Chapter 25 Valve & Digital Electronic 1 free study material by TEACHING CARE online tuition and coaching classes
Do You Know.
- Electronics can be devided in two categories
- Valve electronics (ii) Semiconductor electronics
- Free electron in metal experiences a barrier on surface due to attractive Coulombian
- When kinetic energy of electron becomes greater than barrier potential energy (or binding energy Eb ) then electron can come out of the surface of
(4) Fermi energy (Ef)
Is the maximum possible energy possessed by free electron in metal at 0K temperature
- In this energy level, probability of finding electron is 50%.
- This is a reference level and it is different for different
(5) Threshold energy (or work function W0)
Is the minimum energy required to take out an electron from the surface of metal. Also W0 = Eb – Ef
Work function for different materials
= 4.5 eV
(W0)Throated tungsten = 2.6 eV
(W0)Oxide coated tungsten = 1 eV
(6) Electron emission
Four process of electron emission from a metal are
- Thermionic emission (ii) Photoelectric emission (iii) Field emission (iv) Secondary emission
Thermionic Emission and Emitters.
(1) Thermionic emission
- The phenomenon of ejection of electrons from a metal surface by the application of heat is called thermionic emission and emitted electrons are called thermions and current flowing is called thermion
- Thermions have different
- This was discovered by Edison
- Richardson – Dushman equation for current density (e. electric current emitted per unit area of metal
surface) is given as
J = AT 2e
– W0 / kT
= AT 2e
- qV kT
= AT 2e
– 11600V T
where A = emission constant =
W0 = work function.
12 ´ 104 amp/ m2–K2 , k = Boltzmann’s constant, T = Absolute temp and
- The number of thermions emitted per second per unit area (J) depends upon following :
J µ T 2
J µ e –W0
(2) Thermionic emitters
The electron emitters are of two types
Note : @A good emitter should have low work function, high melting point, high working temperature, high electrical and mechanical strength.
Vacuum Tubes and Thermionic Valves.
- Those tubes in which electrons flows in vacuum are called vacuum
- These are also called valves because current flow in them is
- Vacuum in vacuum tubes prevents the emission of secondary
- Every vacuum tube necessarily contains two electrodes out of which one is always electron emitter (cathode) and another one is electron collector (anode or plate).
- Depending upon the number of electrodes used the vacuum tubes are named as diode, triode, tetrode, pentode…. respectively, if the number of electrodes used are 2, 3, 4, 5…..
Inventor : Fleming
Principle : Thermionic emission Number of electrodes : Two
Working : When plate potential ( Vp ) is positive, plate current ( ip ) flows in the circuit (because some emitted
electrons reaches to plate). If + Vp
also increases and finally becomes maximum (saturation).
Note : @If Vp ®
Negative; No current will flow
@ If Vp ®
(1) Space charge
Zero; current flows due to very less number of highly energised electrons
If Vp is zero or negative, then electrons collect around the plate as a cloud which is called space charge. space
charge decreases the emission of electrons from the cathode.
(2) Characteristic curve of a diode
A graph represents the variation of ip
at a given filament current ( i f ) is known as characteristic curve.
The curve is not linear hence diode valve is known as non- ohmic device.
- Space charge limited region (SCLR) : In this region current is space charge limited Also
ip µ V 3 / 2 Þ
i = kV 3 / 2 ; where k is a constant depending on metal as well as on the shape and area of the
cathode. This is called child’s law.
- Linear region (LR) : ip µ Vp
- Saturated region or temperature limited region : In this part, the current is independent of potential difference applied between the cathode and
ip ¹ f(Vp )
ip = f
The saturation current follows Richardson Dushman equation i.e. i = AT 2e –f / kT
Note : @ The small increase in ip
(iv) Diode resistance
after saturation stage due to field emission is known as Shottkey effect.
- Static plate resistance or dc plate resistance :
R = Vp .
- Dynamic or ac plate resistance : If at constant filament current, a small change ∆VP in the plate potential
produces a small change
in the plate current, then the ratio
DVp / Dip
is called the dynamic resistance, or the
‘plate resistance’ of the diode rp
= DVp .
Note : @In SCLR rp < Rp , In TLR Rp < rp
(3) Uses of diode valve
and rp = ¥ .
- As a rectifier (ii) As a detector (iii) As a transmitter (iv) As a modulator
(4) Diode valve as a rectifier
Rectifier is a device which is used to convert ac into dc
(5) Filter circuit
Filter circuits smooth out the fluctuations in amplitude of ac ripple of the output voltage obtained from a rectifier.
- Filter circuit consists of capacitors or/ and choke
- A capacitor offers a high resistance to low frequency ac ripple (infinite resistance to dc) and a low resistance to high frequency ac Therefore, it is always used as a shunt to the load.
- A choke coil offers high resistance to high frequency ac, and almost zero resistance to It is used in series.
- p – Filter is best for ripple
- For voltage regulation choke input filter (L-filter) is
Inventor : Dr. Lee De Forest Principle : Thermionic emission Number of electrodes : Three
Grid : Is a third electrode, also known as control grid, which controls the electrons going from cathode to plate.
It is kept near the cathode with low negative potential.
Working : Plate of triode valve is always kept at positive potential w.r.t. cathode. The potential of plate is more
æ V ö3 / 2
than that of grid. The variation of plate potential affects the plate current as follows
i = k ç V + p ÷
; where m =
Amplification factor of triode valve, k = Constant of triode valve.
è m ø
When grid is given positive potential then plate current increases but in this case triode cannot be used for amplifier and therefore grid is normally not given positive potential.
When grid is given negative potential then plate current decreases but in this case grid controls plate current most effectively.
- Cut off grid voltage : The valve of VG for which the plate current becomes zero is known as the cut off
voltage. For a given V , it is given by V = – Vp .
p G m
- Characteristic of triode : These are of two types
Note : @Both static and dynamic characteristics are again of two types-plate characteristics and mutual characteristic
- It is a straight line joining the points ( Vpp , 0) on plate voltage axis and (0, Vpp / RL ) on plate current axis of plate characteristics of
- In graph, AB is a load line and the equation of load line is :
Vpp = ip RL + Vp or
i = – 1 V
- The slope of load line
AB = dV = – R
- In graph, OA = Vpp= intercept of load line on VP axis and OB = Vpp / RL = intercept of load line on i p
(3)Constant of triode valve
- Plate or dynamic resistance (rP) : The slope of plate characteristic curve is equal to 1 or It
is the ratio of small change in plate voltage to the change in plate current produced by it, the
grid voltage remaining constant. That is, rp =
, VG = constant .
It is expressed in kilo ohms (KW). Typically, it ranges from about 8 KW to 40 KW. The
rp can be determined from plate characteristics. It represents the reciprocal of the slope of the plate characteristic curve.
If the distance between plate and cathode is increased the rp
of cut off bias or saturation state.
(ii) Mutual conductance (or trans conductance) (gm)
increases. The value of
rp is infinity in the state
- It is defined as the ratio of small change in plate current (Dip) to the corresponding small change in grid potential
æ Dip ö
(DVg ) when plate potential Vp is kept constant i.e. gm = ç ÷
è ø V is constant
- The value of gm is equal to the slope of mutual characteristics of
- The value of gm depends upon the separation between grid and The smaller is this separation, the larger is the value of gm and vice versa.
- In the saturation state, the value of Dip= 0 , gm = 0
- Amplification factor (m) : It is defined as the ratio of change in plate potential (DVp ) to produce certain
change in plate current
æ DVp ö
(Dip ) to the change in grid potential
(DVg ) for the same change in plate current
(Dip ) i.e.
m = -ç ÷
ç DVg ÷
; negative sign indicates that Vp
and Vg are in opposite phase.
è øDIp = a constant
- Amplification factor depends upon the distance between :
- Plate and cathode (d pk )
- Plate and grid (d pg )
- Grid and cathode (dgk )
Also m µ dpg µ dpk µ d
- The value of m is greater than
- Amplification factor is unitless and
Note : @ The triode constants are not independent of each other. They are related by the relation.
m = rp ´ gm
gm depends on ip in the following manner. m
rp µ ip -1 / 3 , gm µ ip1 / 3
m does not depend on ip. The variation of triode parameters with ip are shown in figure.
@ Above three constant may be determined from any one set of characteristic curves.
VP1 – VP 2
I PA – I PB
I PA – I PB
gm = ,
V – V
m = – P1 P 2
VG2 – VG1
(4) Triode as an Amplifiers
Amplifier is a device by which the amplitude of variation of ac signal voltage / current/ power can be increased
- Principle and circuit diagram : The amplifying action of the triode is based on the fact that small change in grid voltage produces the same change in the grid voltage as due to a large change in the plate voltage. A circuit for triode as an amplifier
- Working : First of all the mutual characteristic curves of a triode to be used as an amplifier are plotted and the grid potential – Vgb corresponding to the mid-point of straight portion of characteristic curve is
This negative grid potential is applied on grid and is known as grid bias. The AC signal to be amplified is
connected in series with this grid bias (-Vgb ) . Let the input signal be represented as eg = e0 sinw t .
The net input grid voltage = –Vgb + e0 sinw t , varies between
- Vgb + e0
- Vgb – e0 . The corresponding
amplified output current shown in fig. The output voltage is taken across load resistance RL . If
DVg ) is the
input signal voltage and DVL = RLip (= RL Dip ) is the consequent voltage change across load RL , then
Voltage gain = output voltage = DVL
= DVp =
or A = m
1 + Rp
Rp + RL
The maximum voltage gain is obviously equal to m for RL = ¥ .
Example: 1 The peak voltage in the output of a half-wave diode rectifier fed with a sinusoidal signal without filter is 10V. The d.c. component of the output voltage is [CBSE PMT/PDT (Screening) 2004]
(a) 20/pV (b)
10 / V
(c) 10/pV (d) 10V
Solution : (c) In half wave rectifier V
= V0 = 10 volt
dc p p
Example: 2 When plate voltage of diode increased from 100 V to 150 V then plate current increases from 7.5mA to 12mA the AC plate resistance will be [RPMT 2000]
(a) 10 kW (b) 11 kW (c) 15 kW (d) 11.1kW
Solution : (d) ac plate resistance rP
= 150 – 100 = 11.1kW (12 – 7.5) ´ 10 –3
Example: 3 In the grid circuit of the triode a signal E = 2
cos w t is applied if m = 14, rp = 10KW
then the current [RPMT 1992]
(a) 1.27 mA (b) 10 mA (c) 1.5 mA (d) 12.4 mA
m ´ Vg
Solution : (a)
; From voltage applied across grid, peak voltage V0 = Vg = 2
iP = 14 ´ 2 2 = 1.27 mA.
(10 + 12) ´ 103
Example: 4 A triode having
m = 18
and rp = 8000 ohm is used as an amplifier with a load resistance of 10 kilo ohm in the
plate circuit. The voltage amplification is, then [MP PMT 1991]
(a) 1 (b) 10 (c) 20 (d) 30
m RL 18 ´ 10 ´ 103
Solution : (b) From AV
= r + R =
18 ´ 103
Example: 5 Keeping the grid voltage constant, a change in the plate potential of 50 V, changes the plate current by 10 mA.
And keeping the plate potential constant, a change in the grid potential of 2 V, changes the plate current by 10
mA again. The amplification factor of the triode will be [CBSE 1991]
(a) 100 (b) 25 (c) 5 (d) 20
æ DVP ö 50 3
æ Dip ö
10 ´ 10 –3
Solution : (b)
rP = ç ÷
è P ø
10 ´ 10 –3
= 5 ´ 10
W and gm = ç ÷ =
2 = 5 ´ 10 W
\ m = rP ´ gm = 5 ´ 103 ´ 5 ´ 10 –3 = 25
è ø VP
Example: 6 A diode valve works in the region of space charge limited current. If the voltage is increased four times, how many times the space charge limited current will increase [CBSE 1991]
- Will remain unchanged (b) 2 (c) 8 (d) 4
i æ V
ö 3 / 2
æ 4 ö 3 / 2
Solution : (c) From i µ V 3 / 2
Þ 2 = ç 2 ÷
= ç ÷ = 8
i1 è V1 ø
è 1 ø
Example: 7 A triode whose mutual conductance is 2.5 m A/volt and anode resistance is 20 kilo ohm, is used as an amplifier whose amplification is 10. The resistance connected in plate circuit will be
(a) 1 kW (b) 5 kW (c) 10 kW (d) 20 kW
Solution : (b)
A = m RL
rP + RL
= m RL
m = rP ´ gm = 20 ´ 22.5 = 50
A = m RL
rP + RL
= m RL
\ 4 RL
= 20 = 5kW 5