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Bruce Mayer, PE Registered Electrical & Mechanical Engineer [email protected]

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Engineering 43

Chp 6.4RC OpAmps Ckts

Bruce Mayer, PE

Registered Electrical & Mechanical [email protected]

- Introduce Two Very Important Practical Circuits Based On Operational Amplifiers
- Recall the OpAmp

- The “Ideal” Model That we Use
- RO = 0
- Ri = ∞
- Av = ∞

- Consequences of Ideality
- RO = 0 vO = Av(v+−v−)
- Ri = ∞ i+ = i− = 0
- Av = ∞ v+ = v−

=

v

0

+

- KCL At v- node

- By Ideal OpAmp
- Ri = ∞ i+ = i- = 0
- Av = ∞ v+ = v- = 0

- Separating the Variables and Integrating Yields the Solution for vo(t)

- By the Ideal OpAmp Assumptions

- Thus the Output is a (negative) SCALED TIME INTEGRAL of the input Signal

- A simple Differential Eqn

KVL

=

v

0

+

- By Ideal OpAmp
- v- = GND = 0V
- i- = 0

- KCL at v-

- Now the KVL

- Recall the Capacitor Integral Law

- Recall Ideal OpAmp Assumptions
- Ri = ∞ i+ = i- = 0
- Av = ∞ v+ = v- = 0

- Then the KCL

- Thus the KVL

- Taking the Time Derivative of the above

- Examination of this Eqn Reveals That if R1 were ZERO, Then vO would be Proportional to the TIME DERIVATIVE of the input Signal
- in Practice An Ideal Differentiator Amplifies Electrical Noise And Does Not Operate
- The Resistor R1 Introduces A Filtering Action.
- Its Value Is Kept As Small As Possible To Approximate A Differentiator

- In the Previous Differential Eqn use KCL to sub vO for i1
- Using

ALL electrical signals are corrupted by external, uncontrollable and often unmeasurable, signals. These undesired signals are referred to as NOISE

Signal

Signal

Noise

Noise

- The Signal-To-Noise Ratio

- Use an Ideal Differentiator

- Simple Model For A Noisy 1V, 60Hz Sinusoid Corrupted With One MicroVolt of 1GHz Interference

- The SN is Degraded Due to Hi-Frequency Noise

- Let’s Turn on the Lites for 10 minutes for YOU to Differentiate
- Given the IDEAL Differentiator Ckt and INPUT Signal
- Find vo(t) over 0-10 ms

- Given Input v1(t)
- SAWTOOTH Wave

- Recall the Differentiator Eqn

R1 = 0; Ideal ckt

- The Slope from 0-5 mS

- Given Input v1(t)

- For the Ideal Differentiator

- Units Analysis

- Derivative Scalar PreFactor

- A Similar Analysis for 5-10 mS yields the Complete vO

OutPut

InPut

- Apply the Prefactor Against the INput Signal Time-Derivative (slope)

- For the Ideal Integrator

- Given Input v1(t)
- SQUARE Wave

- Units Analysis Again

- 0<t<0.1 S
- v1(t) = 20 mV (Const)

- The Integration PreFactor

- 0.1t<0.2 S
- v1(t) = –20 mV (Const)

- Next Calculate the Area Under the Curve to Determine the Voltage Level At the Break Points

- Integrate In Similar Fashion over
- 0.2t<0.3 S
- 0.3t<0.4 S

- Apply the 1000/S PreFactor and Plot Piece-Wise

- Simple Circuit Model For a Dynamic Random Access Memory Cell (DRAM)

- Note How Undesired Current Leakage is Modeled as an I-Src

- Also Note the TINY Value of the Cell-State Capacitance (50x10-15 F)

- During a WRITE Cycle the Cell Cap is Charged to 3V for a Logic-1
- Thus The TIME PERIOD that the cell can HOLD the Logic-1 value

- The Criteria for a Logic “1”
- Vcell >1.5 V

- Now Recall that V = Q/C
- Or in terms of Current

- Now Can Calculate the DRAM “Refresh Rate”

- Consider the Cell at the Beginning of a READ Operation

- When the Switch is Connected Have Caps in Parallel

- Then The Output

- Calc the Change in VI/O at the READ

Design an OpAmp ckt to implement in HARDWARE this Math Relation

- Examine the Reln to find an

Integrator

Adder

The Proposed Solution

- The by Ideal OpAmps & KCL & KVL &Superposition

- The Ckt Eqn

- Then the Design Eqns

- This means that we, as ckt designers, get to PICK 3 values
- For 1st Cut Choose
- C = 20 μF
- R1 = 100 kΩ
- R4 = 20 kΩ

- TWO Eqns in FIVE unknowns

In the Design Eqns

20μ

20k

20k

100k

10k

- If the voltages are <10V, then all currents should be the in mA range, which should prevent over-heating

- Then the DESIGN

- Let’s Work These Probs

80k

choose C such that

Find Energy Stored on Cx

APPENDIX

IC GROUND BOUNCE

LEARNING EXAMPLE

FLIP CHIP MOUNTING

IC WITH WIREBONDS TO THE OUTSIDE

GOAL: REDUCE INDUCTANCE IN

THE WIRING AND REDUCE THE

“GROUND BOUNCE” EFFECT

A SIMPLE MODEL CAN BE USED TO

DESCRIBE GROUND BOUNCE

MODELING THE GROUND BOUNCE EFFECT

IF ALL GATES IN A CHIP ARE CONNECTED TO A SINGLE GROUND THE CURRENT

CAN BE QUITE HIGH AND THE BOUNCE MAY BECOME UNACCEPTABLE

USE SEVERAL GROUND CONNECTIONS (BALLS) AND ALLOCATE A FRACTION OF

THE GATES TO EACH BALL