# EC301 Electronics Devices & Circuits-1 tutorial: 1 Junction Diode Characteristics

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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

1. 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

1. 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

1. 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….

1. 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

1. 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…

1. (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)

1. 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=……

1. (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=------ 1. 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

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