# Electricity _ Magnetism

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```					Electricity & Magnetism
Electronics

By YL Siu

Introduction
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Knowledge from CE is assumed
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diode, LED, LDR, thermistor, reed switch, reed relay, electromagnetic relay & potential divider

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Diode - Half-wave & Full-wave Rectification

Transistor
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Input characteristics Current transfer characteristics Collector characteristics Input/Output voltage characteristics
By YL Siu intrinsic properties, -ve feedback & applications

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Operational Amplifier
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Diodes
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Diode - an uni-directional device

Forward bias – small resistance, I is large when
Vin > 0.7 V

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Reverse bias – large resistance, I is small unless
Vin > 80 V

Direction of current flow

Circuit symbol

By YL Siu

Diodes - Applications I
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Half-wave rectification

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Diode only allows current to flow in 1 direction By YL Siu

Diodes - Applications II
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Full-wave rectification I - Centre-tapped transformer First half-cycle, current flows through D1, I flows from right to left Second half-cycle, current flows through D2, I flows from right to left

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Direction of current remains unchanged
a.c. has been "rectified" (changed into d.c.) Sketch the output voltage across R
By YL Siu

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Diodes - Applications III
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Full-wave rectification II - Bridge rectifier First half-cycle, I passes through D2 , R & D4 Second half-cycle, I passes through D3 , R & D1 Direction of I remains unchanged Sketch the output voltage across resistor R How do you compare the two full-wave rectification circuits? By YL Siu

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Diodes - Applications IV
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a.c. voltmeter / ammeter is obtained by replacing resistor with a moving-coil galvanometer
Current passing through the milliammeter is unidirectional but fluctuating Usually, it is calibrated to measure the average (r.m.s) value How to connect this a.c. voltmeter / ammeter?
By YL Siu

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Inductor-Capacitor Smoothing
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Smoothing of fluctuating d.c. by L & C

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C1: storage capacitor, C2: filter capacitor L & C2: filtering the ripple
By YL Siu

L-C Smoothing
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C1 changes voltage across XY to a ripple
The ripple is fed into LC in series L takes up alternating component C2 takes up constant component Voltage across R becomes constant d.c.

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By YL Siu

Transistor
It can be used as a voltage amplifier B or switches in logic circuits  It is a semiconductor device with 3 terminals:
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C

Emitter E, Base B & Collector C
similar to 2 diodes connected together transistor can be of NPN or PNP type only NPN transistor in common emitter configuration will be studied

E circuit symbol

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Common Emitter Configuration:
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IE = IB + IC

By YL Siu

Characteristics of Transistor
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Characteristics to be studied in common emitter configuration includes:

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Input characteristics
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Relationship between IB and VBE Relationship between IC and IB Relationship between IC and VCE

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Current transfer characteristics
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Collector characteristics
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Input/Output voltage characteristics
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Relationship between Vout and Vin

By YL Siu

Input Characteristics (IB & VBE)

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Variation of base current IB vs base-emitter voltage VBE is the input characteristic of transistor Note that VPotential Divider = IBRB + VBE IB is close to zero if VBE is less than 0.7 V By YL Siu

Input Characteristics (IB & VBE)
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IB is close to zero if VBE is less than 0.7 V Input characteristic of transistor is very similar to the I-V characteristic of a diode in forward bias VBE Input resistance ri = IB ri is large initially & becomes small when VBE > 0.7 V RB limits the base current and prevent damages caused to transistor RB avoids large fluctuations in base current
By YL Siu

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Current Transfer Characteristics (IC & IB)

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Variation of the collector current IC vs base current IB is called the current transfer characteristic IB is varied by adjusting R1 while IC is found to be almost directly proportional to IB By YL Siu

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Current Transfer Characteristics (IC & IB)

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Without RL, IC saturates when IB > 200 mA but with RL, IC saturates when IB > 20 mA

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Current is linearly amplified before saturation I C I C  Current amplification factor   By YL Siu IB IB

Current Transfer Characteristics (IC & IB)
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Transistor is a linear current amplifier if IB is not too large

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Practically, resistor RL is added to limit the current flowing into the transistor
Maximum IC = 6 / (2.2 x 103) = 2.73 mA

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Typical  = 130,
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so, maximum IB = (2.73 x 10-3) / 130 = 21 mA

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Note that both the collector current IC and base current IB flow into the transistor while the emitter current IE flows out of the transistor IE ~ IC as IB is very small (IC ~ 98% of IE)
By YL Siu

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Collector Characteristics (IC & VCE)

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Variation of collector current IC vs collector-emitter voltage VCE is called the collector characteristic IB is kept constant by R1 & VCE is varied by R2

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IC depends entirely on IB and hardly at all onBy YL Siu VCE

Collector Characteristics (IC & VCE)
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For VCE > 0.2 V, IC is almost a constant if IB is kept constant Variation of IC vs VCE does not follow Ohm's law
Each increment of IB (IB = 20 mA) causes a constant increment of IC (IC ~ 2.6 mA)

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Consistent with current transfer characteristic as  = IC/IB = (2.6 x 10-3)/(20 x 10-6) = 130
Transistor fails to work properly if VCE < 0.2 V By YL Siu

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Input/Output Voltage Characteristics (Vout & Vin)

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From Kirchhoff's second law:

Vin = IBRB + VBE , Vout = 6 - ICRL
If Vin < 0.7 V, Vin < barrier potential, IB = IC = 0 Vout = 6 V -> transistor is in cut-off state
By YL Siu

Input/Output Voltage Characteristics (Vout & Vin)
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For Vin < 0.7 V, Vout = 6 V -> cut-off For 0.7 V < Vin < 1 V:  Vout = 6 - ICRL = 6 - IBRL = 6 - (Vin - 0.7)RL/RB
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If RL = 2.2 kW, RB = 15 kW &  = 130, we have, Vout = -19Vin + 19 -> linear region Max. IC = (6 - 0.2)/RL = 5.8 / 2200 = 2.64 mA If  = 130, min. value of Vin for saturation is,

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For Vin > 1 V, transistor is saturated because:
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Vin

= IBRB + VBE = ICRB/ + VBE = 2.64 x 10-3 x 15000 / 130 + 0.7 = 1 V
By YL Siu

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Vout = VCE = 0.2 V -> saturation

Input/Output Voltage Characteristics (Vout & Vin)
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Vin < 0.7 V, 0.7 V < Vin < 1 V, Vin > 1 V,

Vout = 6 V -> cut-off
Vout = -19Vin + 19 -> linear region Vout = 0.2 V -> saturation

 For
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 For
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linear region, Vout = 6 - ICRL = 6 - IBRL
 Vin - 0.7    RL  RL  R L or Vout      Vin   6  0.7   6 -   R   R   RB  B B      
By YL Siu

Vout

Vout RL  Voltage gain defined as: G   - Vin RB

Linear Voltage Amplification
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G = -(RL/RB)

[Derivation required]
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Input signal is amplified G comes from

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dynamic ratio
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For undistorted amplification,
a constant d.c. (base bias voltage) is required to overcome the barrier potential, VBE input voltage is restricted to the linear regionYL Siu By

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Exercise
In the circuit shown, if  = 100, sketch the Vout vs Vin characteristic curve
Vout/V 12

0

0.7

1.9

Vin/V

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What is the slope of the linear region? How to get the value of 1.9?

By YL Siu

Example
Refer to the previous question, find the voltage gain. What is Vout when Vin = 1.5 V? Solution: Voltage gain G = -RL/RB = 100 x 2000 / 20000 = 10 When Vin = 1.5 V, IB = (Vin - 0.7)/RB = (1.5 - 0.7)/20000 = 40 mA Vout = 12 - ICRL = 12 - IBRL = 12 - 100 x 40 x 10-6 x 2000 =4V By YL Siu

Operational Amplifier - Basic
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Operational amplifier (op-amp) can be used as
comparator, inverting circuit, summing circuit, mathematical calculation, voltage follower

Symbol for op amp
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2 inputs and 1 output Op amp works under a d.c. power supply VS ~ 15 V

Vo = Ao (V+ - V-), Ao: open-loop voltage gain Op amp - differential voltage amplifier as it amplifies
the differences between V+ and VBy YL Siu

Operational Amplifier - Properties
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High open-loop voltage gain Ao which is about
105 at low frequency -> so accept V+ ~ V-

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High input impedance ri ~ 106 - 1012 W -> so that
negligible current is drawn from the inputs signal

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Low output impedance ro ~ 102 W -> so, most of
the output voltage is transferred to the load resistor which is about a few kW

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High slew rate ~ 106 – 107 V s-1, the rate at

which the output voltage can change when the input voltages are changed -> op amp responds quickly to a.c. inputs up to several MHz so that frequency of Vout = frequency of Vin By YL Siu

Open-loop (without feedback)
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Vo can never exceeds VS, op-amp saturates when Vo = +13 V or when Vo = -13 V

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Typically, Ao = 2 x 105, op-amp will saturate when |V+ - V-| = 13 / 2 x 105 = 65 mV

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Linear region AOB is too small, Vo = +13 V / -13 V
very easily
By YL Siu

Open-loop (without feedback)
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Limitations of op-amp in open loop mode:
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linear region AOB is too small that it tends to
saturates permanently

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voltage gain is sensitive to temperature change voltage gain Ao decreases with increasing frequency
Ao

frequency

By YL Siu

Operational Amplifier
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Assumptions for an ideal op amp:
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Input impedance is infinite -> input current
is zero

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Output impedance is zero -> voltage across
a load = output voltage Vo

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Bandwidth is infinite -> voltage gain is
independent of frequency

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For most cases:
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(V+ - V-) is too small (~65 mV) that we can assumed that V+ = Vop-amp tends to saturate permanently
By YL Siu

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Applications (without Feedback)
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(A) Voltage comparator

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Different input voltages can be compared using the above circuit When V+ > V-, Vo is positive and D1 is lit Conversely, if V+ < V-, Vo is negative and D2 is lit Results of comparison are indicated by diodesBy YL Siu

Applications (without Feedback)
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(B) Conversion of sine wave to square wave

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A small V+ causes Vo to be saturated easily, i.e. 13 V
Note that the power supply VS is often omitted
By YL Siu

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Negative Feedback
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op-amp tends to saturate permanently due to high voltage gain Ao Ao varies with temperature & falls off at high frequencies

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To avoid saturation, part of the output voltage is fed back to the input
Negative feedback - output voltage is being fed back to the negative (inverting) input Closed-loop gain A -> gain with negative feedback (Closed-loop gain A is less than open loop gain Ao) By YL Siu

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Negative Feedback
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Voltage gain is predictable and independent of
the op-amp's intrinsic properties Greater stability - independent of temperature The linear region for amplification is wider less distortion The gain is constant over a wider range of input frequencies (bandwidth) - negative feedback can improve the frequency response Decrease in voltage amplification as A < Ao

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By YL Siu

Applications (with Feedback)
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(A) Inverting amplifier
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Ri: input resistor Rf: feedback resistor a fraction of Vo is being fed back to the input

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Assume op-amp is not saturated, so V- ≈ V+ Since V+ = 0, V- = 0 -> V- is called virtual earth
I = (Vi - V-)/Ri = (V- - Vo)/Rf Vo Rf Closed-loop gain A  Vi Ri

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What would happen if Rf = Ri?

By YL Siu

Applications (with Feedback)
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(B) Summing amplifier

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I = (0 - V)/Rf = I1 + I2 + I3

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-V/Rf = V1/R + V2/R + V3/R

Rf With an inverter, we have Vo  V1  V2  V3  R Vo is summation of Vi By YL Siu

Applications (with Feedback)
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(C) Subtractor

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V1 is first inverted V2 is then summed with -V1

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Rf So, Vo  V1 - V2  R

In particular, if Rf = R, Vo = (V1 - V2)

By YL Siu

Applications (with Feedback)
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(D) Integrator
Replace feedback resistor by capacitor C I = Vi/R = dQ/dt Q = -CVo

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dVo dQ Vi  dVo   -C or  - C  dt dt R  dt  -1 Vo   Vi dt RC How to get an differentiator circuit?

By YL Siu

Non-inverting input with negative feedback
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Vi is applied to the

non-inverting terminal

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feedback factor 
Note  = Ri/(Ri + Rf) Vo = Ao(Vi - Vo) Rearrange the terms, we have
Ao 1 1   Closed-loop gain A = Vo/Vi = 1  A o 1/A o     Rf  i.e. A = 1 + Rf/Ri & Vo  1  Vi  Ri    V+ ≈ V- but they may not close to 0 By YL Siu

Voltage follower
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The output is connected directly to the inverting terminal. i.e. 100%

negative feedback
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So,  = 1, A ≈ 1/ = 1 i.e. Vo = Vi It is called voltage follower as the output voltage Vo just equals to the input voltage Vi Voltage follower can be used as a voltmeter because of its high input impedance

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By YL Siu

Op-amp as High Impedance Voltmeter
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Voltage follower is used to measure the voltage across a charged capacitor The capacitor will discharge extremely slowly during measurement because of the high input impedance of the operational amplifier

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Variation of p.d. across C during measurement can be neglected
Can we use voltmeter or CRO to measure the voltage across capacitor? Why? By YL Siu

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Exercise
In the circuit shown, Vo = 2.0 V, calculate the input voltage Vi and the voltage gain.

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What would happen if the resistors are replaced By YL Siu by 160 kW & 20 kW? 16 MW & 2 MW?

~ END ~

By YL Siu

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