LinkedIn Insight Summer 2017 - Q.1 [Preview] - Grad Plus

Summer 2017 – Q.1 [Preview]

1. (a) Draw the construction diagram and explain working of LED. [6M]

Construction Of LED:-

Construction LED

i) Light Emitting diode (LED) is two lead semiconductor light source.
ii) Basically it is p-n junction diode that emits light when activated.
iii) When a suitable voltage is applied to the leads, electrons are able to recombine with electron holes within the device, releasing energy in the form of photons. This effect is called electroluminescence.
iv) One of the methods used for the LED construction is to deposit three semiconductor layers on substrate
v) In between p-type and n-type there exists an active region
vi) This active region emits light, when an electron and hole recombine.
vii) The basic layered structure is placed in a tiny reflective cup so that the light from thee achive layers will be reflected towards the desired exit direction.

Basic operation:-

i) Whenever a p-n junction is forward biased, electrons cross the p-n junction from the n-type semiconductor material and recombine with the holes in the p-type semiconductor material.

LED forward Biasedii) Free electrons are at higher energy level with respect to the holes.
iii) When a free electron recombines with holes, if falls from conduction band to a valence band
iv) The energy level associated with it changes from higher value to lower value.
v) The energy corresponding to the difference between higher level and lower level is released by an electron while travelling from conduction band to valence band.
vi) This energy is released in the form of photons i.e. in form of light energy.
vii)The energy released in the form of light depends on energy corresponding to forbidden gap which determines the wavelength of light.
Let ‘Eg‘ denotes the energy gap, then Eg=hν, where ν is the frequency of the emitted light. Then

$latex E_g=h\nu=\frac{hc}\lambda$
where λ is the wavelength of emitted light.

(b) Explain with a neat circuit diagram, function of each component in single stage CE amplifier. [6M]

Common Emitter Configuration:-
Following diagram shows is a single stage CE configuration using n-p-n transistor.

CE amplifier

i) The amplifier consists of three main components that are a) transistor biased in active region b) Coupling and bypass capacitor and c) ac or small signal input.
ii) For transistor to be in active region of operation , Emitter – Base junction should be in forward bias and Colector-Base junction should be in reverse bias. This biasing is provided by the network of resistances and Vcc as shown in figure.
iii) So resistances R1, R2, RE and RC along with DC voltage VCC provide proper DC biasing to operate the transistor in active region. Such type of biasing arrangement is called as Emitter bias or Self Bias.
iv) The capacitors C1, C2 are called as coupling capacitors. As these capacitors allow only ac voltage (or current) while blocking the DC voltages (or currents). These are very useful for coupling the ac input and taking out the ac output and hence called as coupling capacitors.
v) In amplifier circuit the transistor is already in the active region by DC biasing (VCC and Resistance).  For amplification, we apply ac input and take only ac output. So these coupling capacitors play very important role in coupling ac at input side as well as blocking DC at output side.
vi) In Common Emitter amplifier ac input is applied between the base and Emitter (Ground) and ac output is taken across the Collector and Emitter (Ground). Hence this configuration is called as Common emitter configuration. For common emitter configuration Base current IB is input current while IC is output current and VCE is the output voltage.
vii) Consider the DC view of the circuit which keeps the circuit in active region i.e. E-B forward biased and C-B reversed biased. After applying thevenin’s equivalent and applying KVL in the input and output loops, we have currents IB, IC and  voltage VCE as below.

dc biasing$latex {\mathrm I}_\mathrm B=\frac{{\mathrm V}_\mathrm{BB}-{\mathrm V}_\mathrm{BE}}{{\mathrm R}_\mathrm B+\left(1+\mathrm\beta\right){\mathrm R}_\mathrm E}$

$latex {\mathrm I}_\mathrm B=\frac{{\mathrm V}_\mathrm{BB}-{\mathrm V}_\mathrm{BE}}{{\mathrm R}_\mathrm B+\left(1+\mathrm\beta\right){\mathrm R}_\mathrm E}$

$latex {\mathrm V}_\mathrm{CE}={\mathrm V}_\mathrm{CC}-{\mathrm I}_\mathrm C{\mathrm R}_\mathrm C-\left({\mathrm I}_\mathrm B+{\mathrm I}_\mathrm C\right){\mathrm R}_\mathrm E$

viii) All these are DC values and define operating point or Q point of the circuit to ensure the circuit in active region. This is the significance of biasing network i.e. R1, R2, RC, RE,and VCC.
ix) We apply ac signal at the base, due to which the base current varies and the variations are amplified in collector current or output current. The process is called as amplification and we get amplified ac at collector.
x) Resistance RE also provides stabilization of the DC biasing; lowers the ac amplification gain. Hence a capacitor is connected across it called as bypass capacitor shown in the diagram as C3.
xi) This capacitor C3, as like other capacitors C1 and C2, acts as an open circuit for DC and short circuit for ac. Hence bypasses the resistance RE in ac analysis.
xii) RC is the load resistance. The current flowing through the RC produces the voltage output of the amplifier. The value of RC is chosen so that at the amplifiers operating point (Q-point) this output voltage lies half way along the transistor load line.

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