CP208 Digital Electronics

Class Lecture 3

February 11, 2009

2

In Today’s Class

We Will Continue to Discuss:

Chapter 1: Introduction to Electronics

1.4.6 Amplifiers( Cont …)

1.7 Logic Inverters

But First … Home Work No. 1

HW #1 Problem (points ??)HW #1 Problem (points ??)

In above ADC, Input changes from 0 to VMAX.(a) Show that the LSB corresponds to a change of

VMAX / (2N – 1) and it is a resolution of the ADC.(b) Show that Max Quantization Error in conversion

= 0.5 x VMAX / (2N – 1)

(c) If VMAX = 5 V, how many bits are needed for a resolution of 1 m V or better? Calculate the Actual resolution? What is Quantization error?

HW Problem SolutionHW Problem Solution

(a) For N bits, there are: 2N Levels from 0 to VMAX, and

2N – 1 Discrete Steps from 0 to VMAX.And, the Step Size is:

= (Total Length) / (Total # of Steps) = VMAX / (2N – 1)

When only (LSB) b0 = 1, it will add a step VMAX / (2N – 1) to Total. That is, the LSB Corresponds to change of VMAX / (2N – 1).

It is also the smallest value possible (Resolution) of the ADC.

0V VMAX V

b0 b1 b2 bN – 1…Step

Level

HW Problem Solution (Cont …)HW Problem Solution (Cont …)

(b)

The Error is Maximum (Highest) when the Analog value Falls in Middle of the Discrete Step.And, the Middle of Step is:

= (Step Length) / 2 OR 0.5 x (Step Length)

Max Quantization Error = 0.5 x VMAX / (2N – 1)

0V

b0 b1

Middle of StepError is Max

HW Problem Solution (Cont …)HW Problem Solution (Cont …)(C) Now, if VMAX = 5 V, For a Resolution = 1 m V or better: Number of bits, N = ??? Resolution = VMAX / (2N – 1) , Therefore,(5) / (2N – 1) ≤ 1 m V 2N – 1 ≥ 5 / (1x10-3) 2N ≥ 5 x103 + 1 2N ≥ 5001 Taking Log on both Sides, we getN x Log (2) ≥ Log (5001) N = Log (5001) / Log (2) N ≥ 12.288 N = 13 (But N =12 for little less resolution than 1 mV)Calculate the Actual resolution? Resolution = VMAX / (2N – 1) = 5 / (213 - 1)

= 0.6 m V ( = 1.22 mV When N = 12)

What is Quantization error? Error is ½ of the Resolution = 0.5 x 0.6 = 0.3 mV or = 0.61 mV When N = 12

1.4.9 Symbol Convention1.4.9 Symbol Convention

• Instantaneous Quantities Lower Case Symbol with Uppercase Sub iA(t), vC(t)

• DC Quantities Upper Case Symbol Uppercase Sub IA, VC

• Sine Wave Signal Amplitude Uppercase Letter with Lowercase Sub Ia, Vc

1.4.9 Symbol Convention1.4.9 Symbol Convention

• PS (dc) Voltages Uppercase V with Double-letter Uppercase Sub, VDD Similar notation for Current from PS

• Incremental Signal Quantities Lowercase Symbol with Lowercase Sub ia(t), vc(t)

1.4.6 Amp Power Supplies1.4.6 Amp Power Supplies

• Power supplied to Load by Amp is Greater than Power Drawn from Input Signal

• To supply that extra power the Amp Need DC Power Supplies for their Operation

• In addition the DC PS supply power that might be Dissipated in Internal Amp Ckt

Power Deliverd to Amp Pdc = V1I1 + V2I2

Amp Power-balance Eq. Pdc+ PI = PL + Pdissipated

• Amp Efficiency η ≡ (PL / Pdc) x 100

• What about PI and Pdissipated ??

• Power Efficiency is Important Performance Parameter for Amps that Handle Large Amounts of Power …

• … and Such Power Amplifiers are Used as Output Amps of Stereo Systems

Simple Ckt Diagram – We shall adopt the convention illustrated and will not Explicitly show connections of Amp to DC PS

Example 1.1 (Page 17)

Data Given;

V1=V2= 10V

I1 = I2 = 9.5 mA

vI = 1 V, vO = 9V

RL = 1kΩ

iI = 0.1 mA

Find:

Av, Ai, Ap,Pdc, Pdissipated, η

1.4.7 Amp Saturation1.4.7 Amp Saturation

• Practically Amp Transfer Characteristics Only Remain Linear for Limited Range of I/O Voltages

• Amp Operated from 2 PS the Output Voltage can not exceed specified positive limit and can not decrease below specified negative limit

An Amplifier Transfer Characteristic. vI Must be kept within Linear Range of Operation.

1.4.8 Nonlinear TC and Biasing1.4.8 Nonlinear TC and Biasing

• Except from output Saturation Effect the Amp TC have been assumed linear

• In Practical Amps the TC may exhibit nonlinearities of various magnitudes

• Biasing is a Simple Technique to Obtain Linear Amplification From Amp Having Nonlinear TC

Considerable Nonlinear Amp TC.Amp is Biased to Obtain Linear Operation and the Signal Amplitude is Kept Small

• Time-varying Signal to be Amplified is Superimposed on the DC Bias VI and Total Instantaneous Input vI(t) = VI + vi(t)

– And vO(t) = VO + vo(t)

– With vo(t) = Av vi(t)

– Where Av = dvO / dvI |at Q is the Slope of Almost Linear Segment of TC

• This Way Linear Amplification is Achieved with a Limitation of Keeping Input Signal Sufficiently Small

Example 1.2Example 1.2 Transfer Characteristic of Amplifier. Note That Amplifier is Inverting (Negative Gain)

Example 1.2(Page 21)

V

OI

I

OI

vo

A

VV

vLL

vv

ev I

gain Voltage Also,

5V that so bias DC

ingcorrespond and and

:Find

V 0.3 and V 0

10 10

_

40-11

1.7 Digital Logic Inverters1.7 Digital Logic Inverters

• Logic Inverter is a Most Basic Element in Digital Ckt Design

• Plays a Role Parallel to the Amp in analog Ckt

• We will get Introduced to Logic Inverter in This Section

1.7.1 Function of the Inverters1.7.1 Function of the Inverters

• Logic Inverter INVERTS the Logic Value of its Input Signal

• That is for 0 input, out put will be 1, and vice versa

In Voltage Level Terms

When vI is Low (close

to 0) the vO will be

high VDD

1.7.2 The Voltage Transfer Characteristic1.7.2 The Voltage Transfer Characteristic

• VTC to Quantify Operation of Inverter• Observe the TC of Amplifier in Ex 1.2

This Inverting Amp can

be used as Inverter

when we use its

Extreme Regions

of Operation –

Opposite to Amp

Dig App Make Use of

Gross nonlinearity exhibited by VTC

VTC of an inverter. VOH Does Not Depend on Exact Value of vI as long as vI ≤ VIL. When vI > VIL Inverter in Amp mode or Transition Region. VIL is Max Value That vI Can Have While Being Interpreted by Inverter as Representing Logic 0.VOL Does Not Depend on Exact Value of vI as long as vI ≥ VIH. VIH is Min Value That vI Can Have While Being Interpreted by Inverter as Representing Logic 1.

1.7.3 Noise Margins1.7.3 Noise Margins• Insensitivity of Inverter

Output to Exact Value of vI within allowed regions is great advantage over analog ckts.

• To Quantify Insensitivity Consider One Inverter Driving Similar Inverter

• Noise Margin for High Input NMH = VOH – VIH

• Noise Margin for Low Input NML = VIL - VOL

Four Parameters VOH,VIH,VIL, and VOL Define VTC of Inverter and Determine its Noise Margin, Which in Turn Measures its Ability to Tolerate Input Signal Variations

• VOL: Output Low Level• VOH: Output High Level• VIL: Max Input as Logic 0• VIH: Min Input as Logic 1• NML: Noise Margin for Low

Input = VIL - VOL

• NMH: Noise Margin for High Input = VOH – VIH

1.7.4 The Ideal VTC1.7.4 The Ideal VTC

• What Makes an Ideal VTC for Inverter?

• Ideal VTC Maximizes the Noise margins and Distributes Them Equally B/W the Low and High Regions

• VOH is at Max Possible Value VDD

• VOL is at Min possible Value 0 V

The Ideal VTC (Cont…)The Ideal VTC (Cont…)

• VIL and VIH are equal at Mid of (VDD/2)

• Width of Transition Region (Imp for Amp) is Zero

• Steep Transition at Threshold Voltage (VDD/2) – Gain infinite

• Noise Margins are Equal

NMH = NML = VDD/2• CMOS Inverter Has

Close to Ideal VTC

1.7.5 Inverter Implementation1.7.5 Inverter Implementation

Implemented using Transistors Operating as Voltage Controlled Switches

• Transistor Switches are NOT Perfect• Their OFF Resistance is High, But• ON Resistance is not Zero and Some

BJTs Exhibit Offset Voltage as well

• The Result is That When vI is High VOL Is not Ideally Zero

More Elaborate Implementation Exists Utilizing Pair of Complementary Switches (CMOS Based). When I/P Low, VOH = VDD, No Current FlowsWhen I/P High, Ron Connect Ground, VOL = 0

1.7.6 Power Dissipation1.7.6 Power Dissipation• Digital Systems use large number of Logic Gates

• Space and Economy Require as Few IC as Possible

• Hence, As Many Logic gates As Possible on IC Chip

• In Present VLSI 100K+ gates on IC

• To Keep Acceptable limit of Power Dissipation in Chip, Power Dissipation/Gate Must be Minimum

• Power Dissipation is Very Important Performance Measure of Logic Inverter

• When vI Low no Power Dissipated

• In other State Dissipation is V2

DD/R and is Substantial

• This Dissipation Occurs even When Inverter not Switching -- Static Power Dissipation

• This Inverter Exhibit No Static Power Dissipation

• BUT …

• There is Always A component of Power Dissipation Due to Capacitance

• Cap Exists between Output Node of Inverter and Ground

• Internal Cap of Switches

• Wires Connecting Output to Other Ckts have Cap

• Input Cap of Any CKt Driven by Inverter

• When Inverter is Switching from One State to Another, Current must Flow thru Switches to Charge and Discharge the Load Capacitance

• The Current Give Rise to dissipation Called Dynamic Power Dissipation (Chap 4)

2DDdynamic fCVP

1.7.7 Propagation Delay 1.7.7 Propagation Delay • Inverters are Characterized in Terms of The Time

Delay Between Switching of vI (Low to High) and Corresponding Change Appearing at the Output

• 2 Reasons for Propagation Delay:– Transistors (Switches) Exhibit Finite (nonzero)

Switching Time– The Cap needs to Charge/Discharge before

Output Change• To Analyze Inverter Switching Need to Understand

Time Response of Single-Time-Constant Ckts (STC) [Appendix D.4.1]

• Step Function Applied to an STC Network with Time Constant τ, Out put at any time:

y(t) = Y∞ - ( Y∞ - Y0+)e-t/τ

Y∞ is final value where response is heading

Y0+ value of response immediately after t=0

• Output at any time t is difference between final value and gap whose initial value is Y∞ - Y0+ and shrinking exponentially

• Y∞= S and Y0+ = 0, Thus• y (t) = S(1- e-t/τ)

Example 1.6Example 1.6

Consider the Inverter in Fig. with a C=10 pF connected between the output and ground. Let VDD=5V, R=1kΩ, Ron=100Ω and Voffset=0.1V. If at t=0, vI goes low and neglecting the delay time of the switch, that is, assuming that it opens Immediately, Find the time for the Output to Reach ½ (VOH + VOL). The Time to this 50% point on the out put waveform is defined as the Low-to-High Propagation Delay, tPLH.

Example 1.6Example 1.6

Before t=0

vI is high, and

vO = VOL = Voffset + V in Ron

V in Ron = (VDD – Voffset)xRon/(R+Ron)

VOL = 0.1 + (5 – 0.1)x0.1/(1.1)

vO = VOL = 0.55 V

Example 1.6Example 1.6At t=0 SW opens, V across Cap cannot change instantaneously.

at t=0+ O/P vO(t0+) = 0.55.

Cap charges thru R and O/P

rises Exponentially to VDD.

vO(∞) = VDD.

Using the Output Eqn of STC Network for Step-function as I/P:

y(t) = Y∞ - ( Y∞ - Y0+)e-t/τ

vO(t) = vO(∞) – [v(∞) - vO(t0+) ]e-t/τ

Using vO(∞) = 5 V and vO(0+) = 0.55V in vO(t) = vO(∞) – [v(∞) - vO(t0+) ]e-t/τ

We get:vO(t) = 5 – (5 – 0.55) e-t/τ

vO(t) = 5 – 4.45 e-t/τ _____ (1)

τ = RC. To Find tPLH … when t= tPLH

vO(t=tPLH) = 0.5(5+0.55) = 2.78Eq (1) becomes:2.78 = 5 – 4.45 e-tPLH /RC

4.45 e-tPLH /RC = 5 – 2.78 = 2.23 etPLH /RC = 4.45/2.23 = 1.9955

Taking ln on both sides:tPLH / RC = ln (1.9955)tPLH = 0.69xRxC

= 0.69x1000x10-11

= 6.9 n Sec

Formal Definitions of Propagation Delay (PD) of Inverter.

I/P and Inverted O/P have Finite Rise and Fall Times.

Also, Delay Time between I/P and O/P Waveforms.

Usually PD is Specified by Average of tPHL and tPLH.

And these measured at 50% of I/P and O/P waveforms.

Transition Times are Specified using 10% and 90% points of (VOH – VOL)

Home WorkHome Work

All Problems of Section 1.7

Specially;

Problem Nos.: 1.86 and 1.89

In Next Class

We Will Discuss:

Chap 3 Diodes Topics:

3.1 The Ideal Diode

Chap 5 Bipolar Junction Diodes (BJTs)

5.10 Basic BJT Digital Logic Inverter