electronics experiments

8/7/2019 electronics Experiments

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Experiment 1 Oscilloscope input resistance.

This section has two purposes. It should help you get used to your equipment. It also

helps you understand the concept of input and output resistances.

Any piece of equipment which accepts input signals will require both a voltage and a

current to make it work. This is because every signal must convey some energy/power except the trivial case of the signal 0 Volts. When you apply an input voltage to,

say, an oscilloscope, it must also draw a small current to make it recognise that a

signal has arrived.

The amount of current required by something to make it respond to a given voltage

depends upon how it has been designed and built. We don't need to bother about these

details, instead we can pretend that a resistor has been connected between its input

terminal and earth, somewhere inside its box. The better a 'scope or voltmeter is, the

smaller the current it needs to register a given voltage - i.e. the higher its input

resistance. The 'scope will probably have an AC/DC/Ground switch for each input.You can force the 'scope to show where 0V is on the screen by setting this to

Ground. Then set it back to DC to use the 'scope - just measure the number of

divisions between where Ground is and the point on the trace whose voltage you want

to measure. For most measurements, these controls should be left on DC. The AC

setting is useful when you want to watch small variations of a relatively large voltage

level, but it tends to alter the shape of some a.c. waves.


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Build the circuit shown in diagram 3 and connect it between the power supply and

'scope as shown. By adjusting the potentiometer you can apply any D.C. voltage from

0 to +15V to the scope. Set the voltage initially to 0 and adjust the vertical position of

the trace to sit on a convenient line.

Measure the input current into the scope for three or four different input voltages.

You can use the 'scope itself to measure the voltage. Use these values to calculate

the 'scope input resistance.

What is the significance of the order of magnitude of the 'scope resistance, ?

Experiment 2 The RC Low-Pass Filter

This experiment shows the main properties of capacitors and how they can be used

with resistors to make filters that pass some frequencies and block others. In this case

the capacitor and a resistor are used to make a Low Pass Filter.

You should build your Low Pass filter circuit on one of the pieces of

Tracked Boardyou have been given. To see photographic images ofwhat your circuit should look like, just click on the image of a camera

near this text.

The circuit diagram for this circuit is

shown to the left.

As is common in circuit diagrams, the

bottom line in this diagram is assumed

to be the earth or 0 Volts wire.

Build this circuit on a board using a


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22k resistor and a 01 F capacitor. Try to make it look as much like the photo as

you can. Use different colour wires for the input, output and ground leads to make it

easier to tell which is which. It is usually a good idea to adopt a favourite set of

usual colours for leads as this will help you recognise what you have built. In most

cases, we would recommend using a green-covered wire for earth or 0 Volts, andcolours like blue for signal input, yellow for signal output, etc. The precise colours

arent important, though. as long as they are consistent and recognisable.

Make a note in your labscript of the wire colours you have used for the circuit.

Connect the signal generator to the input leads ( ). Use both inputs of your scope

so you can observe both the input and output voltages at the same time. In each case

remember that the earth lead of each pair (i.e. the outer of the co-axial cables)

should be connected to the earth-line of the circuit, shown as the bottom line of the

diagram. If you cant see which wire of the coaxial cable is which, remember that the

live wire usually has a red coloured terminal, and the ground wire usually has a

green or black coloured one. If you are not sure, ask a demonstrator to check your

leads.

Apply an input sinewave of approximately 1V peak-to-peak, and use the scope to

note the values of and at a series of frequencies from about 100 Hz up to 5kHz. Use the scope to measure the frequency by observing the period of the

waveform. Then plot a graph of against frequency.

At what frequency does have the value ?

Draw a line on the graph at the filters turn over frequency


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Is this reasonably close to the frequency where the output/input ratio is ?

Remember to label your circuit with your name and keep it to hand in with your

script.

Experiment 3 Resonant Tuned Circuit.

This experiment shows you some of the properties of circuits which contain resistors,

capacitors, and inductors.

Resonance is an important physical phenomenon. It can occur in all sorts of systems,

from a swinging mechanical pendulum to an optical cavity. In each case it requires a

situation where energy is periodically transferred back and forth between two possible

reservoirs. In the case of a pendulum, energy is transferred from gravitational

potential (i.e. the height of the pendulum mass) to kinetic and back again. In an optical

cavity the transfer is between the electric and magnetic fields inside the cavity.

When processing electronic signals in analogue form, we often need to use a filter toselect (or reject) a specific band of frequencies. One of the easiest ways of making a

filter for this sort of task is to combine a resistance, capacitance, and inductance.

Diagram 4 shows a typical arrangement. You should assemble this circuit for

measurement in this experiment. Use a 1,000 pF capacitor forC, a 22 mF inductor

forL, and a 10 resistor forR. Use the signal generator to provide the input

signal, .

As with the earlier experiments, your circuit should be laid out in a

similar way to the circuit shown in the photographs which you can seeby clicking on the image of a camera.

You should use the generator output which is typically labelled 50 or 600 .


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In this case, the lower wire of the circuit has an earth symbol attached to remind you

that this is meant to be the earth/zero-volts line. In practice you will connect the line

to earth via the outer leads of the co-axial cables used for the signal generator and

scope. Use both scope leads and channels to observe both and at the

same time.

Although this circuit doesn't look anything like an optical cavity, it is working in a

similar way. The capacitor can store energy in the form of an electric field in between

its plates. The inductor can store energy in the form of a magnetic field around its coil.

If you put some energy into the circuit it will tend to be moved back and forth

between these two components at a frequency which depends upon their values.

Start by applying a large square-wave input signal with a frequency of a few hundred

hertz. You should see the output voltage ringafter the abrupt input voltage changeswhich occur at each square wave edge. This ringing is a damped resonance which

occurs whenever you abruptly try to alter the state of a resonant system. The time

taken to settle down depends upon the amount of damping which, here, depends upon

the resistance,R, in the circuit. Note that the frequency of the ringing doesn't dependupon the input square wave frequency. It is characteristic only of the resonantfrequency of the circuit.

Sketch the output waveform and use the 'scope to estimate the ringing frequency,

, by timing each cycle.


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(Caution: the time-base readings will be only be correct if the scope display is

correctly calibrated. Check to see if there is a time knob or switch setting marked

something calibrate and ensure it is set to the calibrated position before making any

timing measurements.)

Now switch over to using sine waves. You should find that the ratio of

depends upon the sinewave frequency

Find the frequency, , where is a maximum.

Note this frequency.

How does compare with ?

The circuit can be thought of as being in two parts:

Part 1: a resonant circuit made with theL, C, and 22 resistor

Part 2: an input series 100k resistor.

The properties of the resonant circuit can be examined using this arrangement because

the impedance of a resonant circuit is frequency dependent. In effect, you have made a

potential divider using a resistor (the 100k ) whose resistance doesn'tdepend upon

the signal frequency, and a resonant circuit whose impedance does depend upon the

frequency. As a result varies with the input frequency in a way which

reveals the frequency dependent behaviour of the resonant circuit.

Take the data to form a table of and for a range of frequencies,f, from

about to .


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used as band pass filterwhich will let through signal frequencies , but reject

frequencies which are very different to . The band width of the filter i.e. the

width of the frequency range passed by the filter depends upon its Q.

In principle, the quality factor of your resonant circuit can be calculated in two ways

whereR is the dissipation resistance of the resonant circuit, and is the measured

frequency width of the resonant peak (at the points below its peak).

Take the measured and values from your graph and use expression 3 to

calculate a value ofQ.

Compare this with the value you get if you use expression 4 and the values of the

components you are using. You may well find that these results forQ aren't the same!

Part of the reason for this difference is the fact that the inductor also has a resistance,

which you haven't taken into account. The other resistances (the 100k , and the input

resistance of the 'scope) also have some effect even through they look as if they're

outside the resonant circuit. However, the main problem is one called the skin

effect. This makes a.c. signals tend to prefer to flow in the outer skin of a

conductor. The higher the frequency, the thinner the skin the current is confined to. In

effect, for an a.c. signal you could remove the metal inside the wire just leaving a

hollow tube of metal. As a result the wire behaves as if it is becoming thinner (and

hence more resistive) as you increase the frequency. This means that the behaviour of

an inductor which contains a long wire thin wire wound into a coil can be verydifferent to a plain inductance.

Many textbooks will leave you with the impression that you can calculate Q just from

knowingL andR. The above comparison should serve as a warning that the actualvalue of the dissipation resistance of a circuit isn't always obvious. This is because the

resistance actually experienced by the a.c. signals may not have the value you expect.


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In practice, it is better to discover the Q by measuring and and then calculating

.

ii) The peak impedance.

At any particular frequency,f, the resonant circuit will have an impedance which we

can call . This reaches its maximum value, , at the resonant frequency.

As your circuit is a sort of potential divider you can expect that

where is the input 100k series resistor.

Use your (un-normalised) measurements to calculate a value for at the

resonant frequency, .

Note for those who know something about a.c. circuit theory. When a circuit contains

inductors or capacitors its impedance, , is generally complex. This means that thealternating currents and voltages in it don't always share the same phase. When using the

'scope to measure and you may have noticed their relative phases as well as

sizes changing when you altered the signal frequency. This means that, strictly speaking, in

the above equation , , and should all be considered as complex numbers. Atresonance, however, the impedance of a circuit always becomes real i.e. purely resistive

so we don't need to worry about this complication.

Experiment 4 Characteristics of a Silicon Transistor.

Given that a typical home computer contains around a hundred million transistors (or

more!) and ordinary things like TV's and radios can contain hundreds it's likely that

there are many more transistors on the Earth than people! It's probably a good idea to

understand them...


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There are all sorts, shapes, and sizes of transistor. In this lab we will only consider one

basic type, the bipolartransistor. This comes in two flavours calledPNPandNPN.For the following experiments you should use the BC184L NPN transistors which are

available.

When a theoretician presents a series of lectures about bipolar transistors he or she can

usually make them sound very complex! The good news is that in practice you usually

only have to know a few of the many properties of a transistor. All the other details

only become necessary in that one time in a hundred when you build an unusual

circuit. The basic properties of a BC184L are:

Maximum allowed power dissipation,P= 350 mW

Max. allowed collector current, = 100 mA

Max. allowed collector-emitter voltage, = 30 V

Typical current gain, = 250 to 800

In practice, the transistor has many more properties. Worse still, many of them vary

from transistor to transistor, and may change with temperature, the applied voltages,

etc. Fortunately, we can often ignore these complications!

The BC184L is built into a standard TO-92 package with three leads. The diagrambelow shows what the package looks like and identifies the leads where B = Base, C =

Collector, and E = Emitter.


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Connect up the circuit shown in diagram 5 and use it for the following experiment.

For this experiment, just put the transistor on the circuit board and use the resistors as

part of the leads as shown in the photograph. Once this experiment is complete, you

will use the same transistor and board and add new components to make an amplifier.

As with previous experiments there are some photos to show you what

your circuit should look like. Click on the image of a camera to see the


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photos.

Electronic engineers often adopt the convention that upper case letters, like or

, are used to signify steady or d.c. values and lower case ones, like and ,

are used to represent small changes or a.c. quantities. This convention will be used forthe following explanations.

e.g. signifies the DC voltage as measured between the base and the emitter of the

transistor, whereas signifies the AC voltage fluctuations between the collector

and the emitter.

Note. When you have finished all these measurements keep your transistor on its

board. You will need it for the next section!

Use your 'scope to measure and . Use the Avometer and DVM (Digital

Volt Meter) to measure the currents, and .

Adjust the 1M pot to set the base current, , to 2 A. Setting of the 2.5k

potentiometer to 5 Volts. Make a note of the values of and . Use the 1M

pot to increase in 2 A steps, each time using the 25k pot to set back to

five volts and then noting the new values of and . Stop when you either can't

make equal 5 volts or when mA.

Reduce back down to 2 A and repeat the process but with set to 10

volts.

Plot two graphs of your results. One showing how varies with for both

choices of collector-emitter voltage. The other showing how varies with

for both collector-emitter voltages.

You should find that the V and V curves are fairly similar on

each graph.


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Experiment 5 The Transistor Amplifier.

Transistors are used in a great variety of circuits. Fortunately, we can divide the waysin which they are used into two fairly simple classes: amplifiers and switches.

Transistors switches form the basis of all modern electronic digital computers. Thisparticular lab doesn't deal with digital electronics. Here we will look at an example of

using a bipolar transistor in an amplifier.

Figure 6 illustrates a typical single-transistor amplifier circuit. This arrangement is

often called the common emitteramplifier because the input voltage to the transistor

appears between the base & emitter, and the output voltage appears between the

collector & emitter i.e. the emitter terminal is shared by (or common to) the inputand output.

Note. , , and are the voltages between each of the transistor base,

collector, and emitter terminals and the ground (zero volts). They aren't the same

thing as or which are the voltages from base-to-emitter and collector-to-


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emitter! The diagram also shows the input and output signal AC voltages, and

. These aren't equal to and because the 01 F capacitors block any d.c.

connection between these potentials. (If you're puzzled by all this, ask a

demonstrator.)

In order to build a working amplifier you have to choose suitable values for resistors,

, , , and . For now, assume that (i.e. it is a piece of wire). We

will want to choose a value for later, but for now well worry about everything

else.

Anyone who has been confused by reading an electronics textbook will suspect that

choosing the right values for the resistors is quite complicated. However, it is

possible to select satisfactory values using some simple rules. It is worth bearing in

mind again that electronics is a practical subject which shares some things withcookery! (Transistors can get hot, too...) In particular, there are situations (and this is

one) where there isn't always a single correct solution for the resistor values you

need. It is possible to make a working amplifier using a wide range of resistor values.

For a theorist or mathematician this can be depressing there isn't one right

answer. For the rest of us it's good news as it means there are a wide range of values

which are OK. It also means that some simple approximations aren't likely to lead to

serious problems.

Experience with bipolar transistors has taught engineers that 9 times out of 10 a

good start is to make just three assumptions and use them as rules unless we knowbetter:

1. The base-emitter voltage will always be about 06 Volts (or 06 for a PNP

transistor).

2. The current gain (the value) will be a few hundred.

3. The large value means that , so we can assume that

If you look at your transistor's characteristic curves you should see that, although

does depend upon , over most of the measured range it is around 06 Volts or

so. The of your transistor will probably be somewhere in the 200 600 range. So

these approximations are a moderately good place to start in the absence of any better


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information.

The resistors in the amplifier circuit will determine the steady bias voltages and

currents, , , etc. The capacitors in the circuit are used to control the effects of

a.c. signals. Start off by ignoring the capacitors as they don't affect the way the actual

transistor operates. We can therefore work out all the resistor values, etc, without

bothering about them.

There are various ways to decide what values to choose for the bias resistors. They all

give roughly similar results, and the following simple argument is about as good as

any other.

For the circuit to work as an amplifier we need to make the collector voltage, ,

move up and down in response to any input signal variations. These changes in

collector voltage are coupled out through the capacitor to provide the output voltage

signals, . This means that in the absence of any input signal the transistor

should have a moderate set of applied bias voltages/currents to give room to

move up and down under the influence of any input.

The circuit is driven by a +15V power line and the collector-emitter voltage is applied

via the two series resistors, & . In the absence of any good reason for making

some other choice we might just as well assume that the available voltage should be

shared equally between , , and the transistor. We therefore want about 5 volts

across , 5 volts across , and 5 volts between the collector and emitter. This

means that the amplifier should have, V, V, and V.

The Transistor Amplifier

Choosing Component Values.


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From rule 1 we can nowsay that we want the base

voltage, , to be around

56 volts. The two base

resistors act as a sort of

potential divider and we

can choose their values to

set the voltage we require.

To do this we need to useOhms Law and recognise

that the current through

provides the base

current and the current which goes on through .

From rule 3 we can also say that we require since thecurrents in these resistors will be almost exactly the same and we

want to have 5 volts across each of them (Ohm's Law).

In the previous section you measured your transistor's value at a

particular point on its curves ( mA, V). So let's

choose to try and set the amplifier up with a collector current of

about 2mA. We therefore want the currents passing through andto be 2mA. You now know the current in each of these resistors

and the voltage across each of them. Using Ohm's Law, what values

do you calculate are required for and ? What is the closest

E12 series value available in the lab? Use this value for the emitter

and collector resistor in your circuit.


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Part of the current flowing through will continue on through

and part will enter the transistor to provide its base current. .

Using Ohm's Law again we can say that

where we know that, for to be 5 volts, we want volts, so

we can say that

This gives us two equations but we have three unknowns, , ,

and . To proceed any further we have to choose a sensible value

for one of these.

The best way to proceed is to choose a value for the current, ,

which passes through both resistors. In theory, we can choose any

value we like. However, in practice it turns out to be a good idea to

choose a value since this means that the voltages across the

resistors are largely determined by . This means that any slight

changes in won't mean we've got the wrong results. However, we

don't want to be too big. The reason for this is that we would get a

large current by using very low resistance values. These would

make it difficult to apply an input ac voltage when using the

amplifier.

In practice the simplest convenient choice is to pick something like

so I suggest you choose that. Note, however, that you

could choose almost anything from up to and it

would still probably be possible to make the amplifier work despite

having chosen very different currents and resistor values!


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Note. Here I will assume you found that (rule 2). You can

follow the argument I describe below, but substitute the you

measured to get the correct results for your transistor.

A current gain of 400 means that at mA the current entering

the transistor's base is A. Multiplying this by 25

we get 125 A. Putting this into the above equations we get

k and k . What values do you get for your

transistor? What are the closest E12 series values available to use inyour circuit?

You should now have values for , , , and . However, we

now need to decide what to do with ...

is actually quite important as it turns out to control the voltage

gain of the amplifier. To understand why this is true, have another

look at figure 6 and consider what happens when we quickly waggle

the input voltage up an down with an ac signal. In order to change

the voltage across we also have to change the voltage across

as they are connected in parallel. To change the voltage across

we have to move charge in or out of the capacitor. This takes time.

So if we keep changing our mind and waggling the input voltage up

and down quickly we dont give this a chance to happen.

As a result, for quick variations effectively clamps the voltage

at the top of and wont allow it to change. The transistors base-

emitter voltage remains about 06V. Hence the changes in input

voltage mostly appear as changes in the voltage across .


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An input ac voltage, , therefore tends to produce an ac current

variation in of

Since is relatively tiny (hundreds of times smaller than or )

we now expect the same current fluctuation to appear in . So the

voltage across the collector resistor will vary by an amount

So it is the ratio of these two resistors that tends to control the

voltage amplification factor (gain) of the circuit.

Now, provided we choose a value for which reasonably small

compared to , we can leave the other resistor values alone and

not worry that we have changed the DC levels very much. A small

value will also mean a high gain.

What value of will give your amplifier a voltage gain of

around 20?

Choose the nearest E12 resistor value for your circuit. What

value is this, and what value of gain do you it expect it to

provide?

The Transistor Amplifier

Measuring your amplifier


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The capacitors in the circuit affect how it responds to a.c. signals.

The input and output capacitors, and act as d.c. blocks. They

tend to pass through any voltage fluctuations, but stop any external

connections (to signal generators, 'scopes, etc) from affecting the

d.c. levels in the circuit. Without them, the circuit would be liable to

stop working as soon as you connect anything to it!

The 22 F, , is a shunt capacitor. In principle, we could omit ,

but if we did the amplifier voltage gain would be low. Can you

explain why this would be the case?

Build your amplifier using the values you have been given for

the capacitors and the values you have worked out for the

resistors. Build the circuit using the transistor and board from

the previous experiment.

To see photographs of what your circuit should look like,

click on the image of a camera. Make your circuit assimilar as you can to the amplifier shown in the

photographs, but remember that your resistor values

may be different!

As with earlier experiments, you will need to hand in this circuit

when you are finished to get the experiment marked, so make sure

you also put your name on it.

Switch the amplifier on (i.e. connect the power supply +15V & 0V

lines and turn on the power!) and use the DVM to measure , ,

and . You should find that volts and volts. If

they're more than a volt or so away from these values, check you've

built the circuit correctly. If not sure, ask a demonstrator.


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Make a note on Diagram 6 of the voltages you measure and

indicate the resistance values in your circuit.

You can now measure the a.c. properties of your circuit. Use your

'scope to observe the input and output voltages, & . Connect

the signal generator to provide an input sinewave signal.

The voltage gain, G, of the circuit can be defined as the ratio

. Plot a graph showing how the gain varies with sinewave

frequency. Plot a couple of dozen values over the range from 10Hzup to 50 kHz.

Note. The amplifier gain only means something when the amplifier

behaves in a fairly linear manner. If the amplifier's operating point is

very wrong or if you use too large an input the amplifier will

visibly distort the signal. Watch out for this on the 'scope trace. The

input should look like a good sinewave. The output should also look

like a sinewave. If the output is visibly flattened or clipped then

the amplifier is distorting the signal. Reduce the input level until theoutput looks OK.

You should find that the gain is quite frequency dependent, so the

size of input you can use without distortion will also depend upon

signal frequency.

Why is the gain frequency dependent? In particular, why does

the gain fall away at low frequencies? What could be done to

improve this?


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You should also find that the amplifier tends to invert the signal

i.e. the output appears 'upside-down'. Why is this?

(If you don't know the answers to these questions, ask a

demonstrator.)

Experiment 6 The Op Amp & IC Amplifier

Although transistor amplifiers made with discrete components (i.e. individually

packaged) are still used for some special purposes like high-quality Hi-Fi, most

modern signal processing systems use Integrated Circuits (ICs). The one of the oldest,

most commonly used and cheapest! IC Operational Amplifiers is the SN741. Thisexperiment uses a 741 as a simple audio-frequency amplifier.

741 Op Amps come in a variety of packages. One of the most common is an 8-pin

Dual-In-Line (DIL) or Dual In-line Plastic (DIP) package of the kind shown below


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The 741 has two signal inputs called inverting and non-inverting. It also must be

powered using two voltage lines that provide 15V.

For this experiment, build the circuit shown in figure 7. As with earlier

circuits, make your circuit look similar to the one in the photographs.

Click on the picture of a camera if you want to see the photos.Remember to label your circuit and hand it in with your results. You

should be able to work out which pin to connect to what by comparing this diagram

with those for the 741s package and the wires shown in the photos. If not sure, ask a

demonstrator.


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The circuit shown in diagram 7 can be used as either an inverting or a non-

inverting voltage amplifier depending on how you apply an input signal. This is

because the Op Amp has the property that its output depends on the difference in thevoltages applied to thepairof pins, 2 & 3. First, use it as an inverting amplifier by

connecting it as shown below.

The earth symbol shows where we connect 0V (earth) from the power supply. We

also connect the earth leads (outer wires of the co-axial cables) to this point. The live

input lead is connected to the inverting input resistor (shown as A in figure 7).


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You should find that both the sign and the value of the gain of the two types of

amplifier differ. Say why you think this is the case. (If unsure, ask a

demonstrator.)

Say what change you would make to the circuit you have built if you wanted to

increase the voltage gain of the inverting amplifier to .

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