Im a new member of this forum and im not familiar with electronics so could you help me on some of my questions?
They say that a BJT transistor has a 3 regions of operation namely saturation, active and cut-off. when im going to use my transistor as a switch i should operate them under saturation and cut-off region. My question is how can i saturate an NPN and a PNP transistor?
thanks..
Hi Demeanor,
(If you are a female, can we call you 'Miss Demeanor'?)
Yes, those are basically the three states for a transistor. NPN and PNP are opposites (mostly) so 'cutoff' and 'saturation' mean the opposite things for them.
I should probably say that you really, really want to learn Ohm's Law if you're going to mess with this stuff. I'm just a hobbyist but I use it every time I have to think about electronics. Really. It's that useful. Nothing else will make sense unless you know that V = IR (V means voltage, I means current, and R means resistance). If you already know that, forgive me. If you don't, check out (actually, check out that whole online book--it's a great resource and easy to follow).
A transistor uses a small amount of current flowing into its base to control a larger current flowing between its collector and emitter. Typically you arrange resistors to provide the needed amounts of current into each terminal. The ratio between the base current and the collector-emitter current depends on the transistor's 'beta', or gain (also called its hfe--imagine the 'fe' is written in subscript).
If you provide no (or very little) current to the base compared to the collector-emitter, then the transistor will be in cutoff. For an NPN, this means that no current will flow across the collector-emitter. For a PNP, it means that as much current as is available will flow across the collector-emitter.
If you provide some more current to the base, then at some point more current will be able to flow on the collector-emitter. You are now in the active region, which means that the current flow on the collector-emitter is an amplified version of the base current.
As you keep raising the current into the base, the collector-emitter rises (falls, for a PNP) even more rapidly, until at some point the transistor can no longer allow more current to flow and the collector-emitter current levels off. The transistor is now in saturation.
For switching, you're not interested in the active region (amplification), you just want to slam the transistor between saturation and cutoff. To turn an NPN off, you want to put very little (preferably no) current into its base. To turn it on, you want to put enough current into its base to get its collector-emitter current as high as you need. Typically this means making the base current about 1/10 or so (as a rule of thumb) of the collector-emitter current. The exact values depend on what you're trying to switch and what transistors you're using.
For a PNP, it's the opposite.
Read this for more information, more lucidly written than what I have put here:
Here's another good bit of reading:
https://books.google.ca/books?id=bk...srWpDQ&sig=ACfU3U0ky8qeaVhCStEUsltLpBbFbsneVQ
I highly recommend buying that book if you intend to learn any electronics, by the way. It covers the basics in an easy to follow way but also addresses a lot of other things, and walks you through a lot of neat stuff. It can take some patience, though.
Good luck,
Torben