SI UnitsMega 1000,000 1x106 M Ohmskilo 1000 1x103 k OhmsUnitsmilli 0.001 1x10-3 m Ampsmicro 0.000,001 1x10-6 μ Faradsnano 0.000,000,001 1x10-9 n FaradsPico 0.000,000,000,001 1x10-12 p Farads

p n μ m 1 k M-12 -9 -6 -3 0 +3 +6

SI UnitsSome shortcuts

k x m = cancel out M x μ = cancel out

K x μ = m1/k = m1/M = μM/k = kk/m = M

am x an = am+n

am / an = am-n

Worked Examples

What current flows through a 1MΩ Resistor when a voltage of 9V is applied across it?

I = V/R = 9V/1MΩ= 9/(1x106) Amps= 9 x 10-6 Amps= 9 μA

am x an = am+n

am / an = am-n

Worked Examples

What is the time constant for an RC network of a 220μf capacitor and 3k3 resistor?

t = CxR = 220μf x 3k3= 220x3.3 x (10-6x103)= 726 x (10-3)= 0.73s

am x an = am+n

am / an = am-n

Practice Questions

a)R=180kΩ V=9V I=?b)R=3M3Ω V=6V I=?c)R=100Ω V=3V I=?d)R=1MΩ C=220μf t=?e)R=56kΩ C=330pf t=?f)R=390kΩ C=1000μf t=?

0.05mA1.82μA30mA220s18.5ms390s

I = V/R t = CxR

Practice QuestionsI = V / R I = V / R I = V / R = 9 / 180k = 6 / 3M3 = 3 / 100 = 9 / 180 x 10-3 = 6 / 3.3 x 10-6 = 0.03A = 0.05 x 10-3 = 1.82 x 10-6 = 30mA = 0.05mA = 1.82 μA

t = CxR t = CxR t = CxR = 220μf x 1M = 330pf x 56k = 1000μf x

390k = 220 x 1 x (10-6x106) = 330 x 56 x (10-12x103) = 1000 x

390 x (10-6x103) = 220 x (100) = 18480 x (10-9) = 390000 x

(10-3) = 220s = 18.5 x(10-6) = 390 x (103x10-3)

= 18.5 μs = 390s

Resistors in Series

R total = R1 + R2 + etc

Resistors in ParallelTwo ResistorsR total = R1 x R2 = Product

R1 + R2 Sum

Three or more Resistors 1 = 1 + 1 + 1 etcR total R1 R2 R3

Resistors

Resistors

a) Rtotal = 100+100=200

b) Rtotal = 100x100 = 10,000 = 50100+100 200

c) 1 = 1 + 1 + 1 Rtotal 100 100 100 1 = 3 Rtotal 100

Rtotal = 100 = 33.3 3

Ohms Law

Worked Examples

What current flows through a 1MΩ Resistor when a voltage of 9V is applied across it?

I = V/R = 9V/1MΩ= 9/(1x106) Amps= 9 x 10-6 Amps= 9 μA

am x an = am+n

am / an = am-n

Voltage Divider

When resistors are in series voltage is split in the same

ratio as the resistance.

A voltage divider uses this to give a specified output (Vs). This equates to the value of

R2 divided by the total resistance,

times by the supply voltage.Worked Example, V=9V, R1=3k3, R2=6k9

Vout = R2 x Vsupply R1+R2

Worked Example

(In the exam, copy out the equation first)

Vs = 6k9 x 9V (3k3 + 6k9)= 6,900 x 9V

10,200= 6.08 V

Two ResistorsR total = R1 x R2 = Product

R1 + R2 Sum

Vout = R2 x Vsupply R1+R2

Copy down these formulae:

470 2k2

2k2

1k

9v

10

02

00

3v

Vs = 200 x 9V (200 + 100) = 200 x 9V 300 = 6V

10

02

00

9v

3v

Rt = R1 + R2 = 470 + 2k2 = 470 + 2200 = 2670 = 2k67 ohms

Rt = R1 x R2 R1 + R2

= 1k x 2k2 = 2k2 1k + 2k2 3k2

= 2.2 x k = 0.687k 3.2 = 687ohms

R = 100I = 0.5AV = 50V

V = 9vI = 1mAR = 9M ohm

Capacitors

Capacitor Charging++

++

+++

+Volts

Time

++

++

+++

+

Time

Capacitance = Farads

Capacitors store and hold electric charge

Keywords:Electrolytic, non-electrolytic, metal plates separated by dielectric,

Capacitor Charging++

++

+++

+Volts

Time

++

++

+++

+

Time

t=RCTime constant – the rate at which a capacitor charges through a resistor

After one time constant, capacitor is at 0.6 of its full charge, and fully charged after 5 time constants

Capacitor Discharging

Time

Volts

t=RCTime constant – the rate at which a capacitor discharges through a resistor

Transistor as a switch

In order to switch on the transistor the voltage at the base must be 1.2V or above

Sensors

ΩΩ

Sensors

Voltage Dividers as Sensors

Vout = R2 x Vsupply R1+R2

So, if R2 >> R1, Vout is close to Vsupply

Transistor plus sensor (voltage divider)

In coldVbase = 1/11 x 9V

= 0.81VTransistor is off, bulb off

In warmVbase =2/12 x 9V

=1.5VTransistor on, bulb on

Vout = R2 x Vsupply R1+R2

System Diagram

Systems Electronics

input process output

Systems Electronics

input process output

SwitchLDRThermistorMoisture Sensor Variable ResistorMicrophonePiezo

TransistorDelayOscillatorCounterLatchAmplifier ComparatorLogic GatesPIC

BuzzerSpeakerBulbLEDMotorRelaySolenoidPiezo

Transistor Amplifier

ib

Ice

The current entering the base controls the current that flows through the collector and emitter, with a fixed relationship called the gain (hfe)

Gain (hfe) = Ice

IbOr Ib x Gain = Ice

Eg Gain 100, Ib 1maIce = ?

Thyristor Latch

Similar to a transistor, but here a signal/current at the gate latches the thyristor on for as long as current flows through it, interrupting this resets it to off.

[once ON, it stays on until reset: e.g. car alarm]

741 Op-Amp

Croc Clips link – LDR circuit

IC = integrated circuitDIL = Dual in line

741 op amp = 8 pin DIL IC

OP AMP as a comparator—the OP AMP compares the inverting input voltage to the non-inverting input voltage, and gives a HIGH or LOW output depending upon which is the greater input voltage.

The OP AMP detects very small changes in voltage multiplies the difference by the GAIN (typically 100,000). Because the output is HIGH or LOW, it is used as an analogue to digital converter (ADC) so is suitable for connecting analogue sensors (E.g LDR, THERMISTOR) to logic circuits.

 

ICs have three big advantages over conventional circuits with discrete components:• they take up very little space • they are extremely reliable, and • they are extremely cheap to make

IC = integrated circuitDIL = Dual in line

555 timer = 8 pin DIL IC

14

8

555 timers

Recognising it = pins 6 and 7 are connected,

through R to +V

PIN 3 = output pin

555 timers - MONOSTABLERC timing:

t=RC

t (seconds)R (resistance)C (capacitance)

Careful with units!!!

R

C Monostable state = pin 2 high, pin 3 low

Pin 2 then triggered (taken low), so pin 3 goes high for the timing period, then goes low.[10k pull-up resistor keeps pin 2 high]

555 timers - ASTABLE

Recognising it = pins 6 and 2 are connected,

through C to 0V

PIN 3 = output pinMark (time on) = 0.7 x (R1 + R2) x C1Space (time off) = 0.7 x R1 x C1

C1 charges through R1 and R2, until voltage across C1 is > 2/3 supply voltage. At this point pin 3 goes from high to low.

C1 then discharges into pin 7, until voltage across C1 is < 1/3 supply voltage. At this point pin 3 goes from low to high.

Pin 3 = low = current flows into it (sinking current)Pin 3 = high = current flows out (sourcing current)

NAND gate – ASTABLE

IC = 4011

R and C control frequencyVariable R = adjustment

Typically C is small (say 100nF)And R is large (1M)

NAND gate – MONOSTABLE

IC = 4528

NAND gates with inputs connected – operating as inverters

Timing depends on C, when the voltage across C reaches a threshold level (say 2/3rds of supply) the logic level switches from O to 1

Counters4017 and 4026 data

Some positives… can you say YES to these…

Last years group achieved average 81% in their coursework, and 83% A*-C overall

YOU have achieved average 81% in their coursework

WE have nearly at the end of the revision and course

YOU already know enough to achieve A* - C overall

This morning YOU will know what to expect in next months exam,and know what to do to perform your best

This morning we will study the remaining OUTPUT components and LOGIC

Outputs: relays + motors

The current output from a 4017 won’t drive a motor (too small) – use a transistor or Darlington pair to drive a motor

Relay – small current through the coil causes magnetic field that moves an armature that switches ON the relay – keeps low and high current systems apart

Switch bounce Use a 555 monostable with ~ 1s delay to clean the input

Circuit diagram link

Use a SCHMITT TRIGGER

Analogue and Digital Electronics

Analogue signals are constantly variable, for example temperature, light intensity, sound waves etc

Digital Electronics converts signals into numerical values, using the binary number system based on 1’s and 0’s (on and off)

Advantages of Digital•Can be more reliably reproduced and transmitted•Can be processed

Binary NumbersA single binary number is called a bitAn 8 digit binary number is called a byte, and can represent a decimal number from 0 to 255

128s 64s 32s 16s 8s 4s 2s units decimal

1 0 1 0 1 0 1 0 170

0 0 0 0 1 1 1 1 15

0 1 1 1 1 1 1 1 127

1 1 1 1 1 1 1 1 255

Binary Numbers

Logic Gates

A Q

0 1

1 0

A B Q

0 0 0

0 1 1

1 0 1

1 1 1

A B Q

0 0 0

0 1 0

1 0 0

1 1 1

Logic Gates

A B Q

0 0 0

0 1 1

1 0 1

1 1 1

A B Q

0 0 0

0 1 0

1 0 0

1 1 1

A B Q

0 0 1

0 1 0

1 0 0

1 1 0

A B Q

0 0 1

0 1 1

1 0 1

1 1 0

Logic Gates

A B Q

0 0 0

0 1 1

1 0 1

1 1 0

A B Q

0 0 1

0 1 0

1 0 0

1 1 1

Logic Gates

A Q

0 1

1 0

A B Q

0 0 0

0 1 1

1 0 1

1 1 1

A B Q

0 0 0

0 1 0

1 0 0

1 1 1

A B Q

0 0 1

0 1 0

1 0 0

1 1 0

A B Q

0 0 1

0 1 1

1 0 1

1 1 0

A B Q

0 0 0

0 1 1

1 0 1

1 1 0

A B Q

0 0 1

0 1 0

1 0 0

1 1 1

Using Logic GatesInputsSwitch in Gatehouse ASwitch under barrier (on when barrier closed) BSwitch (pressure pad) before barrier on road CSwitch (pressure pad) under barrier DSwitch (pressure pad) after barrier on road E

OutputsMotor (on barrier raised, off barrier lowered)Red LightGreen Light

Design using Logic Gates systems to; - show a green light when the barrier opens and red when closes - automatically open the gate when the car approaches, then close it when it has passed - allow the gatehouse switch to close the gate unless a car is under it.

PIC ChipsProgrammable Integrated Circuits

Come in various numbers of pins which limit the number of inputs and outputs

Easiest way to imagine a pic is like a programmable ‘process block’

Writing ProgrammesProgrammes can be written as flow charts

Start/End Process

Decision Output

Flashing Light

Input 1 On?

Wait 1

o/p 1 on

Wait 1

o/p 1 off

Circuit modelling and CAD CAM

Breadboard = know how these are configured. You might get a “complete the connections” or a fault finding question

Advantages = test the components you will use, test in read conditions, can change component values and test very quickly

Disadvantages = tricky to fault find, risk of damaging components

Circuit modelling and CAD CAM

CAD = computer aided design

Advantages = quick to model a circuit, voltages and currents can be measured, no damage to components, design can be exported into a PCB layout design program, use of programmable chips (PICs)

Disadvantages = unable to test the circuit in real conditions so you have to make a pcb to test it properly, software can be expensive

CAM

CAM = computer aided manufacture

• CNC processes to cut and drill PCB’s• pick and place components automatically• digital photography to check (QC) component positions• automated processes e.g wave soldering

PCB making:http://www.youtube.com/watch?v=SKccLhFf1DY http://vixy.net/

Or this

SAMPLE QUESTION – think like a computer…