how does one determine the terminals of a BJT and JFET transistor, pn diode and zener diode practically?

Answer 1

On a BJT transistor one would determine the terminals first by putting the flat side towards them, starting at the left terminal you got the emmiter, the center is the base and the right side is the collector. The JFET transistor is simulator just starting at the left you got the Drain, the center is the Gate and the right pin is your source. To hook up a regular PN diode you know to always hook the side with the line to the negative and the black side to the positive. This holds true for a zener diode to but in a circuit the zener diode is designed for limiting voltage. Typically you hook it up in revers bias in parallel with the load your using, this diode then depending on the voltage it was set to work with ground any extra voltage from the source. the most common voltage drop across a diode is 0.7v since most diodes are silicon.

Answer 2

The zener must be kept in reverse breakdown mode for it to function as required and this requires a minimum current. We assume that it may be used in the worst case scenario, given that we use the zener wattage and its voltage to give, I. W= V x I I = 2/5 = 0.4A as you got and RS = (Voltage drop across RS/current through RS). RS = 12-5/0.4 = 17.5 Ohms. However, having said that, your load is quite low (that is, it draws very little current 5mA so a 2W device is a little bit of 'over-kill' for this project, -Anyway! The second part requires a diagram too. But I'll do my best. Typically, a transistor can be connected in one of three modes: Common Emitter,(CE), Common Collector (CC) and Common Base (CB); each having its own advantages and disadvantages when used in circuits. Transistors come in two types: PNP and NPN. My example here will use the more common NPN device. For the NPN device, the ‘N’, ‘P’ and ‘N’ are the emitter, base and collector respectively. Basically, the NPN transistor will 'turn on' when the base is 0.6V more positive than the emitter. So, if the emitter is connected to 0V then the base will need to be +0.6V. If the emitter is at +2V then the transistor will turn on when the base is +2.6V. In this condition the BE junction is said to be forward biased. Just to complicate things a little, the base collector voltage also needs to be reversed biased for the transistor to work. But don’t worry, provided you connect the emitter to 0V and the collector to +12V via a load such as a relay or resistor you will be fine. Typically, the base current is provided by having two resistors in series between the +12V and 0V of the supply and then connecting the base at the intersection point; a arrangment known as a potential divider. If now the transistor is turned 'on' (base 0.6V more positive than the base) then the current that flows into the base will be magnified by the current gain of the transistor and this will appear as a very much larger collector current. If now a load such as a relay is placed between the collector terminal and the positive supply this tiny base current can be used to close the relay and then, if we wish operate a very device such as a powerful motor say. As you will see the transistor is a current operated device. The voltages at the base collector are all relative to one another but the rules remain the same for transistor action: emitter base must be forward biased and base collector reverse biased. With this in mind you could have the emitter at -20V, the base (0.6V more positive ) at -19.4V and the collector at -8V (12V less than 20) if you want . You only need to make sure that the differential voltages are within the spec for the device being used.

Answer 3

The manufacturer can package them any old way they want. The diodes are pretty easy if you use an ohm meter. The others are tough and you need a component diagram.

Answer 4

Fet Terminals

Answer 5

If you do not have a data sheet for the devices, you should realize that they are all basically pn junctions and you can measure them with an ohm meter.

I hope this helps.