Transistor as a switch

[lowell]

I need some help clearing this up please. My question is purely theoretical. I'm not designing anything at the moment.

What I do understand thus far is that a transistor must be in saturation (on) or cutoff (off) to be a switch. Therefore it is not used in it's linear region as in an amplifier stage. I'm also not sure as to why I'd use a jfet vs a bjt.

If I were to use a 2n3904 transistor as a switch would this be correct? My V+ is 9v. http://www.centralsemi.com/PDFs/products/2n3903.pdf

The collector would be tied to the V+ and the emitter to ground or whatever my load would be (maybe a relay). Where I'm confused is where to find in the datasheet what the base voltage should be. I see 2 numbers for Vbe SAT and it's unclear as to whether Vbe should be .65-.95v. Also there's 2 numbers for Ib: 1ma or 5ma.

If someone could clear the ratings up for me that'd be great. Otherwise here's what I think I would do next. And obviously I'd make a voltage divider from V+ to the base to get the .7v.

To find the series base resistance value and to protect the transistor I would do this:

If Vbe SAT is .7v and Ib is 1ma then
.7/.001=700ohms

Is that right?

[Steve Conner]

Therefore it is not used in it's linear region as in an amplifier stage. I'm also not sure as to why I'd use a jfet vs a bjt.

BJTs for switching power (operating LEDs, relays, etc) and JFETs for analog signals.

There are two ways to use the transistor as a switch: an emitter follower or a saturated common-emitter stage.

You are suggesting the emitter follower way, but it's rarely used because the transistor can't saturate, which makes it less efficient. If you want to design it this way, consider that all you're doing is buffering the control voltage so it can supply enough current to drive the load.

To do saturated switching, you need to use a PNP transistor for high-side switching (positive supply to load switched) and an NPN one for low-side switching (ground switched)

[lowell]

Thanks Steve that's great info. Could you point me to a schematics of these? "To do saturated switching, you need to use a PNP transistor for high-side switching (positive supply to load switched) and an NPN one for low-side switching (ground switched)"

[Jazz P Bass]

Keep in mind the base current requirements.
Good reading @:Transistor Circuits

[lowell]

Yes that was what my math in the initial post was for:

If Vbe SAT is .7v and Ib is 1ma then
.7/.001=700ohms

So if figure a 700ohm series base resistor would be suitable?

[Steve Conner]

Keep in mind the base current requirements.
Good reading @:Transistor Circuits

Yes, in particular here is the NPN low-side switch:


and here's the PNP high-side one:


I usually use 4.7k when driving the transistor from 5V logic, 10k from 12V. And sometimes a 3.3k resistor from the base to ground (or positive supply in the PNP circuit) to improve the noise margin and make sure the transistor stays off. The idea is to get about 1mA of base current, which is about right for the usual loads.

If you want the transistor fully saturated, the standard criterion for saturation is Ic/Ib = 10. So if you're switching a 100mA load, you need 10mA base current. So from 5V logic, you'd need a resistor of (5-0.7)/0.01 = 430 ohms. But not all logic chips can supply 10mA.

[lowell]

Okay one last question. I do need to take into account the datasheet for the transistor too right? I'd need to be sure that in your example above "... switching a 100mA load, you need 10mA base current." I'd need to be sure the transistor can handle 100ma of Ic and 10ma if Ib right?

[Steve Conner]

Yes, or it will be fried transistor for lunch.

[dmsherazi]

it would be good for you if u go through this post Transistor as a Switch