Transistor Terminology

A transistor has three legs: a collector, an emitter and a base. Below is the symbols for an NPN and a PNP transistor.


NPN transistor


PNP transistor

Transistors as switches

The easiest way to understand transistors is to think of them as switches. You can switch a big current (between the collector and emitter) with a much smaller current (in the base). Lets look at an example:


NPN transistor as a switch (on)


NPN transistor as a switch (off)

Also note that the emitter is always tied to the fixed voltage (+12V or GND) when using transistors as switches.
In the following tutorials (coming soon) we will show you how to do calculations with transistors.

Diodes have the following properties:

* They only allow current to flow in one direction
* They have a specific voltage drop, which can be taken as a constant for most circuits

A diode is represented with the following symbol:



The diode above will pass current flowing from left to right, but block current trying to flow from the right to the left-hand side. The voltage drop over diodes will vary from diode to diode, the most common ones are:

"Normal" diodes: Have a voltage drop between 0.6V and 0.7V
Schottky diodes: They have a low voltage drop (about 0.1V to 0.2V)
Light emitting diodes (LED): These diodes will emit a light when passing current. Their voltage drop will depend on the colour of the LED, but is around 2V to 3V.
Zener diodes: They behave as a normal diode for current flowing from the left to right, but will allow current to flow from the right to left if the voltage is high enough. You get a lot of different Zener voltage diodes, e.g. 4.7V and 12V.

Using diodes

Below is a simple circuit with a diode. You must always place a resistor in series with a diode to limit the current flow, if you don't do that the diode will act as a short circuit and draw current until the supply or the diode breaks.



The voltage drop across the diode is 0.6V (this is a normal diode and 0.6V is close enough.) The diode and resistor is in series, thus the current through the diode is equal to the current through the resistor.
IR1 = V / R
= (5V - 0.7V) / 470R
= 9.1mA

Thus the current through the diode equals 9.1mA

Here is an example with an LED (you will notice the symbol of an LED has arrows pointing away from it, this is to show that it emits light)



Let's say the LED is a red LED with a voltage drop of 2V, thus:
IR1 = V / R
= (5V - 2V) / 220R
= 13.6mA

The more current an LED's conducts, the brighter it will be. To increase the current you can either increase the voltage or decrease the value of the resistor. Just remember that there is a limit to the amount of current a diode can conduct before it breaks and for LED's this is rather low (about 20mA typically)

Zener diodes are mostly used to create a constant voltage. Say we want an LED to be the same brightness, no matter what our input voltage is (the input voltage could be from a battery that starts out at 14V and goes down to 9V before it is charged again)
The circuit below is one way of doing it:



R1 will determine the amount of current flowing through Z1 (for Zener diodes to work correctly, you need a minimum current flowing through them, about 10mA is a good value)
If the input voltage is 9V:
Id1 = V / R
= (9V - 4.7V) / 380R
= 11.3mA

If the input voltage is 14V:
Id1 = V / R
= (14V - 4.7V) / 380R
= 24.4mA

Because the Zener diode will always have a constant voltage drop over it, the potential at Vzener will always be 4.7V. The current through D2 is (assuming the voltage drop over D2 is 2V):
Id2 = V / R
= (4.7V - 2V) / 220R
= 12.3mA

And this current will always be the same, whether the input voltage is 14V, 9V or anything in between. Of course when the input voltage drops below 4.7V, the current through D2 will drop.

Resistor Tutorial

A basic circuit

Here is an example of using Ohm's Law for a basic circuit (see the electronics introduction if you're not familiar with Ohm's law.)

We have a resistor of 4k7 (4700 Ohm) connected to a 5V supply, what is the current flowing through it?



Answer
I = V/R
= 5V/4k7
= 1.06mA

Resistors in series

When resistors are connected as shown in the example below, we say that they are connected in serial. Serial resistors have the following properties:

* The same current flow flows through them (Itotal = I1 = I2 = I3 ....)
* The total voltage will be a sum of the voltages across them (Vtotal = V1 + V2 + V3 ...)
* The total resistance equals the sum of the individual resistors (Rtotal = R1 + R2 + R3 ...)

Example
A resistors of 10k is connected in series with a resistor of 4k7, what is the current through them and the voltage across each one individually?



Answer
The voltage across both resistors is 5V, thus to calculate the current:
I = V/R
= 5V/(4k7 + 10k)
= 0.34mA

The voltage over R1:
V1 = I*R
= 0.34mA*4k7
= 1.6V

The voltage over R2:
V2 = I*R
= 0.34mA*10k
= 3.4V

And V1 + V2 = 5V!

Resistors in Parallel

When resistors are connected as shown in the example below, we say that they are connected in parallel. Resistors in parallel have the following properties:

* The voltage across them are the same (Vtotal = V1 = V2 =V3 ...)
* The total current flowing through them will be the sum of the individual currents. (Itotal = I1 + I2 + I3 ....)
* The total resistance can be found by adding up the reciprocals of the resistors and then taking the reciprocal of the total (1/Rtotal = 1/R1+ 1/R2 + 1/R3 ...)

There is an easy way to calculate the resistance of 2 resistors in parallel:

Rtotal = (R1 * R2) / (R1 + R2)

Here is the proof, for those interested:
1/Rt = 1/R1 + 1/R2
1/Rt = (R2 + R1) / (R1 * R2)
1 = ((R2 + R1) / (R1 * R2))*Rt
(R1 * R2) / (R1 + R2) = Rt

Example

A resistors of 10k is connected in parallel with a resistor of 4k7, what is the current through them and the voltage across each one individually?



Answer

Because R1 and R2 are connected in parallel, the voltage over R1 equals the voltage over R2 namely 3.3V.
Let's calculate the rest:
I1 = V*R1
= 3.3V/4k7
= 0.7mA

I2 = V*R2
= 3.3V/10k
= 0.33mA

Rt = (R1*R2)/(R1+R2)
= 47k/14k7
= 3k2

Resistors as voltage dividers

In many circuits you'll see resistors used as voltage dividers. In the circuit below we have a 3.3V input voltage and want to calculate the voltage at point Vout. Because the resistors are in series, we know that the total voltage (3.3V in this example) will be divided between R1 and R2. A simple formula to calculate the potential (voltage) at point Vout is:

Vout = Vin * R2 / (R1+R2)

Here is the proof:
(1) Itotal = V1 / (R1 + R2)
(2) Vout = Itotal * R2
Substituting (1) in (2):
Vout = (V1/ (R1 + R2)) * R2
= V1 * R2 / (R1 + R2)
= Vin * R2 / (R1 + R2)



For the above circuit:
Vout = Vin * R2 / (R1 + R2)
= 3.3V * 10K / (10K + 4K7)
= 2.24V

Inductors are coils of wires, wound around some magnetic material and then packaged. The final configuration take a wide variety of forms and appearances. Values may range from fractions of milli-henrys to fractions of henrys.

Coils are the only passive components in common use that may display significant non-linear characteristics. The behavior of inductors is based in phenomena associated with magnetic fields. The source of the magnetic field is charge in motion, or current. If the current is varying with time, the magnetic field is varying with time. A time-varying magnetic field induces a voltage in a by conductor linked by the field. The circuit parameter of inductance relates the induced voltage to the current.

Energy can be stored in magnetic fields and inductors are capable of storing energy. For example, energy can be stored in an inductor an then released to fire a spark plug. Inductors are passive devices because they can not generate energy.

Inductance is the circuit parameter to describe an inductor. Inductance is symbolized by the letter L, measured in henrys (H), and it is represented graphically as a coiled wire.

Inductors with Sine Wave Voltage and Current

1. Recall that an inductor opposes any change in current. When a sine wave voltage is applied to an inductor, the current through the inductor is constantly changing.

2. Also recall that an inductor opposes changes in current. The inductor accomplishes this by developing a voltage across the coil with a polarity that would oppose the cause of the changing current.