Characteristic Curves• To desribe the behavior of a three terminal devices

requires two sets of characteristics- input or driving point characteristics- output characteristic

• The input set show relationship between input current and input voltage for various levels of output voltage.

• The output set relates an output current to an output voltage for various levels of input current.

• Example : Common emitter configuration

Input : IB, VBE

Output : IC , VCE

Input or driving point characteristics (common emitter configuration)

Example : Common emitter configuration

Output characteristics (common emitter configuration)

• The output characteristics has 3 basic regions : active/linear; cutoff and saturation region.

• Active/linear region : BE forward biased, BC reverse biased

• Cutoff region : BE and BC reverse biased (define as the region where the IB = 0A).

• Saturation region : BE and BC forward biased (define as the region of the characteristics to the left of VCE = 0 V).

Collector characteristics curve

• To show how the collector current, IC varies with with VCE for specified values of IB. (VBB and VCC are variable source of voltage).

• Assume VBB is set to produce certain value of IB and VCC = 0V. BE and BC junction are forward-bias because B = 0.7V, E and C = 0V. IB is through BE because low impedance to ground, IC = 0. Transistor is in saturation region.

• As VCC increase, VCE increase as IC increase shown by point A to B. IC increase as VCC increase because VCE < 0.7 V due to BC forward-bias.

• When VCE > 0.7V, BC become reverse-bias and transistor goes into active or linear region. IC remain constant for a given value of IB as VCE continue increase. Actually IC increase (IC = β IB) slightly as VCE increase due to widen BC depletion region, because few hole for recombination, so β increase slightly.

• Relationship IC and VCE with IB. With greater increases of VCC , VCE continues to increase until it reaches breakdown (point C), but IC about the same in the linear region from 0.7V to the breakdown voltage.

DC load line

Cutoff and saturation can be illustrated in relation to collector curve by the use of a load line. Ideal cutoff is IC = 0 and VCE = VCC and saturation when IC = IC(sat) = and VCE = VCE(sat).. In between cutoff and saturation along the load line is active region of transistor operation.

Exercise

a. Assume, VCE(sat) = 0.2V, determine whether or not transistor in saturation region.

Beta, ßDC

• ßDC or hFE for a transistor is not always constant. Temperature and collector current both affect beta.

• Keep junction temperature constant and increase IC causes ßDC to increases to maximum. Further increase IC beyond this maximum point cause ßDC to decrease. If IC is held constant and temperature is varied, ßDC change directly with temperature. If temperature inc, ßDC increase and vice versa.

• Data sheet usually specifies ßDC at specific IC value and it varies from device to device due to inconsistencies in the manufacturing process.

Maximum Transistor Rating • BJT has limitations on its operation.• Limitations are stated in form of maximum power ratings. The data sheet

provides information on these characteristics. • Typically, maximum rating are given for VCE, VCB, VEB, IC and power

dissipation.• Power dissipation, PD(max) = IC x VCE (the product of VCE and IC must not

exceed the maximum power dissipation).

Maximum power dissipation curve and tabulated values

Data Sheet• Maximum collector-emitter voltage with base open, VCEO = 40V.• Maximum collector current, ICmax = 200 mA, ß or hFE is specified at several IC.• The collector-emitter voltage VCEsat is 0.2V maximum for ICsat = 10 mA and

increase with the current.

Transistor Applications

BJT as an Amplifier• Amplification is the process of linearity increasing the amplitude of

an electrical signal.• dc current

- VBE, VCB, VCE are dc voltages from one transistor terminal to another.

- VB, VC, VE are dc voltages from transistor terminal to ground.

- IB, IC and IE are the dc transistor currents.

- RE is an external dc emitter resistance • ac current

- Vbe, Vcb, Vce are ac voltages from one transistor terminal to another.

- Vb, Vc, Ve are ac voltages from transistor terminal to ground.

- Ib, Ic and Ie are the ac transistor currents.- Internal ac resistances, r’

- Internal ac emitter resistances, r’e.

- Re is an external ac emitter resistance

Amplification of a relatively small ac voltage can be had by placing the ac signal source in the base circuit.

Recall that small changes in the base current (Ib) circuit causes large changes in collector current (Ic) circuit.

The small ac voltage causes the Ib to increase and decrease accordingly and with this small change in current the Ic will mimic the input only with greater amplitude.

• Forward bias b-e junction present low r’e (internal ac emitter resistance) and appears in series with RB, so

Vb = Ie/r’eIe = Ic = Vb/r’eVc = IcRC

= IeRC

Av = Vc/Vb = IeRC/(Ier’e) (voltage gain define by Vout to Vin)

= RC/r’eExercise : Calculate Av and ac output voltag if r’e = 70 Ω

BJT as Switch

• A transistor when used as a switch is simply being biased so that it is in cutoff (switched off) or saturation (switched on). Remember that the VCE in cutoff is VCC and 0 V in saturation.

• In cutoff condition, BE junction is not forward-bias, all current is zero and VCE(cutoff) = VCC.

• In saturation condition, BE junction is forward-bias, so enough IB to produce max IC, and transistor is saturated.IC(sat) = (VCC – VCE(sat)) /RC, VCE(sat) << VCC, so neglected.IB(min) for saturation = IC(sat) / ß

Exercise : a. Calculate VCE when Vin = 0V.b. Calculate IB(min) to saturated if ß= 100, neglect VCE(sat) c. Calculate RB max when Vin = 5V

A simple application of a transistor switch

• A square wave input voltage with a period of 2 s is applied to the input.

• When 0 V, the transistor is cutoff, no Ic, then the LED does not emit light.

• When input is high level, the transistor saturates, LED is forward biased, Ic flow though the LED and causes it to emit light.

Transistor Categories and Packaging-General Purpose/Small Signal-

General Purpose transistor is used for low or medium power amplifier or switching circuit. Packages are either plastic or metal.

Typical multiple-transistor packages.

Power TransistorsUsed to handle large current more 1A and/or large voltage. Package are metal tab or case and a heat sink for heat dissipation

. RF Transistors

Design to operate at extremely high frequencies and used in communication system. The shape and lead configuration are designed to optimize certain high-freq parameters.

Typically Faults

Troubleshooting

Testing a transistor can be viewed more simply if you view it as testing two diode junctions. Forward bias having low resistance and reverse bias

having infinite resistance.

The diode test function of a multimeter is more reliable than using an ohmmeter. Make sure to note whether it is an npn or pnp and polarize the test leads accordingly.

Testing a defective npn transistor. Leads are reversed for a pnp transistor.

DC Operating Point• The goal of amplification in most cases is to increase the amplitude of

an ac signal without altering it. If not correctly bias, output can goes into saturation or cutoff.

• In (a), output signal is an amplified replica of the input signal (it is inverted which mean 1800 out of phase with the input).

• In (b), illustrates limiting of the positive portion of the output voltage as a result of a Q-point (dc operating point) being too close to cutoff.

• In (c), illustrates limiting of the negative portion of the output voltage as a result of a Q-point (dc operating point) being too close to saturation.

DC Operating Point-Graphical Analysis-

• How to draw a dc load line for a given biased transistor circuit?

• Figure 5.2(a) shows a transistor that is biased with VCC and VBB to obtain IB, IC, IE and VCE. Figure 5.2(b) shows the characteristic curves. These curve will be used to graphically illustrate the effects of dc bias.

Figure 5 : A dc-biased transistor circuit with variable bias voltage (VBB) for generating the collector characteristics curves shown in part (b)

• For a transistor circuit to amplify it must be properly biased with dc voltages.

• The dc operating point between saturation and cutoff is called the Q-point.

• The goal is to set the Q-point such that it does not go into saturation or cutoff when an ac signal is applied. Therefore assign 3 values of IB and observe what happens to IC and VCE.

VBB is adjusted to produce IB = 200µA, so IC = 20mA and VCE = VCC – ICRC = 10 – (20mA)(220Ω) = 5.6V, Q1

If IB = 300µA, so IC = 30mA and VCE = VCC – ICRC = 10 – (30mA)(220Ω) = 3.4V, Q2

If IB = 400µA, so IC = 40mA and VCE = VCC – ICRC = 10 – (40mA)(220Ω) = 1.2V, Q3

Q1, Q2 and Q3 is shown by diagram below.VCC – ICRC – VCE = 0

IC = -(1/RC)VCE + VCC/RC y = mx + b stright line eq.

Illustration of Q-point adjustments.

• By joining the 3 points, the straight line established and can be drawn

(known as load line)

• Recall that the collector characteristic curves graphically show the relationship of IC and VCE for different IB. With the DC load line superimposed across the collector curves for this particular transistor we see that 30 mA of collector current is best for maximum amplification, giving equal amount above and below the Q-point. Note that this is three different scenarios of IC being viewed simultaneously.

• Linear operation : the region along the load line including all points between saturation and cutoff generally known as linear region.

• As long as the transistor is operated in this region, the output voltage is ideally a linear reproduction of the input.

• Q-point correspond to the zero values of the sine wave as indicated (VCEQ, ICQ, IBQ are dc Q-point values with no input sinusoidal voltage applied).

Transistor being driven into saturation and/or cutoff.

Exercise : Assume ßDC = 200, find IBmax for linear operation.

IBQ = (VBB – VBE)/RB = (10 – 0.7)/47KΩ = 198µA

ICQ = ß IB = 200 x 198 µA = 39.6 mA

VCE = VCC – ICRC = 20V – 13.07V = 6.93V

Ideal saturation, IC(sat) = VCC/RC = 20V/330Ω = 60.6 mA

ICpeak = IC(sat) – ICQ = 60.6mA – 39.6mA = 21mA (Q-point is closer to saturation than to cutoff)

IBpeak = ICpeak/ßDC = 21mA/200 = 105µA