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30

Analysis and Application of Analog Electronic Circuits

This is a relatively enormous small-signal capacitance that appears in parallel with the diode model. The capacitance, CD, affects the speed at which the diode can turn off. However, the forward-biased diode has the time constant:

τd = CD gd = 1.92 E-7 (1 E-2 (2 ∞ 0.026)) 1 μs

(2.11)

which is basically the mean minority carrier lifetime, τ. Thus, CD has little effect on mid-frequency, small-signal performance, but the stored charge does affect the time to stop conducting when vD goes step negative.

Figure 2.5 illustrates the current, voltage, and excess minority charge waveforms in the n-material for a pn junction diode in which a step of forward voltage is followed by a step of reverse voltage. Note that VD remains at the forward value following switching to VR until all the excess minority carriers have been discharged from CD. Then VD VR and depletion capacitance dominates the junction behavior. The storage time is on the order of nanoseconds for most small-signal Si, pn diodes.

2.2.4Schottky Diodes

The Schottky barrier diode (SBD), also called a hot carrier diode, consists of a rectifying, metal insulator–semiconductor (MIS) junction. SBDs actually predate pn junction diodes by about 50 years. The first detectors in early crystal set radios were “cat’s whisker” diodes, in which a fine lead wire made contact with a metallic mineral crystal, such as galena, silicon, or germanium. Such cat’s whisker detectors were, in fact, primitive SBDs. Because their rectifier characteristics were not reproducible, cat’s whisker SBDs were only suited for early radio experimenters and were replaced by pn junction diodes in the 1950s.

The presence of a very thin, insulating interfacial oxide layer from 0.5 to 1.5 nm thick between the metal and the semicon makes a modern SBD different from an ordinary ohmic, metal–semiconductor connection. Figure 2.6 shows the static I–V characteristics of a pn diode and an SBD. Note that the SBD has a lower cut-in voltage (approximately 0.3 V) than the pn diode (approximately 0.55 V) as well as about 1000 times the reverse leakage current (0.1 μA vs. 1 nA), which does not saturate. A schematic cross section of an SBD (after Boylstead and Nashelsky, 1987) is shown on the figure.

The SBD is a majority carrier device; a pnD is a minority carrier device. In order for a pnD to block reverse current, the stored minority carriers must be swept out of the junction by the reverse electric field. This discharging limits the turn-off time to microseconds. On the other hand, the SBD has no minority carrier storage and can switch off in tens of nanoseconds, meaning the SBD can rectify currents in the hundreds of megahertz. Figure 2.7 shows a high-frequency SSM for the SBD when conducting. Note that the SS forward

© 2004 by CRC Press LLC

Models for Semiconductor Devices Used in Analog Electronic Systems

31

 

 

G

 

 

 

F

R

 

 

 

R

 

+

 

 

 

 

 

 

+

ID(t)

VD(t)

 

 

VF

 

 

VR

 

 

 

 

+

 

 

Qx(t)

 

 

 

 

IF τp

 

 

 

 

τp

ID(t)

 

 

 

VF /R

 

τpQx(t)

 

 

 

 

Irs

t

 

 

 

 

 

ID(t)

 

 

 

 

VR /R

 

 

VD(t)

 

Storage

 

 

 

 

time

 

 

 

VDSS

 

 

t

 

 

tr

 

 

VR

 

 

 

 

FIGURE 2.5

Schematic showing the effect of switching a diode from vD = 0, iD = 0, to forward conduction. With VF applied, vD quickly rises to the steady-state forward drop, vDSS 0.7 V, and the XS minority carriers build up to a charge, Qx. When the applied voltage is switched to −VR, a finite time is required for Qx to be dissipated before the diode can block current. During this storage time, vD remains at vDSS.

resistance, rd, and the junction capacitance, CJ, are dependent on iD and vD, respectively. Hewlett–Packard hot carrier diodes of the 5082-2300 series have reverse currents of approximately 50 to 100 nA at vD = 10 V, and CJ = 0.3

pF at vD = 10 V.

Besides high-frequency rectification and detection, SBDs can be used to speed up transistor switching in logic gates. Figure 2.8 illustrates a SBD placed in parallel with the pn base-to-collector diode in an npn BJT. In normal

© 2004 by CRC Press LLC

32

Analysis and Application of Analog Electronic Circuits

 

Anode (+)

iD

 

iD

 

Metal

 

 

 

(mA)

 

 

 

 

2.0

 

 

 

 

 

 

Oxide

 

 

 

 

 

insulator

 

 

 

SBD

pnD

Thin oxide

n-Si

 

SBD

 

 

film

 

 

 

 

 

 

Cathode ()

 

Metal

1.0

 

 

 

 

 

 

pnD

 

 

 

 

≈ 1 nA

vD < 0

 

 

vD

 

0.2

0.4

0.6

(V)

≈ 0.1 A

 

 

 

 

SBD

 

 

 

 

FIGURE 2.6

The static I–V curves of a Schottky barrier diode and a pn junction diode.

 

3 nH

Ls

iD

 

 

+

7

rs

 

 

Cp

vD

CJ = f(vD)

0.15 pF

 

 

 

rd = f(iD)

A

CJ(0) = 1pF

rd 26/iD

 

 

 

B

FIGURE 2.7

(A) Symbol for SDB (not to be confused with that for the zener diode). (B) High-frequency equivalent circuit for a SBD.

linear operation, the CB junction is reverse-biased. When the transistor is saturated, the BE and the CB junctions are forward-biased. Without the SBD, there is large minority charge storage in the CB junction. With the SBD, much of the excess base current will flow through the SBD, preventing large excess charge storage in the CB junction and allowing the BJT to turn

© 2004 by CRC Press LLC

Models for Semiconductor Devices Used in Analog Electronic Systems

33

 

 

 

C

 

iC

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

SBD

 

 

+

 

 

 

 

 

 

 

 

 

 

vCB

+

 

 

 

 

 

 

B

 

 

vCE

 

 

 

 

 

 

 

 

 

 

+ iB

 

 

 

 

vBE

 

 

 

 

 

iE

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

E

 

FIGURE 2.8

Use of an SBD to make a Schottky transistor. The SBD prevents XS stored charge in the BJT’s CB junction.

off much more quickly (Yang, 1988). SBD thin film technology is also used in the fabrication of certain MOS transistors (Nanavati, 1975).

2.3Mid-Frequency Models for BJT Behavior

2.3.1Introduction

Most BJTs found on ICs are silicon pnp or npn. (Other fabrication materials, such as gallium arsenide, exist but will not be treated here.) A heuristic view of BJTs is shown by the “layer cake” models shown in Figure 2.9(B). BJTs are generally viewed as current-controlled current sources in their “linear” operating regions. As in the case of pn diodes, the large-signal, dc behavior of these three terminal devices will first be examined. Figure 2.9(C) illustrates typical volt–ampere curves for a small-signal (as opposed to power) npn BJT.

For purposes of dc biasing and crude circuit calculations on paper, an npn BJT in its active region can be represented by the simple circuit shown in Figure 2.10(A). The voltage drop of the forward-biased BE junction is rep-

resented by VBEQ on the order of 0.6 V. The dc collector current is approximated by the current-controlled current source (CCCS) with gain βo. βo is the

transistor’s nominal (forward) dc current gain. Figure 2.10(B) illustrates the dc model for the transistor when it is saturated. Saturation occurs for iB > iC/βo, i.e., for high base currents. In saturation, the BE and CB junctions

are forward-biased, producing a VCEsat that can range from approximately 0.2 to 0.8 V for high collector currents. The collector-emitter saturation curve

can be approximated by the simple Thevenin circuit shown: rcsat = ∂vCE/∂iC

at the operating point on the saturation curve; VCEsat is used to make the model fit the published curves.

© 2004 by CRC Press LLC

34

 

Analysis and Application of Analog Electronic Circuits

 

 

+

+

 

 

 

IC

 

+

VCB

 

 

IC

 

 

 

 

VCB

+

 

 

 

 

n

 

 

 

IB

 

 

 

 

 

 

 

 

IB

 

 

p

VCE

 

 

+

 

VCE

 

 

+

 

 

n

 

 

 

 

 

VBE

VBE

 

 

 

 

 

 

A

IE

IE

 

B

regionActive

iC

 

 

iB = 100 µA

 

 

 

 

 

 

 

 

 

iB

 

10 mA

 

 

iB = 80

µA

@ VCEQ

 

 

 

 

Saturation

 

 

iB = 60

µA

 

region

 

 

 

 

 

 

 

 

iB > iC /βF

 

 

 

IBQ

Q point

ICQ 5 mA

 

 

iB = 40 µA

 

 

 

 

 

 

 

 

Q point

 

 

 

 

 

iB = 20 µA

 

0

 

 

iB = 0

 

vBE

 

 

 

 

 

 

1

VCEQ

3

0

VBEQ

 

vCE

 

 

“Grounded emitter” operation

 

(Reverse

5 mA

C

 

 

D

operation)

 

 

 

 

 

 

 

 

 

FIGURE 2.9

(A) Symbol for an npn BJT. (B) “Layer cake” cross section of a forward-biased npn BJT. The cross-hatched layer represents the depletion region of the normally reverse-biased CB junction.

(C) Collector–base I–V curves as a function of iB for the npn BJT. (D) Base–emitter I–V curve at constant VCEQ. The BE junction is normally forward biased.

When a pnp BJT’s large-signal behavior is considered, the actual dc currents leave the base and collector nodes and enter the emitter node. Consequently,

in the iC = f(vCE, iB) and iB = g(vBE, vCE) curves shown in Figure 2.11, all currents and voltages are taken as negative (with respect to the directions and signs

of the npn BJT). Figure 2.10(C) and Figure 2.10(D) illustrate the dc biasing models for a pnp BJT. Note that the sign of the CCCS is not changed; a negative IB times a positive β yields a negative IC, as intended.

Figure 2.12 illustrates how the dc biasing model of Figure 2.10(A) is used to set an npn transistor’s operating point in its active region. As an example, the required base-biasing resistor, RB, given the transistor’s βo, and the

desired Q-point (ICQ, VCEQ) for the device will be found, as well as the necessary RC and RE. VCC and the requirement that VCEQ = VCC/2 and VE =

© 2004 by CRC Press LLC

Models for Semiconductor Devices Used in Analog Electronic Systems

35

 

 

C

 

 

 

C

 

 

IC > 0

 

 

ICsat

> 0

+

 

 

 

 

 

 

 

 

 

 

 

 

 

 

VCEsat

 

 

 

 

βoIB

 

 

 

 

IB > 0

 

IBsat > 0

 

rcsat

 

 

 

 

 

B

+ VBE

 

B

+

 

 

E

VBEsat

E

 

 

 

 

 

 

 

 

A

 

 

 

B

 

 

 

 

C

 

 

 

C

 

 

IC < 0

 

 

ICsat

< 0

VCEsat

 

 

 

 

 

 

 

+

 

 

 

 

βIB

 

 

rcsat

 

IB < 0

 

IBsat < 0

 

 

 

 

 

 

B

+

 

B

 

+

 

 

 

VBE

E

 

VBEsat

E

 

 

 

 

 

 

 

 

C

 

 

 

D

 

 

FIGURE 2.10

(A) Simple dc biasing model for an npn BJT in its forward linear operating region (see text for examples). (B) dc biasing model for an npn BJT in forward saturation. (C) Simple dc biasing model for a pnp BJT in its forward linear operating region. (D) dc biasing model for a pnp BJT in forward saturation.

0.5 V are given. Pick numbers: VCC 12 V at the Q-point, VCEQ 6 V, and the desired ICQ 1 mA. βo 80 and VBEQ 0.5 V. It is known that VC = VE + VCEQ = 6.5 V; thus the voltage across RC is 5.5 V. From Ohm’s law, RC = 5.5/0.001 = 5.5 kΩ. Furthermore, IBQ = ICQ/βo = 0.001/80 = 12.5 μA and VB =

VBEQ + VE = 1 V, so RB = (12 – 1)/12.5 = 880 kΩ. Note that the emitter current is IE = (β + 1)IB = IC(1 + 1/β). Thus, RE = 0.5/(1 + 1/80) = 494 Ω.

2.3.2Mid-Frequency Small-Signal Models for BJTs

To find the voltage amplification factor or gain of a simple npn BJT amplifier, we use what is called a two-port, mid-frequency, small-signal model (MFSSM) for the transistor. The MFSSM also allows calculation of the amplifier’s input and output resistances. The most common MFSSM for BJTs is the common-emitter h-parameter model shown in Figure 2.13. The two-input h-parameters define a Thevenin input circuit between the base and emitter, and the output h- parameter pair defines a Norton equivalent circuit between the collector and emitter. The input resistance, hie, is derived operationally from the slope of the nonlinear IB = f (VBE, VCE) curves at the Q-point, as shown in Figure 2.11. That is:

hie

=

vBE

 

Ohms

(2.12)

iB

 

 

 

 

Q,Vce = const.

 

 

 

 

 

© 2004 by CRC Press LLC

36

Analysis and Application of Analog Electronic Circuits

iC

iC @ saturation

 

 

 

 

 

 

 

 

iB =50 A

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

iB

 

 

∆ iC

 

iB = IBQ

iB =40 A

 

 

 

 

 

ICQ

 

∆ iC

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Q point

∆vCE

 

 

 

 

 

 

 

 

 

iB = 30 A

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

iB = 20 A

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

iB = 0

 

 

0

 

 

 

 

 

 

 

 

0

 

VCEQ

vCE

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

iB

@ VCEQ

 

Q point

 

 

 

 

IBQ

 

 

 

∆ iB

 

 

∆ vBE

 

v

 

 

 

0

 

 

 

BE

0

VBEQ

 

 

 

 

 

 

 

 

 

 

 

 

FIGURE 2.11

Top: a large-scale plot of a typical BJT’s iC vs. vCE curves. The Q point is the transistor’s quiescent operating point. The small-signal conductance looking into the collector–emitter nodes is approximated by go = iC/ vCE siemens. The small-signal collector current gain is β = iC/ iB. Bottom: the iB vs. vBE curve at the Q point. The small-signal resistance looking into the BJT’s base is rb = vBE/ iB ohms.

© 2004 by CRC Press LLC

Models for Semiconductor Devices Used in Analog Electronic Systems

37

VCC

 

RC

VC

C

RB

βoIB

IB

 

VB + VBE

VE

E

 

 

RE

 

IE

FIGURE 2.12

Model for dc biasing of an npn BJT with collector, base, and emitter resistors. See text for analysis.

 

 

 

C

 

B

 

 

 

E

 

E

ib B

hie

C

ic

+

+

+

 

 

 

 

vbe

hrevce

hfe ib

hoe vce

 

 

 

E

 

E

 

FIGURE 2.13

Top: an npn BJT viewed as a two-port circuit. Bottom: the linear, common-emitter, two-port, small-signal, h-parameter model for the BJT operating around some Q in its linear region. Note that the input circuit is a Thevenin model and the output is a Norton model.

© 2004 by CRC Press LLC

38

Analysis and Application of Analog Electronic Circuits

hre is the gain of a voltage-controlled voltage source (VCVS), which models how the base voltage is modulated by changes in vCE. In other terms, hre is the open-circuit reverse transfer voltage gain.

hre =

vBE

 

(2.13)

vCE

 

 

 

Q,iB = const.

 

 

For most pencil-and-paper transistor gain calculations using the C–E SMFSSM, hre is set to zero because it (1) has a second-order contribution to calculations and (2) simplifies calculations. For the collector–base Norton model, hoe is the output conductance that results from the iC = f (vCE, iB) curves having a finite upward slope. Figure 2.11 shows that hoe is given by:

hoe =

iC

 

 

siemens

(2.14)

vCE

 

Q,iB = const.

 

 

 

 

 

 

 

 

Finally, the forward, current-controlled current source gain, hfe, is found:

hfe =

iC

 

 

(2.15)

iB

 

Q,Vce = const.

 

 

 

 

 

 

hfe is also called the transistor’s β.

One question that invariably arises concerns the MFSSM to be used when a grounded emitter pnp transistor is considered. The answer is simple: the same model as for an npn BJT. Whether a BJT is npn or pnp affects the dc biasing of the transistors, not their gains and small-signal input and output impedances. The quiescent operating point of pnp and npn transistors affects the numerical values of the CE h-parameters. For example, hre can be approximated by:

hre = rx + rπ ohms

(2.16)

where rx = the base spreading resistance, a fixed resistor with a range of 15 to 150 Ω, and rπ = the dynamic, SS resistance of the forward-biased base–emitter junction.

The value of rπ can be estimated from the basic diode equation for the BE junction:

 

 

IB = Ibs exp(VBEQVT) amps

 

(2.17)

The resistance, rπ, is given by:

 

 

 

 

 

 

 

r

=

1

=

1

=

ηV

=

ηVT hfe

ohms

(2.18)

 

 

T

 

 

 

 

 

 

 

π

 

gπ

 

iB vBE

 

IBQ

 

ICQ

 

 

 

 

 

 

 

 

 

© 2004 by CRC Press LLC

Models for Semiconductor Devices Used in Analog Electronic Systems

39

50

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

VCEQ = 5 V

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

T = 25oC

 

 

 

 

 

 

 

 

 

 

 

 

 

20

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

hoe

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

10

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

hre

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

hxe @ IC 5

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

hie

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

hxe @ 1 mA

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Q

 

 

 

 

 

 

 

hfe

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1.0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

hoe, hfe

 

 

 

 

 

 

 

 

 

 

hre

 

 

 

0.5

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0.2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

hie

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0.1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0.1

0.2

0.5

1.0

2

5

10

20

50

100

 

 

 

 

 

 

 

 

IC

(mA)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ICQ

 

 

 

 

 

 

 

 

 

 

 

FIGURE 2.14

A normalized plot of how the four common-emitter, small-signal h-parameters vary with collector current at constant VCEQ. Note that the output conductance, hoe, increases markedly with increasing collector current, while the input resistance, hie, decreases linearly. (Note that the same sort of plot can be made of normalized, CE h-parameters vs. at constant ICQ.)

For a silicon BJT with IBQ = 5 105 A, rπ = 1.04 kΩ.

Figure 2.14 illustrates how the four CE h-parameters vary with IC/ICQ at constant VCEQ. Note that the forward current gain, hfe, is the most constant with IC/ICQ. The collector–emitter Norton conductance, hoe, increases monotonically with IC/ICQ because, as IC increases, the IC = f (VCE, IB) curves tip upward. hie decreases monotonically with increasing IC, as predicted by Equation 2.18. All four hxe parameters increase with temperature.

Another MFSSM used to analyze transistor amplifier behavior on paper is the grounded-base, h-parameter model, shown in Figure 2.15. Clearly, this MFSSM is useful in analyzing grounded-base BJT amplifiers; however, the same results can be obtained using the more common grounded-emitter, h-parameter MFSSM. The GB h-parameters are found experimentally from:

hib

=

vEB

 

 

 

ohms

(2.19A)

iE

 

 

 

 

 

 

Q, vCB = const.

 

 

 

 

 

 

 

 

 

 

 

 

 

hrb =

 

 

vEB

 

 

(2.19B)

 

 

 

iCB

 

Q,ie = const.

 

 

 

 

 

 

 

© 2004 by CRC Press LLC

40

Analysis and Application of Analog Electronic Circuits

ie E

hib

 

 

C ic

 

+

hfbie

 

+

 

 

 

 

 

hrbvcb

 

hob

vcb

 

 

 

 

 

 

 

B

 

 

B

 

FIGURE 2.15

The common-base h-parameter SSM for npn and pnp BJTs. Using linear algebra, it is possible to express any set of four linear, two-port parameters in terms of any others. (Northrop, R.B., 1990, Analog Electronic Circuits: Analysis and Applications, Addison–Wesley, Reading, MA.)

 

hfb

=

 

 

iC

 

 

(2.19C)

 

 

 

iE

 

Q, vCB = const.

 

=

 

iC

 

 

 

 

hob

 

 

 

 

 

siemens

(2.19D)

 

 

 

 

 

vCB

 

 

 

 

 

 

Q,iE = const.

 

 

 

 

Most manufacturers of discrete BJTs do not give GB h-parameters in their specification data sheets.

2.3.3Amplifiers Using One BJT

Three basic amplifier configurations can be made with one BJT: (1) the grounded-emitter; (2) the emitter-follower (grounded collector); and (3) the grounded-base amplifier. These three configurations are described and analyzed next.

The first example illustrates how to use the CE h-parameter MFSSM to find expressions for the voltage gain, KV, and the input and output resistance of a grounded emitter amplifier, shown in Figure 2.16(A). This figure illustrates a basic grounded emitter amplifier driven by a small ac signal, vs, through a coupling capacitor, Cc. In the MFSSM amplifier model, only smallsignal variations around the BJT’s operating (Q) point are considered; thus all dc voltage sources are set to zero and replaced with short circuits to ground. Figure 2.16(B) illustrates the complete amplifier model (hre = 0 for simplicity). The SS output voltage is, by inspection:

vo

=

hfeib

(2.20)

Gc

+ hoe

 

 

 

The base current is given by Ohm’s law, assuming negligible SS current flows in RB (RB hie).

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Models for Semiconductor Devices Used in Analog Electronic Systems

41

 

 

VCC

 

 

IBQ

 

 

 

 

 

RB

RC

 

 

 

vo

 

 

 

 

 

 

Rs

 

Rs

vb B hie

C vo

 

 

Cc

 

 

Cc

 

npn BJT

hfe ib

 

 

 

 

 

 

 

ib

 

vs

 

vs

RB

hoe GC

 

 

 

E

 

A

 

 

B

 

FIGURE 2.16

(A) Schematic of a simple, capacitively coupled, grounded-emitter BJT amplifier. (B) Linear, mid-frequency, small-signal model (MFSSM) of the grounded-emitter amplifier. Note that, at mid-frequencies, capacitors are treated as short circuits and DC source voltages are small-signal grounds. The mid-frequency gain, vo /vs, can be found from the model; see text.

ib

 

 

vs

(2.21)

Rs

+ hie

 

 

 

Thus, the amplifier’s SS voltage gain is found by substituting Equation 2.21 into Equation 2.20:

vo

=

−hfeRC

= K

 

(2.22)

vs

(hoeRC + 1)(Rs + hie )

V

 

 

 

 

 

 

 

By inspection, its SS input resistance is Rin Rs + hie and its (Thevenin)

output resistance is Rout = 1/(hoe + GC).

In a second example, a BJT emitter-follower amplifier is shown in Figure 2.17. Figure 2.17(B) illustrates the MFSSM for the emitter follower. To find an expression for the EF gain, write the node equations for the vb and vo nodes:

vb [Gs + GB + gie] − vo [gie] = vs Gs

(2.23A)

−vb gie + vo [GE + hoe + gie] − hfe ib = 0

(2.23B)

where gie = 1/hie and ib = (vb − vo)gie. When the expression for ib is substituted into Equation 2.23B,

−vb[gie (1+ hfe )]+ vo[GE + hoe + gie (1+ hfe )]= 0

(2.24)

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42

Analysis and Application of Analog Electronic Circuits

VCC

 

 

 

 

 

npn BJT

Rs

vb

hie

C

RB

 

 

B

 

hfe ib

Cc

 

 

 

ib

 

 

vs

RB

 

hoe

Rs

 

 

Cc

vo

 

 

 

E

 

 

 

 

vo

 

 

 

 

 

vs

 

 

 

 

 

RE

 

 

 

RE

(hfe + 1)ib

A

 

B

 

 

 

FIGURE 2.17

(A) Schematic of a simple, capacitively coupled, emitter-follower amplifier. (B) MFSSM of the EF amplifier.

Now Equation 2.23A and Equation 2.24 are solved using Cramer’s rule, yielding the voltage gain:

vo

=

 

Gs gie (1+ hfe )

 

 

= K

 

(2.25)

 

[

 

]

 

V

vs

 

(Gs + GB ) GE + hoe + gie (1+ hfe )

+ gie (GE + hoe )

 

 

 

 

 

 

This rather unwieldy expression can be simplified using the a priori knowledge that, usually, GE hoe and Gs GB. Equation 2.25 is multiplied top and bottom by hie RE Rs. After some algebra, the simplified EF gain is found to be:

KV

 

 

RE (1+ hfe )

 

(2.26)

RE (1

+ hfe )+ hie

 

 

 

+ Rs

Note that Kv for the EF is positive (noninverting) and slightly less than 1. What then is the point of an amplifier with no gain? The answer lies in the EF’s high input impedance and low output impedance. The EF is used to buffer sources and drive coaxial cables and transmission lines with low characteristic impedances. It is easy to show that the EF’s input impedance

at the vb node is simply RB in parallel with [hie + RE (1 + hfe)] (hoe is neglected). The Thevenin output impedance can be found for the linear MFSSM by

taking the ratio of the open circuit voltage (OCV) to its short-circuit current (SCC). The OCV is simply vs KV:

 

 

vsRE (1+ hfe )

VOC

 

 

 

(2.27)

RE (1

+ hfe )+ hie

 

 

 

+ Rs

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Models for Semiconductor Devices Used in Analog Electronic Systems

43

 

VCC

 

 

 

 

 

 

 

 

hoe

 

 

IBQ

RC

 

hfe ib

 

 

 

 

 

 

 

Rs

vo

Rs ve

 

vo

 

 

 

E

C

 

 

 

 

 

 

 

 

 

 

ib

 

vs

RB

 

vs

 

RC

 

 

 

 

B

 

 

 

 

 

hie

 

 

Cb

 

 

 

 

 

A

 

B

 

 

FIGURE 2.18

(A) Schematic of a simple, capacitively coupled, grounded-base amplifier. (B) MFSSM of the GB amplifier.

The SCC is simply ibsc(1 + hfe). When the output is short-circuited, vo = 0 and ibsc = vs/(Rs + hie). Thus, Rout is:

 

 

vsRE (1+ hfe )

R =

 

RE (1+ hfe )+ hie + Rs

 

 

vs (1

+ hfe )

out

 

 

 

 

 

 

 

(Rs

+ hie )

 

 

 

(Rs + hie )

 

 

=

 

RE

(1+ hfe )

 

(2.28)

 

(Rs

+ hie )

+ R

 

 

(1+ hfe )

E

 

 

 

 

 

Equation 2.28 shows that Rout is basically RE in parallel with (Rs + hie)/(1 + hfe).

In general, Rout will be on the order of 10 to 30 Ω.

The third single BJT circuit of interest is the grounded-base amplifier, shown in Figure 2.18. Again, to find the GB amp’s gain, the node equations are written on the ve and vo nodes:

ve [Gs + gie + hoe] − vo hoe − hfe ib = vs Gs

(2.29A)

−ve hoe + vo [hoe + GC] + hfe ib = 0

(2.29B)

Now ib = −ve gie is substituted into the preceding equations to get the final form for the node equations:

ve[Gs + hoe + gie (1+ hfe )]vo[hoe ] = vsGs

(2.30A)

ve[hoe + hfe gie ]+ vo[hoe + GC ]= 0

(2.30B)

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44

Analysis and Application of Analog Electronic Circuits

These equations are solved using Cramer’s rule. After some algebra, the GB amplifier gain is obtained:

Kv

=

 

RC[hfe

+ hoehie ]

(2.31)

RC[hoe (hie

+ Rs )]+ Rs[1+ hfe + hoehie ]+ hie

 

 

 

It is instructive to substitute typical numerical values into Equation 2.31

and evaluate the gain. Let RC = 5 k; hie = 1 k; hfe = 100; Rs = 100; and hoe = 10− 5 S. Kv = 44.8. Note that the gain is noninverting. From the numerical eval-

uation, certain terms in Equation 2.31 are found to be small compared with others (i.e., hoe 0), which leads to the approximation:

Kv

 

 

RChfe

(2.32)

Rs (1

+ hfe )+ hie

 

 

 

Next an expression will be found for and Rin and Rout evaluated for the GB amplifier. Rin is the resistance the Thevenin source (vs, Rs) “sees” looking into

the emitter of the GB amplifier. Assume hoe 0. The resistance looking into the emitter is approximately:

R

ve

ve

=

ve

=

hie

(2.33)

 

ib (hfe + 1)

(ve hie )(hfe + 1)

(hfe + 1)

inem

ie

 

 

 

 

 

 

 

To get a feeling for the size of Rin, substitute realistic circuit parameter

values. Let RC = 5000 Ω; Rs = 100 Ω; hfe = 100; hoe = 10−5 S; and hie = 1000 Ω. Thus, Rinem = 9.9 Ω. The output resistance is the Thevenin resistance seen

looking into the collector node with RC attached. If hoe 0 is set again, Rout RC, by inspection.

2.3.4Simple Amplifiers Using Two Transistors at Mid-Frequencies

Figure 2.19 illustrates four common two-BJT amplifier architectures. An npn Darlington pair is shown in Figure 2.19(A). Darlington pairs can also use pnp BJTs. These pairs are often used as power amplifiers in the emitterfollower architecture or as grounded-base amplifiers. Figure 2.19(B) is called a feedback pair; it is not a Darlington. It, too, is often used as a power amplifier in the EF or GE modes and will be analyzed later. Figure 2.19(C) is an EF–GB pair, used for small-signal, high-frequency amplification. Figure 2.19(D) is called a cascode pair; it is also used as a small-signal, high-frequency amplifier.

In the first example, an expression for the mid-frequency, small-signal voltage gain, as well as input and output resistances for a grounded-emitter Darlington amplifier, are found. The amplifier circuit and its MFSS CE

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Models for Semiconductor Devices Used in Analog Electronic Systems

45

 

C1

E1

C2

Q1

C2

Q1

 

 

 

B1

Q2

B1

Q2

 

 

E1 B2

C1

B2

A E2 B E2

 

VCC

 

 

 

VCC

 

RB

 

RC

RB

RB

RC

RB

 

 

 

vo

 

vo

 

 

Q1

 

Q2

 

 

 

 

 

 

 

 

 

vs

 

 

 

 

Q2

 

 

 

 

 

 

 

 

 

 

 

Q1

 

 

 

RE

C

 

vs

D

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

RE

 

FIGURE 2.19

Four common two-BJT amplifier configurations: (A) The Darlington pair. Note that Q1’s emitter is connected directly to Q2’s base. (B) The feedback pair. The collector of the pnp Q1 is coupled directly to the npn Q2’s base. (C) The emitter-follower–grounded-base (EF–GB) BJT pair. (D) The cascode pair.

h-parameter model are shown in Figure 2.20. To simplify analysis, assume

RB , hoe1 0, and hoe2 0. From Ohm’s law, ib1 = (vs vb2)/(Rs + hie1),

ib2 = vb2/hie2.

Now write node equations for the vb2 and the vo nodes:

vb2 [1 hie2 + 1 (Rs + hie1)]− hfe1ib1 = vs (Rs + hie1)

(2.34A)

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46

Analysis and Application of Analog Electronic Circuits

VCC

RB

RC

 

Q1

Rs

vo

 

vs

Q2

 

Rs

vb1 hie1

C1

 

ib1

hfe1ib1

 

 

vs

RB

hoe1

 

vb2 hie2

C2 vo

E1

B2

ib2

hfe2ib2

 

 

 

hoe2 RC

FIGURE 2.20

Top: a common-emitter Darlington amplifier. Bottom: CE h-parameter MFSSM of the Darlington amplifier. Analysis is in the text.

 

 

vo [GC] + hfe1ib1 + hfe2ib2 = 0

 

 

(2.34B)

and then substitute the relations for ib1 and ib2 and find:

 

 

vo

 

+ (1+ hfe1) (Rs + hie1) = vs[(1+ hfe1)

(Rs

˘

(2.35A)

[0] + vb2 1 hie2

+ hie1)]˙

 

 

 

 

˚

 

 

vo[GC ] + vb2[hfe2 hie2 hfe1 (Rs + hie1)]= −vshfe1

(Rs + hie1)

(2.35B)

Using Cramer’s rule, find the GE Darlington amplifier’s voltage gain:

vo

 

RC (1+ hfe1 + hfe1hfe2 )

 

 

(Rs + hie1)+ hie2 (1+ hfe1)

= Kv

(2.36)

vs

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Models for Semiconductor Devices Used in Analog Electronic Systems

47

VCC

RB RC

Q1 vo

Rs

Q2

vs

E1

 

hoe1

 

 

 

hfe1ib1

vb2

hie2

C2

vo

Rs B1 hie1

C1

B2

ib2

hfe2ib2

 

ib1

 

 

 

vs

 

 

hoe2

GC

E2

FIGURE 2.21

Top: feedback pair amplifier with load resistor, RC, attached to common Q1 emitter + Q2 collector. Bottom: MFSSM of the feedback pair amplifier above. Analysis in text shows this circuit behaves like a super emitter follower.

By inspection, the Thevenin Rout RC and Rin hie1 + (1 + hfe1)hie2. As the section on BJT amplifier frequency response demonstrates, the GE Darlington

has relatively poor high-frequency response, trading off gain for bandwidth. This example considers the feedback pair amplifier shown in Figure 2.19(B) and Figure 2.21. Note that in this amplifier, it is the collector of the pnp Q1 that drives the base of the npn Q2, and Q2’s collector is tied to Q1’s emitter.

The auxiliary equations are:

 

ib1 = (vs vo)/(Rs + hie1)

(2.37A)

ib2 = vb2/hie2

(2.37B)

© 2004 by CRC Press LLC

48 Analysis and Application of Analog Electronic Circuits

Again, for mathematical convenience, use the approximations hoe1 0 and

hoe2 0, and write the node equations for vo and vb2:

 

vo[GC + 1 (Rs + hie

1)]+ hfe2ib2 hfe1ib1 = vs (Rs + hie1)

(2.38A)

vb2

[1 hie1]+ hfe1 ib1 = 0

(2.38B)

Now substitute the auxiliary relations for ib1 and ib2 into Equations 2.38A

and Equation 2.38B:

 

 

 

 

 

 

 

 

+

1+ hfe1

˘

+ vb2[hfe2

hie2 ]= vs

1+ hfe1

 

vo GC

 

˙

 

(2.39A)

R + h

R + h

 

 

s ie1

˚

 

 

s ie1

 

vo

hfe1

+ vb2[1 hie2 ]= −

vs

(2.39B)

R + h

R + h

 

s ie1

 

s ie1

 

The two preceding equations are solved by Cramer’s rule to find Kv.

 

 

 

1+ h

fe1

˘

 

 

h

fe1

h

fe2

= (1 hie1) GC

+

 

˙

+

 

 

 

 

 

hie2

(Rs + hie1)

 

 

 

Rs + hie1 ˚

 

v

 

=

(1+ hfe1)hie2 + hfe2hie1

 

 

 

 

o

 

 

hie1hie2 (Rs + hie1)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Thus, the voltage gain is:

vo

 

 

RC

(1+ hfe1 + hfe

2hfe1)

 

 

 

 

[

 

]

= K

 

 

[

 

]

 

v

vs

 

RC (1+ hfe1 + hfe1hfe2 )

+ (Rs + hie1)

 

 

 

 

(2.40)

(2.41)

(2.42)

Taking “typical” numerical values, Rs + hie1 = 1050 Ω; hfe1 = 100; hfe2 = 20; and RC = 5000 Ω, then Kv = 0.999895. It appears that the feedback pair amplifier of Figure 2.21 behaves like a super emitter follower with (almost) unity gain.

Next, examine Rin and Rout of the FBP amplifier. To find Rout, use the opencircuit voltage/short-circuit current method. The OCV approximated by

vs voc because of the near unity gain of the amplifier. The short-circuit current at the C2E1 node is:

isc = (1 + hfe1) ib1 hfe2 ib2

(2.43)

With the output short-circuited, vo = 0 and ib1 = vs/(Rs + hie1). Neglecting hoe1, it is possible to write:

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Models for Semiconductor Devices Used in Analog Electronic Systems

ib2 = −hfe1 ib1 = −hfe1 vs/(Rs + hie1)

Thus:

isc = (1 + hfe1)vs/(Rs + hie1) + hfe2 hfe1 vs/(Rs + hie1)

Clearly, Rout = voc/isc, yielding:

Rout

 

 

Rs + hie1

 

+ hfe1 + hfe1hfe2

 

1

49

(2.44)

(2.45)

(2.46)

If the numbers used previously are substituted, Rout = 0.5 Ω — significantly smaller than a single BJT’s emitter follower. Rin looking into the B1 node is just vs/ib1. That is:

Rin = vshie1 (vs − vo ) = hie1

(1− vo

vs ) =

 

hie1

 

1

RC[1+ hfe1 + hfe1hfe2 ]

 

 

 

¬

 

 

RC[1+ hfe1 + hfe1hfe2 ]+ hie1

 

 

 

 

(2.47)

Rin = RC[1+ hfe1 + hfe1hfe2 ]

+ hie1

 

 

 

 

 

 

Rin 10.51 MΩ, using the previous parameters.

Next, examine the behavior of the feedback pair when the load resistor is connected to E2 and C2E1 is connected to Vcc. The circuit is shown in Figure

2.22. It is not necessary to solve two simultaneous node equations to find Kv; by Ohm’s law, ib1 = vs/(Rs + hie1) and ib2 = −ib1hfe1. Now it is clear that:

vo

= ib2 (1+ hfe2 )RE

= −ib1hfe1

(1+ hfe2 )RE =

−vshfe1(1+ hfe2 )RE

= voOC

(2.48)

 

 

 

 

 

 

 

 

 

 

Rs + hie1

 

Thus,

 

 

 

 

 

 

 

 

 

 

 

 

K

 

= −

REhfe1(1+ hfe2 )

 

(2.49)

 

 

v

(Rs + hie1)

 

 

 

 

 

 

 

 

 

 

 

 

 

Rin for this amplifier is found by inspection; Rin = hie1. Rout is found by voOC/ioSC:

 

 

−vshfe1(1+ hfe2 )RE

 

 

Rout

=

(Rs

+ hie1)

= RE

(2.50)

−vshfe1(1+ hfe2 )

 

 

(Rs

+ hie1)

 

 

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50

Analysis and Application of Analog Electronic Circuits

VCC

RB

Q1

Rs

Cc

Q2

vs

+

RE vo

Rs B1 hie1

C1 vb2 hie2

C2

hfe1ib1

B2

ib2

hfe2ib2

ib1

 

 

vs

hoe1

 

hoe2

 

 

 

E2

E1 vo

RE

FIGURE 2.22

Top: feedback pair amplifier with load resistor in Q2’s emitter. Bottom: MFSSM of the amplifier shown above. Analysis in text shows this amplifier has a high inverting mid-frequency gain; it does not emulate an emitter follower.

Next, examine the gain, Rin and Rout, of the emitter-follower-grounded base amplifier shown in Figure 2.19(C). Figure 2.23 illustrates the MFSSM for this amplifier. Again, for algebraic simplicity, assume that hoe1 and hoe2 are negligible; also, R1 = Rs + hie1, all capacitors behave as short circuits at mid-frequencies, and RB •. Two node equations can be written for the MFSSM:

© 2004 by CRC Press LLC

Models for Semiconductor Devices Used in Analog Electronic Systems

51

 

 

 

 

 

RE

 

Rs

B1

hie1

E1

E2

ve

hie2

 

 

 

 

 

 

B2

 

 

ib1

 

 

 

 

vs

 

 

hoe1

 

 

hoe2

 

 

 

 

 

 

 

hfe1ib1

 

hfe2 ib2

 

 

 

 

 

C1

 

C2

 

 

 

 

 

 

vo

 

 

 

 

 

 

RC

FIGURE 2.23

MFSSM for the EF–GB amplifier of Figure 2.19(C). See text for analysis.

vo [GC] + hfe2 ib2 = 0

(2.51A)

ve [GE + 1/hie2 + G1] − hfe2 ib2 − hfe1ib1 = vs G1

(2.51B)

where:

 

ib2 = −ve/hie2

(2.51C)

ib1 = (vs ve)G1

(2.51D)

Substituting Equation 2.51C and Equation 2.51D into the node equations yields:

vo [GC] − ve [hfe2/hie2] = 0

(2.52A)

[

]

G1(1+ hfe1)

 

vo[0]+ ve GE

+ (1+ hfe2 ) hie2 + G1(1+ hfe1)

= vs

(2.52B)

From Equation 2.52A and Equation 2.52B,

 

 

 

 

vo = ve RC [hfe2/hie2]

 

 

(2.53)

can be written and

ve =

[

vsG1

(1

+ hfe1)

]

(2.54)

+ (1+ hfe2 )

 

 

GE

hie2 + G1(1+ hfe1)

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52

Analysis and Application of Analog Electronic Circuits

 

 

 

 

C2

vo

 

 

 

hfe2ib2

 

 

 

 

 

 

hoe2

 

 

 

 

E2

 

 

Rs

B1

ib1

ve

 

 

C1

 

 

 

 

 

hfe1ib1

hie2

RC

 

 

 

ib2

 

vs

 

hie1

hoe1

B2

 

 

 

 

E1

 

 

FIGURE 2.24

MFSSM for the cascode amplifier of Figure 2.19(D). See text for analysis.

Now Equation 2.54 is substituted into Equation 2.53 to find Kv:

K

 

=

vo

=

RChfe2 (1+ hfe1)

 

(2.55)

v

vs

hie2 (R1 RE )+ R1(1+ hfe2 )+ hie2 (1

+ hfe1)

 

 

 

 

 

 

 

 

 

Substitute typical values RC = 5000 Ω; R1 = 1050 Ω; hfe1 = hfe2 = 100; and hie2 = 1000 Ω, and find Kv = +244, i.e., the EF–GB amplifier is noninverting. The EF–GB amplifier will be shown to have excellent high-frequency response. It is left as an exercise for the reader to find expressions for the mid-frequency Rin and Rout for this amplifier.

The final two-BJT amplifier stage to be analyzed using the CE MFSSM is the cascode amplifier, shown in Figure 2.19(D). The MFSSM for this amplifier is shown in Figure 2.24. Again, assume capacitors behave as short circuits

at mid-frequencies, hoe1 = hoe2 = 0, and RB •. Clearly, vo = −hfe2 ib2 RC and ib2 = −ve/hie2, so vo = ve hfe2 RC/hie2 and ib1 = vs/R1. To find ve, write the node equation:

ve [1/hie2] + hfe1ib1 hfe2 ib2 = 0

(2.56)

Substituting for ib1 and ib2 into Equation 2.56 yields:

ve

=

vshfe1hie2G1

(2.57)

1

+ hfe2

 

 

 

© 2004 by CRC Press LLC

Models for Semiconductor Devices Used in Analog Electronic Systems

53

Equation 2.57 is substituted into the preceding equation for vo to find Kv:

K

 

=

vo

=

−hfe1RChfe2

−hfe1RC

(2.58)

v

vs

R1(1+ hfe2 )

R1

 

 

 

 

 

 

 

 

 

 

 

By inspection, Rin = R1, Rout RC. Like the EF–GB amplifier, the cascode amplifier has a high-frequency bandwidth. It also can be described as a GE–GB amplifier.

2.3.5The Use of Transistor Dynamic Loads To Improve Amplifier Performance

In many analog IC designs, including those with differential amplifiers, transistors and small transistor circuits are used to create small-signal, dynamic high impedances that, when used in amplifiers to replace conventional resistors, increase amplifier gain or differential amplifier common-mode rejection ratio. Figure 2.25 illustrates four BJT circuits that act as dc current sources or sinks with high, small-signal, Norton resistances. For example, if 2 mA dc flows through a potential difference of 5 V in an electronic circuit, a 2.5-k resistor is needed. If a dynamic current source is used, the equivalent small-signal (Norton) resistance can range from 100 k to 10 M, depending on the circuit. A good way to view such dynamic resistances is to treat the circuit as a dc current source in parallel with a Norton resistance of 100 k to 10 M.

The most basic dynamic resistance is realized by looking into the collector of a BJT with an emitter resistance that is not bypassed, as shown in Figure 2.25(A). Figure 2.26 illustrates how the MFSSM is used to find the resistance looking into the collector. Note that the base is bypassed to ground at signal

frequencies. Clearly, from Ohm’s law, Rincol = vt/it. (vt is a small-signal test source.)

Begin analysis by finding an expression for ve = f (vt). Writing the node equation:

ve[(1 hie )+ GE + hoe ]− βib = vt hoe

(2.59A)

ib = −ve

hie

(2.59B)

¬

 

 

ve[(1 hie )(1+ β) + Ge

+ hoe ]= vt hoe

(2.59C)

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54

Analysis and Application of Analog Electronic Circuits

 

VCC > 0

 

 

VBB > 0

 

 

 

IS

 

 

IS

 

RC

 

 

RB

IC1

IB1

IB2

 

 

 

Q1

+

Q2

 

 

 

 

 

VBE

 

RE

 

 

 

 

 

B

A

VEE < 0

VCC < 0

 

 

VCC > 0

 

 

IS

 

IS

RC

 

 

IRc

RC

 

 

 

 

IB3

 

 

 

 

Q3

Q1

 

Q2

IC1

 

 

+

 

 

 

 

VBE1

 

Q1

Q2

 

RE

 

+

 

 

 

 

 

 

VBE

 

 

 

 

 

 

 

C

VEE > 0

 

D

 

FIGURE 2.25

Four BJT circuits used for high-impedance current sources and sinks in the design of differential amplifiers and other analog ICs. They can serve as high-impedance active loads. (A) Collector of simple BJT with unbypassed emitter resistance. (B) Basic two-BJT current sink. (C) Widlar current source (pnp BJTs are used). (D) Wilson current sink.

Thus, ve is:

ve =

vthoe

(2.60)

(1 hie )(1+ β) + GE + hoe

and the test current is:

 

it = ve[(1 hie )+ GE ]

(2.61)

© 2004 by CRC Press LLC

Models for Semiconductor Devices Used in Analog Electronic Systems

55

VBB

 

 

 

 

it

 

 

 

 

 

 

 

β ib

 

RB

 

 

 

 

+

B

hie ve

hoe

it

 

 

vt

 

 

 

 

 

 

E

C

+

Cb

 

ib

 

 

 

 

 

vt

RE

 

RE

 

 

 

 

 

 

FIGURE 2.26

Left: test circuit for calculating small-signal resistance looking into BJT’s collector. An ac test source, vt, is used; it is measured. Right: MFSSM of the test circuit used to calculate the expression for the small-signal resistance looking into collector.

When Equation 2.60 is substituted into Equation 2.61, it is possible to solve

for Rincol:

 

 

 

1+

β RE + hoehieRE

 

R =

vt

=

RE

+ hie

 

 

(2.62)

 

 

 

 

incol

it

 

 

hoe

 

 

 

 

 

 

 

Looking into the collector of a BJT with no RE, Rincol hoe1 Ω can be seen. To see how much RE increases Rincol, substitute some typical parameter values

into Equation 2.62: RE = 5k; hie = 1k; β = 100; and hoe = 105 S. Crunching the

numbers, Rincol = 8.4 MΩ. Note that the BJT’s base was bypassed to ground in the preceding development. If the base is not bypassed, hie hie + RB in

Equation 2.62, where RB is the Thevenin resistance that the base “sees.” The

unbypassed base causes Rincol to decrease markedly, but it still remains above 1/hoe.

More sophisticated, high dynamic resistance dc current sources are shown in Figure 2.25(B) and Figure 2.25(C). Figure 2.25(B) is the basic, two-BJT current sink; its dynamic input resistance is about 1/hoe Ω. Figure 2.25(C) is the Widlar current source (note pnp BJTs are used). Figure 2.25(D) is a Wilson current sink. Note that a collector–base junction in all three current source/active load circuits is short-circuited (Q1 in Figure 2.25(B) and Figure 2.25(C), Q2 in Figure 2.25(D)). The transistor with the shorted collector–base junction then behaves as a forward-biased, base–emitter diode junction.

Analysis of the basic two-BJT current sink (Figure 2.25(B)) is made easy when it is assumed that Q1 and Q2 have identical parameters. Because both transistors have the same VBE, their collector currents are equal, i.e., IC1 = IC2. Summing currents at Q1’s collector,

IRc − IC1 = 2IC1/β

(2.63)

© 2004 by CRC Press LLC