TRANSISTOR PINOUTS:

Transistor Pinouts

Just some of the pinouts for a transistor. You need to refer to a data sheet or test the device to determine the pins as there are NO standard pin-outs.

A "STAGE" A "Stage" is a set of components with a capacitor at the input and a capacitor on the output. We have already seen the fact that the capacitor only has an effect on the circuit during the time when it gets charged. It also has an effect when it gets discharged. But when the voltage on either lead does not rise or fall, NO CURRENT flows through the capacitor. When a capacitor is placed between two stages, it gradually charges. When it is charged, the voltage on one stage does not affect the voltage on the next stage. That's why the capacitor is drawn as two lines with a gap. A capacitor is like putting a magnet on one side of a door and a metal sheet on the other. Moving the magnet up and down will move the metal up and down but the two items never touch. Only a rising and falling voltage is able to pass through the capacitor.

Fig 10.  This is a STAGE. A transistor, with a capacitor on the input and output.

Fig 10 has a capacitor on the input and output. This means the stage is separated from anything before it and anything after it as far as the DC voltages are concerned and the transistor will produce its own operating point via the base resistor and LOAD resistor. We have already explained that the value of the two resistors should be chosen so the voltage on the collector should be half-rail voltage and this is called the "idle" or "standing" or "quiescent" conditions. It is the condition when no signal is being processed. When the voltage on the collector is mid-rail, the transistor can be turned off a small amount and turned on a small amount and the voltage on the collector will fall and rise. (note the FALL and RISE).

Fig 11. The Input and output waveforms

Fig 11  shows a small waveform on the input and a large waveform on the output. The increase in size is due to the amplification of the transistor. A stage like this will have an amplification of about 70. This is called "Stage Gain" or "Amplification factor" and consists of two things. The output voltage will be higher than the input voltage and the output current will be higher than the input current. We will discuss the increase in current in a moment. We will firstly cover the voltage increase.

Fig 11a.  Fixed Base Bias Fig 11b.  Fixed Base Bias

Fig 11a and 11b shows a Common Emitter stage with fixed base-bias. This stage produces the maximum voltage amplification but it is very difficult to "set-up" because the value of the base resistor will either make the collector voltage nearly zero or full rail voltage. It is very difficult to get the collector to sit at mid rail. If the base resistor is a high value, the collector will sit at rail voltage.  If the base resistor is a low value, the collector will sit a 0v. If a transistor with a different gain is fitted, the collector voltage will change completely. If it sits at mid-rail, the noise produced by the transistor will make the collector voltage rise and fall and produce a lot of noise. It all revolves around the actual gain of the transistor and this requires a TRANSISTOR TESTER to determine the gain. However, this circuit can be used as an output stage and has some advantages. It is a "Class-C" stage and means it is just at the point of being turned on via the base-bias resistor. It consumes the least current when "sitting around" and is the most efficient stage. Energy from a previous stage provides base current via the coupling capacitor and the base-bias resistor assists too. The output waveform will be distorted at the top or bottom, depending on the biasing and an inductor in the collector can reduce the distortion. See the article on FM Bugs (SPY BUGS) for a Class-C output stage.

Fig 12.

Fig 12  shows the signal (the voltage waveform) as it passes through 2 stages. Note the loss in amplitude as the signal passes through capacitor C2.

CONNECTING 2 STAGES There are 3 ways to connect two stages: 1. direct coupling  - also called DC coupling (not the coupling shown in fig 12.      Fig 12 is AC coupling). DC stands for Direct Current. I know this sounds unusual, but it is the way to explain the circuit will pass (amplify) DC voltages. This type of coupling will pass both AC signals and DC voltages. When the DC voltage moves up and down (even at a slow rate) we call it an AC voltage or AC signal or a rising and falling voltage and when it rises and falls faster, we call it a "signal" or waveform. 2. via a capacitor - this is also called RC coupling (Resistor-Capacitor coupling) - only passes AC signals - fluctuating signals - rising and falling signals. 3. via a transformer - called Transformer Coupling or Impedance Coupling or Impedance Matching - only passes AC signals. Fig 12 shows two stages with a capacitor coupling the output of the first to the input of the second. This is called Capacitor Coupling or Resistor-Capacitor Coupling (RC Coupling). The increase in the size of the waveform at three points in the circuit is also shown.  The waveform is inverted as it passes through each transistor and this simply means a rising voltage will appear as a falling voltage and after two inversions, the output is in-phase with the input. We have already explained the fact that a capacitor only works once and has to be discharged before it works again. When the first transistor turns off a little, the voltage on the collector rises and the resistor pulls the left lead of C2 UP. The right-hand lead can only rise to 0.7v as the base-emitter voltage does not rise above 0.7v.  This means C2 charges and during its charging, it delivers current to the second transistor. When the first transistor turns ON, the collector voltage drops and C2 passes this voltage-drop to the base of the second transistor. But the transistor does not provide a path to discharge the capacitor fully so that when the capacitor gets charged again, it is already partially charged and it cannot activate the base of the second transistor to the same extent as the first cycle. This means a lot of the energy available at the collector of the first transistor is not delivered to the second stage. That's why capacitors produce losses between stages. They are simply an inefficient way to transfer energy. To make them efficient, they must be discharged fully during the "discharge-part" of the cycle. However enough is delivered to produce a gain in the second stage to get an overall gain of about 70 x 70 for the two stages. The value of C2 will be from 10n to 10u, and the larger capacitance will allow low frequencies to be passed from one stage to the other.

Fig 13.

Fig 13  provides a guide to the values of current that will be flowing at 3 important sections of the circuit. The input current to operate the first transistor will be about 3uA. This is worked out on the basis of the current required to saturate the transistor with a 22k load. The collector-emitter current equals 5/22,000 = 200uA. If the gain of the transistor is 70, the input current is 3uA. The only time when energy passes from the first stage to the second is when transistor turns OFF. The collector voltage rises and the 22k pull the 100n HIGH. The maximum current that can be delivered by the 22k is 5v/22,000= 200uA. This is the absolute maximum for a very small portion of the cycle. However it is important to realise it is not the transistor that passes the current to the next stage but the load resistor. The gain of the second stage is not the deciding factor for the output current but the value of the 2k2 load resistor. This resistor will deliver a maximum of 2,000uA (2mA) and that is how a 3uA requirement at the input of the circuit will deliver 2mA at the output.

You can see it is not the gain of the transistors that produce the output current but the value of the load resistors. The transistors play a part but the limiting factor is the load resistors (and the transfer of energy via the capacitor). This is not always the case but applies in the above circuit.

We will now explain an emitter-follower stage and show how it works.
An EMITTER-FOLLOWER is an NPN transistor with the collector connected to the positive rail. (You can also get PNP EMITTER-FOLLOWER stages - see below). Both can be called a COMMON COLLECTOR stage.

Fig 14.  An Emitter-Follower or Common Collector. The names are the SAME

Fig 14  shows an Emitter-Follower. The load is in the emitter and as the base is taken higher, the emitter follows. But the input and output voltage signals are the SAME amplitude! You would ask: "What is the advantage of this?" Answer: You only need a small amount of "lifting power" to raise the base and the emitter rises with 100 times more strength. The voltage waveform stays the same but the CURRENT waveform increases 100 times. The voltage on the emitter is always 0.7v lower than the base and the base can be as low as 0.8v and as high as 0.5v less than the supply voltage. This gives the possibilities of producing an enormous "swing." In the common-emitter stage the transistor is only active when the base rises from 0.55v to about 0.7v  but in the Emitter-Follower stage it rises from 0.8v to nearly rail voltage. This means the stage does not produce a higher output voltage but it does produce a higher output CURRENT.  We mentioned before the current amplification of a stage was not dependent on the transistor characteristics but the value of the load resistor. In an Emitter-Follower stage we can quite easily get a current gain of 100 or more. Why do we want "Current Gain?"    We need current to drive a low resistance load such as a speaker.

Fig 15. A transistor driving a speaker

Fig 15 shows an 8 ohm speaker as the load in the emitter. If the gain of the transistor is 100, the 8R speaker becomes 8x100 = 800 ohms on the base lead. In other words we see the circuit as "800 ohms."

1. For an emitter-follower circuit, we  know the base can rise and fall by an amount equal to about rail voltage. 2. For a common-emitter stage the collector rises and falls by an amount equal to rail voltage. So, why not connect the two stages together without a capacitor? We know a capacitor has considerable losses in transferring energy from one stage to another and removing it will improve the transfer of energy.

Fig 16. Two directly coupled stages

Fig 16. We now have two stages directly connected together. The first transistor does not deliver energy to the second stage but the LOAD RESISTOR does. The value of the load resistor pulls the base of the second transistor UP and this delivers current to the second transistor and the transistor amplifies this 100 times to drive the speaker.

Fig 17. The load resistor and the effective load of the speaker

Fig 17.   Using mathematics we can work out the effective load of the 8 ohm speaker as 8 x 100 = 800 ohms. To put at least half rail voltage into the speaker, (so the speaker can get the maximum  higher voltage and the maximum lower voltage without distorting) the LOAD resistor has to be the same value as the "emitter follower."  This is a simple voltage-divider calculation where two equal value resistors produce a voltage of 50% at their mid-point. This means the LOAD resistor for the first stage has to be 800 ohms.

Fig 18. The load resistor is 800 ohms

Fig 18 shows the circuit with 800R load resistor in the collector of the first transistor. The final requirement is to select a base-bias resistor for the first stage to produce approx mid-rail voltage on the collector. This is generally done by experimentation.

We mentioned the capacitor separating two stages cannot be discharged fully and thus it does not provide very good transfer of energy from one stage to the other. An improved concept is to directly couple two stages - and remove the coupling capacitor. This is called DIRECT COUPLING or DC coupling and the circuit will process DC voltages (the press of your finger as shown above) and AC voltages (as shown by the sine-wave signal shown above). When a capacitor connects two stages they will only amplify AC signals. There are many ways to directly connect two transistors and we will cover the simplest arrangement. It is an extension of Fig 18 above, because this arrangement has very good characteristics as the two stages transfer 100% of the energy due to the absence of a capacitor.

Fig 19.

Fig 19  shows the previous directly-coupled circuit with a load resistor replacing the speaker. We have already learnt the common-emitter stage provides a voltage gain of about 70 but the emitter-follower stage has a voltage gain of only 1. We can improve this by putting two resistors on the second transistor and changing the stage into a common emitter arrangement.

Fig 20.

Fig 20.  This time we get the advantage of the base being able to move up and down so it matches the collector of the first transistor. It also provides a higher voltage gain by adding a collector resistor and taking the output from the collector. The voltage gain of the second transistor will not be as high as the first stage but we have added the advantage of direct coupling (called DC coupling). The voltage gain of the second stage is the ratio of resistor A divided by resistor B.    If resistor A is 10k and resistor B is 1k, the voltage gain is 10,000/1,000 = 10.

Fig 21.

Fig 21  shows biasing of the first transistor has been taken from the emitter of the second transistor. This does not save any components but introduces a new term: FEEDBACK (actually NEGATIVE FEEDBACK). Negative feedback provides stability to a circuit. Transistors have a very wide range of values (called parameters) such as gain and when two transistors are placed in a circuit, the gain of each transistor can produce an enormous final result when the two values are multiplied together. To control this we can directly couple two transistors and take the output of the second to the input of the first.

Fig 22.

Fig 22.  When the voltage on the base of the first transistor rises, the voltage on the collector drops and this is transferred to the second transistor. The voltage on the emitter of the second transistor drops and this is fed back to the base of the first transistor to oppose the rise. Obviously this arrangement will not work as the voltage being fed back is HIGHER than the signal we are inputting, but if we add a 220k resistor we can force against the feedback signal and produce an output.

Fig 23.

Fig 23.  We have added a capacitor (electrolytic) to the emitter of the second transistor. Let's explain how this electrolytic works. An electrolytic is like a miniature rechargeable battery. It charges very slowly because it is a large value.  Initially it has 0v. The circuit starts to turn ON by current flowing through the load resistor and this turns on the second transistor. (The first transistor is not turned on AT ALL at the moment).  The base rises and pulls the emitter up too. And when the emitter is about 0.7v, this voltage is passed to the first transistor via the 220k and the first transistor starts to turn on. This causes current to flow through the collector-emitter leads and pulls the voltage on the base of the second transistor down to about 1.4v

This is how the two transistors settle, with the voltages shown in Fig 23. The electrolytic has 0.7v on it and when a signal is delivered to the base of the first transistor, it is amplified and passed to the emitter of the second transistor. Normally the emitter would rise and fall as explained in the above circuits and the result would be heard in the speaker. But the electrolytic takes a long time to charge (and discharge) and it resists the rise and fall of the signal. This means the signal cannot rise and fall at the emitter. In other words we have placed the second transistor in a stage very similar to the first stage we described a COMMON EMITTER. Since the emitter voltage does not rise and fall, it does not pass a signal through the 220k to the base of the first transistor. This means our input signal is not fighting against the feedback signal and it has a larger effect on controlling the first transistor. This gives the first transistor a bigger gain. A common emitter stage has a voltage gain of about 70-100 and we now have one of the best designs. Two common-emitter stages, directly-coupled (DC) and with very HIGH GAIN. The feedback only controls the DC voltages on the two transistors and does not have an effect on the AC (signals).

Fig 24.

Fig 24 shows typical values for biasing the two transistors.

From what you have learnt, you can see the mistakes and/or the voltages in the following circuit:

Fig 25.

Fig 25.  The two joined transistors create a Darlington transistor and this is just a normal transistor with a large gain. The 330R discharges the 100u and it will only discharge it a very small amount. This means the electro can only be charged a very small amount during the next cycle and the output will be very weak. It is the 330R that determines how much (little) energy gets delivered to the speaker. The 330R has to be 15R to nearly fully discharge the 100u.

Fig 26.

Fig 26.  You can work out the voltage on the various points in this circuit by referring to the examples we have already covered.

Fig 27.

Fig 27. This is a practical example of the circuit we have discussed. It is a MICROPHONE AMPLIFIER (also called a pre-amplifier stage).

Fig 27a.

Fig 27a. Here is the same circuit used as a POWER AMPLIFIER.   Both transistors are common-emitter configurations and the circuit produces high gain due to the DC (direct) coupling.

USING PNP TRANSISTORS A PNP transistor can be used in the 2-Transistor DC amplifier studied above. It does not produce a higher gain or change the output features of the circuit in any way but you may see an NPN and PNP used in this configuration and need to know how they work. Firstly we will discus how a PNP transistor works. All those things you learnt in the first set of diagrams can be repeated with a PNP transistor. The circuits are just a mirror-image of each other and  the transistor is simply "turned-over" and connected to the supply rail. Study the following circuits to understand how a PNP transistor is TURNED ON.

Fig 28. PNP Transistor Symbol

Fig 28.  The symbol for a PNP transistor has the arrow pointing towards the BASE.

Fig 29.  PNP "Water Valve"

Fig 29  shows the equivalent of a PNP transistor as a water valve. As more current (water) is released from the base, more water flows from the emitter to the collector. When no water exits the base, no water flows through the emitter-collector.

Fig 30. PNP connected to the power rails

Fig 30  shows a PNP transistor with the emitter lead connected to the power rail. The collector connects to a resistor called a LOAD RESISTOR and the other end connects to the 0v rail or "earth" or "ground."  The input is the base and the output is the collector.

Fig 31.  PNP Transistor biased with a "base bias" resistor and a LOAD resistor

Fig 31  shows a PNP transistor in SELF BIAS mode. This is called a COMMON EMITTER stage and the resistance of the BASE BIAS RESISTOR is selected so the voltage on the collector is half-rail voltage.  In this case it is 2.5v. Here's how you do it. Use 22k as the load resistance. Select the base bias resistor until the measured voltage on the collector is 2.5v. The base bias resistor will be about 2M2. This is how the transistor gets turned on by the base bias resistor: The base bias resistor allows a small current to pass from the emitter to the base and this makes the transistor turn on and create a current-flow though the emitter-collector leads. This causes the same current to flow through the load resistor and a voltage-drop is created across this resistor. This raises the voltage on the collector. This causes a lower current to flow from the emitter to the base, via the base-bias resistor, and the transistor stops turning on a slight amount. The transistor very quickly settles down to allowing a certain current to flow through the emitter-collector and produces a voltage at the collector that is just sufficient to allow the right amount of current to flow from the base. That's why it is called SELF BIAS.

Fig 32. Turning ON an PNP transistor

Fig 32  shows the transistor being turned on via a finger. Press hard on the two wires and the LED will illuminate brighter. As you press harder, the resistance of your finger decreases. This allows more current to flow from the emitter to the base and the transistor turns on harder.

Fig 33. Two transistors  turning ON

Fig 33  shows a second transistor to "amplify the effect of your finger" and the LED illuminates about 100 times brighter.

Fig 34. Adding a capacitor

Fig 34  shows the effect of putting a capacitor on the base lead. The capacitor must be uncharged and when you apply pressure, the LED will flash brightly then go off. This is because the capacitor gets charged when you touch the wires. As soon as it is charged, NO MORE CURRENT flows though it. The first transistor stops receiving current and the circuit does not keep the LED illuminated. To get the circuit to work again, the capacitor must be discharged. A large-value capacitor will keep the LED illuminated for a longer period of time as it will take longer to charge

Fig 35. Adding a capacitor to the output

Fig 35  shows the effect of putting a capacitor on the output. It must be uncharged for this effect to work. We know from Fig 33 that the circuit will stay on constantly when the wires are touched but when a capacitor is placed in the OUTPUT, it gets charged when the circuit turns ON and only allows the LED to flash.

THE NPN/PNP AMPLIFIER
A 2-Transistor DC amplifier can be constructed using an NPN and PNP set of transistors.

Fig 36.

Fig 36  shows how an NPN-PNP set of transistor is turned on. You can think of the "turning ON" this way:  The base of the NPN get "Pulled UP" and the base of the PNP gets "Pulled DOWN." It does not matter how you refer to the operation of the circuit, you must be able to "SEE" how the circuit works so you can see a more-complex circuit working too!

Fig 37.

Fig 37 shows biasing on the base of the first transistor and the "in" and "out" leads have been identified. This circuit has a very high gain and if "general purpose" transistors are used with a very high spread of gain for each transistor, the result will be a very wide range of voltages on the output terminal.  If each transistor has a gain of 100, a change of 1mV on the input will result is a voltage change of 0.001 x 100 x 100 = 10v.  We don't have a 10v supply so, this type of circuit is very UNSTABLE! We need to design a circuit that has FEEDBACK so the output voltage will remain within the voltage of the supply. This feedback is called NEGATIVE FEEDBACK as it opposes an input signal to provide correction or stability. Later we will talk about POSITIVE FEEDBACK and show what an amazing difference it creates - the circuit behaves totally differently.

Fig 38. This circuit does not work

Fig 38  will not work because the base of the NPN transistor is not turned on when the circuit is switched on. This is one of the things you have to look for when designing a circuit.

Fig 39. The voltages

Fig 39  has a voltage-divider network on the base of the NPN transistor. It turns the first transistor ON and this turns the PNP transistor ON until the voltage at the join of the 3k3 and 1k puts a voltage on the emitter of the first transistor to start turning it OFF. This is a point we have to explain. There are two ways to turn ON an NPN transistor. 1. Hold the emitter fixed and RAISE the base voltage. 2. Hold the base fixed and LOWER the emitter voltage.

In Fig 39  the base is weakly fixed by the voltage divider made up of the 1M and 220k and even though the base can move up and down a little bit, we will assume the voltage is constant. If we raise the emitter voltage, the transistor will be turned off. This is what the FEEDBACK voltage via the 3k3 does. It raises the emitter voltage and turns the NPN transistor OFF slightly so an equilibrium point is reached where the two transistors are turned on a small amount and if one gets turned on a little more, the other sends signal to turn it OFF. This is not a practical circuit as an increase of 1mV on the input will produce a large change on the output and this will be reflected back to the emitter of the first transistor to cancel the input voltage.

Fig 40.  A practical example

Fig 40. By changing the value of the feedback resistors we get Fig 40. The values are now 10k and 100R. This gives a ratio of 10,000:100 or 100:1 and it means the output can rise 100mV before the emitter gets 1mv to cancel the input voltage. This means the amplifier will have a gain less than 100 but provides a very stable set of voltages.

Fig 40a.  Another practical example

Fig 40a. Here is an amplifier with the same DC biasing as Fig 40 but with a lower overall gain (2,200:100 or 22:1) and high-frequency feedback (attenuation) via the 2n2 capacitor.

MEASURING THE VOLTAGE(S) The voltage on each line (connection) of a circuit can be measured with a multimeter. To help you take (make) a reading, we have written an eBook titled: Testing Electronic Components. There is a certain amount of skill required to take a reading and this eBook will help you enormously. OSCILLATORS If we remove some of the components from Fig 39 and put a LED on the emitter of the PNP transistor we have a circuit that will illuminate the LED. We have already talked about FEEDBACK in terns of NEGATIVE FEEDBACK to stabilize a circuit. We will now cover a new term called POSITIVE FEEDBACK - it changes the performance of circuit completely. It makes the circuit OSCILLATE. Negative feedback "kills" a circuits performance - positive feedback makes it oscillate. It increases the signal so much that the circuit becomes unstable. This is called oscillation.

Fig 41.

Fig 41 shows a circuit using an NPN and PNP connected via a 1k resistor and turned ON via a 330k base resistor. The LED will illuminate. There is nothing magic about this circuit. It is simply a HIGH-GAIN, DC-AMPLIFIER using two transistors.   The values of current are only approximate and show how each section allows an increasing amount of current to flow. A current of 100mA is too high for a LED and it will be damaged. This circuit demonstrates the possible current-flow. If this current flows for a very short period of time, the LED will not be damaged. Fig 42 shows how the circuit is converted to an oscillator or "flasher."

Fig 42.	Fig 42.  When we connect a capacitor as shown, an amazing thing happens. The high-gain amplifier turns into an OSCILLATOR. When the voltage on point "X" is rising, the voltage on point "Y" is rising TOO. But point "Y" rises much higher than point "X." This means that if we DIRECTLY join points X and Y, the voltage-rise from point Y will push point X higher and turn the circuit ON more. This will continue until the circuit is fully turned ON and the two transistors are SATURATED.
This effect is called POSITIVE FEEDBACK and the circuit will get turned ON until it cannot turn on any more. But we haven't joined points "X" and "Y" DIRECTLY (we have used a capacitor) so we have to start again and explain how the circuit works. When the power is applied, the 10u gradually charges and allows a voltage to develop on the base of the NPN transistor. When the voltage reaches 0.6v, the transistor turns ON and this turns on the PNP transistor. The voltage on the collector of the PNP transistor increases and this raises the right side of the 10u electrolytic and it firstly pushes its charge into the base of the NPN transistor. Then the 330k takes over then it continues to charge in the opposite direction via the base-emitter junction of the NPN transistor. This causes the two transistors to turn ON more. This keeps happening until both transistors cannot turn ON any more and the 10u keeps charging. But as it continues to charge, the charging current eventually drops slightly and this turns off the first transistor slightly. This gets passed to the PNP transistor and it also turns off slightly. This instantly lowers both leads of the 10u and both transistors turn OFF. The 10u is partially charged and it gets discharged over a long period of time by the 330k resistor and when it starts to charge in the opposite direction, the base of the first transistor sees 0.6v and the cycle starts again. The end result is a very brief flash and a very long pause (while the capacitor starts to charge again). As you can see, there is very little difference between the high-gain DC amplifier we discussed above and the oscillator circuit just described. That's why you have to be very careful when looking at a circuit, to make sure you are identifying it correctly.

Fig 43.

Fig 43  is the same circuit with the components re-arranged. It is a high-frequency oscillator with an inductor as the load and when the circuit turns off, the inductor produces a high voltage in the opposite direction to the supply voltage and this is high enough to illuminate a LED. The LED will not illuminate on the 1.5v supply so when the LED illuminates, you know the circuit is working.

Fig 44.

Fig 44 is the same arrangement of the two transistors we have just studied,  but with a third transistor above the two. We have already seen the importance of charging a capacitor (and then it must be discharged so that the re-charge will produce a "current-flow.") That's what the two transistors in the output are doing. The top transistor charges the electrolytic and the bottom transistor discharges it. In the process, the charging and discharging current flows through the speaker to produce audio. We have already studied the two lower transistors. The BC327 turns ON and allows current to pass through the emitter-collector leads and this discharges the electrolytic.

The top transistor is an emitter-follower and it turns ON when the bottom two transistors are effectively "out of circuit." The base is pulled to the supply rail by the 1k and the emitter follows. In other words the collector-emitter leads allow current to flow and this charges the electrolytic. The charging current flows through the speaker.

CURRENT GAIN OF AN EMITTER FOLLOWER STAGE We have seen the need to provide current into and out of a speaker to move the cone. This is because current produces magnetic flux and many items work on magnetic flux,  such as: motors, relays and speakers. And some items need a lot of current to be activated - especially globes.   Most transistors will provide a CURRENT GAIN of 100 when up to 25% of their rated current flows, but only a gain of 50 for the next 25% increase in current and a gain of 30 for the next 25% increase in current and a gain of only about 10 when the maximum allowable current flows. That's why you have to understand transistor data-sheets. The gain of a transistor is very low when maximum current flows. There is a hidden factor with motors and globes. They take 6 TIMES more current for a globe to start glowing or to start a motor revolving. This is because the resistance of a cold globe is only one sixth of its glowing resistance and a motor has a very low resistance until the back emf (electro-motive force - another name for voltage)  produced by the armature, reduces the current-flow. This means you have to design a circuit that will deliver up to 6 times the operating current, so these items will turn on. We explained the 800R LOAD resistor provides the turn-on current for the speaker in the following circuit. When the BC547 turns off, the current through the 800R is amplified by the emitter-follower transistor to drive the speaker. This is a very wasteful way of operating a circuit as current is always flowing through the 800R and during part of the cycle, this current is not achieving any result. We can design a circuit where this current is provided by a transistor. This is important when we are providing high currents as a transistor can be turned on to deliver the current and turned off when the current is not required,. This saves energy and prevents over-heating. We will look at the following 2-Transistor DC amplifier driving a speaker (taken from Fig 18) and modify the circuit.

Fig 45.  An emitter-follower driving a speaker

Fig 45. The EMITTER FOLLOWER drives a speaker.

Fig 46.

Fig 46. We replace the speaker with a motor.

Fig 47.

Fig 47. We replace the LOAD resistor with a transistor and add a resistor called a: Current Limiting Resistor. It is designed to limit the current between the first and second transistors as these will turn ON and allow a very high current to flow if the resistor is not included.

Fig 48.

Fig 48. The current required by the motor is 300mA. The emitter-follower will have a gain of 10 and the gain of the other two transistors produces the set of conditions shown on the diagram. You can see that very little input current is required to activate the motor when 3 transistors are used.

Fig 49.

Fig 49. The input current can be supplied from a voltage-divider using a pot (to adjust the setting) and a Light Dependent Resistor. We cannot use only 2 transistors as the LDR cannot supply 1mA under low-level light conditions and that's why 3 transistors are needed.

Fig 50. Dancing Flower

Fig 50. Here is a commercial version of a 3-transistor circuit. This circuit was taken from a dancing flower. A motor at the base of the flower has a bent shaft up the stem and when the microphone detects music, the shaft makes the flower wiggle and move. The circuit will respond to a whistle, music or noise. The circuit uses a different arrangement to our 3-transistor design and we will discuss the differences.

It is very easy to get a change in voltage from an input device such as an LDR or electret microphone. Simply add a LOAD resistor and "tap off" the change in voltage at the join of the two components. There is also a very small change in CURRENT at the join of the two components (but we normally refer to the change in voltage).  We can amplify this voltage via two transistors to get a voltage equal to rail voltage. This is not a problem for 2 transistors. But we also need to amplify the CURRENT to operate a motor. We cannot get enough CURRENT GAIN with 2 transistors and that's why we need 3 transistors.  The change in voltage must be passed through 3 transistors to get the CURRENT GAIN required by a motor. Both circuits (Figs 49 and 50) appear to perform the same but you need to look at the voltage drop across the leads of  the output transistors to see how the two circuits compare. There are two important values for a FULLY-TURNED-ON transistor: Fig 51. The characteristic voltage drops across a fully-turned-ON transistor

Fig 52. The voltage losses across the output transistor

The emitter-follower design (the first circuit) has a total voltage drop of 0.8v and the motor will see a maximum of 2.2v.    The motor in the common-emitter design will see a maximum of 2.8v. SUMMARY You can see the advantages and disadvantage of each design. Because the emitter-follower has a 0.6v drop between base and emitter, it is generally used in a PUSH-PULL arrangement as we will see in Fig 53, to charge and discharge the electrolytic or in an H-Bridge to drive a motor forward and reverse as shown in Fig 54. But when a common-emitter stage is used, the output voltage  increases 0.6v. THE TRANSISTOR as a LINEAR AMPLIFIER The EMITTER FOLLOWER stage can also be called a LINEAR AMPLIFIER as the output follows the input voltage EXACTLY except it is about 0.6v lower than the input. The output has about 100 times more current capability than the input and this gives it the name AMPLIFIER. See Emitter-Follower for circuits. A Linear Amplifier can amplify the current from a pot to create a very simple Motor Speed Controller or LED Illuminator: The actual result in increasing the speed of the motor or the brightness of the LED will not seem to be linear because they do not respond in a linear way to an increase in voltage. The pot also has to be linear to produce a linear output. Motor Speed Controller LED Illuminator THE PUSH-PULL STAGE also called  PUSH-PULL AMPLIFIER We have studied the emitter-follower in Figs 45 to 49. We have also shown how to connect a PNP transistor to the power rails. (It is basically a mirror-image of the NPN transistor.) Combining these facts we can produce a circuit consisting of two emitter-followers as shown in Fig 52a. The top emitter follower is an NPN transistor and the lower emitter-follower is a PNP transistor. The is called a PUSH PULL output stage or PUSH PULL AMPLIFIER or Complementary-Symmetry output stage. Fig 52a. Push-Pull Output Fig 52b. Push-Pull Current Dumping Fig 52b shows a very clever variation on the Push-Pull circuit described above. It uses a low-value resistor between the collector of the driver transistor and output. This resistor transfers the low-level signals directly to the speaker. As the signal-level increases, the output transistors come into operation. This arrangement removes cross-over distortion and uses less parts. It is called CURRENT DUMPING.   Lifting the Input line will raise the output line and it will have "100 times more strength." Lowering the input line will make the output line go down with "100 times more strength." In other words this circuit turns a "weak line into a strong line." This feature is also called IMPEDANCE MATCHING. The circuit is also called a PUSH PULL OUTPUT as one transistor "pushes energy" into a device (connected to the output) during one half of a cycle while the other transistor will "pull energy" out of a device. This is one of the ways to charge and discharge a capacitor on the output and any device connected to the other side of the capacitor will see the AC waveform and become active. This is shown in Fig 53:

Fig 53 PUSH-PULL to charge/discharge the 100u electrolytic Fig 54  PUSH-PULL driving the motor forward/reverse

 

Fig 54aa  PUSH-PULL Amplifier

Fig 54aa is a 3-Transistor Push-Pull amplifier. When the supply is turned on, current flows though the 8R speaker and through R4 to the base of T2. This pulls the base of T2 towards the 9v rail and the transistor rises to nearly the 9v rail. The voltage on the emitter of T2 is 0.6v lower than the base and this pulls the emitter of T3 towards the 9v rail. The base of T3 is 0.6v lower than the emitter. This is as far as we can go with the current-path at the moment and we now have to go to T1. The join of the two emitters has a voltage near the 9v rail and this voltage is passed to the base of T1 via the 82k resistor. The 82k resistor forms a voltage divider with 12k and the resulting voltage at their join is sufficient to put 0.6v on the base of T1. This turns ON T1 and the voltage between collector and emitter drops to a low value. The exact value will be shown in a moment. We can now go back to the base of T3 and continue the current-path (also called the voltage path) from the 9v rail to the 0v rail. T1 pulls the base of T3 towards the 0v rail. We now have three transistor that all turn on.  They are not fully turned on but partially turn on. The exact amount of “turn-on” for each of the transistors is due to the 83k and 12k biasing components and diodes D1 and D2. Here’s how the DC coupled amplifier self-adjusts to a state called the QUIESCENT STATE. This is the state where some of the components adjust the “turn-on” of other components and the circuit reaches a point where the voltages settle down and reach a stable value and the current is a constant minimum value. The voltage at the midpoint of the two output transistors is fairly high and this creates a slightly higher voltage on the base of T1. This turns on T1 slightly more and the voltage on the collector drops. This lowers the voltage on the base of T3 and the emitter voltage drops. This lower voltage is passed to the base of T1 and the transistor turns OFF slightly. This is how the three transistors adjust themselves to a final value. The exact final voltage is called a DESIGN VOLTAGE and designer of the circuit want the voltage on the join of the two emitters to be half-rail-voltage. This allows the circuit to rise and fall and reproduce a waveform without clipping or cutting off the top or bottom of the wave. To get the  circuit to sit with the output (the join of the two emitters) at 4.5v, the values of R2 and R3 have been selected.    We now have the circuit sitting, ready to amplify a signal. The output stage is called PUSH PULL because one transistor pushes current through the winding of the speaker via the 100u electrolytic and the other transistor pulls current through the speaker via the electrolytic. You could connect the speaker directly to the output of the stage and remove the electrolytic. The circuit would work just the same. However if the speaker is connected directly, a voltage of 4.5v will be paced across the speaker and this voltage will cause a current to flow in the winding of the peaker (the voice coil) and the cone will be pulled in. If we try to reproduce a waveform, the cone is already partially pulled-in and it will not reproduce half of the waveform. In addition, this constant current will heat up the voice coil.    By adding the 100u, we remove the Dc component of the output and only the AC (waveform) will be passed to the speaker. Now we have to understand how an electrolytic passes energy (current) to the speaker. If you connect an electrolytic and speaker directly to a supply, you will hear a “plop” This is the electrolytic charging and the charging current flows through the speaker and produces the noise. But after a very short time the electrolytic is charged and no ore current flows. Even if you remove the supply and connect it again, no sound will be reproduced because the electrolytic is already charged. The only way to hear another plop, is to remove the components and short between the power leads. When the supply is re-applied, you will hear another plop. To get sound from the circuit, this is what it has to do. Firstly it has to charge the electrolytic. Then it has to discharge the electrolytic. As you can see from the circuit, the lower transistor charges the electrolytic and the top output transistor discharges the electrolytic. Now we have to drive the two transistors so that they charge and discharge the electrolytic. To charge the electrolytic, T1 turns ON and pulls T3 towards the 0v rail. This is the easy part. How do you pull T2 UP so that it discharges the electrolytic? This is how it is done. It is very clever. Connected between T2 and T3 are two diodes. Each if these diodes has a voltage drop of 0.6v. This voltage drop is exactly the same voltage as between the base and emitter of the two transistors in the output. This means we can directly pull on the base of the top transistor, just like we are directly pulling on the base of the lower output transistor. Now we have a situation where we can pull down on both transistors and this will turn ON the lower transistor and turn OFF the upper transistor. This is done when T1 turns ON. When T1 turns OFF, the top transistor is pulled HIGH via the 1k8. That’s how it works.

Fig 54a  Two Push-Pull circuits driving the primary of a transformer

Fig 54b

Fig 54b shows a free-running multivibrator configured so the transistors drive a transformer in Push-Pull

THE TOTEM POLE OUTPUT STAGE
A slightly different push-pull output stage can be created with two NPN transistors. It is called a Totem Pole Output stage.

Fig 55a

Fig 55a. When the input is less than 1v, the output is pulled high via the 1k resistor and the "strength" of the "pull-up" will be 1,000/100 = approx 10 ohms. When the input reaches 1.4v, the output is pulled low via the lower transistor and will about 0.2v from the 0v rail. The "strength" of the "pull-down will be about equivalent to a 10 ohm resistor. This is about the same as the output driving capability of a normal Push-Pull arrangement, however there is a mid-point where both transistors are turned on at the same time and this produces a large current that can overheat the transistors or damage them.

THE BRIDGE
Another way to connect a transistor to produce a "stage" is called a BRIDGE. It consists of 4 resistors: