New old question about biasing.

[mjarus]

O.k. I have a few questions about biasing.  I have searched as much as I can think of, but there is something that just has to be eluding me...

My questions will concern BJT transistors.  Particularly the 2N3904.  Datasheet is here: http://www.jameco.com/wcsstore/Jameco/Products/ProdDS/178597.pdf

I was interested in putting this in a Green Ringer and also wanted to change the resistor values that are currently used for biasing it to values more easily obtainable (i.e. the common ones that I have).  Green ringer page is here:
http://www.generalguitargadgets.com/index.php?option=content&task=view&id=59&Itemid=85

So here we go...
1)  Everyone always says to put the voltage divider on the base @ 4.5v, but they have it @ 2V.  Why?  What should the base voltage be at?  Shouldn't it just be 0.7v over the emitter?

2)  How do you decide what voltages to put onto the emitter and collector when designing a circuit?  I assume that you want the largest area before saturation right?  How is this figured out?

3) Everywhere I go on the 'net there is some plot of base current vs. collector current and voltage.  I have YET to see one like they show on any datasheet.  Where are the getting these things from?!  An example is here: http://www.mitedu.freeserve.co.uk/Design/bjtbias.htm#Self%20Stabilizing in the first picture.  There is an example at the bottom, but how did he get 10V @ the collector?  Every example works with like 20Vcc.  The datasheet works with 5V, I want to use 9V... I know there is a trick I am missing somewhere... but where?!

I have read almost all of RG's stuff @ GEO, and I am sure the answer is probably there somewhere, but I sure can't find it!

For extra credit... using the datasheet for the 2N3904 how do you come up with the resistor values for biasing with a 9Vcc supply?

Many, many thanks in advance!

mjarus.

[George Giblet]

What you are asking isn't easily answered because there's no unique solution.  You want to know how to design transistor circuits.  There's many trade-offs and that's why you see different designs.

1)  Not true.  That is generally what's chosen for emitter followers but beyond that there is no requirement for the base to be at half the supply.  The actually base voltage is below 4.5V with equal base resistors because the transistor base current make is drop a little.  The base is approximately 0.6 to 0.7 above the emitter.

2)  You often start off with a fixed supply voltage, say 9vDC for effects. Then perhaps some other requirements like how much gain you want, the output impedance you need to drive the next stage, and the input impedance.  There's usually a bit of slack in choosing the output impedance.

The output impedance pretty much determines the collector resistor.  The gain and the collector resistor determines the emitter resistor.  Once that is chosen you juggle the collector current to get the best swing; and that is done by juggling the base voltage.  You pick the size of the base resistor to ensure your input impedance is OK.  In high gain circuits you will need a low emitter resistor and this makes the input impedance drop.  It might drop too far regardless of how big you make the base resistors.

That's where you start.  From that point you see if all your requirements are met.  If you have less than about 0.6V across the emitter resistor your biasing might not be stable enough with temperature. Here you can use a larger emitter resistor then use a series cap and resistor across the emitter resistor (but you trade the bias stability with loss of swing).  If your base biasing resistors are too large you circuit's bias point might be too dependent on the transistors gain; in this case consider lower bias resistors and buffer before that stage.

Low output impedance means low collector values and that means more current drain from the battery.  You might consider large collector resistors and a buffer following that stage.

The choices go on and we haven't even touched on noise issues.

3)  That comes from days when there was a piece of test equipment called a curve tracer.  The idea was useful for valves but not that great for semiconductors.  Those graphs generally aren't that useful and that's why you don't see them in data sheets much.  The slope of the flat part is useful but it's too flat on the scale of the graphs; sometime manufactures give a hoe or ro value for that slope which is more accurate/useful.

[R.G.]

Let me amplify on George's comments.

Biasing bipolars is fairly simple, but it's confusing because the idea is to make the transistor not matter.

There are two tricks in biasing a bipolar as a linear amplifier.
(1) the emitter always goes to 0.6V below the base if it's possible for it to do so.
(2) the gain is determined by the collector and emitter resistors.

To do a bias from scratch, you first know your supply voltage. We'll pick 9V for everyone here. Then you decide what you want this stage to do, and that largely determines the collector and emitter resistors. In general, for biggest linear swing (which is sometimes not what effects designers want) you put the output pin in the middle of the available power supply.

So if you're doing an emitter follower, you want the emitter at 4.5V to get the biggest possible swing. The base then must be 0.6V higher, and you then want it at 5.1V. Since you want all the voltage swing across the emitter resistor, you make the collector resistor be zero so it doesn't eat up any of the available voltage. If you already have a 4.5V bias point that youuse for other things, then it's acceptable to put the base at 4.5V and the emitter will be at 3.9-4.0V and you'll have only lost a little linear swing.

If you want two outputs, collector and emitter, you put a resistor in both leads and take an output from each. This is the classical phase splitter from things like the Univibe and the envelope detector in the dynacomp. You'll want equal swings from each one, and since essentially the same current flows in both collector and emitter, you make the resistors equal. When the transistor is off, the voltage between them is 9V and when the transistor is fully on, the voltage is 0V. So in the middle, no signal, the voltage across each resistor must be 1/4 of the power supply. So that makes the static voltage be 1/2 the supply across the transistor. The base sits at 1/4 of the powser supply plus 0.6V, or 2.25V + 0.6  = 2.85V.

If you want an amplifier: the same current flows in the collector and emitter resistors. Since the emitter follows the base, the "gain" at the emitter is one. The signal voltage at the collector will then be the collector resistor divided by the emitter resistor.  Once again, you have to decide what portion of the power supply you spend on the transistor at zero signal compared to the collector resistor. Since the biggest linear swing is when the transistor and collector resistor have equal voltages, but the emitter resistor eats some as well, the transistor voltage is then picked to be equal to the collector resistor voltage but subtracting off the emitter voltage. The base is the placed at a voltage to make that emitter voltage come true.

So how do you place the base voltage (which is where you started). Cheapest is with a single resistor from the power supply which eats up just the right amount of voltage with the base current needed. This is a bad, bad way to do it because each transistor is different, and the base current is fairly unpredictable.

Next cheapest is with two resistors in a voltage divider across the power supply. You have picked a base voltage in the collector/emitter biasing above. So now you have to make that come true. If the upper resistor is R1 and the lower resistor is R2, then the base voltage is
   Vsupply* R2/(R1+R2)
But that assumes the base current is insignificant compared to the resistors' current. We make that true by calculating roughly the base current. The transistor's emitter current is the emitter voltage divided by the emitter resistor. The base MUST have a current equal to the emitter current divided by HFE going into it. For, say, a 1ma emitter current, the base current in a 2N3904 with a minimum HFE of 200 will be 5uA.

So we have to make the resistor divider current be much larger than 5uA. We can easily do that by making it at least ten and preferably more times greater. So - PICK THE RESISTOR STRING CURRENT to be more than ten times the bias current. In this case, twenty times is good, just picking a number out of the air, and the resistor string needs to be about 100uA.

So 9V/(R1+R2) = 100uA and 9V*R2/(R1+R2) = base voltage. And when you solve that set of two equations in two unknowns, you have your resistor values. They will NOT in general be standard values.

GREAT. NOW WHAT?

You go easter egging for standard values close to the values you came up with and see how bad it is with near standard values. Generally you can come close. If not, you pick a different current for the string. 9V/(R1+R2) could also be 200uA if that makes standard values come out better.

And that's how it's done. There's a lot more on AC versus DC gain, emitter resistor bypassing, etc. But then I took a 3 credit hour course in this.

[mjarus]

Thanks a lot guys!  I had a feeling it came down to a bit of fiddling, but I didn't know where to even start.  I appreciate the help!  I know that in depth answers take time, and I really appreciate the time that both of you took to help me get started.  

R.G... thanks again.  It was GEO that originally got me started into deciding that this was something that I could do.  I must have read about %70 of the stuff on your site... and that is ALOT.  BTW, before I saw your remote bypass board and I was planning on an AVR controlled board.  I think that analog effects with digital controls would rock.  Have you looked into using the digital pot chips?  Are they as good as real pots?  I was thinking about making an entire control board, not just bypass like my Korg AX1500 but using analog effects.

George, that is one heck of a third post.  Very helpful and insightful, I will have more questions for ya in the future... so please stick around!

mjarus.

[mjarus]

I noticed that neither of your analysis used anything from the datasheet.  For stompboxes, does none of the information on the datasheet really matter?  I would find it hard to believe that the answer is no... but is it?!

In this example I found on another site (common emitter amplifier):

Vcc=20V, Vc=10V, Ic = 1mA, transistor is BC107A with hFE=195

Rc= Vc /Ic = 10 / 1m = 10k
Ve = 10% * 20 = 2V
Re = Ve / Ie= 2 / 1= 2k
Vb = 2+ 0.6 = 2.6V
Ib = Ic / hFE = 1 / 195 =0.005128mA
R2 = Vb / 10* Ib = 2.6 / 0.05128 = 50.7k (use 47K)
R1 = Vcc-Vb / 10 * Ib = (20-2.6) / 0.05128 = 339.3k (use 330K)

1) Is this a valid way to bias?  They basically set the Ve= 10% of Vcc and calculate the rest from there.  How did they get Vc of 10?  Just 1/2 of 20?

2) Where did they get the Ic=1ma?  From the datasheet?  If so would the 2N3904 do well with 10ma?

For what R.G. said... I guess I got the base biasing part... but what about the collector/emitter voltages?  That part was a little confusing... sorry.

I know I may be being a PITA, but I would really like to learn this stuff rather than just soldering stuff to a pcb.

mjarus

[R.G.]

I noticed that neither of your analysis used anything from the datasheet. For stompboxes, does none of the information on the datasheet really matter? I would find it hard to believe that the answer is no... but is it?!

You do use the datasheet - but you use it like an adversarial set of evidence in court. The info on the datasheet is the minimum the maker can tell you and still get you to buy the thing. Most of the data is couched as "typical" values - which can be anything in the real devices you get - or as only a minimum or maximum. F'rinstance, Hfe is ofted specified as something like a minimum of 100. What maximum? Do I sometimes get one with Hfe of 10000?

The things to consult the datasheet for are the guaranteed minima and maxima. Vceo is where the voltage breaks down the collector-base diode, so if that's less than your power supply voltage, the power supply is gonna kill the transistor. So you pick one with Vceo bigger than your power supply. This mostly doesn't matter with 9V batteries, but it does for some things. Maximum collector current had better be bigger than you want to use, and the maximum power dissipation needs to be bigger than you'll dissipate. Gain has to be... enough. Generally that's 100 or more.

I keep saying this, but I don't think it soaks in to a lot of people: The point of good transistor design is to MAKE THE TRANSISTOR SPECIFICATIONS NOT MATTER.  Design so a transistor with a gain of 100, 200, 500 all work.
Design so that any device with ft of over 100MHz works. Design so any device that will dissipate over 100mW works. A lot of high powered time went into thinking about how to make the individual device not matter so you don't have to select for the few perfect devices.


In this example I found on another site (common emitter amplifier):

Vcc=20V, Vc=10V, Ic = 1mA, transistor is BC107A with hFE=195

Rc= Vc /Ic = 10 / 1m = 10k
Ve = 10% * 20 = 2V
Re = Ve / Ie= 2 / 1= 2k
Vb = 2+ 0.6 = 2.6V
Ib = Ic / hFE = 1 / 195 =0.005128mA
R2 = Vb / 10* Ib = 2.6 / 0.05128 = 50.7k (use 47K)
R1 = Vcc-Vb / 10 * Ib = (20-2.6) / 0.05128 = 339.3k (use 330K)

1) Is this a valid way to bias? They basically set the Ve= 10% of Vcc and calculate the rest from there. How did they get Vc of 10? Just 1/2 of 20?

Yes. You apportion how much voltage you allow to collector resistor, transistor, and emitter resistor, then calculate from there. It's quite similar to what I typed. I personally would quibble that they've not done a good job, since they should have set the collector at half of whatever was left over after subtracting the emitter voltage, but they're close.

2) Where did they get the Ic=1ma? From the datasheet? If so would the 2N3904 do well with 10ma?

They decided "Hmmm, I think I'll run this baby at about 1ma." They then checked the datasheet (or should have) to see that it would do 1ma and that 1ma times Vce would be under the power dissipation limit. It probably will, as 1ma and 20V is only 20mW. The 2N3904 is specified for Icmax of 100ma, and a power dissipation of 100mW in the TO-92 package as I recall. So with 10ma and 10V across the device, it's only dissipating 10mW, and it's OK. Doing that means that the collector and emitter resistors have to drop by a factor of ten to keep the same voltages with ten times the current, of course. And the base current comes up by about the same factor of ten, but yes, it will work.

For what R.G. said... I guess I got the base biasing part... but what about the collector/emitter voltages? That part was a little confusing... sorry.

I said about what your other source said: Pick where you want your voltages to be on collector and emitter depending you your application; then pick the current you want through the device. When you have done that, the resistors are determined. For an emitter follower, you want half the power supply across the emitter resistor, none across the collector resistor, and enough current to drive your load. For small signal cases, you can often just pick a value of current. The range of 10uA to 10ma covers almost all cases, and 100uA to 1ma is most common. Say we want an emitter voltage of 10V for a 20V supply. The collector voltage is 20V, and if we pick 10ma, then the emitter resistor is 10v/10ma = 1K. The collector resistor is 0V/10ma - 0 ohms.

If we want a gain of 10 at 1ma, and biggest voltage swing, we know that we'll need a collector resistor 10 times the emitter resistor, and the voltage across the transistor should be about half the voltage supply. For a 10V supply, the voltage on the collector is ten parts, the transistor ten parts, and the emitter one part, so the emitter voltage is 1/21st of 10V; the collector resistor has 10/21 of the supply as does the transistor. So the emitter resistor is (10V/21)/0.001 = 476 ohms. The collector resistor is 4760 ohms. Picking nearest standard values of 470 and 4.7K gets you very close.

You pick the operating current, power supply voltage, and apportioned voltages on collector resistor, transistor, and emitter resistor, and then just do Ohm's law to get the values. You'll then have to diddle to get it in standard values, but you can get really close.

Once you know the emitter voltage, you know the base voltage, because the base voltage on a bipolar ***MUST*** be about one diode drop higher or the transistor won't make your careful planning come true. So you set the base voltage to be that, and using some guess at Hfe, you allow enough current to flow into the base to make the base current satisfy the base voltage.

I know I may be being a PITA, but I would really like to learn this stuff rather than just soldering stuff to a pcb.

PITA? Are you kidding? You're the first person in a long time to want to really know how this works. I wish I had one like you every day.

[aron]

ARCHIVE

[mjarus]

I wish I had one like you every day.

Thanks!  Coming from someone of your stature in the community that means alot!

I personally would quibble that they've not done a good job, since they should have set the collector at half of whatever was left over after subtracting the emitter voltage, but they're close.

So in the example the collector should REALLY have been set at approx. 9V, right? Vc=(Vcc-Ve)/2 or 9=(20-2)/2   And the fact that they chose 10% of Vcc for Ve is a good rule of thumb?  They explain it as the amount of drift to allow for temperature fluctuations that could otherwise effect bias.

If all of this is true... then for any small signal amplifier BJT (for 9V stompboxes, with "reasonable" hfe, etc...) I can just start with Vcc=9.00;Ve=0.90;Vc=4.05 and Vb=1.5... before fiddling, right?  I know currents are important too, but those will be on a case by case basis depending on the amount of current I want to flow on the device, right?  After all that I think I got it...

I assume that the lower you go on Vc below (Vcc-Ve)/2 the more you will clip with a bottoming out at Vb, at which point you will already be saturated right?

A lot to digest, and I appreciate all of your time.  You should really get your posts together from this discussion and put them in a tech note on GEO.  I know you talk about these different things at different places on there, but seeing it all together has REALLY made the 'ole lightbulb come on for me!  This has to be some of the best information I have seen on the subject in one place.

mjarus.

edit:  looks like aron already beat me to saying it! (archive)

[vortex]

This is an interesting thread. I'm trying to get my head around biasing and have just cracked Jack Darr's Transistor Audio Amplifiers book. I'm still getting a toehold on this. In the past I have built Rangemaster and Fuzzface type effects and have tuned in the collector voltages by ear and by meter.

So at the moment, I have built the Interfax Harmonic Perculator and have set up a breadboard with pots to tweak all of the resistor values. I am two days into it and there are quite a few qualities to the circuit that I have dialed in. Once in a while, when I like what I am hearing I check the voltages on the transistors and it seems that there are times where the voltages look "sensible" other times they don't conform to the standard rules of biasing. Either way, I am finding that the sound is dynamic and pleasing to my ear. It's quite exciting to have the control of "designing" even if it is not based an technical data.

I look forward to having a better technical grasp on the situation, but the "tuning by ear" is a great experience.

[R.G.]

So in the example the collector should REALLY have been set at approx. 9V, right?Vc=(Vcc-Ve)/2 or 9=(20-2)/2

That's correct. Well, it's correct in a picky way.

Getting the biggest linear swing out of any amplifier implies that you bias it in the dead middle of the available voltage so it can swing equal amounts positive and negative. Otherwise, one polarity bangs into the closest power supply first. This may or may not be an advantage. If your output signal is always 100mv or less, then it doesn't matter much that you have 3volts one way and 4 volts the other before distortion.

Or you may want to limit first in one direction for the distortion, the classical example in effects being the Rangemaster.

And the fact that they chose 10% of Vcc for Ve is a good rule of thumb? They explain it as the amount of drift to allow for temperature fluctuations that could otherwise effect bias.

I've been giving you the simplistic explanation. Let's go deeper.

The bias point really depends on both the base current and Hfe. Hfe increases with temperature, leakage increases with temperature, and base-emitter junction voltage decreases with temperature, all of which add up to more current as the temperature goes up. Bipolar transistors that dissipate any significant power are always in serious danger of thermal drift if not outright thermal runaway and death if the power supply allows that. Of the three positive temperature coefficient effects, all three get ameliorated by emitter resistor feedback.

The base-emitter junction voltage is somewhere between 0.5 and 0.7, with 0.5 to 0.6V being most common in small signal amplifiers. Some very low current bias schemes can get Vbe as low as 0.4V. But it drifts smaller with temperatures. If you bias the base with no emitter resistor, then the only thing that limits current is the exponential function of base current with base voltage and the internal Shockley emitter resistance. This is one reason the first transistor in a Fuzz Face drifts - it has no emitter ballasting.

If you put a resistor in the emitter, it will drop a (relatively) constant voltage with temperature. And the biasing point is then determined by Vbe  PLUS the emitter resistor voltage. So if the emitter voltage is big compared to the Vbe drift, the drift doesn't much matter and the bias point will not drift significantly. So the question is - how much is enough?

In general, things that are one order of magnitude larger than other things make the smaller things insignificant. Vbe drift is about 2mV/C, so with a 25C change in temperature, the Vbe can move 50mV. And current is an exponential function of that Vbe, so current changes a lot. But if the emitter voltage is in series with that Vbe, and it's as much as 0.5V, then 50mV of drift doesn't change things much.

For instance, assume you have biased an NPN with Vbe = 0.5V and no emitter resistor and Ic = 1ma. If Vbe drifts 50mV, then the Ic increases  by 30% to 1.3ma. If you have a 0.5V emitter ballast, then the base is biased at 1.0V, and when the temperature drift happens, the emitter voltage changes from 0.5V to 0.55V, and the current has increased only 10%. If Ve is 1V, then the current increases only 5%. That's the thermal stabilization.

If you also want to get a specified gain, not just as much as you can, the emitter resistor does that too. If the transistor's gain is as much as 100, then the emitter current is equal to the collector current plus the base current, and the base current is only 1% of the collector current, so we can in general ignore it. That says that the emitter current equals the collector current. If that's true, and it's also true that the emitter voltage always must follow the base voltage (it does) then the gain of the transistor stage simply must be the ratio of the collector resistor and the emitter resistor. This is always lower than the no-emitter-resistor gain of the device. In effect, the emitter resistor is providing feedback to stabilize the Vbe temperature variations, and loss of gain is the price we pay for stability.

If you want your gain back, you can get that by placing a big capacitor across the emitter resistor, or across part of it. The capacitor has a low impedance at signal frequencies, and so to AC the emitter looks shorted to ground, and the gain goes back up to the raw device gain. If you split the emitter resistor into two parts, then they both contribute to DC stability and only the unbypassed one determines AC gain with Rc.

There are other refinements.

[mjarus]

Thanks again R.G.!  Another great post.  I will hold the rest of my questions while I breadboard some of this stuff to let it "set" in my mind.  I am sure that I will wind up with a couple of blown transistors :shock: and a few more questions in a couple of days... :roll:

mjarus.

[seanm]

Ok, I want to make sure I have a couple of things straight.

As I see it, R.G. presented two methods of biasing transistor amps (I am ignoring emittor followers): setting for maximum voltage swing or setting for gain.

So let's take a 2N5088 Vcc=9, Ic=1mA, min hfe=350, Vbe=0.8 (I got this from Vbe(on) in the data sheet).

I end up with Rc 4k, Re 1k, R1 252k, R2 63k (no attempt to standardize)

This means a gain of 4. That means if I wanted a gain of 10, I would have asymetrical clipping, correct?

Also, I finally had a lightbulb moment with my bass preamp problems. If I want a gain of 10 (20dB) and I have a 2 V p-p input signal, I have a 20 V p-p output signal. This cannot be handled with a 12V power supply. If everything is balanced properly, I assume I need at 24V power supply to handle 20V p-p. Am I correct in thinking that for biasing, the peak to peak values are more useful?

P.S. I should be picking up a scope tomorrow  So things should get a bit clearer.

[R.G.]

This means a gain of 4. That means if I wanted a gain of 10, I would have asymetrical clipping, correct?

Good thinking - but no, not correct. The part that tripped you up was that gain has no relation to clipping. In fact, you have to turn the input signal level down so it does not clip to measure gain. If you have 1ma of collector current, 4K collector resistor and 1K emitter resistor, then the emitter voltage is 1V, the collector voltage is 4V down from 9V or 5V, and you have 4V from collector to emitter. In theory, this will clip symmetrically, as there is the same voltage between the static bias and saturation as there is cutoff.

And the gain is only 4 if you leave the emitter resistor un-bypassed. If you clip a big capacitor around the emitter resistor, then for all frequencies above the one where the cap's impedance is 1K, the gain is determined by the capacitor impedance (which goes to nearly zero) and the internal Shockley resistance, which is 26mV/0.001A, or 26 ohms. The gain will be quite high - and highly dependent on the gain of the transistor. If you make that emitter resistor be a pot and put the BFC across the wiper to ground, you can turn the gain up and down with the pot.

Lessee... did you do the other stuff right?  Base has to be at 1.6V about, so 9V*(63/(63+252)) = 1.8V. Ok, pretty close. The emitter will be a bit higher than 1V, the collector a bit lower than 5V, but not too shabby.

More important perhaps is your lightbulb moment.  YES!! That's it!  If the gain times the input voltage is more than the power supply, the transistor bangs into the power supply limits and can't do any more - it clips. It may bang into the power supply or saturation where Vce is approaching zero, or if you have it biased just right for biggest undistorted signal, it hits cutoff and saturation at the same time and clips symmetrically. If it hits either cutoff or saturation first, clipping is asymmetrical.

For small enough signals, everything will quit clipping. And yes, it's all about peak to peak signal volts. Once you choose the power supply voltage and the biasing point, the peak to peak output voltage is fixed. That's all there is and there ain't no more. So you can then independently set gain for your signals.

Good insights!!!

I honestly don't know how everyone lives without a scope.

[Gus]

I would like to add that sometimes you might not want the collector or emitter at 1/2 the power supply depending on the load.
  R.G. that was a lot of typing.
  Thank you for posting the point "make the transistor not matter".    I have tried to do that with the stuff I build( the 100 ohm and 10k resistors in the FF type circuits for first stage open loop gain and temp stablity).  In some of my schematics I wrote tune the bias by ear IMO 1/2 the power supply is not always what one wants.

A grounded emitter makes me wonder what the "designer" was thinking.

[R.G.]

I would like to add that sometimes you might not want the collector or emitter at 1/2 the power supply depending on the load.

You are of course perfectly correct. I just didn't want to get into the differences between AC and DC load lines. My fingers already have calluses on the ends, even my non-fretting hand.

I had to no-fooling sit through a 4 credit hour sophomore course which had this kind of stuff as its primary meat. We did get up to stagger tuning for FM receiver tuners, but most of the course and all of the labs was biasing, predictable gain, cascaded frequency response, loop stability, and so on. Dr. Allison was pretty clear - if the transistor's characteristics matter much in the circuit, you've done a really poor job of design. He graded accordingly on the class projects. I got the message.

Tuning audio by ear is always going to be preferable - if you can afford it. It's not affordable if you have a production line turning out 10K of these a day. That's the kind of situation that will make you understand the true meaning of one of my favorite quips - if you can't get your work done in 24 hours a day, work nights. 

[seanm]

Ok, here is my first attempt at an all transistor, non-fet, preamp. The Vcc is 24V and all the transistors are MPSA18s. I cannot take credit for the high impedance input buffer, it comes from the AMP 360 bass head. I want two stages of gain since I plan to put a tone stack between the stages.



In the breadboarded prototype I measured about 700k input impendance and 20dB of gain. The output sounded clean, but I didn't really push it. No Neil Young tonight

I may add an output buffer, since it will be running into 10k load.

Comments welcome, especially any issues with the input buffer. I think the 100pF cap is for noise reduction, but I am not sure.