How to saturate a transistor?

[demeanor]

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..

 

[Torben]

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

 

[Pommie]

Torben,

Your description is correct except for the PNP transistor which is off with no current and on when a current flows out of the base.

Mike.

 

[Torben]

Torben,

Your description is correct except for the PNP transistor which is off with no current and on when a current flows out of the base.

Mike.

Mike,

You're right. I always get that mixed up with the PNP. I had a brainfart.

I think it's from playing with simulators and misinterpreting what I'm seeing. Need to get back to the calculator and pencil.


Torben

 

[demeanor]

Thanks for the help....if i keep raising my base current does that mean i am decreasing the beta of the transistor?

 

[Torben]

Thanks for the help....if i keep raising my base current does that mean i am decreasing the beta of the transistor?

No, the beta of the transistor stays the same.


Torben

 

[Roff]

Saturation has exactly the same meaning for NPNs and PNPs. In either case, the base-emitter and base-collector junctions will both be forward biased. Cutoff, of course also has the same meaning in both cases. Collector current is zero (except for leakage).
Check this Wikipedia article if you don't believe me.
EDIT: I guess Pommie made the same point.

 

[Torben]

Hi again all,

I did some more reading and have figured out where my confusion comes from. I think. If any of the following is wrong please correct me.

I was associating "saturation" of a transistor with "having the base pulled high". That's the case with NPNs (and "cutoff" is pulling base low), but for PNPs, the current is not flowing into the base but rather out of it. So to saturate a PNP you pull the base low (to maximize current flow out of the base) and to cut it off you pull the base high.

Am I right on that?

Hehe. I just noticed that if I wanted to defend my original post I'd note that I only referred to current flowing *into* the base, so if I wanted to be ornery I'd argue on that point--but both Pommie and Roff know what I meant by what I wrote, and what I meant was wrong.


Torben

 

[Pommie]

Hehe. I just noticed that if I wanted to defend my original post I'd note that I only referred to current flowing *into* the base, so if I wanted to be ornery I'd argue on that point--but both Pommie and Roff know what I meant by what I wrote, and what I meant was wrong.

You are, as you know, completely correct. If you minimise (make negative) the current flowing into the base then it will turn on. I like arguments like that as they make you think (and laugh).

Mike.

 

[Willbe]

I thought the beta in saturation goes down to about 10. It'll be on the datasheet.

 

[audioguru]

When a transistor is poorly saturated without enough base current then the collector current doesn't change much when it has much more base current so it saturates much better. So the beta doesn't apply.

The datasheet for most transistors shows a max saturation voltage loss when the base current is 1/10th the collector current.
Some low current but very high beta transistors show a max saturation voltage loss when the base current is 1/20th the collector current.
Some power transistors show a max saturation voltage loss when the base current is very high. A lousy old 2N3055 transistor has a max saturation voltage loss of 3V when its base current is 3.3A and its collector current is 10A.

 

[Hero999]

At a minimum, I'd keep the current at the base >Ic/(2*Beta) whichis generaly good enough.

 

[kchriste]

I thought the beta in saturation goes down to about 10. It'll be on the datasheet.

What you are thinking of is that the beta will decrease for higher collector currents. ie: A transistors collector may pass 100ma for a base current of 1ma, but only pass 500ma for a base current of 10ma.
For saturation, the general rule of thumb is to design the circuit so that the base current is 1/10th of the required collector current.

At a minimum, I'd keep the current at the base >Ic/(2*Beta) whichis generaly good enough.

Surely you meant 2*Ic/Beta??? Where Beta is the MINIMUM beta on the datasheet.