Book 2 Module 4 ‘TRANSISTORS GRATED CIRCUITS,; IRCUIT BOAR: Licence By Post B 4.1.2 to 4.2 ISSUE 262AUTHORITY It is IMPORTANT to note that the information in this book is for study training purposes only. When carrying out a procedure/work on aircraft/aircraft equipment you MUST always refer to the relevant aircraft maintenance manual or equipment manufacturer's handbook. You should also follow the requirements of your national regulatory authority (the CAA in the UK) and laid down company policy as regards local procedures, recording, report writing, documentation etc For health and safety in the workplace you should follow the regulations/ guidelines as specified by the equipment manufacturer, your company, national safety authorities and national governments.CONTENTS Transistor theory & construction Transistor testing Transistor as an amplifier Directly coupled amplifiers Classes of amplifiers Multivibrators Flip-flops JUGFETs MOSFETs cmos Feedback in amplifiers Oscillators The transistor as a switch Integrated circuits Operational amplifiers Logic circuits SR flip-flops Clocks, JK flip-flops PCBs Page aoe 15 17 21 24 26 28 30 31 33 34 35 38 45 50 51 52 54HOW TO TACKLE THIS BOOK The same applies to this book as applies to book 1. This means that it is written to the B2 level, and B1 people should check the syllabus to see what subjects should be learnt and to what depth. None of the contents applies to the A Line Mechanic. For category B engineers the book will probably need at least 2 read throughs to get the information to sink in ~ in some cases it may need moreTRANSISTORS, Construction and Theory of Operation The bi-polar or junction transistor consists of two P-N junctions in the same crystal. If two P-N junctions were fused together so that the two 'N’ regions form a very thin (0.1 to Imm thick) lightly doped layer between the two more heavily doped ‘P’ regions a PN P transistor is formed. Figure 1 shows the layout of the transistor and its symbol. Note the electrodes are called COLLECTOR, BASE and EMITTER (emitter - the one with the arrow in the symbol). The emitter is more heavily doped than the collector. IC Collector. «= [P| 8 Base B E Emits = € Fig. 1 PNP TRANSISTOR Similarly if two heavily doped 'N’ regions are separated by a very thin lightly doped 'P’ region then an N PN transistor is formed. Figure 2 shows the layout and its symbol. The emitter is again more heavily doped than the collector. Note. For both the PNP and NPN transistors the arrows show the direction of conventional current flow. ic Collector c| ("| [P| 8 Base B Ie] E Emitter E Fig. 2. NPN TRANSISTORAction of NPN Transistor For transistor action to occur the BASE-EMITTER junction must be forward biased (POSITIVE to 'P’, NEGATIVE to 'N’) and the COLLECTOR-BASE junction must be reverse biased (POSITIVE to ‘N’, NEGATIVE to | It should be noted that the battery Es is much smaller than the battery Ez, it must also be of sufficient voltage to overcome the barrier potential of 0.6v for silicon. n poo Electrons 4 iz ae A emitter [ Collector ¥ © Jy Y siectrons ow ot Ee Ec SGA Forward bias Reverse bias Je ARROWS INDICATE ELECTRON FLOW CONVENTIONAL FLOW IS IN THE OPPOSITE DIRECTION Fig. 3 NPN OPERATION Under the influence of the electric field due to battery E- electrons cross the junction into the base. Only a small proportion (about 1 to 2%) of the electrons combine with holes in the base due to it being very thin and lightly doped. Most of the electrons (98 to 99%), under the very strong positive influence of the battery E., are swept through the base to the collector to E¢ to form the collector current in the external circuit Electrons are the majority carriers in the N P N transistor. The small amount of electron-hole combination in the base gives it a momentary negative charge, which is immediately corrected by battery E- supply holes, or can be considered as electron flow. Remember conventional current flow is in the opposite direction. So transistor action is the controlling of a large current in the high resistance (reverse biased) collector-base junction by a small current through the low resistance (forward biased) base-emitter junctionCollector Fig. 4 CONVENTIONAL CURRENT FLOW NPN TRANSISTOR Action of PN P Transistor Again the base-emitter junction is forward biased and the collector-base junction is reverse biased. Under the influence of the electric field due to battery Ee, holes cross the junction into the base. Only 1 to 2% of holes recombine with free electrons in the base due . to it being very thin and lightly doped. The majority of the holes 98 to 99% are aecelerated towards the very strong negative influence of battery Ex. Holes are the majority carriers in the P N P transistor. pon p Collector A © Ty proeceors ove Ee Ec tit Forward bias Reverse bias [ARROWS INDICATE ELECTRON FLOW CONVENTIONAL FLOW IS IN THE OPPOSITE DIRECTION Fig. 5 PNP OPERATION Due to recombination of holes and electrons in the base, the base loses free electrons and will therefore exhibit a positive charge. The electrons will be lattracted by battery Ex into the base to ‘make-up’ for those lost by recombining with holes. Figure 6 shows the conventional current flow through the transistor.of Collector Fig. 6 CONVENTIONAL CURRENT FLOW PNP TRANSISTOR Since the carriers in the N PN and PN P transistors originate at the emitter and distribute themselves between base and collector, the sum of the base and collector currents must always be equal to the emitter current, therefore: le = le + bb Type BFI94 = BCI08.—=—«ZTK300.-—«2N370S.-——-2NSOSS & BFYSI ‘ne . ja prc cBe cee Bce cBeE MMIOB TOI. “Eline (X59) T0928 TOS (metal can (Glascicy ——fmeta) (plant) (plaete} connected to C) Fig. 7 TYPICAL TRANSISTORS - The transistor can be used as an AMPLIFIER circuit and also as a SWITCH. The amplifier action is based on applying a low current to the base-emitter with a higher current flowing through the collector-emitter. The switching action is the effect of applying a small current to the base for the unit (NPN) to ‘switch on’ allowing current to flow between the collector-emitter. Removing the base-emitter current will cause the unit to switch off. These switching times can be very fast (say 2ns or 2 x 10 seconds or 0.000 000-002 seconds} (ns = nano seconds). Fast switching times are needed in computing.Testing Transistors Using an analogue multimeter switched to the ohms range. On most analogue multimeters on the ohms range the negative (-) terminal has a positive polarity and the positive terminal (+) has a negative polarity. This is an important point with regards to identifying N P N and PN P transistors. If a digital multimeter is used then check the polarities of the terminals on the ohms range. Figure 8 shows the readings you would expect using an analogue multimeter. LOW RESISTANCE HIGH RESISTANCE LOWRESISTANCE HIGH RESISTANCE MULTIMETER. alos eee NPN Fig. 8 TESTING TRANSISTORS USING A MULTIMETER LINEAR circuits are amplifying-type circuits. They will have analogue inputs and the output will vary continuously and be more or less an exact but amplified copy of the input, ie the output is a linear representation of the input. Many class A transistor amplifiers, eg audio frequency and radio frequency amplifiers, are linear circuits. TRANSISTOR AS AN AMPLIFIER First of all we need to look at how the bias is applied in a practical circuit. In our previous discussions batteries were used for the bias. Fig. 9 AMPLIFIER CIRCUIT - 1If de only is applied to the circuit shown figure 10 then Ri and Re will divide the supply voltage into the same ratio as that of the resistors. So if the resistor values were 80k and 20k0 then with a supply voltage of 10v the voltages across Ri and R; would be 8v and 2v respectively Fig. 10 AMPLIFIER CIRCUIT - 2 ‘The voltage across must be 0.6V to overcome the barrier potential. This could be achieved by removing Re and making R2 of such a value so that 0.6V is dropped across it, however, the problem here would be Re would have to be quite low and the amplification would ve restricted ‘The voltage across the base emitter junction (Vse) must be 0.6V and is the difference between the voltage across Rz and Re. Var = Veo - Vre. TRY, t ow IR: 2 Lav 4 Fig. 11 AMPLIFIER CIRCUIT - 3 So Re must be of a value that when the standing de current is flowing 1.4 will be dropped across Re leaving Vpe to be 0.6v. So in the static condition, ie de only applied, a standing current (quiescent current] flows through the circuit and TRI, Ri, Rz and Re provide the bias necessary to operate TR1 and allow current to flow.