Upcoming SlideShare
×

# Unit 3

1,009

Published on

Semiconductor devices

1 Like
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

Views
Total Views
1,009
On Slideshare
0
From Embeds
0
Number of Embeds
1
Actions
Shares
0
92
0
Likes
1
Embeds 0
No embeds

No notes for slide
• Sdaa
• ### Unit 3

1. 1. BIPOLAR JUNCTION TRANSISTORS (BJT)
2. 2. INTRODUCTION – BASIC BJT A Transistor is often called a Bipolar Junction (BJT) or bipolar transistor. The name bipolar is adopted from its structure whereby it has ‘two junction’ and consist o three doped region either ‘n-p-n’ or ‘p-n-p’ combination.
3. 3. BIPOLAR JUNCTION TRANSISTOR The bipolar junction transistor is a semiconductor device constructed with three doped regions. These regions essentially form two ‘back- to-back’ p-n junctionsin the same block of semiconductor material (silicon). The most common use of the BJT is in linear amplifier circuits (linear means that the output is proportional to input). It can also be used as a switched (in, for example, logic circuits).
4. 4. STRUCTURE AND SYMBOLS FORBJTa. 3 Layer semiconductor device consisting: - 2 n – and 1 p-type layers of material >NPN transistor - 2 p- and 1 n-type layers of material > PNP transistorb. The term bipolar reflects the fact that holes and electrons participate in the injection process into the oppositely polarized material.c. A single PN junction has two different types of bias: - forward bias - reverse bias
5. 5. STRUCTURE AND SYMBOLS FORBJT N P N P N P
6. 6. PHYSICAL STRUCTURE FOR BJT Consist of 3 alternate layers of n- and p-type semiconductor is called emitter (E) , base (B) and collector (C). Majority of current enters collectors , crosses base region and exits through emitter. A small current also enters base terminal, crosses base emitter junction and exits through emitter. Carrier transport in the active base region directly beneath the heavily doped(n) emitter dominates i-v characteristics of BJT.
7. 7. TRANSISTOR OPERATION The basic operation will be described using the pnp transistor. The operation of the npn transistor is exactly the same if the roles played by the electron and hole are interchanged. One p-n junction of a transistor is reverse- biased,whereas the other is forward-biased.
8. 8. TRNSISTOR OPERATION Majority carriers can cross the reverse-biased junction because the injected majority carriers in the n-type material. Applying KCL to the transistor : IE = IC + IB The comprises of two component – the majority carriers IC = Icmajority + ICOminority ICO – IC current with emitter terminal open and is called leakage.
9. 9. EXAMPLE CONNECTION
10. 10. TRANSISTOR OPERATION We will consider npn transistors Pnp devices are similar but with different polarity of voltage and currents. When using npn transistors. Collecter is normally more positive than the emitter VCE might be a few volts Devices resembles two back – to – back diodes – but has very different characteristics With the base open – circuit negligilbe current flows from the collector to the emitter.
11. 11. BJT OPERATES AN AMPLIFIER ANDSWITCH A simple transistor amplifier - Rb is used to ‘bias’ the transistor by injecting an appropriate base current C is a coupling capacitor and is used to couple the AC signal while preventing external circuits from affecting the bias
12. 12.  AC-coupled amplifier - Vb is set by the conduction voltage of the base-emitter junction and so is about 0.7V - voltage across Rb is thus Vcc- 0.7 - this voltage divided by by Rb gives the base current Ib - the collector current is then given by Ic= hFE Ib - the voltage drop across Rc is given by IcRc - the quiescent output voltage is therefore Vo= Voc- IcRc
13. 13. BJT OPERATES AS AN AMPLIFIER AND SWITCH
14. 14.  Base current is small, so Vb= Vcc R2/R1+R2= 10 10kΩ/27kΩ+10kΩ =2.7V Emitter voltage Ve= Vb- Vbe =2.7-0.7 =2.0VSince Ib is small,collector current Ic=Ie= 2mAOutput voltage =Vcc-IcRc = 10- 2mA x 2.2kΩ= 5.6V
15. 15.  A common-collector amplifier - unity gain - high input resistance - low output resistance - a very good buffer amplifier
16. 16.  Regions of BJT operations - Cut-off region: The transistor is off. There is no conduction between the collector and the emitter. (Ib= 0 therefore Ic=0) - Active region: The transistor is on. The output current(Ic) is relatively insentive to Vce. In this region the transistor can be an amplifier. - Saturation region: The transistor is on. The collector current varies very little with a change in the base current in the saturation region. The collector crrent is strongly dependt on Vce unlike the active region. It is desirable to operate transistor switches in or near the saturation region when in their state.
17. 17.  A common –emitter configuration - emitter is a common or reference to both input and output terminals - emitter is usually the terminal closest to or at ground potential almost amplifier design is using connection of CE due to the high gain for current and voltage Two set of characteristics are necessary to describe the behaviour for CE: input(base terminal) and ouptut (collector terminal) parameters.
18. 18. Common-emitter configuration Active region Saturation region Cut-off region• B-E junction is forward •B-E and C-B junction is •Region below Ib= 0µA isbias forward bias, thus the to be avoided if an•C-B junction is reverse value of Ib and Ic is too undistorted o/p signal isbias big. required•Can be employed for •The value of Vce is so •B-E junction and C-Bvoltage, current, and small junction is reverse biaspower amplification •Suitable region when the •Ib= Iceo where is this transistor as a logic switch current flow when B-E is •NOT and avoid this reverse bias. region when the transistor as an amplifier.
19. 19. Common-base configuration Active region Saturation region Cut-off region•Ie increased,Ic increased •BE and CB junction is •Region below the line of•BE junction forward bias forward bias Ie= 0Aand CB junction reverse •Small changes in Vcb •BE and CB is reverse biasbias will cause big different to •No current flow at•Ic is not depends on Vcb Ic collector,only leakage•Suitable region for the •The allogation for this currenttransistor working as region is to the left ofamplifier Vcb= 0V
20. 20. Common-collector configuration also called emitter-follower (EF) Is called common-emitter configuration since both signal source and the load share the collector terminal as a common connection point Output voltage is obtained at emitter terminal Input characteristic is similar with common-emitter configuration Provided with the load resistor connected from emitter to ground Primarily for impendance-matching purpose since it has high impedance an low output impedance
21. 21. DEFINE – β DC AND β AC Ratio of dc collector current(IC) to the dc base current (IB) is dc beta(βdc) which is dc current gain where IC and IB are determined at a particular operating point, Q-point(quiescent point). Ic 30< βdc<300 2N3904 βdc IB On data sheet, βDC = hFE with h is derived from ac hybrid equivalent cct.FE are derived from forward- current amplification and common-emitter configuration respectively.
22. 22.  For ac conditions an ac beta has been defined as the changes of collector current(Ic) compared to the changes of base current6(IB). On data sheet, βac=hFE
23. 23. Beta(β) In the dc mode the level of Ic and Ib are related by a quantity called beta and defined by the following equation: βdc =Ic/IB For the ac situation an ac beta has been defined as follows: βac=∆ Ic/∆IB β indicates the amplification factor of transistor. (β is sometimes refer as hfe,a term used in transistor modeling calculations)
24. 24. Relationship between β and α Both indicate an amplification factor. α= β/ β +1 β=α/α-1 β provides a Relationship between Currents Ic= β*IB IE=(β+1)IB
25. 25. Voltage-divider bias Because the base current is small, the approximatioon V =[R2/R1+R2] Vcc is useful for calculating the base voltage. B After calculating VB, you can find VE by subtracting 0.7V for VBE. Next, calculate IE by applying Ohm’s law to RE : IE=VE/RE Then apply the approximation Ic=IE Finally, you can find the collector voltage from Vc=Vcc-IcRc
1. #### A particular slide catching your eye?

Clipping is a handy way to collect important slides you want to go back to later.