BJT Emitter Resistors

[Faber]

I've been studying BJTs for a while now and I've been looking at this:  http://www.diystompboxes.com/wiki/index.php?title=Debugging

I have some questions:

1.  The power-collector voltage is the difference between between the battery and the collector, across Rc right?  From what I read in the Debugging the answer is yes, but I just want to make sure.
2.  Do the positive and negative voltage swings need to be equal?  From what I read, the answer is no.
3.  How can you calculate Ve without a voltmeter?

Thanks!

[SISKO]

1. Dont understand
2. Most of the time, yes. Thats why some collectors are biased at half the power supply voltage, (Fuzz Face Q2 collector voltage = 4.5V).
3. There are may ways.  Have a look here http://pisotones.com/RangeMaster/RainMaster/Rainmaster.htm

[zachomega]

For silicon transistor:
Ve = Vb - .6V

For germanium transistor:
Ve = Vb - .25V

-Zach Omega

[Sir H C]

1.  Yes, the voltage across the resistor is the supply voltage minus that on the collector.

2.  No.  Usually if you are wanting to maximize your swing then you would want the swings to be equal, but for some low swing situations not having equal swings works better for other reasons (noise, supply variance, etc.).  Also for asymmetrical clipping for fuzzes, uncentered output swings can be a benefit.

3.  Well you know with high beta transistors, collector current is about equal to emitter current.  Therefore if you know the drop across the collector resistor, then you can calculate the Ve voltage by knowing the current in the collector resistor (Ic = Vc / Rc), and then Ve = Ic * Re.  Without knowing any voltages other than the supply, you have to go through a bunch of math and then get the value and this requires you to know the base biasing scheme and the beta for the transistor (importance of this depends on the biasing method).

[brett]

Hi
I use this a lot.  It saves me time and headaches.  Most of the time you'll want to change the bias type from "series resistor" to "voltage divider" for stompboxes.

http://www.daycounter.com/Calculators/Transistor-Bias/NPN-Transistor-Bias-Calculator.phtml

Brett

[R.G.]

I've been studying BJTs for a while now and I've been looking at this: 

But have you read the article on how to bias a BJT at GEO or the several times I've typed it in here in the forum?

1.  The power-collector voltage is the difference between between the battery and the collector, across Rc right?  From what I read in the Debugging the answer is yes, but I just want to make sure.

Yes.

2.  Do the positive and negative voltage swings need to be equal?  From what I read, the answer is no.

The answer is maybe. And the answer to the next question (that is, how do you tell how big to make them) is YOUR job as the designer to answer. You have to know whether equal positive and negative swings are necessary in your application or desirable. The only thing that equal positive and negative gives you is the biggest signal possible from that power supply voltage before clipping. Normal EE practice is that equal and the biggest possible swings are the right way. In musical effects, we know the answer is not always according to normal practice. Equal swings is only good for getting biggest possible undistorted output.

3.  How can you calculate Ve without a voltmeter?

Ohm's law.

If you know the power supply voltage and the collector voltage as well as the collector and emitter resistances, the collector current by ohm's law must be (Vpower-Vc)/Rc.

That same current (plus the 1% or less added base current) must flow through the emitter. So you can calculate Ve as Ic*Re.

In designing a BJT circuit, you, the designer, first DECIDE what proportion of your available power supply you want to assign to the collector resistor, the Vce, and the emitter resistor. The sum of those three is the power supply. Once you have decided the voltage apportionment, you know what voltage the collector and emitter should be for your application. Then you DECIDE how much current you want to flow in the stage. BJTs work as amplifiers from microamperes to tens of amperes. As the designer, you pick. Many amplifier transistors are specified for hfe at 100uA and 1ma. 1ma makes it easy, if a bit high current. 100-200uA is common for small signal circuits.

So you decide the stage current. Once you do that, the value of the collector and emitter resistors are fixed because both their voltage (from the apportionment) and their currents (which are the same) are decided, and Ohm's law tells you the resistance.

When you have decided the emitter voltage, you have also decided the base voltage to a high degree. This is a BJT, and the base must and always will be one diode drop higher than the emitter (NB except for very high current applications), so the base voltage to bias properly must and always will be Ve+Vbe. It's now up to YOU the designer to make this come true by picking resistors to bias the base to force that voltage onto the base and provide enough current into the base to keep the collector current happy at those voltages.

And that's how you bias a four- or five-resistor BJT.

[R.G.]

Hi i use this a lot.  It saves me time and headaches.  Most of the time you'll want to change the bias type from "series resistor" to "voltage divider" for stompboxes.

Each time you use a special purpose calculator to do the work for you, you put off by one more time the work you would need to do to really understand and be familiar with whatever you are calculating. This calculator, all by itself, helps keep many, many people from understanding how to bias a BJT. But it sure keeps the traffic up to that web site, yes? 

[kvb]

Sometimes when I look through GEOFEX I have trouble locating certain things. So I took a quick look and found the article about biasing under "electronics" in the "how it works" article.

Just a quick note for anyone like me (I'm pretty much an E hack) who might actually start doing the math some day.

[Faber]

Let this be a credit to RG's genius.  He answered half of another question while answering this one!

A huge thanks to all of you!!!

[brett]

Hi

Each time you use a special purpose calculator to do the work for you, you put off by one more time the work you would need to do to really understand and be familiar with whatever you are calculating.

Oops! You are so right, RG.  Shortcuts can leave you shortchanged.  I should have said something like - I did the calculations a hundred times by hand, and used some not very helpful calculators before I found this site. 

Users should be aware that there's a difference between (i) punching in "dumb" numbers and getting estimated results, and (ii) the way the circuit will actually work.  A good example of this in the current ( ) example is increasing the collector resistance.  Change Rc and the calculator won't ask for a new value for Beta, even though Beta declines at high Rc (low Ic), and it might be important.  Any calculator is only as useful (or useless) as the user.