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Transcript
Bipolar Junction Transistors (BJTs) It is a three terminal semiconductor device consists of two p-n junctions. The basic principle involved is to use the voltage between two terminals to control the current flowing in the third. There are two types of BJT (npn and pnp) * The transistor is design to have heavily doped Emitter and very thin and lightly doped base. Depending on the bias condition (forward or reverse) of the junctions, different modes of operation are obtained. BJT modes of operation: Mode Cutof EBJ CBJ Reverse Reverse Saturation Forward Forward Active Application Switching Applications logic circuit Forward Reverse Amplifier pnp transistor in the active mode 1) The forward bias voltage on Emitter –base junction causes current to flow across this junction. Two components: holes injected from emitter into the base (Ip) and electron injected from base into the emitter (In). Ip is much higher than In (base is thin and lightly doped). 2) Most of the holes ( I P ) injected into the base will diffuse (continue their way) through the base toward the collector. This is because of that the base is thin and attracted by the large negative of the battery (VBC). Some of the holes (1 ) I P will attracted by the negative of the battery (VBE ). (i.e. some of the holes will combine with the base electrons and disappear). These electrons will be substituted by electrons from the battery) 3) Collector-Base reverse current (ICBo) Minority carrier base holes (Ipo) and collector electrons (Ino) generates reverse current across the collector base junction (Ico) or leakage current. 4) Transistor terminal currents: IE I p In I C I p I po I no I B (1 ) I p I n I po I no Note that I C I B I p I po I no I p I p I n I po I no I p I n I E (i.e. The current enters the transistor should be equal current leaving it ) Summary pnp transistor in active Current Change carrier IP Majority holes injected from emitter into base In Majority electrons injected from base into emitter I P Most of IP that go to collector (1 ) I P Some of IP that do not go to collector Ipo Minority holes injected from base to collector Ino Minority electrons injected from collector to base 5) IB is very small and usually of the order of A 's or nA 's for most practical cases with (collector to emitter-current gain) 1 I C I E by making the deflection area very small.(Base very thin) 6) Current relations in active mode I p I co IC 100 (Base collector current gain) I B (1 ) I p I n I co 1 I E IC I B I C I E I C I B 7) The emitter current IE can be expressed as I E I SE (eVBE / VT 1) (Forward bias pn junction current) I C I E I SC (eVBE / VT 1) ISC: constant called saturation current or current scale factor npn transistor IE I p In I B (1 ) I n I p I po I no I C I n I po I no conclusion In the active mode, the forward-bias voltage VBE cause an exponentially related current iC to flow in the collector terminal. The collector current iC is almost independent of the value of the collector voltage as long as the collector –base junction remains reversebiased. In the active mode, the collector terminal behaves as a current source and its value is determined by VBE . ic –VCE characteristics VCE VBC VBE 1) when VCE 0 VBC 0.7 (Forward) IC 0 2) VCE 0.1 VBC 0.6 (Forward) I C Start appearing 3) VCE 0.2 VBC 0.5 (Forward) 4) VCE 0.3 VBC 0.4 (in between) 5) VCE 0.4 VBC 0.3 (Revere) Summary Cutoff Active Saturation Both junction Reverse BEJ Forward Both junction Forward biased BCJ Reverse biased All current practically zero IC I B V BE 0.7 V CE unknown I E IC I B Symbol I B IC V BE 0.7V V CE 0 0.2V or max 0.3 I E IC I B BJT For npn transistor VBE VB VE 0.7 to be ON 1) If VBE 0.7 cutoff I B I E I C 0 2) To operate in active region BCJ should be reverse in active region only: I C I B ; I E (1 ) I B ; I C I E 3) To operate in saturation region BCJ should be forward ( I B VCE 0.2V only in saturation region For pnp transistor VBE and VCE are negative and in saturation only VEC 0.2V IC ) DC analysis Example: Determine the region of operation and the values of I B , I C and VCE for the circuit shown in the figure a) RB 300 k b) RB 150 k F 100 Assume the transistor is working Saturation VCE 0.2 Check is ( I B IC ) ? 10 0.7 0.031mA 300k I 10 0.2 IC 4.9mA C 0.049mA I B 2k IB This means that the assumption is wrong. Assume the transistor is working in active mode IC I B but check BCJ is reverse I B 0.031mA I C 3.1mA VC 10 RC I C 10 (2)(3.1) 3.8V VCE VC VE 3.8 0 3.8V VBC VB VC 0.7 3.8 3.1V BCJ is reverse which confirms the assumption or since VCE>0.3 the assumption is verified. Find the smallest value of RB for which the transistor still operate in the active mode. At the edge of saturation and active. VCE 0.3V 10 0.3 4.85mA 2k I I B C 0.0485mA IC IB 10 0.7 0.0485mA RB RB 9.3 191.8k 0.0485mA For the circuit shown find VB, I B , I E , I C Assume the transistor is working in active mode IC I B but check BCJ is reverse 20 I B 0.7 5 I E 5 I E (1 ) I B 53.3I B 286.5 I B 4.3 I B 0.015mA I E I C I B (1 ) I B 0.8mA VB 20 I B 0.3V VE VB 0.7 1V VC 5 5I C 1.08V VBC VB VC 0.3 1.08 1.38V Re verse Active VEC VE VC 1 (1.08) 2.08V >0.3V Active Note: β is very large I B 0 and IE=IC 52.3 Find all the currents and voltages? 100 Assume the transistor working in the active I C 100 I B check BCJ is reverse 10 10 I E 0.7 10 I B 10 I E 10 I E ( 1) I B I B 9.5 A I C 0.95mA I E 0.96mA VC 10 10 I E 0.5V VB VC 10 I B 0.405V VE VB 0.7 0.3V VBC 0.405 0.5 0.1V VCE 0.5 (0.3) 0.8V 0.3 (Re verse active) (active) Small signal equivalent circuit Amplifier IB IS q VBE kT (e 1) vbe VBE v BE iB I S q vBE kT (e 1) q q q (VBE vbe) vbe VBE kT kT kT .e ISe I B ib I S e IS IS q VBE kT e q VBE kT e IB IS vbe ib IS q vbe ) kT q VBE kT e q vbe kT q VBE e kT q VBE kT e q vbe kT kT / q 1 q IB IB. kT 1 25 1000 I B 40 I B ib I S vbe ib (1 I C I B iC i B I C iC ( I B i B ) I C I B iC i B Small signal ac equivalent circuit of the BJT Transconductance g m ib IC VT vbe 40 I B vbe 1 40 I B 40 I C vbe g m vbe g m 40 I C g m r 40 I C vbe 1 g m r 40 I B