electronic devices - utcluj.ro. fullwave dr rectifiers. dc... · laura-nicoleta ivanciu, electronic...

ELECTRONIC DEVICESAssist. prof. Laura-Nicoleta IVANCIU, Ph.D.

C4 – Full-wave DR rectifiers.DC switching circuits. DRC rectifiers. LEDs.

2Laura-Nicoleta IVANCIU, Electronic devices

Contents

Full-wave DR rectifiers Two-port DC networks

Positive/negative peak detector Translation networks Voltage doubler

DRC rectifiers (power supply filtering) LEDs and photodiodes



Full-wave DR rectifiers

Full-wave rectifier – diode bridge

positive half, vI >0D1, D3 – (on) D2, D4 – (off)

negative half, vI <0D1, D3 – (off) D2, D4 – (on)

~

~

-+

neglecting 0.7 V



Full-wave rectifier – center-tapped transformer

neglecting 0.7 V

C4 – Full-wave DR rectifiers.DC switching circuits. DRC rectifiers. LEDs. Full-wave DR rectifiers


Example

vI

tVtv II ωsinˆ)( =

For the circuit in the figure, RL=50Ω. Assumea) Plot vO(t) and iO(t)b) Compute the values of the maximum reverse voltage vDR across each

diode and the maximum forward current through each diode. c) Repeat a) and b) assuming

.7ˆ VVI =

.100ˆ VVI =

C4 – Full-wave DR rectifiers.DC switching circuits. DRC rectifiers. LEDs. Full-wave DR rectifiers


Two-port DC networks

Preamble

The steady-state properties (i.e. the VTC) of two-port DC networks are of no interest in applications.The focus is on the time-domain behavior – how does vO look for variable vI?

The analysis of switching two-port DC networks works with the constant voltage drop model of the diode.

D, C




Temporal extreme two-port DC networks

vD (t) = vI (t) – vO(t)

vO(t) = vC(t)

1. If vD tends to be > 0.7 V, D – (on)

vD = 0.7 V, iC(t) > 0, vC(t) increases (C charges)

2. If vD< 0.7 V, D – (off)

iC(t)=0 vC (t)= constant

Based on the states of the diodes, there are two possible cases:


-10

-5

0

5

10

vI [

V]

t

0

5

10

vO

= v

C

[V]

t

-15

-10

-5

0

vD

[V]

t

9.3

4.3

-9.3

0.7

-19.3



Temporal extreme two-port DC networks

vD (t) = vI (t) – vO(t)

vO(t) = vC(t)

Application:

Positive peak detector

Plot vO(t), vD(t) if vI(t) is a 6 V amplitude sine wave with a dc component of:

i) 0 V; ii) +2 V; iii) -7 V

Change the circuit, in order to obtain a negative peak detector.


-10

0

10

vI [

V]

t

0

5

10

vC

[V]

t

-20

-10

0

10 v

O =

vD

[V

]

t

9.3

-9.3

-19.3

4.3

0.7



Translation two-port DC networks

vD (t) = vI (t) – vC(t)

vO(t) = vD(t)

Application:

Translation towards negative values (downward)


-5

0

5

vI [

V]

t

012345

t

vC

[V]

0

5

10

t

vO

= -

vD

[V

] 9.3

4.3



Translation two-port DC networks

vD (t) = -vI (t) – vC(t)

vO(t) = -vD(t)

Application:

Translation towards positive values (upward)

steady-statetransient state




Voltage doubler

• variable (ac) input voltage• dc output voltage• output voltage is (almost) twice the amplitude of the input voltage

Solution 1: series connection of a positive peak detector and a negativepeak detector




Voltage doubler

Solution 2: chain connection of an upward translation circuit and a positivepeak detector

Translation toward positive values

Positive peak detector

V4.1ˆ2 −= IO VvIO Vv ˆ2≈ neglecting 0.7 V

steady-state regime waveforms




Voltage doublerV01

V-

V02

D1D1N4448

Vi

PER = 1ms

V1 = +5VV2 = -5V

D2

D1N4448

V+

C222n

0

V V

C1

22n

V




Voltage triplerOPTIONAL



DRC rectifiers (power supply filtering)

Half-wave rectifier w/ capacitive filter (loaded positive peak detector)

• vI is the voltage in a secondary winding of astep-down line transformer.

• It is required to obtain an almost dc voltage(on a load resistor)

-10

0

10

t

-10

0

10

t

-10

0

10

t

T

td tc

Δv

vI

vO (without C in the circuit)

vO

OV

OV

How can the voltage ripple Δv be computed?




Half-wave rectifier w/ capacitive filter (loaded positive peak detector)Between two successive voltage peaks, D – (off) and Cdischarges through R

RCt

RCt

c eVVetv−

∞

−+−= )0()1()(

If RC >>T => the capacitor discharge during td can be approximated with a linear variation of the output voltage (across the capacitor)

tIvC ∆=∆TtTtt cd ≈−==∆

RVI O=

The discharging current is assumed to be constant

RCV

fRCTVI

CTv O

O

ˆ1ˆ ===∆




Example

V7.10ˆ =IV f=50Hz Ω= 100LR V5.1<∆v C = ?

Solution: electrolytic capacitor, C =1500 µF/25 V

What is the actual value of the output ripple?

What should be a new value of C, in order to reduce the output ripple to half of theinitial value?

Solve again, assuming full-wave rectification.

V5.1ˆ

<=∆fRCVv O

μF1333100505.1

105.1

ˆ=

⋅⋅=>

VfRVC O μF1333>C

10VV7.0ˆˆ =−= IO VV




AM demodulator

Envelope detector




DC component restoration

Consider a rectangular wave, transmitted on a line w/ capacitive coupling.

The signal loses its initial dc level.

The upward translation circuit provides a well determined dc component: dc restoration.

OPTIONAL



LEDs and photodiodes

LED – light emitting diode

A

K

vD

iD

behavior similar to conventional diodes 1.5 V to 3 V forward voltage drop in forward bias the LED lights up:

red, yellow, green, blue, white IR – remote control

emits radiation in the visible, infrared, or laser range typically 5 mA to 20 mA power LED: 3.5V @ 500 mA

ID = ?




LED – excerpt from datasheet




LED stripsSingle Color LED Strip

RGB LED Strip

180 180390




Photodiodes

photosensitive pn-junction transforms light into electrical signal (current) used in reverse bias




Photodiodes - excerpt from datasheet



SummaryOnly now we can safely say diodes are not a secret anymore, after studying:

Full-wave DR rectifiers Two-port DC networks DRC rectifiers (power supply filtering) LEDs and photodiodes

Next week: Zener diodes. Operational amplifiers. Simple comparators with OpAmp.

To do: Homework 2


electronic devices - utcluj.ro. fullwave dr rectifiers. dc... · laura-nicoleta ivanciu, electronic...

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