OTHER BIAS METHODS 241

Use Equation 5–7 to determine IE.

From IE, you can determine IC and VCE as follows:

Related Problem What value of is required in this example in order to neglect RIN(BASE) in keepingwith the basic ten-times rule for a stiff voltage divider?

bDC

VCE = VC - VE = -3.79 V + 2.71 V = 1.08 V

VE = -IERE = -(1.23 mA)(2.2 kÆ) = -2.71 V

VC = -VCC + ICRC = -6 V + (1.23 mA)(1.8 kÆ) = -3.79 V

IC = IE = 1.23 mA

=3.15 V

2.57 kÆ= 1.23 mA

IE =-VTH + VBE

RE + RTH>bDC=

2.45 V + 0.7 V

2.2 kÆ + 371 Æ

RTH =R1R2

R1 + R2=

(68 kÆ)(47 kÆ)

(68 kÆ + 47 kÆ)= 27.8 kÆ

1. If the voltage at the base of a transistor is 5 V and the base current is A, what is thedc input resistance at the base?

2. If a transistor has a dc beta of 190, VB 2 V, and IE 2 mA, what is the dc inputresistance at the base?

3. What bias voltage is developed at the base of a transistor if both resistors in a stiffvoltage divider are equal and

4. What are two advantages of voltage-divider bias?

VCC 10 V?

5 mSECTION 5–2CHECKUP

5–3 OTHER BIAS METHODS

In this section, four additional methods for dc biasing a transistor circuit are discussed.Although these methods are not as common as voltage-divider bias, you should be ableto recognize them when you see them and understand the basic differences.

After completing this section, you should be able to

Analyze four more types of bias circuits Discuss emitter bias

Analyze an emitter-biased circuit Discuss base bias

Analyze a base-biased circuit Explain Q-point stability of base bias Discuss emitter-feedback bias

Define negative feedback Analyze an emitter-feedback biased circuit Discuss collector-feedback bias

Analyze a collector-feedback biased circuit Discuss Q-point stability overtemperature

Emitter Bias

Emitter bias provides excellent bias stability in spite of changes in or temperature. It usesboth a positive and a negative supply voltage. To obtain a reasonable estimate of the key dcvalues in an emitter-biased circuit, analysis is quite easy. In an npn circuit, such as shown

b

242 TRANSISTOR BIAS CIRCUITS

in Figure 5–17, the small base current causes the base voltage to be slightly below ground.The emitter voltage is one diode drop less than this. The combination of this small dropacross RB and VBE forces the emitter to be at approximately Using this approxima-tion, you can obtain the emitter current as

VEE is entered as a negative value in this equation.You can apply the approximation that to calculate the collector voltage.

The approximation that is useful for troubleshooting because you won’t needto perform any detailed calculations. As in the case of voltage-divider bias, there is a morerigorous calculation for cases where you need a more exact result.

VE -1 V

VC = VCC - ICRC

IC IE

IE =-VEE - 1 V

RE

-1 V.

Calculate IE and VCE for the circuit in Figure 5–16 using the approximationsVE -1 V and IC IE.

EXAMPLE 5–6

Solution

Related Problem If VEE is changed to what is the new value of VCE?-12 V,

VCE = 8.4 V - (-1) = 9.4 V

VC = VCC - ICRC = +15 V - (1.4 mA)(4.7 kÆ) = 8.4 V

IE =-VEE - 1 V

RE=

-(-15 V) - 1 V

10 kÆ=

14 V

10 kÆ= 1.4 mA

VE -1 V

RE10 k

VCC+15 V

RC4.7 k

RB

47 k

VEE–15 V

FIGURE 5–16

The approximation that and the neglect of may not be accurateenough for design work or detailed analysis. In this case, Kirchhoff’s voltage law can beapplied as follows to develop a more detailed formula for IE. Kirchhoff’s voltage law ap-plied around the base-emitter circuit in Figure 5–17(a), which has been redrawn in part (b)for analysis, gives the following equation:

Substituting, using Ohm’s law,

VEE + IBRB + VBE + IERE = 0

VEE + VRB+ VBE + VRE

= 0

bDCVE -1 V

OTHER BIAS METHODS 243

Substituting for and transposing

Factoring out IE and solving for IE,

IE VEE VBE

RE RB/BDC

a IE

bDCbRB + IERE + VBE = -VEE

VEE,IB IE>bDC

Equation 5–9

Voltages with respect to ground are indicated by a single subscript. The emitter voltagewith respect to ground is

The base voltage with respect to ground is

The collector voltage with respect to ground is

VC = VCC - ICRC

VB = VE + VBE

VE = VEE + IERE

VEE

VCC

RC

RB

REVB

VE VC

RCIC

+

–

RE

+

–

+–VBE

RB

IB

VEE

IE

+ –

(a) (b)

VCC FIGURE 5–17

An npn transistor with emitter bias.Polarities are reversed for a pnp tran-sistor. Single subscripts indicate volt-ages with respect to ground.

Determine how much the Q-point (IC, VCE) for the circuit in Figure 5–18 will changeif increases from 100 to 200 when one transistor is replaced by another.bDC

EXAMPLE 5–7

RE10 k

VCC+15 V

RC4.7 k

RB

47 k

VEE–15 V

FIGURE 5–18

244 TRANSISTOR BIAS CIRCUITS

Solution For

Therefore,

For

Therefore,

The percent change in IC as changes from 100 to 200 is

The percent change in is

Related Problem Determine the Q-point in Figure 5–18 if increases to 300.bDC

%¢VCE = aVCE(2) - VCE(1)

VCE(1)b100% = a9.71 V - 9.83 V

9.83 Vb100% = 1.22%

VCE

%¢IC = a IC(2) - IC(1)

IC(1)b100% = a1.38 mA - 1.37 mA

1.37 mAb100% = 0.730%

bDC

VCE(2) = VC - VE = 8.51 V - (-1.2 V) = 9.71 V

VE = VEE + IERE = -15 V + (1.38 mA)(10 kÆ) = -1.2 V

VC = VCC - IC(2)RC = 15 V - (1.38 mA)(4.7 kÆ) = 8.51 V

IC(2) IE =-VEE - VBE

RE + RB>bDC=

-(-15 V) - 0.7 V

10 kÆ + 47 kÆ>200= 1.38 mA

bDC = 200,

VCE(1) = VC - VE = 8.56 V - (-1.3 V) = 9.83 V

VE = VEE + IERE = -15 V + (1.37 mA)(10 kÆ) = -1.3 V

VC = VCC - IC(1)RC = 15 V - (1.37 mA)(4.7 kÆ) = 8.56 V

IC(1) IE =-VEE - VBE

RE + RB>bDC=

-(-15 V) - 0.7 V

10 kÆ + 47 kÆ>100= 1.37 mA

bDC = 100,

Base Bias

This method of biasing is common in switching circuits. Figure 5–19 shows a base-biasedtransistor. The analysis of this circuit for the linear region shows that it is directly depend-ent on Starting with Kirchhoff’s voltage law around the base circuit,

Substituting IBRB for you get

Then solving for IB,

Kirchhoff’s voltage law applied around the collector circuit in Figure 5–19 gives the fol-lowing equation:

Solving for VCE,

VCE VCC ICRC

VCC - ICRC - VCE = 0

IB =VCC - VBE

RB

VCC - IBRB - VBE = 0

VRB,

VCC - VRB- VBE = 0

bDC.

Equation 5–10

Substituting the expression for IB into the formula yields

IC BDCaVCC VBE

RBb

IC = bDCIB

Equation 5–11

RC

RB

VBE

+

+VCC

VCE

+

––

FIGURE 5–19

Base bias.

OTHER BIAS METHODS 245

Q-Point Stability of Base Bias Notice that Equation 5–11 shows that IC is dependenton The disadvantage of this is that a variation in causes IC and, as a result, VCEto change, thus changing the Q-point of the transistor. This makes the base bias circuitextremely beta-dependent and unpredictable.

Recall that varies with temperature and collector current. In addition, there is alarge spread of values from one transistor to another of the same type due to manu-facturing variations. For these reasons, base bias is rarely used in linear circuits but is dis-cussed here so you will be familiar with it.

bDC

bDC

bDCbDC.

Determine how much the Q-point (IC, VCE) for the circuit in Figure 5–20 will changeover a temperature range where increases from 100 to 200.bDC

EXAMPLE 5–8

RC560

RB

VCC+12 V

330 k

FIGURE 5–20

Solution For

The percent change in IC as changes from 100 to 200 is

The percent change in VCE is

As you can see, the Q-point is very dependent on in this circuit and thereforemakes the base bias arrangement very unreliable. Consequently, base bias is not nor-mally used if linear operation is required. However, it can be used in switchingapplications.

Related Problem Determine IC if increases to 300.bDC

bDC

= a8.17 V - 10.1 V

10.1 Vb100% = 19.1% (a decrease)

%¢VCE = aVCE(2) - VCE(1)

VCE(1)b100%

= a6.84 mA - 3.42 mA

3.42 mAb100% = 100% (an increase)

%¢IC = a IC(2) - IC(1)

IC(1)b100%

bDC

VCE(2) = VCC - IC(2) RC = 12 V - (6.84 mA)(560 Æ) = 8.17 V

IC(2) = bDCaVCC - VBE

RBb = 200a12 V - 0.7 V

330 kÆb = 6.84 mA

For bDC = 200,

VCE(1) = VCC - IC(1)RC = 12 V - (3.42 mA)(560 Æ) = 10.1 V

IC(1) = bDCaVCC - VBE

RBb = 100a12 V - 0.7 V

330 kÆb = 3.42 mA

bDC = 100,

246 TRANSISTOR BIAS CIRCUITS

Emitter-Feedback Bias

If an emitter resistor is added to the base-bias circuit in Figure 5–20, the result is emitter-feedback bias, as shown in Figure 5–21. The idea is to help make base bias more pre-dictable with negative feedback, which negates any attempted change in collector currentwith an opposing change in base voltage. If the collector current tries to increase, the emit-ter voltage increases, causing an increase in base voltage because VB VE + VBE. This in-crease in base voltage reduces the voltage across RB, thus reducing the base current andkeeping the collector current from increasing. A similar action occurs if the collector cur-rent tries to decrease. While this is better for linear circuits than base bias, it is still de-pendent on and is not as predictable as voltage-divider bias. To calculate IE, you canwrite Kirchhoff’s voltage law (KVL) around the base circuit.

Substituting for IB, you can see that IE is still dependent on

IE VCC VBE

RE RB/BDC

bDC.IE>bDC

-VCC + IBRB + VBE + IERE = 0

bDC

=

Open the Multisim file E05-08 in the Examples folder on the companion website.Set and measure IC and VCE. Next, set and measure ICand VCE. Compare results with the calculated values.

bDC = 200bDC = 100

RC

RE

RB

VCC

FIGURE 5–21

Emitter-feedback bias.

Equation 5–12

The base-bias circuit from Example 5–8 is converted to emitter-feedback bias by theaddition of a emitter resistor. All other values are the same, and a transistor witha is used. Determine how much the Q-point will change if the first transis-tor is replaced with one having a Compare the results to those of thebase-bias circuit.

Solution For

For

The percent change in IC is

Although the emitter-feedback bias significantly improved the stability of the bias fora change in compared to base bias, it still does not provide a reliable Q-point.

Related Problem Determine IC if a transistor with is used in the circuit.bDC = 300

bDC

%¢VCE = aVCE(2) - VCE(1)

VCE(1)b100% = a7.90 V - 5.35 V

7.90 Vb100% = 32.3%

%¢IC = a IC(2) - IC(1)

IC(1)b100% = a4.26 mA - 2.63 mA

2.63 mAb100% = 62.0%

VCE(2) = VCC - IC(2)(RC + RE) = 12 V - (4.26 mA)(560 Æ + 1 kÆ) = 5.35 V

IC(2) = IE =VCC - VBE

RE + RB>bDC=

12 V - 0.7 V

1 kÆ + 330 kÆ>200= 4.26 mA

bDC = 200,

VCE(1) = VCC - IC(1)(RC + RE) = 12 V - (2.63 mA)(560 Æ + 1 kÆ) = 7.90 V

IC(1) = IE =VCC - VBE

RE + RB>bDC=

12 V - 0.7 V

1 kÆ + 330 kÆ>100= 2.63 mA

bDC = 100,

bDC = 200.bDC = 100

1 kÆEXAMPLE 5–9

OTHER BIAS METHODS 247

Collector-Feedback Bias

In Figure 5–22, the base resistor RB is connected to the collector rather than to VCC, as itwas in the base bias arrangement discussed earlier. The collector voltage provides the biasfor the base-emitter junction. The negative feedback creates an “offsetting” effect thattends to keep the Q-point stable. If IC tries to increase, it drops more voltage across RC,thereby causing VC to decrease. When VC decreases, there is a decrease in voltage acrossRB, which decreases IB. The decrease in IB produces less IC which, in turn, drops less volt-age across RC and thus offsets the decrease in VC.

Analysis of a Collector-Feedback Bias Circuit By Ohm’s law, the base current can beexpressed as

Let’s assume that The collector voltage is

Also,

Substituting for VC in the equation

The terms can be arranged so that

Then you can solve for IC as follows:

IC VCC VBE

RC RB/BDC

ICaRC +RB

bDCb = VCC - VBE

ICRB

bDC+ ICRC = VCC - VBE

IC

bDC=

VCC - ICRC - VBE

RB

IB = (VC - VBE)>RB,

IB =IC

bDC

VC VCC - ICRC

IC W IB.

IB =VC - VBE

RB

Equation 5–13

Since the emitter is ground, VCE VC.

VCE VCC ICRC

=

Equation 5–14

Q-Point Stability Over Temperature Equation 5–13 shows that the collector current isdependent to some extent on This dependency, of course, can be minimizedby making An important feature of collector-feedbackbias is that it essentially eliminates the dependency even if the stated condi-tions are met.

As you have learned, varies directly with temperature, and VBE varies inverselywith temperature. As the temperature goes up in a collector-feedback circuit, goesup and VBE goes down. The increase in acts to increase IC. The decrease in VBEacts to increase IB which, in turn also acts to increase IC. As IC tries to increase, thevoltage drop across RC also tries to increase. This tends to reduce the collector voltageand therefore the voltage across RB, thus reducing IB and offsetting the attempted in-crease in IC and the attempted decrease in VC. The result is that the collector-feedbackcircuit maintains a relatively stable Q-point. The reverse action occurs when the tem-perature decreases.

bDC

bDC

bDC

bDC and VBE

RC W RB>bDC and VCC W VBE.bDC and VBE.

RC

+VCC

+–VBE

IC + IB

IC

VCRB

IB

FIGURE 5–22

Collector-feedback bias.

248 TRANSISTOR BIAS CIRCUITS

Calculate the Q-point values (IC and VCE) for the circuit in Figure 5–23.EXAMPLE 5–10

RC10 k

VCC+10 V

RB

180 k

+–0.7 V

βDC = 100

FIGURE 5–23

Solution Using Equation 5–13, the collector current is

Using Equation 5–14, the collector-to-emitter voltage is

Related Problem Calculate the Q-point values in Figure 5–23 for and determine the percentchange in the Q-point from

Open the Multisim file E05-10 in the Examples folder on the companion website.Measure IC and VCE. Compare with the calculated values.

bDC = 100 to bDC = 200.bDC = 200

VCE = VCC - ICRC = 10 V - (788 mA)(10 kÆ) = 2.12 V

IC =VCC - VBE

RC + RB>bDC=

10 V - 0.7 V

10 kÆ + 180 kÆ>100= 788 MA

1. Why is emitter bias more stable than base bias?

2. What is the main disadvantage of emitter bias?

3. Explain how an increase in causes a reduction in base current in a collector-feed-back circuit.

4. What is the main disadvantage of the base bias method?

5. Explain why the base bias Q-point changes with temperature.

6. How does emitter-feedback bias improve on base bias?

bDC

SECTION 5–3CHECKUP

5–4 TROUBLESHOOTING

In a biased transistor circuit, the transistor can fail or a resistor in the bias circuit canfail. We will examine several possibilities in this section using the voltage-divider biasarrangement. Many circuit failures result from open resistors, internally open transistorleads and junctions, or shorted junctions. Often, these failures can produce an apparentcutoff or saturation condition when voltage is measured at the collector.

After completing this section, you should be able to

Troubleshoot faults in transistor bias circuits Troubleshoot a voltage-divider biased transistor circuit

Troubleshoot the circuit for several common faults Use voltage measurementto isolate a fault

Bipolar-Junction (BJT) transistors

References:

Barbow (Chapter 7), Hayes & Horowitz (pp 84-141), Rizzoni (Chapters 8 & 9)

A bipolar junction transistor is formed by joining three sections of semiconductors with

alternatively different dopings. The middle section (base) is narrow and one of the other two

regions (emitter) is heavily doped. Two variants of BJT are possible: NPN and PNP.

BB

C

E

C C

E

B

E

NPN Transistor

n

p

+

n

Circuit Symbols

B

C

E

CC

B

E

B

E

n

p

Circuit Symbols

+

PNP Transistor

p

We will focus on NPN BJTs. Operation of a PNP transistor is analogous to that of a NPN

transistor except that the role of “majority” charge carries reversed. In NPN transistors,

electron flow is dominant while PNP transistors rely mostly on the flow of “holes.” Therefore,

to zeroth order, NPN and PNP transistors behave similarly except the sign of current and

voltages are reversed. i.e., PNP = − NPN ! In practice, NPN transistors are much more

popular than PNP transistors because electrons move faster in a semiconductor. As a results,

a NPN transistor has a faster response time compared to a PNP transistor.

At the first glance, a BJT looks like 2 diodes placed back to back.

Indeed this is the case if we apply voltage to only two of the three

terminals, letting the third terminal float. This is also the way that

we check if a transistor is working: use an ohm-meter to ensure both

diodes are in working conditions. (One should also check the resistance

between CE terminals and read a vary high resistance as one may have

a burn through the base connecting collector and emitter.)

The behavior of the BJT is different, however, when voltage sources are

attached to both BE and CE terminals. The BE junction acts like a

diode. When this junction is forward biased, electrons flow from emitter

to the base (and a small current of holes from base to emitter). The

base region is narrow and when a voltage is applied between collector

and emitter, most of the electrons that were flowing from emitter to

base, cross the narrow base region and are collected at the collector

region. So while the BC junction is reversed biased, a large current can

flow through that region and BC junction does not act as a diode.

The amount of the current that crosses from emitter to collector region depends strongly

on the voltage applied to the BE junction, vBE . (It also depends weakly on voltage applied

ECE60L Lecture Notes, Spring 2004 55

between collector and emitter, vCE .) As such, small changes in vBE or iB controls a much

larger collector current iC . Note that the transistor does not generate iC . It acts as a valve

controlling the current that can flow through it. The source of current (and power) is the

power supply that feeds the CE terminals.C

vCE

BE

CB

i

+B

_

+

+_

_

v

vi

i E

A BJT has three terminals. Six parameters; iC , iB, iE, vCE , vBE , and

vCB ; define the state of the transistor. However, because BJT has three

terminals, KVL and KCL should hold for these terminals, i.e.,

iE = iC + iB vBC = vBE − vCE

Thus, only four of these 6 parameters are independent parameters. The relationship among

these four parameters represents the “iv” characteristics of the BJT, usually shown as iB vs

vBE and iC vs vCE graphs.

The above graphs show several characteristics of BJT. First, the BE junction acts likes

a diode. Secondly, BJT has three main states: cut-off, active-linear, and saturation. A

description of these regions are given below. Lastly, The transistor can be damaged if (1) a

large positive voltage is applied across the CE junction (breakdown region), or (2) product

of iCvCE exceed power handling of the transistor, or (3) a large reverse voltage is applied

between any two terminals.

Several “models” available for a BJT. These are typically divided into two general categories:

“large-signal” models that apply to the entire range of values of current and voltages, and

“small-signal” models that apply to AC signals with small amplitudes. “Low-frequency” and

“high-frequency” models also exist (high-frequency models account for capacitance of each

junction). Obviously, the simpler the model, the easier the circuit calculations are. More

complex models describe the behavior of a BJT more accurately but analytical calculations

become difficult. PSpice program uses a high-frequency, Eber-Mos large-signal model which

is a quite accurate representation of BJT. For analytical calculations here, we will discuss a

simple low-frequency, large-signal model (below) and a low-frequency, small-signal model in

the context of BJT amplifiers later.

ECE60L Lecture Notes, Spring 2004 56

A Simple, Low-frequency, Large Signal Model for BJT:

As the BE junction acts like a diode, a simple piece-wise linear model can be used :

BE Junction ON: vBE = vγ, and iB > 0

BE Junction OFF: vBE < vγ, and iB = 0

where vγ is the forward bias voltage (vγ ≈ 0.7 V for Si semiconductors).

When the BE junction is reversed-biased, transistor is OFF as no charge carriers enter the

base and move to the collector. The voltage applied between collector and emitter has not

effect. This region is called the cut-off region:

Cut-Off: vBE < vγ , iB = 0, iC ≈ iE ≈ 0

Since the collector and emitter currents are very small for any vCE , the effective resistance

between collector and emitter is very large (100’s of MΩ) making the transistor behave as

an open circuit in the cut-off region.

When the BE junction is forward-biased, transistor is ON. The behavior of the transistor,

however, depends on how much voltage is applied between collector and emitter. If vCE > vγ ,

the BE junction is forward biased while BC junction is reversed-biased and transistor is in

active-linear region. In this region, iC scales linearly with iB and transistor acts as an

amplifier.

Active-Linear: vBE = vγ , iB > 0,iCiB

= β ≈ constant, vCE ≥ vγ

If vCE < vγ, both BE and BC junctions are forward biased. This region is called the

saturation region. As vCE is small while iC can be substantial, the effective resistance

between collector and emitter in saturation region is small and the BJT acts as a closed-

circuit.

Saturation: vBE = vγ , iB > 0,iCiB

< β, vCE ≈ vsat

Our model specifies vCE ≈ vsat, the saturation voltage. In reality in the saturation region

0 < vCE < vγ . As we are mainly interested in the value of the collector current in this region,

vCE is set to a value in the middle of its range in our simple model: vCE ≈ vsat ∼ 0.5vγ.

Typically a value of vsat ≈ 0.2 − 0.3 V is used for Si semiconductors.

ECE60L Lecture Notes, Spring 2004 57

The above simple, large-signal model is shown below. A comparison of this simple model

with the real BJT characteristics demonstrates the degree of approximation used.

i B

vBEvγ vsat vCE

CiSaturation

Active Linear

Cut Off

BJT ON

BJT OFF

How to Solve BJT Circuits:

The state of a BJT is not known before we solve the circuit, so we do not know which model

to use: cut-off, active-linear, or saturation. To solve BJT circuits, we need assume that

BJT is in a particular state, use BJT model for that state to solve the circuit and check

the validity of our assumptions by checking the inequalities in the model for that state. A

formal procedure will be:

1) Write down a KVL including the BE junction (call it BE-KVL).

2) Write down a KVL including CE terminals (call it CE-KVL).

3) Assume BJT is in cut-off (this is the simplest case). Set iB = 0. Calculate vBE from

BE-KVL.

3a) If vBE < vγ, then BJT is in cut-off, iB = 0 and vBE is what you just calculated. Set

iC = iE = 0, and calculate vCE from CE-KVL. You are done.

3b) If vBE > vγ, then BJT is not in cut-off. Set vBE = vγ . Solve above KVL to find iB. You

should get iB > 0.

4) Assume that BJT is in active linear region. Let iE ≈ iC = βiB. Calculate vCE from

CE-KVL.

4a) If vCE > vγ , then BJT is in active-linear region. You are done.

4b) If vCE < vγ , then BJT is not in active-linear region. It is in saturation. Let vCE = vsat

and compute iC from CE-KVL. You should find that iC < βiB. You are done.

ECE60L Lecture Notes, Spring 2004 58

CE

i C

Ω1 k

BEv

Bi40 kv

Ω +

_+

_

i E

+-

4 V

12 VExample 1: Compute the parameters of this circuit (β = 100).

Following the procedure above:

BE-KVL: 4 = 40 × 103iB + vBE

CE-KVL: 12 = 103iC + vCE,

Assume BJT is in cut-off. Set iB = 0 in BE-KVL:

BE-KVL: 4 = 40 × 103iB + vBE → vBE = 4 > vγ = 0.7 V

So BJT is not in cut off and BJT is ON. Set vBE = 0.7 V and use BE-KVL to find iB.

BE-KVL: 4 = 40 × 103iB + vBE → iB =4 − 0.7

40, 000= 82.5 µA

Assume BJT is in active linear, Find iC = βiB and use CE-KVL to find vCE :

iC = βiB = 100iB = 8.25 mA

CE-KVL: 12 = 1, 000iC + vCE , → vCE = 12 − 8.25 = 3.75 V

As vCE = 3.75 > vγ , the BJT is indeed in active-linear and we have: vBE = 0.7 V, iB =

82.5 µA, iE ≈ iC = 8.25 mA, and vCE = 3.75 V.

CE

i C

i

i B

vBE

1 kΩ

_

+

_

1 kΩ

+v

40 kΩ

E

-+

12 V

4 V

Example 2: Compute the parameters of this circuit (β = 100).

Following the procedure above:

BE-KVL: 4 = 40 × 103iB + vBE + 103iE

CE-KVL: 12 = 1, 000iC + vCE + 1, 000iE

Assume BJT is in cut-off.

Set iB = 0 and iE = iC = 0 in BE-KVL:

BE-KVL: 4 = 40 × 103iB + vBE + 103iE → vBE = 4 > 0.7 V

So BJT is not in cut off and vBE = 0.7 V and iB > 0. Here, we cannot find iB right away

from BE-KVL as it also contains iE.

ECE60L Lecture Notes, Spring 2004 59

Assume BJT is in active linear, iE ≈ iC = βiB:

BE-KVL: 4 = 40 × 103iB + vBE + 103βiB

4 − 0.7 = (40 × 103 + 103× 102)iB

iB = 24 µA → iE ≈ iC = βiB = 2.4 mA

CE-KVL: 12 = 1, 000iC + vCE + 1, 000iE, → vCE = 12 − 4.8 = 7.2 V

As vCE = 7.2 > vγ , the BJT is indeed in active-linear and we have: vBE = 0.7 V, iB = 24 µA,

iE ≈ iC = 2.4 mA, and vCE = 7.2 V.

Load line

The operating point of a BJT can be found graphically using the concept of a load line. A

load line is the relationship between iC and vCE that is imposed on BJT by the external

circuit. For a given value of iB, the iCvCE characteristics curve of a BJT is the relationship

between iC and VCE as is set by BJT internals. The intersection of the load line with the

BJT characteristics represent a pair of iC and vCE values which satisfy both conditions and,

therefore, is the operating point of the BJT (often called the Q point for Quiescent point)

The equation of a load line for a BJT should include only iC and vCE (no other unknowns).

This equation is usually found by writing a KVL around a loop containing vCE. For the

example above, we have (using iE ≈ iC):

KVL: 12 = 1, 000iC + vCE + 1, 000iE → 2, 000iC + vCE = 12

An example of a load line, iCvCE characteristics of a BJT, and the Q-point is shown below.

ECE60L Lecture Notes, Spring 2004 60

BJT Switches and Logic Gates

i C RC

CC

i

o

V

v

v

i B

i C

BR

RC

CC

i

o

V

v

v

The basic element of logic circuits is the transistor switch. A

schematic of such a switch is shown. When the switch is open,

iC = 0 and vo = VCC . When the switch is closed, vo = 0 and

iC = VCC/RC .

In an electronic circuit, mechanical switches are not used. The

switching action is performed by a transistor with an input

voltage switching the circuit, as is shown. When vi = 0,

BJT will be in cut-off, iC = 0, and vo = VCC (open switch).

When vi is in “high” state, BJT can be in saturation with

vo = vCE = Vsat ≈ 0.2 V and iC = (VCC − Vsat)/RC (closed

switch). When Rc is replaced with a load, this circuit can switch

a load ON or OFF (LED and motor drive circuits of ECE20A

Lab).

The above BJT circuit is also an “inverter” or a “NOT” logic gate. Let’s assume that the

“low” states are voltages between 0 to 0.5 V, “high” states voltages are between 4 to 5 V,

and VCC = 5 V. When the input voltage is “low” (vi ≈ 0), BJT will be in cut-off and

vo = VCC = 5 V (“high” state). When input voltage is “high,” with proper choice of RB,

BJT will be in saturation, and vo = vCE = Vsat ≈ 0.2 V (“low” state).

Resistor-Transistor Logic (RTL)

The inverter circuit discussed above is a member of RTL family of logic gates. Plot of vo

as a function of vi is called the transfer characteristics of the gate. To find the transfer

characteristics, we need to find vo for a range of vi values.

When vi < vγ, BJT will be in cut-off, iC = 0 and vo = VCC . Therefore, for input voltages

below certain threshold (denoted by VIL), the gate output is high. For our circuit, VIL = vγ .

When vi exceeds vγ , BE junction will be forward biased and a current iB flows into BJT:

iB =vi − vγ

RB

As BE junction is forward biased, BJT can be either in saturation or active-linear. Let’s

assume BJT is is in saturation. In that case, vo = vCE = Vsat and iC/iB < β. Then:

iC =VCC − Vsat

RC

→ iB >iCβ

=VCC − Vsat

βRC

ECE60L Lecture Notes, Spring 2004 61

Therefore, BJT will be in saturation only if iB exceeds the value given by the formula above.

This ouccrs when vi become large enough:

vi = vγ + RBiB > vγ + RB ×VCC − Vsat

βRC

= VIH

Therefore, for input voltages larger than the a certain value (VIH) , the gate output is low.

For vi values between these two limits, the BE junction is forward biased but the BJT is

NOT in saturation, therefore, it is in active linear. In this case, the output voltage smoothly

changes for its high value to its low value as is shown in the plot of transfer characteristics.

This range of vi is a “forbidden” region and the gate would not work properly in this region.

This behavior can also seen in the plot of the BJT load line. For small values of vi (iB = 0)

BJT is in cut-off. As vi is increased, iB is increased and the operating point moves to the

left and up on the load line and enters the active-linear region. When iB is raised above

certain limit, the operating point enters the saturation region.

i B

i C

BR

R1 i 1

i 2

RC

CC

i

o

V

v

v

A major drawback of the this RTL inverter gate is the limited

input range for the “low” signal (VIL). Our analysis indicated

that VIL = vγ , that is the gate input is low for voltages between 0

and vγ ≈ 0.7 V. For this analysis, we have been using a piecewise

linear model for the BE junction diode. In reality, the BJT

will come out of cut-off (BE junction will conduct) at smaller

voltages (0.4–0.5 V). To resolve this shortcoming, one can add

a resistor between the base and ground (or between base and a

negative power supply) as is shown. (You have seen this circuit

in ECE20A, motor drive circuit.)

To see the impact of this resistor, note that VIL is the input voltage when BJT is just leaving

the cut-off region. At this point, vBE = vγ, and iB is positive but very small (effectively

ECE60L Lecture Notes, Spring 2004 62

zero). Noting that a voltage vBE has appeared across R1, we have:

i1 =vBE

R1

i2 = iB + i1 ≈ i1 =vBE

R1

VIL = vi = RBi2 + vBE = vBE

RB

R1

+ vBE = vγ

(

1 +RB

R1

)

This value should be compared with VIL = vγ in the absence of resistor R1. It can be seen

that for RB = R1, VIL is raised from 0.7 to 1.4 V and for RB = 2R1, VIL is raised to 2.1 V.

R1 does not affect VIH as iB needed to put the BJT in saturation is typically several times

larger than i1.

RC

CC

o

BR BR

1 2

v

v

V

v

RTL NOR Gate

By combining two or more RTL inverters, one obtains

the basic logic gate circuit of RTL family, a “NOR”

gate, as is shown. More BJTs can be added for addi-

tional input signals. (You have seen in 20B that all

higher level logic gates, e.g., flip-flops, can be made

by a combination of NOR gates or NAND gates.)

Exercise: Show that this ia NOR gate, i.e., the gate

output will be low as long as at least one of the inputs

is high.

RTLs were the first digital logic circuits using transistors. They were replaced with other

forms (DDT, TTL, and ECL) with the advent of integrated circuits. The major problem

with these circuits are the use of large resistors that would take large space on an IC chip (in

today’s chip, resistor values are limited to about 20 kΩ and capacitance to about 100 pF).

Before we move on to more modern gates, we consider two important characteristics of a

digital gate.

ECE60L Lecture Notes, Spring 2004 63

Switching Time and Propagation Delay:

Consider the inverter gate with an input voltage close to zero (and/or negative). In this

case, the BJT is in cut-off, iC = 0 and the output of the gate is high. Suppose a “high”

voltage is applied instantaneously to the gate at some point. We expect BJT to enter

saturation with iC = ICsat and output to drop to the “low”state. However, this does not

occur instantaneously.

When the BJT is in cut-off, BE junction

is reversed biased. When a forward volt-

age is applied to the BE junction, it takes

some time for the BE junction transition

capacitance to charge up. Time is also re-

quired for minority carries to diffuse across

the base and enter the collector. This re-

sults in the delay time td, which is of the

order of a nanosecond for a typical BJT.

Before BJT can enter saturation, it should traverse the active-linear region. The rise time,

tr (on the order of 1-10 ns) account for this transition. The time that takes for the gate to

switch “ON” is represented by ton.

Suppose that the input voltage to gate is then reduced instantaneously to low state. BJT

will leave saturation region and go to cut-off. Again, this not occur instantaneously. When

a BJT is in saturation, both BE and BC junctions are forward biased and conducting. As

such, an excess minority charge is stored in the base. For the transistor to leave saturation

and enter active-linear (BC junction to become reversed biased), this excess charge must be

removed. The time required for the removal of excess charge determines the storage time, ts

(order of 100 ns). Then, transistor traverses the active-linear region before entering cut-off.

This account for the fall time tf (1-10 ns). The total time it takes for the gate to switch

“OFF’ is represented by toff . As can be seen, BJT switching is mainly set by the storage

time, ts.

Propagation delays introduced by transistor switching time are important constraints in

designing faster chips. Gate designs try to minimize propagation delays as much as possible.

Fan-out: All digital logic circuits are constructed with cross-coupling of several basic gates

(such as NOR or NAND). As such, a basic gate may be attached to several other gates.

The maximum number of gates that can be attached to a digital gate is called “fan-out.”

Obviously, one would like to have large fan-out.

ECE60L Lecture Notes, Spring 2004 64

Diode-Transistor Logic (DTL)

The basic gate of DTL logic circuits is a NAND gate which is constructed by a combination

of a diode AND gate and a BJT inverter gate.

o

R

CC

A

1

1

A

2

2

1

2

D

i

D

i

i

v

V

v

v

Diode AND Gate: First, let’s consider the diode AND

gate as is shown. To study the behavior of the gate we will

consider the state of the circuit for different values of v1 and

v2 (either 0 or 5 V corresponding to low and high states).

To aid the analysis, let’s assume VCC = 5 V and RA = 1 kΩ.

We note that by KCL, iA = i1 + i2 (assuming that there is

no current drawn from the circuit).

Case 1, v1 = v2 = 0: Since the 5-V supply will tend to forward bias both D1 and D2,

let’s assume that both diodes are forward biased. Thus, vD1 = vD2 = vγ = 0.7 V and i1 > 0,

i2 > 0. In this case:

vo = v1 + vD1 = v2 + vD2 = 0.7 V

iA =VCC − vo

RA

=5 − 0.7

1, 000= 4.3 mA

Current iA will be divided between two diodes by KCL, each carrying one half of iA (because

of symmtery). Thus, i1 = i2 = 2.1 mA. Since diode currents are positive, our assumption of

both diode being forward biased is justified and, therefore, vo = 0.7 V.

So, when v1 and v2 are low, D1 and D2 are ON and vo is low.

Case 2, v1 = 0, v2 = 5 V: Again, we note that the 5-V supply will tend to forward bias

D1. Assume D1 is ON: vD1 = vγ = 0.7 V and i1 > 0. Then:

vo = v1 + vD1 = 0.7 V

vo = v2 + vD2 → vD2 = −4.3 V < vγ

and D2 will be OFF (i2 = 0). Then:

iA =VCC − vo

RA

=5 − 0.7

1, 000= 4.3 mA

i1 = iA − i2 = 4.3 − 0 = 4.3 mA

Since i1 > 0, our assumption of D1 being forward biased is justified and, therefore, vo = 0.7 V.

So, when v1 is low and v2 is high, D1 is ON and D2 is OFF and vo is low.

ECE60L Lecture Notes, Spring 2004 65

Case 3, v1 = 5 V, v2 = 0 V: Because of the symmetry in the circuit, this is exactly the

same as case 2 with roles of D1 and D2 reversed.

So, when v1 is high and v2 is low, D1 is OFF and D2 is ON and vo is low.

Case 4, v1 = v2 = 5 V: Examining the circuit, it appears that the 5-V supply will NOT

be able to forward bias D1 and D2. Assume D1 and D2 are OFF: i1 = i2 = 0, vD1 < vγ and

vD2 < vγ. Then:

iA = i1 + i2 = 0

vo = VCC − i1RA = 5 − 0 = 5 V

vD1 = vo − v1 = 5 − 5 = 0 < vγ and vD2 = vo − v2 = 5 − 5 = 0 < vγ

Thus, our assumption of both diodes being OFF arejustified.

So, when v1 and v2 are high, D1 and D2 are OFF and vo is high.

Overall, the output of this circuit is high only if both inputs are high (Case 4) and the output

is low in all other cases (Cases 1 to 3). Thus, this is an AND gate. This analysis can be

easily extended to cases with three or more diode inputs.

DTL NAND Gate:

The basic gate of DTL logic circuits is a NAND gate which is constructed by a combination

of a diode AND gate and a BJT inverter gate as is shown below (left figure). Because RB is

large, on ICs, this resistor is usually replaced with two diodes. The combination of the two

diodes and the BE junction diode leads to a voltage of 2.1 V for the inverter to switch and

a VIL = 1.4 V for the NAND gate (Why?). Resistor R1 is necessary because without this

resistor, current iB will be too small and the voltage across D3 and D4 will not reach 0.7 V

although they are both forward biased (Recall LED driver circuit of ECE20A in which the

LED started to lit for vin about 0.8 V instead of estimated 1.4 V).

RA

1

1

A

2

2

1

2

i B

i C

BR

RC

CC

o

D

i

D

i

i

v

v

V

vi B

i C RC

CC

o

R1

RA

1

A

2

2

1

2

1 3 4D

i

D

i

i

D D

V

v

v

v

ECE60L Lecture Notes, Spring 2004 66

DTLs were very popular in ICs in 60s and early 70s but are replaced with Transistor-

Transistor Logic (TTL) circuits. TTL are described later, but as TTLs are evolved from

DTLs, some examples of DTL circuits are given below.

i B

i C

1

CC

o

2

2

1

2

1 3 4

5

A

34

D

i

D

i

i

D D

ii

Ω5k

5kΩ

Ω1k

v

V

v

v

v

Example: Verify that the DTL circuit shown is

a NAND gate. Assume that “low”state is 0.2 V,

“high” state is 5 V, and BJT βmin = 40.

Case 1: v1 = v2 = 0.2 V It appears that

the 5-V supply will forward bias D1 and D2.

Assume D1 and D2 are forward biased: vD1 =

vD2 = vγ = 0.7 V and i1 > 0, i2 > 0. In this

case:

v3 = v1 + vD1 = v2 + vD2 = 0.2 + 0.7 = 0.9 V

Voltage v3 = 0.9 V is not sufficient to froward bias D3 and D4 as v3 = vD3 +vD4 +vBE and we

need at least 1.4 V to forward bias the two diodes. So both D3 and D4 are OFF and i4 = 0.

(Note that D3 and D4 can be forward biased without BE junction being forward biased as

long as the current i4 is small enough such that voltage drop across the 5 kΩ resistor parallel

to BE junction is smaller than 0.7 V. In this case, i5 = i4 and iB = 0.) Then:

i1 + i2 = iA =5 − v3

5, 000=

5 − 0.9

5, 000= 0.82 mA

And by symmetry, i1 = i2 = 0.5iA = 0.41 mA. Since both i1 and i2 are positive, our

assumption of D1 and D2 being ON are justified. Since i4 = 0, iB = 0 and BJT will be in

cut-off with iC = 0 and vo = 5 V.

So, in this case, D1 and D2 are ON, D3 and D4 are OFF, BJT is in cut-off, and vo = 5 V.

Case 2: v1 = 0.2 V, v2 = 5 V Following arguments of case 1, assume D1 is ON. Again,

v3 = 0.7 + 0.2 = 0.9 V, and D3 and D4 will be OFF with i4 = 0. We find that voltage across

D2 is vD2 = v3 − v2 = 0.9 − 5 = −4.1 V and, thus, D2 will be OFF and i2 = 0. Then:

i1 = iA =5 − v3

5, 000=

5 − 0.9

5, 000= 0.82 mA

and since i1 > 0, our assumption of D1 ON is justified. Since i4 = 0, iB = 0 and BJT will

be in cut-off with iC = 0 and vo = 5 V.

So, in this case, D1 is ON, D2 is OFF, D3 and D4 are OFF, BJT is in cut-off, and vo = 5 V.

ECE60L Lecture Notes, Spring 2004 67

Case 3: v1 = 5 V, v2 = 0.2 V Because of the symmetry in the circuit, this is exactly

the same as case 2 with roles of D1 and D2 reversed.

So, in this case, D1 is OFF, D2 is ON, D3 and D4 are OFF, BJT is in cut-off, and vo = 5 V.

Case 4: v1 = v2 = 5 V Examining the circuit, it appears that the 5-V supply will NOT

be able to forward bias D1 and D2. Assume D1 and D2 are OFF: i1 = i2 = 0, vD1 < vγ and

vD2 < vγ. On the other hand, it appears that D3 and D4 will be forward biased. Assume D3

and D4 are forward biased: vD3 = vD4 = vγ = 0.7 V and i4 > 0. Further, assume the BJT is

not in cut-off vBE = vγ = 0.7 V and iB > 0. In this case:

v3 = vD3 + vD4 + vBE = 0.7 + 0.7 + 0.7 = 2.1 V

vD1 = v3 − v1 = 2.1 − 5 = −2.9 V < vγ vD2 = v3 − v2 = 2.1 − 5 = −2.9 V < vγ

Thus, our assumption of D1 and D2 being OFF are justified. Furthermore:

i4 = iA =5 − v3

5, 000=

5 − 2.1

5, 000= 0.58 mA

i5 =vBE

5, 000=

0.7

5, 000= 0.14 mA

iB = i4 − i5 = 0.58 − 0.14 = 0.44 mA

and since i4 > 0 our assumption of D3 and D4 being ON are justified and since iB > 0 our

assumption of BJT not in cut-off is justified.

We still do not know if BJT is in active-linear or saturation. Assume BJT is in saturation:

vo = vCE = Vsat = 0.2 V and iC/iB < β. Then, assuming no gate is attached to the circuit,

we have

iC =5 − Vsat

1, 000=

5 − 0.2

1, 000= 4.8 mA

and since iC/iB = 4.8/0.44 = 11 < β = 40, our assumption of BJT in saturation is justified.

So, in this case, D1 and D2 are OFF, D3 and D4 are ON, BJT is in saturation and vo = 0.2 V.

Overall, the output in “low” only if both inputs are “high”, thus, this is a NAND gate.

Note: It is interesting to note that at the input of this gate, the current actually flows out

of the gate. In the example above, when both inputs were high i1 = i2 = 0, when both were

low i1 = i2 = 0.4 mA, and when one input was low, e.g., v1 was low, i1 = 0.8mA. The input

current flowing in (or out of the gate in this case) has implications for the fan-out capability

of logic gates as is shown in the example below.

ECE60L Lecture Notes, Spring 2004 68

Example: Find the fan-out of this NAND DTL gate. Assume that “low”state is 0.2 V,

“high” state is 5 V, and BJT β = 40.

i Bi C

CC

o

1

2

2

1

2

1 3 4

5

A

i R

4

L

D

i

D

i

i

D D

ii

Ω5k

5kΩ

Ω

Other gates

i

i

1k

V

v

v

v

The circuit is the same DTL NAND gate of previous example and we can use results from

previous example here. “N” other NAND gates are attached to the output of this gate.

Fan-out is the maximum value of N . Since we want to make sure that our gate operates

properly under all conditions, we should consider the worst case, when all of the second stage

gates have maximum currents.

For a NAND DTL gate, the maximum current i occurs when all of the inputs are high with

exception of one input. We found this value to be 0.82 mA (Cases 2 & 3 in the previous

example). Therefore, the worst case is when the input of all second stage gates are low (for

the first stage, vo = 0.2 V) and each draw a current 0.82 mA (a total of iL = N × 0.82 mA

is drawn from the first stage gate).

Considering the first stage gate, we had found that vo = 0.2 V only for Case 4. For that

case, we found iB = 0.44 mA. Then:

iR =5 − Vsat

1, 000=

5 − 0.2

1, 000= 4.8 mA

iC = iR + 0.82N = 4.8 + 0.82N

The first stage gate operates properly as long as the BJT is in saturation, i.e.,

iC < βiB → 4.8 + 0.82N < 40 × 0.44 → N < 13.7

As the fan-out should be integer, the fan-out for this gate is 13.

Fan-out of DTL gates can be greatly increased by a small modification. Fan-out can be

increased by increasing the base current of the BJT. iB is, however, limited by the current iA(and i4). Reducing the value of RA in the AND diode part of the circuit will have increase

iB. Unfortunately, as this resistor is reduced, power dissipation in the gate increases and the

fan-out capability decreases dramatically.

ECE60L Lecture Notes, Spring 2004 69

i B

i C

CC

R

1

A

2

2

1

2

1

RC

o

R1

4

A

D

i

D

i

iD

V

v

v

v

A simple solution which keeps current iA small

but increases iB drastically is to replace diode D3

with a BJT as is shown. As can be seen, the DTL

NAND gate is now made of 3 stages: 1) input

stage (diodes), 2) driver stage (first BJT) and 3)

output stage (2nd BJT).

Circuit Symbol

2

1

CC

i B

i C

R1

A

RC

o

R2RA

i

v

v

V

v

Transistor-Transistor Logic (TTL)

A simplified version of an IC-chip NPN transistor is shown.

The device is fabricated on a p-type substrate (or body) in

a vertical manner by embedding alternating layers of N and

P-type semiconductors. By embedding more than one N-

type emitter region, one can obtain a multiple-emitter NPN

transistor as shown. The multiple-emitter NPN transistors

can be used to replace the input diodes of a DTL NAND

gate and arrive at a NAND gate entirely made of transistors,

hence Transistor-Transistor Logic (TTL) gates.

A simple TTL gate is shown with the multiple-emitter BJT

replacing the input diodes. This transistor operates in

“reverse-active” mode, i.e., like a NPN transistor in active-

linear mode but with collector and emitter switched. Oper-

ationally, this BJT acts as two diodes back to back as shown

in the circle at the bottom of the figure. As such the oper-

ation of this gate is essentially similar to the DTL NAND

gate described above (note position of driver transistor and

D4 diode is switched).

Similar to DTL NAND gates, a typical TTL NAND gate

has three stages: 1) Input stage (multi-emitter transistor),

2) driver stage, and 3) output stage. Modern TTL gates

basically have the same configuration as is shown with the

exception that the output stage is replaced with the “Totem-

Pole” output stage to increase switching speed and gate fan-

out. For a detailed description of TTL gate with “Totem-

Pole” output stage, consult, Sedra and Smith (pages 1175

to 1180).

ECE60L Lecture Notes, Spring 2004 70