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Transistors. Eugeniy E. Mikhailov The College of William & Mary Week 6 Eugeniy Mikhailov (W&M) Electronics 1 Week 6 1 / 12 Transistors invented in 1947 amplify current lower power consumption cheap for mass production robust to vibration long working time (decades) when properly used replaced vacuum tube building block of modern electronics Some areas where vacuum tube are still good ultra high voltage applications (more then 1000 V) radiation prone locations Eugeniy Mikhailov (W&M) Electronics 1 Week 6 2 / 12 Bipolar junction Transistor (BJT) NPN-transistor PNP-transistor Collector C N Base Collector C P B Base P N N P E Emitter B E Emitter C C B B E Eugeniy Mikhailov (W&M) E Electronics 1 Week 6 3 / 12 Notation Base-emitter current (Ibe ) C Collector-emitter current (Ice ) Base-emitter voltage difference (Vbe = Vb − Ve ) Collector-emitter voltage difference (Vce = Vc − Ve ) Ice B Ibe E Eugeniy Mikhailov (W&M) Electronics 1 Week 6 4 / 12 Simple NPN-transistor rules To support shown currents direction C Ice B Ibe E C B E Eugeniy Mikhailov (W&M) Electronics 1 Week 6 5 / 12 Simple NPN-transistor rules To support shown currents direction Vce > 0 C Ice B Ibe E C B E Eugeniy Mikhailov (W&M) Electronics 1 Week 6 5 / 12 Simple NPN-transistor rules To support shown currents direction Vce > 0 Vbe > 0 C since, it is forward biased diode Vbe ≈ 0.6 V Ice B Ibe E C B E Eugeniy Mikhailov (W&M) Electronics 1 Week 6 5 / 12 Simple NPN-transistor rules To support shown currents direction Vce > 0 Vbe > 0 C since, it is forward biased diode Vbe ≈ 0.6 V Vcb > 0 since, it is reversed biased diode, no current goes from collector to base, all collector current is directed to emitter if Vcb < 0 transistor goes to saturation and cannot be described by the following simple rule. Ice B Ibe E C B E Eugeniy Mikhailov (W&M) Electronics 1 Week 6 5 / 12 Simple NPN-transistor rules To support shown currents direction Vce > 0 Vbe > 0 C since, it is forward biased diode Vbe ≈ 0.6 V Vcb > 0 since, it is reversed biased diode, no current goes from collector to base, all collector current is directed to emitter if Vcb < 0 transistor goes to saturation and cannot be described by the following simple rule. Ice B Ibe E C B If above holds true then E Eugeniy Mikhailov (W&M) Electronics 1 Week 6 5 / 12 Simple NPN-transistor rules To support shown currents direction Vce > 0 Vbe > 0 C since, it is forward biased diode Vbe ≈ 0.6 V Vcb > 0 since, it is reversed biased diode, no current goes from collector to base, all collector current is directed to emitter if Vcb < 0 transistor goes to saturation and cannot be described by the following simple rule. Ice B Ibe E C B If above holds true then Ice = βIbe thus a BJT is a current amplifier Eugeniy Mikhailov (W&M) Electronics 1 E Week 6 5 / 12 Simple NPN-transistor rules To support shown currents direction Vce > 0 Vbe > 0 C since, it is forward biased diode Vbe ≈ 0.6 V Vcb > 0 since, it is reversed biased diode, no current goes from collector to base, all collector current is directed to emitter if Vcb < 0 transistor goes to saturation and cannot be described by the following simple rule. Ice B Ibe E C B If above holds true then Ice = βIbe thus a BJT is a current amplifier E the static forward current transfer ratio β (or sometimes hfe ) ≈ 100 . . . 200 Eugeniy Mikhailov (W&M) Electronics 1 Week 6 5 / 12 Simple NPN-transistor rules To support shown currents direction Vce > 0 Vbe > 0 C since, it is forward biased diode Vbe ≈ 0.6 V Vcb > 0 since, it is reversed biased diode, no current goes from collector to base, all collector current is directed to emitter if Vcb < 0 transistor goes to saturation and cannot be described by the following simple rule. Ice B Ibe E C B If above holds true then Ice = βIbe thus a BJT is a current amplifier E the static forward current transfer ratio β (or sometimes hfe ) ≈ 100 . . . 200 Ie = Ibe + Ice = (β + 1)Ibe ≈ βIbe Eugeniy Mikhailov (W&M) Electronics 1 Week 6 5 / 12 Simple PNP-transistor rules Apply the same rules as before for NPN BJT but multiply currents and voltages by -1. Hints the arrow indicates the direction in which current is supposed to flow. the arrow always connects the base and emitter. C Ice B Ibe E C B E Eugeniy Mikhailov (W&M) Electronics 1 Week 6 6 / 12 Design considerations for β Remember β is not a constant! It depends on many parameters temperature collector current varies from device to device even in the same batch Good design should not depend on β value. Eugeniy Mikhailov (W&M) Electronics 1 Week 6 7 / 12 Constant current source Current through the load resistor does not depend on the load resistance. Vcc IL = Ic = βIbe = β Vctrl − .6V Rset RL A Vctrl Rset B C E Eugeniy Mikhailov (W&M) Electronics 1 Week 6 8 / 12 Constant current source Current through the load resistor does not depend on the load resistance. Vcc RL IL = Ic = βIbe = β Vctrl − .6V Rset This is actually a sample of bad design since the current through the load depends on β. A Vctrl Rset B C E Eugeniy Mikhailov (W&M) Electronics 1 Week 6 8 / 12 Constant current source Current through the load resistor does not depend on the load resistance. Vcc RL IL = Ic = βIbe = β This is actually a sample of bad design since the current through the load depends on β. A Vctrl Rset B Vctrl − .6V Rset Vc = Vcc − RL IL C E Eugeniy Mikhailov (W&M) Electronics 1 Week 6 8 / 12 Constant current source Current through the load resistor does not depend on the load resistance. Vcc RL IL = Ic = βIbe = β This is actually a sample of bad design since the current through the load depends on β. A Vctrl Rset B Vc = Vcc − RL IL C E remember that Vc must be > Vb thus current cannot be bigger then the saturation current Isat = max(IL ) ≤ Eugeniy Mikhailov (W&M) Vctrl − .6V Rset Electronics 1 Vcc − Vb Vcc ≈ RL RL Week 6 8 / 12 Constant current source (continued) From Vcc point of view, left schematic is equivalent to the right one. Vcc Vcc Rtrans RL A Vctrl Rset B Vcc − IL RL Vc = = IL IL C RL A C Rtrans E E Transistor Tran(sform)-(re)sistor Eugeniy Mikhailov (W&M) Electronics 1 Week 6 9 / 12 Constant current source. Power dissipation. Transistor power dissipation Ptrans = Pbe + Pce = Vbe Ibe + Vce Ice Vcc Since Vbe ≤ Vce , Ibe = Ice /β Ice , and Ice = IL Vcc RL RL A Vctrl Rset B C Ptrans ≈ Vce Ice = Rtrans IL2 A C Rtrans E Eugeniy Mikhailov (W&M) E Electronics 1 Week 6 10 / 12 Constant current source. Power dissipation. Transistor power dissipation Ptrans = Pbe + Pce = Vbe Ibe + Vce Ice Vcc Since Vbe ≤ Vce , Ibe = Ice /β Ice , and Ice = IL Vcc RL RL Ptrans ≈ Vce Ice = A Vctrl Rset B C E Rtrans IL2 Maximum power dissipation in transistor Eugeniy Mikhailov (W&M) Electronics 1 A C Rtrans E Week 6 10 / 12 Constant current source. Power dissipation. Transistor power dissipation Ptrans = Pbe + Pce = Vbe Ibe + Vce Ice Vcc Since Vbe ≤ Vce , Ibe = Ice /β Ice , and Ice = IL Vcc RL RL Ptrans ≈ Vce Ice = A Vctrl Rset B C E Rtrans IL2 Maximum power dissipation in transistor is when Rtrans = RL Eugeniy Mikhailov (W&M) Electronics 1 A C Rtrans E Week 6 10 / 12 Constant current source. Power dissipation. Transistor power dissipation Ptrans = Pbe + Pce = Vbe Ibe + Vce Ice Vcc Since Vbe ≤ Vce , Ibe = Ice /β Ice , and Ice = IL Vcc RL RL Ptrans ≈ Vce Ice = A Vctrl Rset B C E Rtrans IL2 Maximum power dissipation in transistor is when Rtrans = RL max(Ptrans ) = Eugeniy Mikhailov (W&M) A C Rtrans E 2 Vcc Vcc , when IL = 4RL 2RL Electronics 1 Week 6 10 / 12 Voltage controlled switch When properly designed outcome does not depend on reasonable variations of β Recall that typically β = 100 . . . 200 We will assume the worst case scenario β = 10 Vcc Notice that RL limits collector current Bulb RL IL = Vctrl Rb Eugeniy Mikhailov (W&M) Vcc RL Ibe = Vctrl − .6V IL = Rb β Rb ≤ Vctrl − .6V βRL Vcc Electronics 1 Week 6 11 / 12 Emitter follower Vcc Vout = Vin − 0.6V Vin Vout RL Eugeniy Mikhailov (W&M) Electronics 1 Week 6 12 / 12 Emitter follower Vcc Vout = Vin − 0.6V Gain. What gain? Vin Vout RL Eugeniy Mikhailov (W&M) Electronics 1 Week 6 12 / 12 Emitter follower Vcc Vout = Vin − 0.6V Gain. What gain? We achieved the input impedance increase. Vin Vout RL Eugeniy Mikhailov (W&M) Rinput = Vin ≈ RL (β + 1) Ibe As result our Vin source is not overloaded and our load receive all required current (as long as the collector power supply can support it). Electronics 1 Week 6 12 / 12