Electron Devices: Chapter 6: Thyristors UJT and Optoelectronic Devices
Thyristors UJT and Optoelectronic Devices: Important Example Solved Problems
Electron Devices
Electron Devices: Chapter 6: Thyristors UJT and Optoelectronic Devices: Shockley Diode, Light Emitting Diode (LED) - Anna University Solved Problems, Assignment Problems and Important Solved Problems
Electron
Devices
Chapter 6: Thyristors UJT
and Optoelectronic Devices
Shockley
Diode
Ex. 1: A certain Shockley diode is
in its forward blocking region. The d.c. a values for the two transistors are α1
= 0.3 and α2 = 0.4. The leakage currents for both are 125 nA.
Calculate the anode current. If the anode to cathode voltage is 15 V, calculate
forward resistance of the diode.
Solution:
α1= 0.3, α2 = 0.4, ICBO1 = ICBO2 = 125 nA, VAK = 15 V
IA
= [ ICB01 + ICBO2 ]
/ [ 1‒(α1+α2) ]
=
[125×2×10‒9] / [1‒(0.3+0.4)]
=
0.833 μ Α
and
Rf
= VAK / IA
=
15 / 0.833×10‒6
=
18 MΩ
Thus
the forward resistance of shockley diode in the forward blocking region is very
very high and it acts as an open switch.
Ex. 2: Find the value of anode
current for the device shown in the Fig. 6.1.5 which is ON. VBR(F) =
40 V. Assume VBE = 0.71V and VCE(sat) = 0.15 V for the
internal transistor. Also find the forward resistance of the diode.
Solution:
VAK
= VBE + VCE(sat)
=
0.71+ 0.15 = 0.86 V
... Device ON
VRS
= VBIAS ‒VAK
= 55‒0.86 = 54.14 V
IA
= VRS / RS
=
54.14 / 12×103
=
4.511 mA
Rf
= VAK / ΙΑ
=
0.86 / 4.511×10‒3
=
190.64 Ω
So
Rf in ON state is very low.
Light
Emitting Diode (LED)
Ex. 3: What is the current through
LED shown in Fig. 6.6.5.
Solution: :
VS
= 15 V, RS = 2.2 kΩ
Assume
LED voltage drop as VD = 2 V
IS = (VS‒VD) /
RS
= (15‒2) / 2.2×103
= 5.91 mA
…………. LED current
Electron Devices: Chapter 6: Thyristors UJT and Optoelectronic Devices : Tag: electronics : Electron Devices - Thyristors UJT and Optoelectronic Devices: Important Example Solved Problems
Electron Devices: Chapter 6: Thyristors UJT and Optoelectronic Devices
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