transistor gain

Started by Vitrolin, May 26, 2009, 09:04:24 PM

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Vitrolin

i would like to know, though this migt be obvious to some it isn't to me, how do i know the amount of gain available in a transistor, and when it would start to saturate? this question applies to BJTs, MOS- and JFETs.

R.G.

Quote from: Vitrolin on May 26, 2009, 09:04:24 PM
i would like to know, though this migt be obvious to some it isn't to me, how do i know the amount of gain available in a transistor, and when it would start to saturate? this question applies to BJTs, MOS- and JFETs.
The amount of gain available in a transistor, whether BJT, MOSFET, or JFET, is at least as dependent on the circuitry around it as it is the transistor itself, so there is no concise answer to your first question. I've seen gains from less than one up to a couple of thousand depending on the external circuitry and device.  The biggest issues are the external impedances on the base, emitter and collector (or their analogs in FETs). It's not a simple answer.

As to when they start to saturate, it depends on what you mean by "saturate". That word has a specific meaning for transistors, for their operation where increasing the input drive no longer produces concomitant increases in current. Outside electronics, it has several meanings, including in the music world for "distortion". Did you actually mean when would a transistor start to distort?
R.G.

In response to the questions in the forum - PCB Layout for Musical Effects is available from The Book Patch. Search "PCB Layout" and it ought to appear.

Vitrolin

yes, i did mean when it would distort.

the point in all this is to select transistors to saturate/distort/overdrive other transistors.

GibsonGM

It's pretty complicated, Vitrolin, but the characteristic curves that are available in specific transistors' datasheets show how they operate (and take some time to learn to interpret!).

A "properly biased" transistor operates somewhere on the linear portion of its curve for a given input and a given output current. What comes out is a larger and pretty faithfully reproduced (and possibly inverted) version of the input signal. To make it saturate, meaning no longer be able to amplify linearly but to distort (a non-faithful reproduction of the input signal), means to design circuitry around the transistor that will cause it to operate in the the NON-linear part of its curve.  This amounts to starving it for input power, overloading the input, or biasing it to operate way "out of range" (at least for linearity).  Essentially this means "cranking it up", lol.

For what you are talking about doing, you might be operating the 1st transistor well within its linear range, and just pushing a lot of output power at the 2nd transistor.  For the most part, when I do this sort of thing, I play with the gain of both until I end up where I want. You could try 2 Big Muff recovery stages to experiment with for this.... 

There are formulae that will tell you exactly where you'll end up on the curves of both transistors - a lot of Ohm's Law, too - but I like the hands-on approach!  Not too complicated, but the biasing aspect involves making assumptions about the output, and working backwards to get the right voltages/currents at the base & emitter to produce a desired output current at the collector - for common emitter circuits, that is.

Check out "transistor tutorial" on Google and find a good site, and/or get a book on electronics that has some detail about transistors.  Once you get past the unfamiliar and odd concepts, you'll be glad you did, ha ha.  It eventually becomes 2nd nature in a way (until you learn more from a guru!).

And, I bet R.G. will be back, and will have a better explanation for you!!!     
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Vitrolin

what really triggered this is i build a ROG flipster and it works fine, but this the page says as follows:

"Possible Modifications:

As seen on the schematic, you can try any number of N-channel JFETs in the circuit. Some beta testers found the circuit, when fully loaded with J201, had too little headroom. By mixing J201 and with lower gain FETs, such as 2N5457 or MPF102, the gain level can be customized to your taste."

so wondered how do i know higher gain trans from lower gain, without testing, i can get a bit lazy some times. :icon_lol:

but thanks for you, for helping me out.

Ripthorn

Probably what you are looking for is a transistors hfe rating.  This can be measured by some dmm's.
Exact science is not an exact science - Nikola Tesla in The Prestige
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valdiorn

Quote from: Vitrolin on May 30, 2009, 09:52:12 PM
what really triggered this is i build a ROG flipster and it works fine, but this the page says as follows:

"Possible Modifications:

As seen on the schematic, you can try any number of N-channel JFETs in the circuit. Some beta testers found the circuit, when fully loaded with J201, had too little headroom. By mixing J201 and with lower gain FETs, such as 2N5457 or MPF102, the gain level can be customized to your taste."

so wondered how do i know higher gain trans from lower gain, without testing, i can get a bit lazy some times. :icon_lol:

but thanks for you, for helping me out.

JFETs work on a different principle from the NPN transistors. They have something called a Threshold voltage. Remember, when applying 0 volts between gate and source an n-channel (normal) JFET conducts at its best, but apply negative boltage to the gate (compared to ground) and the transistor lets less and less current pass through. At some point there is almost zero current, and that value of gate voltage is called threshold voltage, or pinch-off voltage.

The J201 has a very small pinch-off voltage, around -1.5v. this mean that any signal below -1.5v will not make a change in the conductance of the transistor (it is effectively in cutoff mode = distortion in the audio signal), thus you only need a 1.5v peak-to-peak signal to swing the drain output of a J201 all the way between the power rails.

The MPF102 or 2N5484 have a large threshold voltage, which means you need a larger signal to distort the output signal in the way previously described. Thus, they are lower gain than the J201...

brett

Hi
In the case of JFETs there's an extra factor to consider besides the direct effect of the "pinchoff" voltage (Vgsoff, -volts).  As well as getting more "headroom" in many circuits with low Vgsoff, you'll also get less transconductance (the thing that causes gain).  So the effect of lower Vgsoff is twofold.  In fact, JFETs with low (more negative) Vgsoff are fairly unpopular for this reason (too clean, too low gain).  Examples include the MPF102 and the 2N5486 (Vgsoff about -5 or -6 V).  The 2N5484/5/6 series is convenient, because the XXXX84 has higher Vgsoff than the XXXX85, which is higher than the XXXX86.  Check out their datasheets.  I think you'll find the 2N5484 a "mild", general-purpose device relative to the high gm, high Vgsoff J201. 
cheers
Brett Robinson
Let a hundred flowers bloom, let a hundred schools of thought contend. (Mao Zedong)