SI Units Mega1000,0001x10 6 MOhms kilo10001x10 3 kOhms Units milli0.0011x10 -3 mAmps micro0.000,0011x10 -6 μFarads nano0.000,000,0011x10 -9 nFarads Pico0.000,000,000,0011x pFarads pnμm1kM

SI Units Some shortcuts k x m = cancel out M x μ = cancel out K x μ = m 1/k = m 1/M = μ M/k = k k/m = M a m x a n = a m+n a m / a n = a m-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/(1x10 6 ) Amps = 9 x Amps = 9 μA a m x a n = a m+n a m / a n = a m-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 -6 x10 3 ) = 726 x (10 -3 ) = 0.73s a m x a n = a m+n a m / a n = a m-n

Practice Questions a)R=180kΩV=9V I=? b)R=3M3ΩV=6V I=? c)R=100ΩV=3V I=? d)R=1MΩC=220μf t=? e)R=56kΩC=330pf t=? f)R=390kΩC=1000μf t=? 0.05mA 1.82μA 30mA 220s 18.5ms 390s I = V/R t = CxR

Practice Questions I = V / RI = V / RI = V / R = 9 / 180k = 6 / 3M3 = 3 / 100 = 9 / 180 x = 6 / 3.3 x = 0.03A = 0.05 x = 1.82 x = 30mA = 0.05mA = 1.82 μA t = CxR t = CxRt = CxR = 220μf x 1M = 330pf x 56k = 1000μf x 390k = 220 x 1 x (10 -6 x10 6 ) = 330 x 56 x ( x10 3 ) = 1000 x 390 x (10 -6 x10 3 ) = 220 x (10 0 ) = x (10 -9 ) = x (10 -3 ) = 220s = 18.5 x(10 -6 ) = 390 x (10 3 x10 -3 ) = 18.5 μs = 390s

Resistors in Series R total = R1 + R2 + etc

Resistors in Parallel Two Resistors R total = R1 x R2 = Product R1 + R2 Sum Three or more Resistors 1 = etc R total R1 R2 R3

Resistors

a) Rtotal = =200 b)Rtotal = 100x100 = 10,000 = c) 1 = Rtotal = 3 Rtotal 100 Rtotal = 100 =

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/(1x10 6 ) Amps = 9 x Amps = 9 μA a m x a n = a m+n a m / a n = a m-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 Resistors R total = R1 x R2 = Product R1 + R2 Sum Vout = R2 x Vsupply R1+R2 Copy down these formulae:

4702k2 1k 9v v Vs = 200 x 9V ( ) = 200 x 9V 300 =6V v 3v Rt = R1 + R2 = k2 = = 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 = 100 I = 0.5A V = 50V V = 9v I = 1mA R = 9M ohm

Capacitors

Capacitor Charging Volts Time Capacitance = Farads Capacitors store and hold electric charge Keywords: Electrolytic, non- electrolytic, metal plates separated by dielectric,

Capacitor Charging Volts Time t=RC Time 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=RC Time 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 Ω Ω

Voltage Dividers as Sensors V out = R2 x V supply R1+R2 So, if R2 >> R1, V out is close to V supply

Transistor plus sensor (voltage divider) In cold Vbase= 1/11 x 9V = 0.81V Transistor is off, bulb off In warm Vbase=2/12 x 9V =1.5V Transistor on, bulb on V out = R2 x V supply R1+R2

System Diagram

Systems Electronics inputprocessoutput

Systems Electronics inputprocessoutput Switch LDR Thermistor Moisture Sensor Variable Resistor Microphone Piezo Transistor Delay Oscillator Counter Latch Amplifier Comparator Logic Gates PIC Buzzer Speaker Bulb LED Motor Relay Solenoid Piezo

Transistor Amplifier i b I ce The current entering the base controls the current that flows through the collector and emitter, with a fixed relationship called the gain (h fe ) Gain (hfe)= I ce I b Or I b x Gain= I ce Eg Gain 100, Ib 1ma I ce =?

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 circuit DIL = 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 circuit DIL = Dual in line 555 timer = 8 pin DIL IC timers

Recognising it = pins 6 and 7 are connected, through R to +V PIN 3 = output pin 555 timers - MONOSTABLE RC 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 pin Frequency (Hz) =1.44 (R1 + 2xR2) x C Frequency (Hz) =1.44 (R1 + 2xR2) x C Frequency (Hz) =1.44 (R1 + 2xR2) x C Frequency (Hz) =1.44 (R1 + 2xR2) x C Mark (time on) = 0.7 x (R1 + R2) x C1 Space (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 frequency Variable 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

Counters 4017 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 Numbers A single binary number is called a bit An 8 digit binary number is called a byte, and can represent a decimal number from 0 to s64s32s16s8s4s2sunits decimal

Binary Numbers

Logic Gates AQ ABQ ABQ

ABQ ABQ ABQ ABQ

ABQ ABQ

AQ ABQ ABQ ABQ ABQ ABQ ABQ

Using Logic Gates Inputs Switch in Gatehouse A Switch under barrier (on when barrier closed) B Switch (pressure pad) before barrier on road C Switch (pressure pad) under barrier D Switch (pressure pad) after barrier on road E Outputs Motor (on barrier raised, off barrier lowered) Red Light Green 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 Chips Programmable 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 Programmes Programmes 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: Or this

SAMPLE QUESTION – think like a computer…