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190

 

 

 

 

Analysis and Application of Analog Electronic Circuits

 

(

jωτ

A

K

vo

+ 1

 

 

 

 

Vo

 

a vo

 

)

[Gμ + jω(CT

+ CN )]Vo jωCN

= VbGμ

(4.72)

 

 

Avo

 

 

 

 

 

 

 

 

 

 

 

 

 

Equation 4.72 can be wrestled algebraically into the transfer function of the capacitance-neutralized GME amplifier:

Vo

=

 

 

 

 

 

Avo

(4.73)

 

 

 

 

 

 

 

V

 

jω

2

τ A C

+ C R K

+ jω τ A K + R C R C A − 1 + 1

 

(

 

) a vo ( T

N ) μ vo

[ a vo vo μ T μ N ( vo )]

 

 

b

 

 

 

 

From the standard quadratic form, the closed-loop amplifier’s undamped natural frequency squared is:

ωn2 =

Kvo

(r s)2

(4.74)

τaAvo (CT + CN )Rμ

 

 

 

The amplifier’s damping factor is found from the second term in the denominator of Equation 4.73

ξ =

(

ω

n

2

)[

τ A

K

vo

+ R C R C

A

− 1

(4.75)

 

 

 

a vo

 

μ T

μ N (

vo

)]

 

It is now useful to illustrate how the capacitance-neutralized, PVF amplifier works with typical numbers. Let Kvo = 105; τa = 10−2 sec; Avo = 2; Rμ = 108 Ω; CT = 3.05 pF; and CN = 2.95 pF. From these numbers it can be found that ωn = 9.129 ∞ 104 r/s, and ξ = 0.464 (an underdamped second-order system with good transient response). Note that if CN = CT, the system is highly underdamped and, if CN is slightly greater than CT, the amplifier’s closedloop poles lie in the right-half s-plane, making the amplifier oscillate. The price paid for extending the system’s bandwidth can be shown to be excess noise (see Section 9.8.4 in Chapter 9).

4.6Chapter Summary

Most electronic amplifiers use negative voltage feedback (NVF), which:

1.Reduces mid-band gain

2.Reduces total harmonic distortion at the output at a given signal output power level

3.Reduces Zout at low and mid-frequencies

4.Decreases gain sensitivity to certain circuit parameters

©2004 by CRC Press LLC

General Properties of Electronic Single-Loop Feedback Systems

191

5.Decreases the output signal-to-noise ratio slightly

6.Increases the closed-loop amplifier’s bandwidth

7.Can increase or decrease Rin, depending on the circuit

8.Can make a system unstable if incorrectly applied

Negative current feedback was shown to be useful in making a VCCS out of an amplifier with normally low Rout. NCF is used to raise the Thevenin Rout so that the NCF amplifier appears to be a current source. Otherwise, its properties are similar to those resulting from NVF. Positive voltage feedback reverses properties 1 through 4 and 6 in the preceding list for NVF. It does have use for capacitance neutralization and is used in the design of certain oscillators because it can make systems unstable more easily than NVF.

Home Problems

4.1Negative current feedback is used in a VCCS circuit used to power a laser diode (LAD), as shown in Figure P4.1. The LAD presents a nonlinear load to the VCCS. We wish the diode current to be independent of the LAD’s nonlinear resistance.

a.Assume the op amp is ideal and the differential amplifier (DA) is a VCVS in which V2 = KD (Vo − VL). Derive an expression for the VCVS’s transconductance, GM = IL/Vs.

b.Now assume that the power op amp is nonideal, so Vo = Kvo (Vi − Vi). Derive an expression for the Norton output conductance seen by the LAD. (Hint: set Vs = 0 and replace the LAD by a test voltage source, vT. Find Gout = iT/vT.)

 

Vi

 

 

 

 

Vo RF

VL

+

Vi

POA

 

 

IL

Vs

 

 

 

 

 

R1

R2

 

LAD

 

 

DA

 

 

V2

 

 

FIGURE P4.1

4.2Negative voltage feedback is used to make an electronically regulated dc voltage source, shown in Figure P4.2. The design output voltage is 15 V at

1-A load. The power NPN BJT can be modeled by a mid-frequency

h-parameter model with hfe = 19; hoe = 2 105 S; hre = 0: hie = VT/IBQ; VT =

0.026 V; and IEQ = 1 A. DA gain is KD = 1 104; RL = 15 Ω, neglect current through RF + R1; and VR = 6.2 V, feedback attenuation β = R1/(R1 + R2) = 6.2/15. The raw dc has a ripple voltage, vr, added to it. RS = 50 Ω.

© 2004 by CRC Press LLC

192

 

Analysis and Application of Analog Electronic Circuits

 

 

IEQ

vo

 

 

RS

 

vr

RF

IL = IE

 

βVo

 

 

 

 

 

 

IBQ vb

 

 

Vs

DA

RL

 

R1

 

 

 

 

VR

FIGURE P4.2

a.Find an algebraic expression for and evaluate numerically the

regulator’s ripple gain, AR = vo/vr. Use the MFSSM for the BJT (e.g., dc sources ground).

b.Now let vr = hoe = 0. Find an expression for and evaluate numerically the regulator’s output resistance, Rout, seen by RL. You can find Rout = vt/it by replacing RL with a small-signal test voltage source, vt.

c. Evaluate the regulator’s regulation, ρ = Vo/ RL.

4.3An op amp is ideal except for a finite differential voltage gain: vo = Kvo (vi vi′). In the circuit of Figure P4.3, the op amp is connected to make a VCCS for

RL.

a.What kind of feedback is used: NVFB, PVFB, NCFB, or PCFB?

b.Give an expression for GM = IL/Vs. Show what happens to GM as Kvo •.

c.Find an expression for the Thevenin Rout that RL “sees.”

vi

Vo

 

vi

OA

 

 

 

 

Vs

 

RL

IL

 

 

R1

 

R2

VF

 

βVF

 

 

 

 

0

 

 

 

 

RM

FIGURE P4.3

© 2004 by CRC Press LLC

General Properties of Electronic Single-Loop Feedback Systems

193

VDD

 

RD

 

 

V2

 

D

DA

Vo

 

 

S

V1

RF

RS

VSS

FIGURE P4.4

4.4Figure P4.4 illustrates a source-follower–grounded-gate JFET amplifier with

feedback. The JFETs are identical with the same MFSSM in which gm > 0 and gd = 0. The feedback voltage, Vo = −KDV2, is applied through resistor RF to the common source node.

a.Derive an expression for Av = V2/V1 with no feedback (set RF = ).

b.Find an expression for Av = Vo/V1 with feedback. Let KD ; give Av.

4.5A three-op amp VCCS circuit is shown in Figure P4.5. Assume that op amps OA2 and OA3 are ideal and the power op amp, OA1, is characterized by the transfer function:

 

V − V′ K

vo

V =

(

1 1)

 

(τ s + 1)

 

o

 

 

 

 

 

and zero output resistance.

a.Derive an expression for the VCCS’s output transadmittance,

YM(jω) = IL/Vs, in time-constant form. Sketch and dimension 20 log YM(jω) vs. ω and YM(jω) vs. ω (Bode plot). Show what happens to YM(jω) as Kvo .

b.Derive an expression for the Norton output admittance the non-

linear load sees, Yout(jω) = IL/VL, in time-constant form. Sketch and dimension a Bode plot of Yout(jω) vs. ω. Show what happens to Yout(jω) as Kvo .

© 2004 by CRC Press LLC

194

 

Analysis and Application of Analog Electronic Circuits

 

V1

 

 

IL

 

 

 

 

 

 

 

Vo

RF

 

V1

POA

VL

 

 

 

 

 

 

Vs

 

R

 

 

 

 

 

RL

 

 

 

 

 

R

 

 

 

VL

 

IOA

 

IOA

 

 

R

 

R

 

 

 

 

VL

FIGURE P4.5

4.6A DA is given a form of common-mode negative feedback, as shown in Figure P4.6. The DA has a differential output from which voc is derived by a voltage divider. The amplifier is described by the scalar equations:

vo = AD v1c + AC v1c

vo′ = −AD v1c + AC v1c

Also, it is clear that voc = AC v1c. Define α = R1/(R1 + Rs) and β = Rs/(R1 + Rs).

a.Find expressions for v1c and v1d in terms of vsc and vsd, and α, β, K, AD, and AC.

b.Give an expression for vo in terms of vsc and vsd and circuit parameters.

c.Find an expression for the system’s single-ended CMRR.

d.Let AD = 100; AC = 0.01; α 1; β = 0.01; and K = 104. Evaluate the single-ended CMRR numerically and compare it to the CMRR of the DA.

RS

vs

R1

vo

 

 

v1

R

 

 

RS

DA

voc

v1

R

 

 

vs

vo

R1

 

 

K

FIGURE P4.6

© 2004 by CRC Press LLC

General Properties of Electronic Single-Loop Feedback Systems

195

I1

I2

 

vi

V1 OA

vi

Ve

R1

R2

FIGURE P4.7

4.7The op amp circuit shown in Figure P4.7 is a form of current mirror. The BJT can be modeled by its MFSSM with hre = hoe = 0. The op amp is ideal except for the gain:

v =

Kv (vi − vi)

(τ s

+ 1)

o

 

 

Find an expression for I2/I1(jω) in time constant form.

4.8Negative current feedback is used to make an op amp/FET VCCS, shown in

Figure P4.8. MFSS analysis will be used. The FET is characterized by (gm, gd) and the op amp has a finite voltage gain so vg = KV (v1 − vs). Assume VDS is large enough to keep the FET in channel saturation.

a.Find an expression for the VCCS’s small-signal transconductance, GM = id/v1.

b.Find an expression for the VCCS’s Norton output conductance,

Gout.

c.Let Kv = 104; gm = 103 S; gd = 105 S; and RS = 1 kΩ. Find numerical values for GM and Gout.

id

vi

+

vg

OA

VDS

v1 vi

S

vs

id RS

FIGURE P4.8

© 2004 by CRC Press LLC

196

Analysis and Application of Analog Electronic Circuits

 

RF

 

 

R1

 

 

Vi

Vo

 

OA

 

 

V1

 

 

10

RL

FIGURE P4.9

4.9A certain power op amp (POA) has an open-loop gain of Vo/Vi′ = − 5 ∞ 104. When it is connected as a gain of −250 amplifier, it has a total harmonic distortion (THD) of 0.5% of the RMS fundamental output signal voltage, Vos,

when the RMS Vos = 10.0 V. Find the percent THD when the amplifier is given a gain of −1 (unit inverter) and the Vos is again made 10.0 VRMS. See Figure P4.9.

4.10The circuit of Figure P4.10 is a constant-current regulator used for charging

batteries; NCFB is used. Assume the DA’s gain is KD = 105; hoe = hre = 0; hfe = 19; Rm = 1.0 Ω; VR = 5 V (sets IL); and VBE = 0.7 V. Battery: VB = 12.6 V (nominal); series resistance (nominal); RB = 0.1 Ω; VS = 24 V; and RS = 13 Ω.

a.Find an expression for and evaluate numerically the small-signal transconductance of the regulator, GM = IL/VR.

b.Find an expression for and evaluate numerically the small-signal

Norton conductance the battery sees. (Hint: set VR = 0 and use a test source, vt, in place of the battery. Gout = it/vt.)

RS

 

 

 

Vo

 

 

 

 

 

 

 

 

VBE

IL

RB

 

 

+

B

attery

 

 

 

VS

IB

 

 

VB

 

 

 

 

 

 

 

 

 

 

Vm

 

 

Vb

 

DA

Rm

 

 

 

 

 

 

 

 

 

VR

 

 

FIGURE P4.10

© 2004 by CRC Press LLC

General Properties of Electronic Single-Loop Feedback Systems

197

 

Is

 

 

 

Vi

−βK2

Vo

 

 

 

 

 

+

 

Vs

R1

K1(Vi Vi’)

 

Vi

FIGURE P4.11

4.11We have shown that NVFB reduces amplifier output impedance. In this problem, you will investigate how NVFB affects input impedance. In the schematic of Figure P4.11, an amplifier is given negative feedback as shown. Find the input resistance with feedback, Rin = Vs/Is.

4.12One way of additively combining feedback with the input signal in a SISO

feedback amplifier is to use a difference amplifier, illustrated in Figure P4.12. The output, v2, is given by the relation: v2= KF (vs − vo). Use a simple MFSSM for the matched BJTs in which hre = hoe = 0 and hfe and hie > 0. A dc current source supplies the emitter currents of the two BJTs. Derive an expression for KF in terms of circuit parameters.

VCC

R

R

 

v2

 

ve

+

+

vs

vo

 

2IEQ

FIGURE P4.12

4.13Both positive and negative feedback are used on an ideal op amp, shown in Figure P4.13.

a.Derive an expression for Vo/Vs in terms of the circuit parameters.

b.What vector condition on the impedances will make the system unstable?

© 2004 by CRC Press LLC

198

Analysis and Application of Analog Electronic Circuits

Z4

Z3

vi

IOA

vo

vi

Z1

Z2

vs

FIGURE P4.13

© 2004 by CRC Press LLC