BTEC NC - Analogue Electronics - Design and Testing of Analogue Circuits Using ECAD

8/2/2019 BTEC NC - Analogue Electronics - Design and Testing of Analogue Circuits Using ECAD

1/25

Brendan Burr BTEC National Certificate in ElectronicsDesign and Testing of Analogue Circuits Using ECAD

- 1 -


2/25


- 2 -

Task 1

1.1 a) ELEX 2A21


3/25


- 3 -

b) ELEX 2A22


4/25


- 4 -

c) ELEX 2A24


5/25


- 5 -

ELEX 2A21 Results

R1 = 10K D1 = 1N4001

AC Input

Measured RMS of AC Input Voltage = 12VMeter Range Used = AC VoltsMeasured Peak to Peak of AC Input Voltage = 17V

DC Output

Measured DC of Output Voltage at A = 11.5V

Measured Peak to Peak of DC Output Voltage at A = 16.3VMeasured Periodic Time of Waveform at A = 10mSCalculated Frequency of Waveform at A = 100Hz

ELEX 2A22 Results

R1 = 100 D1 = 1N4001 C1 = 815F

Measured DC Output Voltage at B = 14.6V

Measured Peak to Peak Ripple Voltage at B = 3.2V

ELEX 2A24 Results

R1 = 10K D1 D4 = 1N4001

Measured DC of Output Voltage at D = 15.45VMeasured Peak to Peak of DC Output Voltage at D = 0.3V

Measured Periodic Time of Waveform at D = 10mS

Calculated Frequency if Waveform at D = 100Hz


6/25


- 6 -

Task 2

2.2a

The Transformer will decrease or increase the voltage supply coming from thesignal generator.The Bridge Rectifier produces a lumpy DC Voltage. The Smoothing Capacitorthen charges to full capacity. The Voltage Regulator works with theSmoothing Capacitor to smooth the lumpy DC Voltage and make it more likea true DC Voltage.


7/25


- 7 -

2.2b


8/25


- 8 -

2.2cCalculations:


9/25


- 9 -


10/25


- 10 -


11/25


- 11 -


12/25


- 12 -

2.4

Voltage Current

(a) No Load 12.1V 0A(b) Full Load 12.0V 250mA

b) Full Load


13/25


- 13 -

a) No Load


14/25


- 14 -

Task 3


15/25


- 15 -

Results

R1 = 6K8 R2 = 27K C1 = 220nF C2 = 0.1F

WAVEFORM AT PIN 6 TEST POINT A MEASURED FROM CRO

Maximum Voltage Level (Volts) 6VMinimum Voltage Level (Volts) 3V

Peak to Peak Voltage Level (Volts) 3V

WAVEFORM AT THE OUTPUT PIN 3 MEASURED FROM CROMaximum Voltage Level (Volts) 9VMinimum Voltage Level (Volts) 0V

Time Period in Secs of High Level [Mark] (t1) 5.15mSTime Period in Secs of Low Level [Space] (t2) 4.2mS

Periodic Time in Secs (T) 9.35mS

Frequency in Hz (f) 106.95Hz

MEASURED FROM CROMaximum Voltage at PIN 6 6VMinimum Voltage at PIN 6 3V

CALCULATED FROM ABOVERatio of Maximum Voltage at PIN 6 / Vcc 2:3Ratio of Minimum Voltage at PIN 6 / Vcc 1:3

Calculations

MEASURED CALCULATEDMark Time t1 (Secs) 5.15 x 10^-3 5.2052 x 10^-3

Space Time t2 (Secs) 4.2 x 10^-3 4.158 x 10^-3Periodic Time T (Secs) 9.35 x 10^-3 9.3632 x 10^-3

Mark-Space Ratio 1.223:1 1.252:1Duty Cycle (%) 55.08% 55.592%Frequency (Hz) 105.95 Hz 106.8 Hz


16/25


- 16 -

Questions

1) Explain the difference if any between your calculated values andmeasured values.The tolerance levels of the components can result in discrepancies.

This applies not only to the circuit but also to the measuring equipmentcomponents.

2) What happens when PIN 4 is connected to 0V?

There is an exponential decay in the Voltage level at PIN 6 and 2. Thiswould indicate the capacitor (C1) is discharging due to the IC notresetting continuously, but remaining at a constant 0V as the dischargeis not removed from 0V.

3) What happens to the frequency when C1 is decreased in anastable circuit?The frequency increases. This is because the capacitor (C1) has tocharge to 1/3 of the Vs, before the trigger begins each cycle.When the value of the voltage across the C1 Capacitor reaches 2/3 ofthe supply voltage the capacitor positive is connected to the Discharge.With a lower valued capacitor to fill, the time to fill it decreases,resulting in a quicker cycle.This takes less time for each cycle (Periodic Time) also resulting in aneffect on the frequency.


17/25


- 17 -

Task 4

The 555 IC Timer used for Task 3 is a highly stable device which can be usedfor generating accurate time delays or oscillation.In a timing scenario for the Task 3 Astable Circuit two resistors and two

capacitors are used.

The 555 IC Timer is cheap and perfectly effective for the creation of theAstable Multi-vibrator. This is because with the simple use of two resistorsand two capacitors you can easily calculate and create a timer whichoscillates states, on and off, to a reasonably accurate scale.

The Voltage supply (+Vcc) goes to pin 8 with 0V (Ground) connected to

Pin 1. The Trigger (Pin 2) starts the timing IC by providing a pulse.

Depending on the time interval of the chip the output will vary, it willgive an output voltage level for that time period of the +Vcc SupplyVoltage.

The timing interval can be interrupted by using the reset at pin 4.

At pin 5 there is access to the internal Potential divider.

The threshold determines when the timing interval is complete.

The Discharge (Pin 7) is connected to a capacitor, this will alsodetermine the time interval.

To conclude, it is very simple to design and build an Astable Timer using thisIC. The calculations are straight forward to work out and you are able tocalculate the Mark Space Ratio and therefore work out the Periodic Time ofthe waveform.Other IC devices are much more complex than the 555 IC and thereforeslightly more expensive, however there is clearly no need for them to be used.

These are the reasons for using this IC for Task 3.


18/25


- 18 -

Task 5a) Voltage Ripple on No Load

There is no current being drawn through the Load Resistor when the switch isopen, this puts the circuit in a No Load state.

The 400mA current then finds the easiest available path available and travelsthrough the Zener Diode.The voltage across the Zener Diode then reaches levels of around 12.278V.This is due to the lack of load which would cause a small amount of voltage tobe dropped from the overall voltage if the circuit were in a Loaded State.The Zener Diode only allows current to pass through, when the voltage levelis greater than 12V. At 12V the Zener Diode breaks down and allows currentto pass freely. If the voltage level dropped to 11V then the current would beblocked and the circuit would not work as it did at 12V.The effect of this means the voltage waveform is much flatter and appears tobe an almost perfect DC from a zoomed out scale.

When the waveform is zoomed in, it is apparent that there is definitely aVoltage Ripple. The peak value is 11.634mV this gives me a 23.268mV peakto peak value. The periodic time of the waveform is 1mS.The ripple is a variance of the Direct Current and is there because of theAlternating Current not being suppressed completely by the power supply.


19/25


- 19 -

b) Voltage Ripple on Full Load

When the Switch is closed the Load Resistor is connected to the circuit, thismakes the circuit Fully Loaded.With the Switch closed, 255.616mA is drawn away from the circuit, leaving

only 150mA to pass through the Zener Diode.

There is a voltage peak of 14.572mV, this is according to the graph abovewhen the circuit is under Full Load.

This gives me 29.144mV peak to peak voltage.

The current of 255.616mA flows through the Load Resistor and then toground.The voltage ripple has increased when compared to the No Load graph. Thisis due to the


20/25


- 20 -

Task 6

Analyse the results for Task 2 as follows:-

a) Produce a graph of the waveform of the output voltage ripple v time on Full

Load


21/25


- 21 -

b) Produce a graph of the waveform of the input voltage v time on full load

Input Voltage:


22/25


- 22 -

Voltage After Transformer:


23/25


- 23 -

c) Determine the frequency of the output voltage ripple on Full Load:

31010

1

1

F

PF

T

HzF 100

Compared to the Frequency of the Input Voltage:

31020

1

1

F

PF

T

HzF 50

So the Frequency has appeared to have doubled.


24/25


- 24 -

Task 7a) ECAD

When working with an ECAD package it is possible to add and delete multipletimes with minimal effort. The benefit of this is a major saving in time, which

would be the downfall when physically building the circuit from realcomponents.

The tolerance levels of the components on an ECAD package allow thedeveloper to check their calculations, for the correct design criteria. A 5KResistor would be precisely 5000 Ohms, meaning that calculations madeusing this component are more likely to be accurate.

Failure in the design process could save large expense, when using an ECADpackage. This is because a microprocessor circuit for example may be wiredincorrectly in the design. The consequence could be that the chip will blowalong with multiple other components. If this is the case then the cost couldbe high. When developing the circuit on the ECAD package it is possible toblow components up when the circuit is wired incorrectly, but then they can bereplaced completely free of charge. This would also save a long time in theordering process of components, as the new component may take about aweek to arrive, ceasing the design process.

When using an ECAD package you are also able to view information that youwouldnt be able to do during a test on a real circuit. For example the currentalong a specific wire would not be visible unless you reconnected an ammeter

into the circuit, which would again take time.

Information can also be displayed in the form of a graph. The simulation canbe paused, allowing the designer to analyse individual time periods, of eachcycle of a waveform. This allows further calculations to be carried out to amuch higher level as information can be displayed and paused.


25/25


b) Building and testing a prototype on the bench

When building the circuit for real, it is possible to analyse whether the circuitgives true results. This would be mainly due to the accuracy of thecomponent values. A resistor with a Gold band has a tolerance of 5%,

corresponding to the stated value. So a 100 Ohm resistor could actually be95 Ohms up to 105 Ohms, making current and voltage levels extremelyvariable.

Overall circuit performance can be tested and a real life output can be used tocheck if the circuit does actually do the required thing. For example a DCdevice can be attached to the output to test if the device would functioncorrectly. This wouldnt be possible for a ECAD package to reconstruct.

On Croc Clips there is a function which can make the circuit componentsindestructible. If the whole circuit was indestructible, it would be impossible totell whether the circuit would work in real life, according to the individualcomponents. For example a resistor may have a power rating too small tocope with the current which is passing through it, the result would be that theresistor would simple melt from overheating. This could be monitored on atest rig but not on an ECAD package. The benefit of this would be that thedesigner could not only think about the design calculations but the practicalapplication of the device. This would allow the designer to think of criteriasuch as an inclusion of a heat sink or a fan to lower the temperature of certaincomponents in the circuit, for example the transformer or the load.

Documents

BTEC NC - Analogue Electronics - Design and Testing of Analogue Circuits Using ECAD