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In Forw ard bias:Diode is short circuited i.e. ON

Ideal diode Rf= 0

Practically Rfvalue is less.

In Reverse b ias:Diode is open circuited i.e. OFF

Ideal diode Rr=

Practically Rrvalue is very high.

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Minority-carrier distribution as a function of the distance x from junction.

(a) A reverse biased junction (b) A forward biased junction

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The waveform in (b) is applied

to the diode circuit in (a),

(c) The excess carrier density

at the junction,

(d) The diode current, and

(e) The diode voltage

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The transistor act as a switch, when it is either in cut-off or in saturation region.

When both emitter-base & collector-base junction are reverse biased, transistor operates

in cut-off region & act as open switch. When both emitter-base & collector-base junction

are forward biased, it operates in saturation region & act as closed switch.

ICRL-VCC+VCE=0

VCE=VCCICRL

IC = 0, VCC = VCE

then, VCE =0

VCC = ICRL

IC =VCC/ RL

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When pulse input is given for the transistor then we will find different

ON time and OFF time.

tON=td+ tr (delay time + rise time)

tOFF= ts+tf (storage time + fall time)

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The delay time, td :

When the base current is applied, the transistor does not

switch on instantaneously. Some time delay takes place before the collector

current starts flowing. This time gap between the base current application and

collector current flow termed as delay t ime.

Rise time, tr:Even though the collector current begains to flow,it attains

its maximum value IConly after some time delay tr. This delay is termed as

rise time.

We can define r ise timeas the time required for collector current torise from 10% of its maximum level to 90% of its maximum level.

Storage time, s:The time interval between the instant IB= 0 and the instant

when ICfalls to 90% of its maximum value is termed as storage t ime.

Fall time, tf :

The time required for ICto fall from 90% of its maximum

value to 10% of its maximum value is termed as fa l l t ime.

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When trnsistor is in cut in,

VB = VBB(R1/R1+R2) + Vi(0) (R2/R1+R2)

Vi(0) = V(0) (Vi= 0, in cut in)

Therefore, VB = VBB(R1/R1+R2)

VB< VBE required for transistor, i.e. 0V for Si transistor and -0.1V for Ge

transistor only for cut-off region.

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When Vi = V(1),

RC = VCC-VCE(sat)IC

Current through resistor R1, I1 = ViVBE(sat)

R1

Current through resistor R2, I2 = VBE(sat)VEE

R2

IB = I1 - I2

IB(min) = IC / hFE(min)

IB = ViVBE(sat) _ VBE(sat)VEE

R1 R2

IC = V(1)VBE(sat) _ VBE(sat)VEE

hfe(min) R1 R2

VCC-VCE(sat) = V(1)VBE(sat) _ VBE(sat)VEE

RC.hFE(min) R1 R2

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