What is a Semiconductor

Semiconductors
This section deals with discrete semiconductors.   A later section deals with ‘Integrated Circuits’ which are large-scale semiconductor devices.

The ORP12 Light-dependent resistor.   This device has a high resistance in the dark and a low resistance in bright light.   It can be placed in a circuit to create a switch which operates with an increase in light level or a decrease in light level:

In this version, the voltage at point ‘A’ controls the circuit.  In darkness, the ORP12 has a resistance ten times greater than that of R1 which is 12,000 ohms.   Consequently, the voltage at point ‘A’ will be high.   As the light level increases, the resistance of the ORP12 falls, dragging the voltage at point ‘A’ downwards.   As the variable resistor ‘VR1’ is connected from point ‘A’ to the ground rail (the -ve of the battery), its slider can be moved to select any voltage between 0 Volts and the voltage of ‘A’.   A slider point can be chosen to make the transistor switch off in daylight and on at night.   To make the circuit trigger when the light level increases, just swap the positions of R1 and the ORP12.

The transistor shown is a BC109 although most transistors will work in this circuit.   The BC109 is a cheap, silicon, NPN transistor.   It can handle 100mA and 30V and can switch on and off more than a million times per second.   It has three connections: the Collector, marked ‘c’ in the diagram, the Base, marked ‘b’ in the diagram and the Emitter, marked ‘e’ in the diagram.

As mentioned before, it has a very high resistance between the collector and the emitter when no current flows into the base.   If a small current is fed into the base, the collector/emitter resistance drops to a very low value.   The collector current divided by the base current is called the ‘gain’ of the transistor and is often called ‘hfe’.   A transistor such as a BC109 or a BC108 has a gain of about 200, though this varies from actual transistor to actual transistor.   A gain of 200 means that a current of 200mA passing through the collector requires a current of 1mA through the base to sustain it.   Specific information on the characteristics and connections of semiconductors of all kinds can be obtained free from the excellent website which provides .pdf information files.

The BC109 transistor shown above is an NPN type.   This is indicated by the arrow of the symbol pointing outwards.   You can also tell by the collector pointing to the positive rail.   There are similar silicon transistors constructed as PNP devices.   These have the arrow in the transistor symbol pointing inwards and their collectors get connected, directly or indirectly, to the negative rail.   This family of transistors are the earliest transistor designs and are called ‘bi-polar’ transistors.

These silicon transistors are so efficiently constructed that they can be connected directly together to give greatly increased gain.   This arrangement is called a ‘Darlington pair’.   If each transistor has a gain of 200, then the pair give a gain of 200 x 200 = 40,000.   This has the effect that a very, very small current can be used to power a load.   The following diagram shows a Darlington pair used in a water-level detector.   This type of alarm could be very useful if you are asleep on a boat which starts taking on water.

Here, (when the circuit is switched on), transistor TR1 has so little leakage current that TR2 is starved of base current and is hard off, giving it a high resistance across its collector/emitter junction.   This starves the buzzer of voltage and keeps it powered off.   The sensor is just two probes fixed in place above the acceptable water level.   If the water level rises, the probes get connected via the water.   Pure water has a high electrical resistance but this circuit will still work with pure water.

The odds are that in a practical situation, the water will not be particularly clean.   The resistor R1 is included to limit the base current of TR1 should the sensor probes be short-circuited.   Silicon bi-polar transistors have a base/emitter voltage of about 0.7V when fully switched on.   The Darlington pair will have about 1.4V between the base of TR1 and the emitter of TR2, so if the sensor probes are short-circuited together, resistor R1 will have 6 – 1.4 = 4.6V across it.   Ohms Law gives us the current through it as R = V / A or 47,000 = 4.6 / A or A = 4.6 / 47,000 amps.   This works out at 0.098mA which with a transistor gain of 40,000 would allow up to 3.9A through the buzzer.   As the buzzer takes only 30mA or so, it limits the current passing through it, and TR2 can be considered to be switched hard on with the whole battery voltage across it.

NPN transistors are more common than PNP types but there is almost no practical difference between them.   Here is the previous circuit using PNP transistors:


Not a lot of difference.   Most of the circuit diagrams shown here use NPN types but not only are these not critical, but there are several ways to design any particular circuit.   In general, the semiconductors shown in any circuit are seldom critical.   If you can determine the characteristics of any semiconductor shown, any reasonably similar device can generally be substituted, especially if you have a general understanding of how the circuit works.   Either of the two previous circuits can operate as a rain detector.   A suitable sensor can easily be made from a piece of strip board with alternate strips connected together to form an interlacing grid:

Here, if a raindrop bridges between any two adjacent strips, the circuit will trigger and sound a warning.

The transistors in the circuit above are connected with their emitter(s) connected to the ground rail (the lower battery line shown in any circuit is considered to be “ground” unless it is specifically shown elsewhere).   This connection method is called ‘common emitter’.   The following circuit uses the transistor connected in ‘emitter follower’ mode.   This is where the emitter is left to follow the base voltage – it is always 0.7V below it unless the base itself is driven below 0.7V:

This is almost the same as the light-operated circuit shown earlier.   In this variation, the transistors are wired so that they work as an ‘emitter-follower’ which follows the voltage at point ‘A’ which rises as the light level drops and the resistance of the ORP12 increases.   This causes the voltage across the relay to increase until the relay operates and closes its contacts.   A relay is a voltage-operated mechanical switch which will be described in more detail later on.

The disadvantage of the above circuit is that as the light level decreases, the current through the relay increases and it may be a significant amount of current for some considerable time.   If it was intended to power the unit with a battery then the battery life would be far shorter than it need be.   What we would like, is a circuit which switched rapidly from the Off state to the On state even though the triggering input varied only slowly.   There are several ways to achieve this, one of them being to modify the circuit to become a ‘Schmitt Trigger’:

Here, an additional transistor (‘TR2’) has changed the circuit operation significantly, with transistor TR3 switching fully on and fully off, rapidly.   This results in the current through the relay being very low until the circuit triggers.

The circuit operates as follows.   When the voltage at the base of TR1 is high enough, TR1 switches on, which causes the resistance between its collector and emitter to be so low that we can treat it as a short circuit (which is a nearly-zero resistance connection).   This effectively connects the 10K and 1K8 resistors in series across the battery.   The voltage at their connecting point (both the collector and emitter of TR1) will then be about 1.8 Volts.   The two 18K resistors are in series across that voltage so the voltage at their junction will be half that; 0.9 Volts.

This puts the Base of TR2 at about 0.9 Volts and its emitter at 1.8 Volts.   The base of TR2 is therefore not 0.7 Volts above its emitter, so no base/emitter current will flow in TR2, which means that TR2 is switched hard off.   This means that the TR2 collector/emitter resistance will be very high.   The voltage at the base of TR3 is controlled by the 1K8 resistor, the TR2 collector/emitter resistance (very high) and the 3K9 resistor.   This pushes the base voltage of TR3 up to near the full battery voltage and as it is wired as an emitter-follower, its emitter voltage will be about 0.7 Volts below that.   This means that the relay will have most of the battery voltage across it and so will switch hard on.

Some practical points:   The current flowing into the base of TR3 comes via the 3K9 resistor.   A 3K9 resistor needs 3.9 Volts across it for every 1 mA which flows through it.   If the relay needs 150 mA to operate and TR3 has a gain of 300, then TR3 will need a base current of 0.5 mA to provide 150 mA of current through its collector/emitter junction.   If 0.5 mA flows through the 3K9 resistor, there will be a voltage drop across it of some 2 Volts.   The TR3 base/emitter voltage will be a further 0.7 Volts, so the voltage across the relay will be about 12.0 – 2.0 – 0.7 = 9.3 Volts, so you need to be sure that the relay will work reliably at 9 Volts.

If you used a Darlington pair of transistors, each with a gain of 300, instead of TR3, then their combined base/emitter voltage drop would be 1.4 Volts, but they would only need a base current of 150 mA / (300 x 300) = 1/600 mA.   That current would only drop 0.007 Volts across the 3K9 resistor, so the relay would receive 10.6 Volts.

So, how do you work out the gain of any particular transistor?   The main working tool for electronics is a multimeter.   This is a digital or analogue meter which can measure a wide range of things: voltage, current, resistance, …   The more expensive the meter, generally, the greater the number of ranges provided.   The more expensive meters offer transistor testing.   Personally, I prefer the older, passive multimeters.   These are looked down on because they draw current from the circuit to which they are attached, but, because they do, they give reliable readings all the time.   The more modern battery-operated digital multimeters will happily give incorrect readings as their battery runs down.   I wasted two whole days, testing rechargeable batteries which appeared to be giving impossible performances.   Eventually, I discovered that it was a failing multimeter battery which was causing false multimeter readings.

For the moment, let us assume that no commercial transistor tester is to hand and we will build our own (or at least, discover how to build our own).   The gain of a transistor is defined as the collector/emitter current divided by the base/emitter current.   For example, if 1mA is flowing through the collector and 0.01mA is flowing into the base to sustain that collector flow, then the transistor has a gain of 100 times at 1mA.   The transistor gain may vary when it is carrying different current loads.   For the circuits we have been looking at so far, 1mA is a reasonable current at which to measure the transistor gain.   So let’s build a circuit to measure the gain:

With the circuit shown here, the variable resistor is adjusted until a collector current of 1mA is shown on the milliammeter and the gain of the transistor is then read off the scale on the variable resistor knob.   The circuit is built into a small box containing the battery and with a socket into which the transistor can be plugged.   The question then is, what values should be chosen for the resistor R1 and the variable resistor VR1?

Well, we might choose that the minimum gain to be displayed is 10.   This would correspond to where the variable resistor slider is taken all the way up to point ‘A’ in the circuit diagram, effectively taking the variable resistor out of the circuit.   If the transistor gain is 10 and the collector current is 1mA, then the base current will be 0.1mA.   This current has to flow through the resistor R1 and it has a voltage of (9.0 – 0.7) Volts across it as the base/emitter voltage is 0.7 Volts when the transistor is on.   Ohms Law gives us Ohms = Volts / Amps, which for the resistor R1 means Ohms = 8.3 / 0.0001 or 83,000 ohms, or 83K.

Rule of thumb: 1K provides 1mA if it has 1V across it, so 10K will give 0.1mA if it has 1 Volt across it.   With 8.3 Volts across it, it needs to be 8.3 times larger to hold the current down to the required 0.1mA so the resistor should be 83K in size.

As 83K is not a standard size, we need to use two or more standard resistors to give that resistance.   Nearest standard size below 83K is 82K, so we can used one 82K resistor and one 1K resistor in series to give the required 83K.

Suppose that we say that we would like to have 500 as the highest gain shown on our tester, then when VR1 is at its maximum value, it and R1 should provide 1/500 of the collector current of 1mA, i.e. 0.002mA or 0.000002 Amps.   From Ohms Law again we get VR1 + R1 = 4,150,000 ohms or 4M15.   Unfortunately, the largest value variable resistor available is 2M2 so the circuit as it stands, will not be able to cope.

Suppose we were to just use a 2M2 variable resistor for VR1, what transistor gain range could we display?   Well Ohms Law … lets us calculate the base current with 8.3 Volts across (83,000 + 2,200,000) ohms and from that the maximum transistor gain which would be 277.77 (at 1mA).   You would buy a ‘linear’ standard carbon track variable resistor so that the change in resistance is steady as the shaft is rotated.   The scale which you would make up would be in even steps and it would run from 10 at the minimum setting, to 278 at the highest setting.

But that is not what we wanted.   We wanted to measure up to 500.   But they don’t make variable resistors big enough, so what can we do?   Well, if we wanted, we could lower the battery voltage, which in turn would lower the resistor values.   As a 9V battery is very convenient for this kind of circuit, lets not go down that route.   We could add extra circuitry to drop the 9V battery voltage down to a lower value.   The most simple solution is to add an extra resistor and switch to give two ranges.   If we switched in an extra 2M2 resistor above VR1 then the circuit would measure transistor gains from 278 to just over 500 and all we would need to do would be to add a second scale for the VR1 pointer knob to move over.   We could, provide extra ranges which overlap and which have more convenient scales to mark.   The design is up to you.

The design covered above is not the only way to measure the transistor gain.   A second way, which accepts that it is not so accurate, picks a set base current and measures the collector current as a guide to the gain.   In this simple method, one or more resistor values are chosen to give gain ranges, and the milliammeter used to read the corresponding gain:

Here, resistor R1 might be chosen to give a collector current of 1mA (which is a full-scale deflection on the meter) when the transistor gain is 100.   Resistor R2 might be picked to give a full-scale deflection for a gain of 200, R3 for a gain of 400, R4 for a gain of 600, and so on.   Generally speaking, it is not essential to know the exact gain but any reasonable approximation to it is sufficient.   You are normally selecting a transistor where you need a gain of 180, so it is not important if the transistor you pick has a gain of 210 or 215 – you are only avoiding transistors with gains below 180.

How do you work out the values of the resistors R1 to R4?   Well, you probably won’t expect this, but you use Ohms Law.   Voltage drop is 8.3 Volts and the base current is given by the full-scale deflection’s 1mA divided by the transistor gain for each range, i.e. 1/100 mA for R1, 1/200 mA for R2,… 1/600 mA for R4,…

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