KPK Board 12th class Physics Ch 17 Electronics short questions answers

The region just in the vicinity of the PN-junction is called depletion region. There is no mobile charge in depletion region. Formation: Just after the formation of PN-junction the diffusion of holds and free electrons take place across the junction. But soon this action is stopped because when free electrons in N-type section cross the junction and combines with the holes in P-type section. The N-type section becomes positively charge while the P-section becomes negatively charged. A region is formed around the junction in which charge carriers are not present. This region is known as depletion region.

(a) Base Thin: In a transistor, usually the base is very thin of the order of 10 –⁶m and lightly doped so that to increase the current gain. The base is thin so 90% of the charges leaving the emitter pass through the base section and enters the collector region. From here these charges are attracted by the terminal of V_BC. These charges from the collector flow around the circuit causes collector current I_c. The base is very thin, so a few charges flow from base to emitter causing base to emitter current I_B. Here Ic >>1. The amplification factor or current gain is                                                                                                                                                                                     β=I_C/I_B

Thus to increase the current gain the RSI base strip is generally made thin.

(b) Collector Large: The region in the right of the base is called collector. The collector has less concentration of impurity as compared to emitter. The collector is comparatively large in size than the emitter, so that to receive large number of charges emitted by the emitter when the unit is in function.

Normally, when transistor is used as an amplifier, its emitter-base junction is forward biased while base-collector junction is reversed biased. So the emitter injects a large number of electrons in the base section. Since the base section is extremely thin, so nearly all the electrons leaving the emitter passing through base section and enter the collector region. These electrons are attracted by positive terminal of Vbc and forming current Ic, that flows through the external circuit, back to emitter section. A few electrons combines with the holes of the base section, forming current Is, which flows form base to emitter region. Therefore, the emitter current I, is given by.

I_E = I_C +I_B

Here I, is very small as compared to I_C,

Hence.

I_E=I_C

In normal operation of a transistor, the emitter to base junction is forward biased the collector to base junction is reverse biased. This is because in forward biasing the emitter offer low resistance and inject a large number of charges into the base region. These charges entering the base can flow in either of two directions. They can either flow from base to emitter or base to collector. But here base to collector is reverse baised and base section is thin, so most of the charges are attracted by the collector. Therefore the collector current IC, is nearly equal to the emitter current lE.The charges which flows via base to emitter farming base current 13.

Here I_E = Ic +I_B

But I_c >> I_B

SO I_E = I_C

1. NPN transistor: In the fig NPN transistor circuit diagram is shown. In normal operation the two batteries VBE and Voc are so connected that the emitter is made more negative while the collector is made more positive. The Vic has much higher value then Voe. The emitter injects a large of number of electrons into the bused region, out of which 90% cross the base and enter the collector region, forming Current I_C. The rest of the electrons flows from base to emitter creating current I_B. Thus the emitter current le is.

I_E = I_C+I_B

2. PNP Transistors: In the fig PNP transistor circuit diagram is shown where the two batteries are so connected that the emitter is made more positive while the collector is made more negative. The emitter injects a large number of holes into the base region. Nearly 90% holes passes through the base which are attracted by the collector forming current I_c. The remaining holes are passing through base into the emitter causing current Is. The emitter current l_E becomes. I_E=I_C+I_E

N-type material

1. A Ge or Si crystal formed after adding penta-valent impurity is known as N-type semi- conductor.
2. When in a pure Ge or Si, a small amount of penta-valent impurity is added in the form of an element from 5th group such as Arsenic, Antimony or Bismuth. The four valance electrons of the impurity atom (Sb) form covalent bonds with four neighbouring atom of Ge or Si in the crystal, while one electron of the impurity atom is free.
3. In this way one valence electrons of each atom is free in the crystal. These free electrons are responsible for the flow of electric current in the crystal thus formed.
4. Hence these electrons are called majority carriers.

P-type material

1. A crystal of germanium or silicon formed after adding a trivalent impurity is known as P- type semi-conductors.
2. When in a pure germanium or silicon, a small amount of impurity is added in the form of an element from the third group such as aluminum, Boron, Indium. The three valance electrons of the impurity atom forms covalent bond with the neighbouring atoms of Ge or Si. There is no electron for the 4^th atom of Ge or Si.
3. So there is a hole for the 4th atom of Ge or Si, as shown. The hole is free to move from one atom to another in the crystal.
4. In this way the holes are called majority carries.

A germanium or silicon crystal formed after adding a trivalent impurity from 3 group such as Indium is known as P-type, semiconductor. Each impurity atom shares three electrons forming covalent bonds and has a hole for forth atom of Ge or Si. In this way a P-type semiconductor has a large number of holes as positive charge carriers. But it is still electrically neutral, because in P-type or N-type semi-conductor each and every atom in neutral. The total number of electrons in the semi- conductor is equal to the total number of protons. Hence there is no not charge the crystal. Therefore P-type semiconductors are electrically neutral.

The common emitter (CE) configuration is the most popular of the three types I amplifying circuits. In most cases CE mode is preferred because its current gain and voltage gain are high while its power gain is the highest.

Current gain = β=I_C/I_B

Voltage gain=V_out/V_in

Power gain=I²_out/I²_inxR_C/r_ie

At the input the emitter junction is forward biased there input resistance is small while at the output, the collector junction is reversed biased, therefore output resistance is high. Hence current is transferred from low to high resistance circuit. The circuit for simple transistor amplification is the NPN transistor to act an A.C amplifier with ground emitter. Transistors are used as a amplifier in many electronic equipment such as oscillators, radio, CRO, T.V. computer, loudspeakers mobile etc.

Transistor is basically a current amplifying device. It converts low alternating voltage or current signals to high voltage or current signals. The circuit for a simple transistor amplifier is common-emitter circuit where emitter is ground. When a signal is applied to the input which causes small change in the base current of the base current .This causes large changes in collector current. In this way emitter ground CE converts low AC signals to high alternating current signals. Transistor is basically a current amplifier which is the building block of every complex electrical circuit such as radio, CRO, oscillators, T.V, load speaker etc.

Absolute temperature = T = 25°C + 273 = 298 K

Total number of atoms = N = 10¹⁰ + 10′ = 1.0 x 10¹⁰ atoms

Boltzmann Gas Constant = K = 1.38 x 10^-23J/ K

Total Kinetic Energy supplied is given by.

K.E=3/2NKT = 3/2×1.0x10¹⁰x1.38×10^-23x298=6.1686×10^-11

K.E=6.1686×10^-11/1.6×10^-19=3.86×10⁸eV

In case of silicon 1.09 eV energy is required to break a covalent bond of electron and hole. So the number of free electrons is.

No. of free electrons =3.86×10⁸eV/1.09eV = 3.54×10⁸electrons

No. of holes = number of electrons = 3.54 x 10⁸ holes.

In case of germanium 0.72 ev energy is required to break the bond. So the number of electrons in case of germanium is

No. of electrons =3.86×10⁸eV/0.72eV=5.36×10⁸ electrons

No. of holes = No. of electrons = 5.36 x 10⁸ holes.