Recall Lecture 8• Clipper

– Step 1: Find the clip value by doing KVL at the output branch

– Step 2: Set the conditions to know whether diode is on or off – sketch your output waveform

• Clamper– Step 1: Calculate value of VC by knowing which cycle it is

charged to (what is the peak value of that cycle) then confirmed the polarity of the capacitor

– Step 2: performed KVL as diode is now reverse- biased to obtain equation of Vout . Sketch your output waveform

Multiple Diode CircuitsMultiple Diode Circuits

Final Exam SEM I 2013/2014

DIODE ID VD

OFF 0 VD < V

ON ID > 0 VD = V

REMEMBER THAT:

A pn junction diode will conduct when the p-type material is more positive than the n-type material

OR GATE

V1 V2 VO

Vo = voltage across R

D1 and D2 off; no current flow, 0 0 0

D1 off, D2 on, current flow,

Vo – V2 + V = 0

0 5V ( 1 ) 4.3V

D1 on, D2 off, current flow,

Vo – V1 + V = 0

5V ( 1 ) 0 4.3V

Both on, using both loops will give the same equation

5V ( 1 ) 5V ( 1 ) 4.3V

V1 V2 VO

Both on, using both loops will give the same equation

0 0 0.7

D1 on, D2 off 0 5V ( 1 ) 0.7

D1 off, D2 on 5V ( 1 ) 0 0.7V

Both are off; open circuit no current flowing through R since no GND destination

5V ( 1 ) 5V ( 1 ) 5V

AND GATE Vo = node voltage

Chapter 4Chapter 4Bipolar Junction TransistorBipolar Junction Transistor

REMEMBER THIS

Current flow in the opposite direction of the electrons flow; same direction as holes

e e e

I

h h h

Transistor StructuresTransistor Structures

The bipolar junction transistor (BJT) has three separately doped regions and contains two pn junctions.

Bipolar transistor is a 3-terminal device.

Emitter (E)

Base (B)

Collector (C)

The basic transistor principle is that the voltage between two terminals controls the current through the third terminal.

Current in the transistor is due to the flow of both electrons and holes, hence the name bipolar.

Transistor StructuresTransistor Structures There are two types of bipolar junction transistor: npn and

pnp.

The npn bipolar transistor contains a thin p-region between two n-regions.

The pnp bipolar transistor contains a thin n-region sandwiched between two p-regions.

ActiveOperating range of the amplifier.Base-Emitter Junction forward biased.Collector-Base Junction reverse biased

CutoffThe amplifier is basically off. There is voltage but little current.Both junctions reverse biased

SaturationThe amplifier is full on. There is little voltage but lots of current.Both junctions forward biased

3 Regions of Operation3 Regions of Operation

OPERATIONS - npnOPERATIONS - npn

The base-emitter (B-E) junction is forward biased and the base-collector (C-B) junction is reverse-biased,.

Since the B-E junction is forward biased, electrons from the emitter are injected across the B-E junction into the base IE

Once in the base region, the electrons are quickly accelerated through the base due to the reverse-biased C-B region IC

ACTIVE MODE

C

B

E

Some electrons, in passing through the base region, recombine with majority carrier holes in the base. This produces the current IB

+

-VBE

iB

C B E

TO ILLUSTRATE

-

VBE

+

•Imagine the marbles as electrons

•A flat base region with gaps where the marbles may fall/trapped – recombine

•A sloping collector region represents high electric field in the C-B region

•Hence, when enough energy is given to the marbles, they will be accelerated towards to base region with enough momentum to pass the base and straight ‘fly’ to the collector

MATHEMATICAL EXPRESSIONSC

B

E

+

-VBE

IE

IC

IB

IE = IS [ e VBE / VT -1 ] = IS e VBE / VT Based on KCL: IE = IC + IB

No. of electrons crossing the base region and then directly into the collector region is a constant factor of the no. of electrons exiting the base region

IC = IB

No. of electrons reaching the collector region is directly proportional to the no. of electrons injected or crossing the base region.

IC = IEIdeally = 1, but in reality it is between 0.9 and 0.998.

Based on KCL: IE = IC + IB IC = IB IC = IE

IE = IB + IB = IB( + 1)

= [ / + 1 ]

IE = IB( + 1)

NowWith IC = IB IB = IC / Hence, IE = [ IC / ] ( + 1)IC = IE [ / + 1 ]

Comparing with IC = IE

B

C

E

-

+

VEB

OPERATIONS - pnpOPERATIONS - pnpFORWARD ACTIVE MODE

The emitter – base (E- B) junction is forward biased and the base-collector (B- C) junction is reverse-biased,.

IE = IS [ e VEB / VT -1 ] = IS e VEB / VT

**Notice that it is VEB

IE

IC

IB

Based on KCL: IE = IC + IB

pnp Transistor- Active mode

SUMMARY: Circuit Symbols and SUMMARY: Circuit Symbols and ConventionsConventions

npn bipolar transistor simple block diagram and circuit symbol.

Arrow is on the emitter terminal that indicates the direction of emitter current (out of emitter terminal for the npn device)

pnp bipolar transistor simple block diagram and circuit symbol.

Arrow is on the emitter terminal that indicates the direction of emitter current (into of emitter terminal for the pnp device)

Based on KCL: IE = IC + IB

EXAMPLE 4.1 Calculate the collector and emitter currents, given the base current and current gain. Assume a common-base current gain and a base current of . Also assume that the transistor is biased forward in the forward active mode.

Solution: The common-emitter current gain is

The collector current is

And the emitter current is

BJT: Current-Voltage BJT: Current-Voltage CharacteristicCharacteristicIICC versus V versus VCECE

The Emitter is common to both input (base-emitter) and output (collector-emitter).

Since Emitter is grounded, VC = VCE

With decreasing VC (VCE), the junction B-C will become forward biased too. The current IC quickly drops to zero

because electrons are no longer collected by the collector

Common-Emitter Configuration Common-Emitter Configuration - npn- npn

Node B

0V

Characteristics of Common-Emitter Characteristics of Common-Emitter - npn- npn

NOTE: VEC for PNP

Examples

• EXAMPLE 1• Given IB = 6.0A and

IC=510 A

Determine , and IE

• EXAMPLE 2• NPN Transistor • Reverse saturation

current Is = 10-13A with current gain, = 90. Based on VBE = 0.685V, determine IC , IB and IE

EXAMPLE 3 PNP Transistor = 60, IC= 0.85mA Determine , IE and IB