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Basic Electronics
Two main subdivisions:
Digital Electronics .. deals with "ones and noughts" (typically "nought" is 0-1 V
and "one" is 4-5 V, but low-power equipment is often based on "one" being about
3-3.5 V).
Analogue Electronics .. deals with voltages and currents having (in principle) any
values.
Electronics differs from the Electrical work we have done so far in that, in
Electronics, we are dealing with information rather than power. Even when we use
a powerful amplifier to produce a deafeningly loud noise from a loudspeaker, we
do generally want to hear the music and not just a noise !
Analogue Electronics
The main things we may want to do to a current or voltage signal are:
Generate it .. various types of signal generator circuits.
Amplify it .. to produce a larger but similar version of the signal.
Produce a voltage or current whose value depends on that of a non-electrical
quantity for measurement purposes .. with a Transducer.
Use it to modulate a high-frequency signal in order to be able to transmit our signal
by radio.
Demodulate the received signal.
Filter out unwanted frequencies.
We will investigate Amplification and see how we can make a circuit which will
amplify a voltage input. We are handicapped in this respect by the fact that it is
easy to make circuits which can make the signal much larger but more difficult to
make circuits which will make the signal an accurately-defined number of times
larger, so an audio amplifier designed to produce a pleasant level of sound from
the loudspeaker might either be inaudible or deafening !
The main general-purpose amplifying device is called an Operational Amplifier and
this is its symbol.
v+
+
v-
-
Output voltage = Av(v+ - v-)
The quantity Av is called the Voltage Gain and it is very large (anything from
100000 up to several million) for a typical op-amp. The problem is that we do not
know where it is in the range ! But, if we are content with a voltage gain of up to a
few hundred times, we can use one of the following circuits to produce such an
amplifier.
+
Input voltage
vin = v+
Output voltage
vo
R2
R1
It is easy to calculate the voltage gain of this circuit because of two things.
1. The operational amplifier has hardly any current flowing in or out of its inputs, so
the current in R1 is for practical purposes equal to that in R2.
2. So R1 and R2 are effectively in the following circuit:
R2
v- = ?
vo
R1
As the currents in R2 and R1 are the same, the resistors are in series and the total
resistance is therefore R1 + R2. The current can therefore be worked out by Ohm's
Law .. it is vo/(R1 + R2). (If you don't believe this, what will the current be if vo is
6 V, R2 is 1 k and R1 is 2 k ?)
We can now work out the voltage at the top of R1 as R1 times the current in R1
(whose law is this ?). It gives us R1vo/(R1 + R2), and we know that the output is
Av(v+ - v-) = Av(vin -R1vo/(R1 + R2)) = vo. If we divide everything on both sides of the
equation by Av, we get
vin -R1vo/(R1 + R2) = vo/Av
and, as Av is so large, we can neglect vo/Av, as it is so small.
So vin = R1vo/(R1 + R2) and vo/vin = the voltage gain = (R1 + R2)/R1 = 1 + R1/R2.
So we can make a circuit using the operational amplifier which has a voltage gain
which we can set accurately using appropriate values for R1 and R2.
We can also make an amplifier (known as an Inverting Amplifier) which produces a
negative output voltage for a positive input voltage (and vice versa). This is the
circuit:
R2
R1
vin
+
vo
This time we note that v+ is zero as its terminal is connected to earth, and, as vo is
Av(v+ - v-) and Av is very large, must be very small. So v- is approximately 0 V (we
often call it a "virtual earth") and we also remember that the current into the opamp input is negligibly small.
So the current in R1 is (vin - 0)/R1 (to the right) and that in R2 is (0 - vo)/R2, also to
the right. None enters or leaves the "-" input, so (vin - 0)/R1 = (0 - vo)/R2 and the
voltage gain vo/vin is -R2/R1.
(Can we extend the idea to subtracting one voltage from another ?)
When we make such circuits, we design them to use "preferred" resistor values.
The values go as follows:
1 1.1 1.2 1.3 1.5 1.6 1.8 2 2.2 2.4 2.7 3.0 3.3 3.6 3.9 4.3 4.7 5.1 5.6 6.2 6.8 7.5 8.2
9.1 10 ... times an appropriate power of 10 if required, so for example all of 3.9 ,
39 , 390 , 3.9 k, 39 k, 390 k etc. are available.
We normally try to use values between about 1 k and about 100 k where
possible.
Variable resistors (also called potentiometers) are also available.
Where does the power come from ?
Simple (when we know how) ... all operational amplifiers have d.c. power supplies
(often + 10 V and -10 V, but some manage with one supply voltage) from which
the power to make the amplifier work comes.
"Op-amps" can also be used as the basis of circuits to do most of the other tasks
listed at the beginning of the Analogue section.
A good read: Horowitz and Hill, "The Art of Electronics" (Government Health
Warning that it contains sums but it focusses on how to produce good circuits
which work well).
A few examples
1. If the value of the voltage gain Av for a particular operational amplifier is
100000, what will the difference in its input voltages be if its output voltage is 5 V ?
(50 microvolt).
2. If we used this op-amp to make an "inverting" amplifier with a voltage gain of 10,
select suitable resistor values. If we apply an input voltage vin of 1 V, what will the
voltage at the inverting input (v-), which the notes have argued to be zero, actually
be ?
(Any in a ratio of 1:10, so for example R1 = 1 k, R2 = 10 k. We note that
vo = -10 V, and v+ = 0, so v- = (-10)/(-100000) = 10-4 V = 0.1 mV.)
3. Choose resistor values to produce:
a) An inverting amplifier with a voltage gain of 37.5.
b) A non-inverting amplifier with a voltage gain of 3.
c) A non-inverting amplifier with a voltage gain of 101.
(Possibilities: a) 2 k and 75 k; b) 1 k and 2 k; c) 1 k and 100 k ... it is the ratio of the
two which is important)