Electronics Lab (Short Report)

Electronics LaboratoryBrian Bannon(08332321)

April 2013

Abstract

The purpose of this experiment is to provide an introduction to analogue electronicsconsisting of resistor/capacitor circuits, diode circuits, transistor circuits and opera-tional amplifiers.A breadboard was given to the student to construct the named cir-cuits and to evaluate if the circuits were constructed correctly the right componentswere used and that the circuit perform to its theoretical predictions.These circuitswere compared to simulations which were carried out using computer software pro-gram called TINA PRO which carried out analytic and numerical calculations of thecircuits before construction.

Data Analysis

Exercise 1

The first circuit that the student attempted was the simple Resistor/Capacitor Cir-cuit.The RC is the perfect example of how a voltage divider works.The purpose ofthe voltage divider is to obtain a certain desired output voltage from an initial inputvoltage.Most likely it is typical that the output voltage will be lower then the inputvoltage, but in small cases it can be higher, but this case is very rare.The main com-ponents of a RC circuit are voltage source, voltmeter and three resistors two 1kΩ anda load resistor which is inserted and which can be varied. If we look at Theveinin’stheorem, the student can evaluate the output voltage from a simple RC-circuit withno load resistor by the formula,

Vout = VinR2

R1 +R2

(1)

If we introduce a load resistor into the circuit the voltage varies.As the resistanceacross the load resistor increases the voltage across R2 increases and a greater voltage

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is read on the output.This will increase until no voltage crosses the the load and thevoltmeter reads the same as if the load was not attached.

Fig(1).A simple RC-circuit with a load resistor attached.

Thevenins theorem can adjusted to deal with such a circuit.It states that any twoterminal network of resistors and the voltage sources is equivalent to single resistorRth, and a single power source Vth, connected in series.Vth and Rthcan be calculatedusing the following,

Vth = Vin(R2

R1 +R2

) (2)

Rth = (R1R2

R1 +R2

) (3)

Vout = Vth(Rload

Rth +Rload

) (4)

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A plot of the load resistance and output voltage yields a graph and the behaviourof the load resistor and output voltage can be seen clearly.The purpose of the loadresistor is to produce a voltage drop in the circuit and as the load varies so too doesthe voltage.If the resistance of the load resistor was set to 0 then there would be novoltage on the output.

Fig(2).Plot of load resistance and output voltage with the load resistance variedfrom 0Ω to 300kΩ

Exercise 3

In this setup we are using the Thevenin equivalent circuit,

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Fig(3).Thevenin equivalent circuit.

The components used and labelled in this circuit are,the Vth is 30V,Rth is 10kΩ andRload is 20kΩ, the output voltage on this circuit is 20V.Selecting the load resistor forDC analysis in TINA, will produce a plot as follows,

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Fig(4).Plot of Thevenin Output as a function of load resistance over the loadresistor.

Exercise 5

This circuit was the first construction on the breadboard.The construction of thiscircuit was a simple RC-circuit, the circuit consisted of one resistor and one capaci-tor.The voltage source in was a signal generator producing a square wave signal withVin of 5V, the capacitance of the capacitor was 1µF and the resistor had a resistanceof 1kΩ.The circuit was attached to an oscilloscope with a built in signal generator,so we could graph the input signal into the circuit and the output signal.

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Fig(5).TINA simulation of the RC-circuit with oscilloscope attached

Fig(6).Image from the simulation oscilloscope on TINA of the input and output.

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Fig(7).Image from the oscilloscope used in the lab in the students circuit.

Exercise 7

In this exercise the student is using what he had learned before about voltage dividersand applying it now to include impedance, the impedance may be defined as,

Z =V

I(5)

and in general the impedance will be complex for a capacitor

Z =−iwC

(6)

Since the impedance is purely imaginary there is only a phase angle change.SinceEq(5) is a generalisation of Ohm’s law,Thevenin’s theorem may be generalised toinclude impedances and the voltages and the voltage divider equation becomes

Vout = VinZ2

Z1 + Z2

(7)

The circuit for this student had to build a simple RC-circuit with a signal generatorinputting a sinusoidal signal that varies with time, in this case a sine wave.Therewas a capacitor of 1µF and a resistor of 1kΩ used as well as an output in the formvoltmeter.

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Fig(8).Schematic of a simple RC-circuit.

I had difficulty with this exercise, firstly the circuit in the simulation seemed to notgive out a voltage from the out put so I could not plot the amplitude and phase inMathCad.I was able to get a Bode plot but I am pretty sure that it is incorrect.

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Fig(9). Bode plot of a RC-circuit

Exercise 12

A diode in electronics is a PN junction and goes with the direction of current.A diodecan be in tow positions,forward bias when the voltage is connected to the anode andis made positive to the cathode and if the potential difference will let current flow,reverse bias that the voltage is with the cathode with the anode and no current willflow.In this case if there is a large voltage in this direction the diode will breakdown.This voltage depends on factors such as the material,doping level etc.Vf is equal tothe barrier potential produced across the forward bias PN junction.This is equal tothe forward current.

If = Vbias − Vf (8)

In this exercise we measured the I-V curve for a 1N914 diode. IN this circuit wewill be using a voltage source of 5V, a 1k resistor a 2k potentiometer to vary thevoltage across the diode and to ensure that the maximum forward current throughthe diode is not exceeded.In the spec. sheet this number was given as 300mA.And anammeter to measure the the current through the circuit.Unfortunately I was unableto open the .tdr file which I had saved my simulation graph on my laptop,but thisfile is available in my copy of my report saved on the digital version of this reportsaved under ex12.tdr.The circuit in my simulation is of the following.

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Fig(10).The circuit to demonstrate a I-V curve in the lab.

In the lab the student built the circuit and tested it using a voltage source and anoscilloscope to try to capture the characteristics of a I-V curve.The results are asfollows,

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Fig(11). Image from oscilloscope of a segment of the I-V curve.

Fig(12).Image of the graph of an I-V curve and we can see the segment in whichour experiment image came from in the forward characteristic section.

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Exercise 19

In this exercise we are dealing with transistors. Transistor are semiconductors, whichact as current , voltage and power amplifiers.Transistors comes in two types, a bipolarjunction, and unipolar.The bipolar consists of two PN junctions back to back,the p-type layer is very thin and the n-type layers are heavily doped.

Fig(13).npn-transistor and pnp-transistor

The transistors have three legs, the collector, the base, and the emitter.The npntransistor must be operated with the collector and base with respect to the emit-ter.The preferred transistor is the silicon for its low current leakage and high temper-ature acceptance.In the transistor the current flowing through the transistor are thebase-emitter and collector-emitter currents, this gives the ability of the base-emittercurrent to control the collector-emitter current.The D.C. current gain is given by

hfe =ICIB

(9)

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and the sum of the base and emitter current must be equal to the collector current,

Ic = Ib + Ie (10)

For this exercise the student had to construct the circuit,

Fig(14).Circuit layout for a a transistor acting as a current amplifier UnfortunatelyI did not get time take measurement on this circuit on my breadboard in the lab.

Exercise 20

In this exercise, the student learned how to use transistors as a switching mecha-nism.Transistor surpass the electronic switches by being cheaper, smaller more re-liable,less moving parts,and there precision in switching can be anything up to amillion times a second.

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Fig(15).Circuit for using a transistor as a switch

(Fig(16).Graph of the input voltage and of the output voltage when the resistanceon R1 was changed from 1Ω-10kΩ.

The above figure is the response curve of the transistor in its three stages, the firstbeing the transistor is off and no collector current is present and the output is Vin.

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The second the transistor is partly on, the current increases the voltage across theload resistor increases and hence the output drops. And lastly the transistor is fullyon the current across the collector-emitter may not be at it’s maximum, but it can’tgo any higher because that would require a larger input and we say the transistor issaturated.

Exercise 21

In this exercise the student is required to build a light sensitive alarm. In this causethe diode is a lamp which should switch on in darkness and switch off in brightness.In the manual it recommends the red LED for it has a reverse bias voltage of 1.8Vand a current of ∼5mA and would produce reasonably output.

Fig(17). Layout inTINA of a light sensitive switch.

In this circuit we had to measure the resistance of the LDR on its own in light anddark and evaluate our choice of other components. Firstly we needed a LED whichwas the red bulb recommended in the manual, a variable resistor,to fine tine theresistance and control the voltage going through the base-emitter channel.We chose

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a 1kΩ resistor going into the base, 1kΩ would have been sufficient.We put 1kΩ on theload resistor, and the last resistor we chose as 2.6kΩ from our readings of the of theLDR on its own in light and dark.My simulation worked fine but in my breadboardI believe I put in too high of a resistor in were the 2.6kΩ should have been ∼ 10kΩ.And this did not give me back a great result when testing this circuit in the lab.

Exercise 22

In this exercise, the student is using a transistor as a voltage Amplifier.

Fig(18).Layout circuit for using a transistor as an amplifier.

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Fig(19). Graphs of the input and out voltage for different amplitudes and offsets ofthe signal generator.

In this circuit the input signal was a 1kHz sinewave and the load resistor was set to500Ω.

Exercise 25

In this section the student learned about real Operational Amplifiers.”Opamps” areintegrated circuits which contain 20+ transistors in addition to resistors, capacitors

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and diodes.If you are constructing amplifiers or other circuits the preferred choicewould be opamps over transistors.Opamps have two inputs and one output.Morepins are available on the opamps that resistors and capacitors may be connected toimprove the performance of the circuit.Opamps have very high gain Aol, but this fallsas the the frequency of the input increases.They have very high input impedance, sothat little current is drawn from the input source and the input source and the inputvoltage is transferred with little loss. And lastly low output impedance.The opamp acts like a differential amplifier and produces an output

Vout = Aol(V+ − V−) (11)

Opamps have uh high gain the output will most likely be saturated, unless the inputsare within a few µV of each other.The voltage at the inverting input is equal to thevoltage at the non-inverting input,

V− = Vin (12)

The current I through a resistance Ri is

Ii =VinRi

(13)

Also since the inputs have no current all of Ii must go throughRf and Ii = If ,thus

Vout = V− + IfRf (14)

= Vin + IiRf (15)

= Vin +VinRi

Rf (16)

= Vin(1 +Rf

Ri

) (17)

Acl =VoutVin

= (1 +Rf

Ri

) (18)

In the exercise we constructed the circuit given to us in the handout,

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Fig(20). Non-inverting amplifier using negative feedback

The circuit was driven by a 1kHz Sine-wave and we had to compare the input andthe output. IN the simulation the output should be a gain on the input to the orderof the ratio of resistors used in the circuit which was achieved in simulation but wasnot obtained in the experimental process.

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Fig(21). Oscilloscope image of the input and output waves of the Ompamp.As youcan see the output wave is not showing any gain in amplitude to match my

theoretical results.

Exercise 29 In this exercise student learns about current to voltage convert-ers.An op-amp with negative feedback can be used as a current to voltage converterby replacing Ri in the inverting circuit with a current source.Since V− is a groundall the current flows through Rf which can be chosen to give the appropriate outputvoltage.In this circuit we used a BPW34 infra-red photodiode and is operated in reverse biasmode and produces a current proportional to the infra red light falling on it.

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Fig(22). Infra-red photo-diode circuit

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Fig(23). Oscilloscope image of the output of the circuit.

References

[1] 3rd Year Electronics Laboratory,Dr.J. Quinn

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