dynamic resistance is given as,
rₚ =dVₚ/diₚ.
From the Langmuir-Child equation,
iₚ =kVₚ3/2, ----------- (i)
where k is the constant of proportionality. Differentiating both sides,
diₚ =k*(3/2)Vₚ1/2 *dVₚ ------(ii)
Dividing (ii) by (i),
diₚ/iₚ =(3/2)Vₚ⁻¹ *dVₚ
→dVₚ/diₚ =(2/3)Vₚ/iₚ
→rₚ =(2/3)*60/(10x10⁻³) Ω
=4000 Ω
=4 kΩ.
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 ∝ iₚ1/3i.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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