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EC301 Electronics Devices & Circuits-1
TUTORIAL: 1
Junction Diode Characteristics
1. The current of germanium diode is 100uA at a voltage of -1V, at room
temperature. Determine the magnitude of the current for the voltages of ±0.2V at
room temperature. [Ans: 219mA, 99.95uA]
ANS:
I=Io(eV/n.Vt-1)
Io=100uA given
V= +0.2 forward bias>>>>>>find out I=219mA
V= -0.2V reverse bias>>>>>>>>find out I=99.95uA
2. A silicon diode has a reverse saturation current of 60nA. calculate the voltage at
which 1% of the rated current will flow through the diode, at room temperature if
diode is rated for 1A.[Ans:0.6252]
ANS:
Io=60nA
I=0.01A(1% of 1A)
Use
I=Io(eV/n.Vt-1)
Find out V=0.6252Volt
3. For a silicon diode with reverse saturation current of 0.2uA , calculate dynamic
forward and reverse resistance at a voltage of 0.72V and -0.72V respectively,
applied across the diode, at room temperature of 27⁰C.[Ans:0.252Ω, 268.09GΩ].
ANS:
Dynamic resistance: rd= n.Vt/Io. eV/n.Vt
For Forward biased>>>take V=+0.72V..find out rd
For Reversed biased>>>>take V=-0.72V find out rd
4. A silicon diode operates at a forward voltage of 0.4V. Calculate the factor by
which the current will be multiplied when the temperature is increased from 25⁰C
to 150⁰C.
ANS:
I=Io(eV/n.Vt-1)
For I=I1>>>>>Io=Io1>>vt=vt1
I=I2>>>>>Io=Io2>>vt=vt2
now
Io=k.Tm. e(VGO/n.Vt)
For Io=Io1, Io=Io2…….
Find out value of Io1 and Io2…and substitute in equation of I1 and I2…
Then take ratio of I2/I1….
5. For a Silicon diode has an ohmic series resistance of 0.5Ω. Determine the fraction
of applied voltage that falls across the ohmic resistance for diode currents of
10uA, 1uA, 10mA and 1A. Assume I0=10-12A and n=1.5 for the
diode.[Ans:0.001,0.09,0.8,41]
ANS:
Vr=drop across resistance =0.5 I
Drop across diode:
I=Io(eV/n.Vt-1)….Io is given in data
Find out v=diode drop=n.Vt.ln.(1+I/Io)
Fractional Change= V/V+Vr Χ 100%
I
Vr
v
% change
10uA
1uA
10mA
1A
6. For an asymmetrical silicon diode, mean life time of holes be 10ns and η=1. If a
forward current of 0.1mA is flowing in the diode, determine diffusion
capacitance.[Ans:38.5pF]
ANS:
Use formula Cd = г.I/n. Vt
Where r= mean life time of holes, I=current, vt=26mV…
7. (a) For what voltage of the reverse saturation current in Ge diode reach 90% of its
saturation value at room temperature.[-0.059V]
ANS:
I= - 0.09Io
Use I=Io(eV/n.Vt-1) and find out V
(b) What is the ratio of the current for a forward bias of 0.05V to the current for
the same magnitude of reverse bias?[-6.83] [CLO 1 PO1]
ANS: use formula I=Io(eV/n.Vt-1)
(c) If the reverse saturation current is 10uA, calculate the forward currents for
voltages of 0.1V, 0.2V and 0.3V respectively.[0.458mA,0.021A,1A] [CLO 1
PO1]
ANS:
I=Io(eV/n.Vt-1)
8. The diode shown in the circuit is of silicon with Vy=0.6V and Rf=20Ω. Assuming
diode to be ON, show that Vo=0.9V1+0.1V2-0.54.
ANS: I =V1-V2-0.6/180
So V0 = I.180 + V2.. substitute value of I and solve for Vo=……
9. (i)Determine I, V1, V2 and Vo for the circuit shown. Assume Vy=0.7V for the
diode.[Ans:2.072mA,9.74V,4.56V,-0.44V]
ANS: I = 10 – 0.7 + 5/(4.7K+2.2K)
V1= I. 4.7K
V2=I.2.2K
Vo= V2-%
(ii) Find out value of I and Vo if diodes are ideal.[Ans:4mA,+1V]
ANS:
Common anode configuration
D3 is ON state:
I = 5 – 1/ 1K
And Vo= 5 – 1K.I=------
10. Consider an asymmetrical silicon junction with Na=1019 /cm3 and Nd=1017/cm3 If
the cross section area of the junction is 9.6µm2, determine its transition
capacitance with no applied bias. [Ans: 9.2fF]
ANS:
CT= A. E/Wd
Where Wd = √2Eo.Er.Vj/Nd
Where Vj= Vt.ln Na.Nd/ni2