npn fundamentals: biasing emitter follower as switch - Page 1
Forum section: electronics / beginners

author: hirada
author_rank: Regular Contributor
timestamp: February 02, 2018, 12:47:14 pm
content: I am having trouble understanding (as in biasing) the emitter follower, especially when used as a switch. And whenever I think I got it, some other factor pops up and the more I read, the more I get confused. All I just want is a behavior as close to an no switch as possible.The core of my problems I believe is: How can I make sure, that I have enough potential left after the base resistor. Or, if I do not need one, how can I limit the base current.Two examples, more to follow:The BJT archives desired saturation Ibsat=3mA, has an Ube=0.7V and the comparator outputs 4,7V@3mA. So I expect Uout=4V.The LED would be bright enough with ~1mA.Now for my confusion: Any resistor before the base to limit its current would also cause a voltage drop, effectively reducing the basepotential and therefore ruin the output voltage?In example a) one could argue, that Rbmay not be needed, as Ibis also limited by the emitter resistor Re? Meaning, to get the BJT into saturation I would need to increase the diode current to at least Ibsator parallel a resistor. Fair enough.But if that happens to be the case: if the maximum current flowing through the emitter is basically Ib, that would mean Ic=~0A? Makes no sense.Or would it be, by some silicon magic, the base current gets automagically be reduced until hfeequates to Ic=1mA? After all, we do not need saturation for this case. But from what I have read, it is the base current, that determins how far the transistor opens, becomes conductive that is.Then, what with example b)? Next to the LED current we have around 1A flowing through Rl. Without any Rb, how can we limit basecurrent to not exceed it's limits (and the ones from the sourcing IC)?How does the emitter know: "I'll take just Ibsatfrom the base and draw the rest (~1 amp in this example) from the collector?"As you can see, while trying to understand the basics, I have taken a wrong turn at some point in time again and gotten completely off track or am I just missing some npn magic?Before I put things to breadboard, I would like to have at least a very basic understanding, something I am obviously lacking. I am more than happy for pointers to documentation, but most I have come across so far are either very general and are repeating each other more or less or have been written for university students. Maybe it's just my weird way of thinking.

author: bd139
author_rank: Super Contributor
timestamp: February 02, 2018, 01:11:39 pm
content: Emitter follower and switch are two different topologies. You need to separate them.The emitter follower is a common collector amplifier with load connected to the emitter. The switch, the emitter is usually commoned and the load connected to the collector.As you’re talking about saturation, the switch operates entirely in either the saturation and cut off regions. The emitter follower tends to work in the active region (there are some exceptions but don’t worry about those yet)Obtain (wink wink) a copy of The Art of Electronics and read the first few parts of the section on transistors. That has a better explanation than I can give.

author: Wimberleytech
author_rank: Super Contributor
timestamp: February 02, 2018, 01:42:58 pm
content: If you are determined to use an emitter follower, then work the problem like this.Desired LED current = 5mA (to achieve your desired brightness...choose your value)Vdiode from diode spec at desired current = 2V (or whatever it is for your diode)Vbase = 5V (no resistor required)Vbe = 0.7 (more or less)Vemitter = Vbase - 0.7 - 4.3 volts.Voltage across emitter resistor - 4.3 - 2 = 2.3 volts5mA flows through a resistor with 2.3 volts across it, thus the resistor is 2.3/5mA = 460 ohmsYour base current will be approximately 5mA/beta Beta is a function of emitter current, but say it is at least 100 at an emitter current of 5mA.That means your base current is 50uA . Your comparator has no problem providing that current.Your BJT cannot saturate BTW because Vce can never be less than Vbe using the voltages you have chosen.

author: Zero999
author_rank: Super Contributor
timestamp: February 02, 2018, 01:47:29 pm
content: An emitter follower never goes into saturation. The voltage at the emitter always settles below the voltage at the base.There's no need to limit the base current, which settles at a value determined by the Hfe and the emitter current. If the transistor has an Hfe of 100 and the emitter current is 100mA, then the base current will be 1mA.If the device driving the transistor can't supply enough base current, then use a transistor with a higher Hfe or cascade two to make a Darlington pair, but bear in mind the base-emitter voltage will be doubled.

author: Wimberleytech
author_rank: Super Contributor
timestamp: February 02, 2018, 01:49:58 pm
content: LTspice example

author: hirada
author_rank: Regular Contributor
timestamp: February 02, 2018, 04:04:29 pm
content: Thanks very, very much, for your replies and even the Spice effort, I finally may even have gotten it.I guess I was treating a BJT as a mosfet, just with current instead of voltage. So the more current I supply, the less resistance it has. And the base current therefore is determined by the sourcevoltage minus Ubeand the resistors in it's path to ground. Be it Rbor ReBut, unless limited by the source, a BJT is not supplied with a base current, it rather takes it, it sucks in what it needs - therefore the base cannot be overloaded. As long as sticking to the various specs, of course.Just to make sure that I am on the correct path:Sticking with hfeof 100 and taking a common emitter configuration where Icis limited to 100mA, I could do without a base resistor and without the transistor being destroyed, despite the base being shorted to ground by nothing but a diode in between, because thanks to hfeit would only draw 1mA anyway? Not that this is recommended or good practice, just for sake of understanding. And eventually verifying on a breadbord.QuoteIf I bring down Ucbelow Ueby placing two diodes in the collector path, it should saturate? I did not draw these in my sketches, correct, but that should be possible?

author: Wimberleytech
author_rank: Super Contributor
timestamp: February 02, 2018, 05:00:10 pm
content: Quote from: hirada on February 02, 2018, 04:04:29 pmThen it is NOT an emitter follower.

author: rstofer
author_rank: Super Contributor
timestamp: February 02, 2018, 05:02:29 pm
content: Quote from: hirada on February 02, 2018, 04:04:29 pmI don't think I grasp how you plan to get Vc less than Ve on a conducting NPN device. Maybe a sketch would clear it up.

author: Wimberleytech
author_rank: Super Contributor
timestamp: February 02, 2018, 05:08:52 pm
content: Quote from: hirada on February 02, 2018, 04:04:29 pmNo. That is not how bipolar transistors work. For the emitter follower--which you are using as your template for understanding--feedback is at play, causing the emitter voltage to rise until there is equilibrium.

author: Audioguru
author_rank: Super Contributor
timestamp: February 02, 2018, 06:21:52 pm
content: You show a comparator with an output of +4.7V at 3mA which is impossible. A comparator is "open collector output" and has nothing to make its output produce a positive voltage and current.A Mosfet follower is even worse than an emitter follower. It has a gate-source voltage as high as 10V!

author: Audioguru
author_rank: Super Contributor
timestamp: February 02, 2018, 06:58:02 pm
content: Do it like this:

author: floobydust
author_rank: Super Contributor
timestamp: February 02, 2018, 07:30:38 pm
content: Quote from: Audioguru on February 02, 2018, 06:58:02 pm+1but I like less than 47k to drive BJT at 5VMost comparators have open-collector/open-drain outputs and so they cannot source any output current.I'm not sure what part you are using for your simulations, but check.

author: Damianos
author_rank: Frequent Contributor
timestamp: February 02, 2018, 07:38:46 pm
content: Quote from: Audioguru on February 02, 2018, 06:21:52 pmWrong information!Some examples:Push-Pull Output Comparators

author: hirada
author_rank: Regular Contributor
timestamp: February 02, 2018, 09:21:37 pm
content: Sorry, I forgott to mention, that I only use push pull devices, so far open collector/drain does not make any sense for me.QuoteSketch is attached. The two diodes (or a zener) create a voltage drop of ~1.4V, that brings Ucdown to 3.6V. Uehas a value of 4V, so Ucshould be way smaller. From what I've read so far, this is not a problem when the BJT happens to be in the saturation region. But I'll happily accept that I've read bs, if well-founded.QuoteOk, I stand corrected here, did not know that. But it remains a high side switch, what is the relevant part. I just concluded, that it still was an emitter follower due to the location of the load.QuoteWhich makes things complicated again, because, if feedback is at play, where is the loop?I do agree, that "sucks it in" was a really bad wording, I should have said, that the base impedance is changing depending on Icand beta. But with feedback that may be a wrong assumption as well.@Audioguru:I would like to avoid pullup if possible for the same reason I avoid open collector: it wastes energy, as this thing will in total be more idle than active. And since most (all?) push-pull devices also can sink current, I would have a constant flow of current for nothing. And the case is rather small, so heat could become an issue. We'll see, I lack experience here. But no need to push it.

author: rstofer
author_rank: Super Contributor
timestamp: February 02, 2018, 10:05:45 pm
content: Quote from: hirada on February 02, 2018, 09:21:37 pmNo, the base has a value of 4.7V but that doesn't mean that under all conditions the emitter will be 4.0V. I suspect that overdriving the base will get the device into saturation and this will cause Ve = Vc - 0.2V. Vc is 3.6V so Ve should be on the order of 3.4V. I don't think the base current will make up the difference.I tried this with LTspice and the result is Vc > Ve as I had hoped it would be. VceSat seems really low but it is still a drop.Notice that the current from the 5v source is essentially 0. All of the emitter current is coming from the base as near as I can tell.

author: Zero999
author_rank: Super Contributor
timestamp: February 03, 2018, 05:40:50 pm
content: Quote from: Damianos on February 02, 2018, 07:38:46 pmYes, comparators with a push-pull output are available, but the majority of comparators are open collector. Another disadvantage of only using push-pull comparators is it restricts the voltage to under 16V. I've been here before. If there's a push-pull comparator available with a higher voltage rating, which isn't too expensive and is easy to obtain, I'd like to know.

author: hirada
author_rank: Regular Contributor
timestamp: February 04, 2018, 11:49:22 am
content: QuoteThat makes we wonder again. In previous posts I've learned, that Ibis a function of beta and Ic. So, being saturated, let's assume a beta of 10 and said 1mA through the LED, than I would expect Ibto be 100uA?Besides, I did not sketch Rl, that is in parallel to the LED and will vary between 1-50k. So Uoutis basically the terminal for the actual load, the LED is just an indicitator that the switch is active. I have both in my inital sketches. Also there will be an output resistor in front of the load to limit short circuit current.QuoteThanks for the heads up and again for the effort with LTSPice. Had a look into the 2N2222 Datasheet, and in deed, Vbesatcan be as high as 2.6V (even though for way higher Ic), haven't paid any attention to this so far. Will be intersting to see, when bringing Vcefurther down, who wins: Vcesator Vbesat.Anyway, circle is closing, this leads us where bd139 was in the first reply: Common collector (or kind of) is only of limited use as a switch. But now at least I have a basic foundation for experimenting. And a book to investigate into.Again thanks to all for your help!QuoteEven those with 12V are rather scarce, and then additionally also rather slow in comparisson. Still fast enough for me old man, but you are right, air is getting thin above 5V. That is why I originally thought, I'd put a transistor behind, to step up, kind of like one would do with a relay: 5Vb, 12Vc->e. I was a bit naive here and obiously that does not work out and seems harder to archieve than I initially thought.Looking into analog switches may be worthwhile here or perhaps symmetrical jfets, but latter ones will be a different topic alltogether after I got this one halfway sorted out.

author: Wimberleytech
author_rank: Super Contributor
timestamp: February 04, 2018, 09:12:57 pm
content: QuoteBeta depends on terminal conditions of the BJT. If a BJT is in deep saturation beta is much smaller than it is in, say, optimally biased condition.You really need to spend a day and just study the BJT.