EIE209 Basic Electronics Transistor Devices Contents BJT and FET Characteristics Operations 1
What is a transistor? Three-terminal device whose voltage-current relationship is controlled by a third voltage or current We may regard a transistor as a controlled voltage or current source. v c i c i v 2
Types of transistors According to the physics of the device, we can classify transistors into two main classes: 1. Bipolar junction transistors (BJT) 2. Field effect transistors (FET) Conduction is controlled by electric field which is produced by voltage applied to the control terminals. So, the control draws no current and FET is a voltagecontrolled device. Diode-based device which is usually blocked unless the control terminals are forwardbiased. So, the control is a current, and BJT is a current amplifier by nature. 3
Bipolar junction transistor (BJT) 2 types of BJT devices collector collector C C B base B E E emitter emitter npn transistor pnp transistor Basic model C C a bit of physics Consider the npn BJT. The collector-base junction is reverse-biased. So, no current can flow down. But if the base-emitter junction is forwardbiased ( 0.6V), then the diode contact potential barrier can be overcome. Electrons can go to base called base injection. These electrons are minority carriers, which are strongly attracted/captured by the collector. Hence, current flows down from collector to emitter. B I B I C E npn B I B I C THUS, we use a small base current to induce a large collector current. This large collector current is proportional to the base injection. I C = b I B E pnp 4
Simple BJT model Base I B Collector I C Emitter Consider npn transistor. Collector is more positive than Emitter. B-E and B-C junctions are pn junctions, like diodes. In normal operation, B-E is forward biased and B-C is reverse biased. Main relation: Base Collector I B n p n I C I C = bi B b 100 typically Emitter 5
Some properties Base I B Collector I C Emitter V BE 0.6 V when the transistor turns on. Never try to stick a large voltage across V BE because it may produce enormous current or may just kill the device! b is a bad parameter. Don t trust the databook. Its value can vary to ±50% or more. I B I C I C = bi B holds only under some carefully set conditions. We ll look at it later. I E = I B I C 6
Typical operations 1. Cut-off 2. Active operation 10 V 3. Saturation Determining factors: How large is I B or V BE How large is R L I B R L I C V BE 7
Cut-off When the B-E junction is not forward-biased, the transistor is basically not doing anything. This is called CUT-OFF. 10 V 0A V BE = 0 8
Active operation When the following holds: I C = bi B the BJT is said to be in active operation. This is the case of current amplification. But we need I C R L < 10V I B V BE R L 10 V I C 9
Condition for active operation: I C R L <V CC Let b = 100. 10 V 10 V 10 V 1kΩ I C =1mA 5kΩ I C =1mA 10kΩ I C =1mA I B =10µA I B =10µA I B =10µA V CE = 9V V CE = 5V V CE = 0V How about 11kΩ? 10
Saturation When V CE is reduced to 0, the BJT is saturated. 15kΩ I B =10µA 10 V I C =0.6667mA V CE = 0V I C cannot be 1mA!! In fact, it must drop in order to make up for the total voltage. In this case, I C = 10V/15kΩ = 0.6667mA I C = bi B 11
What makes it saturate? Large R L Large I B 10 V 10 V 10kΩ I C =1mA 1kΩ I C =10mA I B =10µA V CE = 0V just saturate! I B =100µA V CE = 0V just saturate! 12
Application: BJT as switch Saturation I B = (10-0.7)V 1kW = 9.3mA I C = 100x9.3 = 0.93A which is too large and surely saturates the BJT!!! So, I C 0.1A. Situation 1 1kW B 10V C 10V 0.1A 100Ω lamp Light bulb turns on. Cut-off Situation 2 E I B = I C = 0. Light bulb turns off. 13
Detailed BJT characteristics Input characteristic (I B versus V BE ) Obviously, V BE and I B are related by diode characteristic. I B Ê I B = I ss e V BE /V T ˆ Á Ë -1 thermal voltage Boltzman s constant absolute temperature V T = kt q 25 mv @room temp 0.6 V BE electronic charge 14
Detailed BJT characteristics Transfer characteristic (I C versus V BE ) Also, I C is just I B multiplied by b. I C = b I B Ê I C = bi ss e V BE /V T ˆ Á Ë -1 = I Ê e V BE /V T -1 ˆ Á s Ë thermal voltage Boltzman s constant same shape as I B V T = kt q absolute temperature 25 mv @room temp 0.6 V BE electronic charge 15
Detailed BJT characteristics Output characteristic (I C versus V CE ) I C is nearly flat unless near saturation. I C I C for one particular choice of I B or V BE V CE Not Ohm s law!! V CE 16
Important small-signal characteristic I B Different I B (or V BE ) has different output characteristic. A range of V BE corresponds to a range of I C. 0.6 V V BE Transconductance: g m = DI C DV BE I I C1 C slope = g = I C1 m 0.025 W1 I C saturation active V =0.68 V BE V =0.65 V BE V =0.60 V BE = slope on the transfer char. 0.6 0.65 0.68 V V BE 0.2 V V =0 (cut-off) BE V CE 17
What is g m? g m = DI C DV BE A simple differentiation gives g m = di C dv BE = d Ê bi ss (e V BE /V T ˆ Á -1) dv Ë BE = bi ss (e V BE /V T ) 1 V T ª I C V T or I C 25mV at room temp 18
A bit more precise At saturation, V CE is not really 0, it is about 0.2 V. I C 10 V for one particular choice of I B or V BE 10kΩ I C =0.98mA not 1mA!! I B =10µA V CE = 0.2V 0.2V V CE 19
A bit more precise In active region, I C is not really flat. It goes up gently! This is called Early Effect! I C I C for one particular choice of I B or V BE slope I C / V A Early voltage typically 100V 0.2V V CE 20
Field Effect Transistors (FET) Drain Gate channel Current goes down from D to S, controlled by the gate voltage at G. Source Two kinds of channels: n-channel FET p-channel FET Two kinds of gate electrodes: Junction FET (JFET) Metal-oxide-semiconductor FET (MOSFET) 21
Terminology confusion Before we move on, it is important to clarify some possible confusions due to terminology difference. BJT saturation region active region cut-off FET triode region saturation region cut-off 22
n-channel MOSFET S G D SiO 2 insulator n p n The channel is not conducting initially when gate is zero volt. body or substrate When gate is ve, electrons are attracted to it and this becomes n-type conducting channel. This action is called channel enhancement. 23
n-channel MOSFET characteristic Drain Characteristics: Gate current = 0 (always) The channel conduction is determined by V GS Gate V G S Source triode V GS =2V saturation V GS =1.9V (like active in BJT) V GS =1.8V Threshold voltage V th = 1.7 V, for example. V GS =1.7V 24
Saturation region Drain So, it looks like the npn BJT!! But if we look closer, we find that the saturation current is proportional to (V GS V th ) 2. Gate V G S Source = K (V GS V th ) 2 for saturation region saturation V GS =2V V GS =1.9V (like active in BJT) V GS =1.8V Threshold voltage V th = 1.7 V, for example. V GS =1.7V 25
Saturation region Drain = K (V GS V th ) 2 for saturation region If we plot the saturation versus V GS, we have a quadratic (parabolic) curve. Gate V G S Source V th V G S 26
Triode region Triode region like a quadratic (parabolic) function K (V GS V th ) 2 So, the equation is: y = a x (2M x) = a (2M ) Obviously, M = V GS V th, a = K, = K [2 (V GS V th ) ] M 2M = V GS V th for triode region 27
n-channel MOSFET characteristic Complete model summary: triode region (quadratic in ) = K [2 (V GS V th ) ] K (V GS V th ) 2 saturation region (flat) = K (V GS V th ) 2 V GS V th 28
Example (biasing in saturation) 10V V th = 3V By using load line 5V 2kΩ = 0.5x2 2 = 2mA K = 0.5 mav 2 load line slope = 1/2k 53 = 2V 6V 10V 29
Example (biasing in saturation) 10V V th = 3V By analysis 2kΩ K = 0.5 mav 2 5V = 0.5x2 2 = 2mA = 10 2x2 = 6V which is > 2 OKAY! What happen if a 4.5kΩ is used? 53 = 2V 6V 30
Example (biasing in triode) 5V 4.5kΩ 10V V th = 3V K = 0.5 mav 2 = 0.5x2 2 = 2mA = 10 4.5x2 = 1V Oops!!? = 1.88mA = 1.54V So, it is in the triode region. = K [2 (V GS V th ) ] = 0.5 (4 ) = 0.5 (104.5 )(410 4.5 ) i.e., 10.125 2 35 30 = 0 = 1.8845 ma or 1.5722 ma 53 = 2V And 1.88mA gives = 104.5x1.88=1.54V. But 1.57mA gives = 104.5x1.57=2.95V!! 31
Enhancement and depletion MOSFET What we have just studied is the enhancement MOSFET. Enhancement the channel is originally not conducting when gate voltage is 0, and we have to apply a positive gate voltage (bigger than a threshold V th ) to make it conduct (enhance it). Depletion mode Enhancement mode Depletion In fact, we also have another kind of MOSFET, in which the channel can conduct even when gate voltage is not applied. Then, we need to apply reverse gate voltage to cut it off. This is called depletion MOSFET. V th V th V G S NOTE THAT DUE TO A SEMICONDUCTOR DOPING PROPERTY, For n-channel MOSFET, both enhancement and depletion types can be made. For p-channel MOSFET, only enhancement type can be made. 32
Junction FET (JFET) Drain Current can flow initially because plenty of electrons are available in the channel. Gate p n p Gate : Apply negative voltage to increase the depletion width, so as to reduce the current. When the gate voltage is negative enough, current will stop. Hence, this is a depletion device. Source depletion region width depends on the magnitude of the gate reverse bias 33
Junction FET (JFET) Drain Drain Gate ve voltage applied to reduce current p n p Source Gate more ve voltage Channel becomes narrower p n p Source 34
Pinch off in JFET Drain Gate more ve voltage p n p Channel pinch off; Current stops V p V G S Source Pinch-off voltage V p surely depletion type 35
n-channel JFET characteristics Drain The characteristics are very similar to those of MOSFET. But, now the threshold is a negative value, which is called the pinch-off voltage V p instead of threshold voltage. Gate V G S Source Pinch-off voltage of this JFET is V p = 2 V triode V GS =2V saturation V GS =1V (like active in BJT) V GS =2V V GS =0V 36
n-channel JFET characteristics Drain Everything is almost the same!! Gate V G S triode region (quadratic in ) = K [2 (V GS V p ) ] Source Pinch-off voltage of this JFET is V p = 2 V K (V GS V p ) 2 saturation region (flat) = K (V GS V p ) 2 Be careful about sign! V GS can be negative or positive, but V p is negative. V GS V p 37
Example (biasing in saturation) 10V 10kΩ = K (V GS V p ) 2 = 0.2(2) 2 = 0.8mA VDS = 10 10x0.8 = 2V just okay in saturation! V p = 2 V K = 0.2mA/V 2 But if the resistor is more than 10kΩ, it will be in triode region! 0(2)=2V 38
Example (biasing in triode) 10V 12kΩ VDS = K (V GS V p ) 2 = 0.2(2) 2 = 0.8mA = 10 12x0.8 = 0.4V < 2V So, it can t be in saturation! V p = 2 V K = 0.2mA/V 2 Recalculate : = K [2 (V GS V p ) ] = 0.2 (1012 )[2x2(1012 )] i.e., 28.8 2 37.4 12 = 0 = 0.7195mA or 0.5791mA And, = 0.7195mA gives = 1.366V ---okay But, = 0.5791mA gives = 3.051V --- reject! 0(2)=2V = 0.7195mA = 1.366V 39
Example (biasing in triode) 10V 12kΩ VDS = K (V GS V p ) 2 = 0.2(2) 2 = 0.8mA Load line V p = 2 V K = 0.2mA/V 2 Of course, you may also solve it by using load line. 0(2)=2V 10V = 0.7195mA = 1.366V 40
Important small-signal characteristic Similar to BJT!!! V GS =1.7V V GS =2V V GS =1.9V V GS =1.8V Consider only the saturation region. If we change V GS in a small range, then also changes in a range. The ratio of the change in to the change in V GS is called transconductance. g m = D DV GS which is the slope of the curve versus V GS, or analytically, g m = d dv GS = d dv GS K (V GS -V th ) 2 = 2K (V GS -V th ) = 2 K K (V GS -V th ) = 2 K 41
Other FETs So far, we have only talked about 1. n-channel MOSFET (enhancement type) 2. n-channel JFET (depletion type) Other FETs: similar to npn BJT enhancement FET MOSFET n-channel MOSFET p-channel MOSFET depletion enhancement JFET n-channel JFET p-channel JFET depletion depletion 42
p-channel FETs Operation is almost the same as n-channel FETs. Voltage polarity and current direction reversed. BUT for p-channel devices, the carriers are holes (not electrons). So, mobility is lower and minority carrier lifetime shorter. Consequence: p-channel devices are usually POORER! higher threshold voltage, higher resistance, and lower current capability. 43