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External Read Cycle How fast does The RAM have to Be? 7 osc. Cycles Recall design Of P0: active pullup And pulldown so Data bus can “float”

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Presentation on theme: "External Read Cycle How fast does The RAM have to Be? 7 osc. Cycles Recall design Of P0: active pullup And pulldown so Data bus can “float”"— Presentation transcript:

1 External Read Cycle How fast does The RAM have to Be? 7 osc. Cycles Recall design Of P0: active pullup And pulldown so Data bus can “float”

2 Memory Mapped System P0 P2 8051 ALE \RD \WD A[14:0] A[7:0] D[7:0] RAM \RE 32Kx8 \WE \E4 ADC1 \E1 D[7:0] ADC2 \E2 D[7:0] A[4:0] D[7:0] LCD \E3 E LATCH D Q PLD

3 Glue Logic Design 1.Define Memory Map Ram: Upper 32K LCD: 0-31 ADC1: 32 ADC2: 64 2.Come up w/ address decoder logic for each case (many to 1 ok, 1 to many not okay!) E4 = A15 covers only upper 32K addresses E3 = A15’A6’A5’ covers 2^12 addresses including 0-31 but excluding upper 32K, 64, 32 E2 = A15’A6’A5 covers 2^13 addresses but excludes upper 32K, 64, 0-31 E1 = A15’A6A5’ covers 2^13 addresses but excludes upper 32K, 32, 0-31 3.Account for inversions, and qualify with RD \E4 = E4’ //RAM already qualified by \RE \E3 = E3’+\RD \E2 = E2’+\RD \E1 = E1’+\RD

4 8051 Why? power supply can’t change instantaneously Power lines have inductance Rapidly reduced R eff at constant I  Voltage drop at the load So what? Could cause processor to reset/go to unknown state Power Supply Noise

5 What to do about it? 8051 How big should cap be? Depends on speed and inductance of the supply, and  I when switched Typical values for digital boards are.1uF/IC placed very close to IC ….then there’s capacitive loads…. Called a “Bypass Cap”

6 Capacitive Loads 8051 Why is this worse than resistive load? Recharge current is only limited by available electrons! Can cause massive voltage drop until battery catches up. So what? Last year’s capstone project: sonar firing caused processor reset C1C1 C2C2 chrg flash chrg flash chrg

7 Charge Sharing 1.Initially Q 1 =V 0 C 1 2.Then close switch, what is V’/V 0 ? 3.V 0 = Q 1 /C 1 (initial condition) 4.Q 1 ’+Q 2 ’ = Q 1 (post condition) 5.Q 1 = V’C 1 +V’C 2 = V’(C 1 +C 2 ) 6.V’ = Q 1 /(C 1 +C 2 ) 7.V’/V0 = Q 1 /(C 1 +C 2 ) * C 1 /Q 1 8.V’/V 0 = C 1 /(C 1 +C 2 ) If C 1 dominates, then V’ ~ V 0 If C 1 = 10C 2 Then V 1 /V 0 = 10/11 C1C1 C2C2 V0V0

8 Other Business Streaming v. Local RAM Streaming Rate of consumption v. Need for buffering (Pilot/PC, 8051) Network Congestion RAM: Laziness Standalone Wiring Next Week’s Lab? Post Meeting Notes Rest of Class Schedule Major Topics: Embedded HW (Mostly Finished) Embedded OS (Next) Embedded Networking Software Engineering Concepts for Embedded/RT


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