Measuring gain of common emiter stage

[Brymus]

In a common emiter gain stage with a standard 4 resistors bias arrangment, could we calculate the gain of an unlnown NPN by:
Measuring the voltage drop across: collector R ,the V drop across base to ground R ,then take vbR *.1 *vcR =Hfe ?
Or would one have to actually solve for I across collector R,and base and emiter R then use,  Hfe = Ic divided by Ib ?
And how does the answer in gain correspond to increase in Db with a 100mv guitar signal ?

[R.G.]

In a common emiter gain stage with a standard 4 resistors bias arrangment, could we calculate the gain of an unlnown NPN by:
Measuring the voltage drop across: collector R ,the V drop across base to ground R ,then take vbR *.1 *vcR =Hfe ?
Or would one have to actually solve for I across collector R,and base and emiter R then use,  Hfe = Ic divided by Ib ?

I wouldn't do it either of those ways. I'd take note of the fact that if the current gain of the transistor is greater than about 100, which it almost always is for small signal NPNs, then the emitter current is less than 101% of the collector current, which we can ignore and say they're the same. Since the same current flows in both collector and emitter resistors, and the emitter resistor follows the base, then the gain is Rc/Re to within a few percent. This is a very decent approximation for all cases where the emitter and collector resistors are not dramatically different, and for which the Shockley resistance in the base-emitter is much smaller than the emitter resistor. The feedback in the emitter resistor forces this to be, just like the feedback resistor in opamps force their gains.

And how does the answer in gain correspond to increase in Db with a 100mv guitar signal ?

Voltage gain expressed in db is always, always,  db = 20*log₁₀(Vout/Vin).

Gain of two is 6db. Gain of ten is 20db. Gain of 20 is 26db.Gain of 100 is 40db. The starting point doesn't matter except as one of the numbers in calculating the gain. +20db based on 100mV is the same gain as +20db based on 100V. Both are gain of ten.

[Brymus]

Thank You for the answer RG
So to see if I understand right,the gain of the BMP's last stage with a 12K collector resistor and 2k7 emiter resistor would be about 4.4 ?

How does one measure gain when bypassing the emiter with a cap ? :What I have read (am reading) states that this forces the stage to the Hfe of the transistor VS what the stage is designed for.
So I am thinking that with a beta of >300 thats a huge difference from 4.4,which doesnt seem right logically to me
Even though I understand its frequency selective based on the caps value
I have done this and am not getting THAT big of a difference as in 4.4 vs 300
Is that because the Hfe is current gain and we are calculating the voltage gain?
Or am I just completely lost still ?

[PRR]

> the gain ... would be about

Yes, if the load impedance is infinite, source impedance is zero, and the emitter resistor RE is large enough for stable bias, the voltage gain is about RC/RE. Taking 5K and 1K, the voltage gain is 5.

> when bypassing the emitter with a cap ?

(Assuming ample-size cap) Then as far as you know so far, "RE" is "zero" and voltage gain is infinite.

But what is the Input Impedance? Neglecting some stuff which we can't neglect, it is the resistance in the emitter path times hFE. Assume hFE is 100. Then for RE=1K the input is about 100K. And for RE=zero then zero times 100 is zero input impedance.

> am not getting THAT big of a difference as in 4.4 vs 300

If the input impedance is zero, and the source impedance is zero, what do we get? Impossible answer.

The simplified "RC/RE" approximation has failed. The emitter resistance is not zero, but very low; and there are no ZERO ohm sources.

The intrinsic emitter impedance is (for linear amp purposes) 30/Ie ohms, where Ie is in milliAmps. So 30 ohms at 1mA, 3 ohms at 10mA, 300 ohms at 0.1mA.

Taking 9V and 5K and 1K we might work at 1mA. So the emitter impedance is 30 ohms. Our bypass cap must be less-than, not the 1K, but the 30 ohms. That leads to values near 100uFd.

Now the input impedance is hFE*Re or 100*30 or 3K ohms.

Your source may be 50K.

Unbypassed, the 100K input is a slight load on the 50K source. You get 0.66.

BYpassed, the 3K input is a heavy load on the 50K source. You get 0.056.

So unbypassed the gain is 0.66 * 5 = 3.3.

And bypassed the gain is 0.056 * 166 = 9.3.

Unlike FETs, wheezy BJT stages can NOT be understood without accounting for the input and output impedances.

In a specific case, you work with what you have.

How to study them in the abstract? The classic way is to imagine an infinite cascade of identical stages. Then all the in/out parameters get accounted.

But we never build real stuff that way. Even two identical stages is unusual. The first one is scaled for low hiss and high gain, the last one is scaled for strong output, and in audio we never get through two stages before either reaching our goal or having to stop for something (volume and tone controls, etc).

There's a big book, and a lot of practical designs, which alternate a "transconductance" stage with a current-to-voltage stage. First transistor has large emitter resistor, high input impedance, and an output current set by the resistor. Second stage has very low input impedance and a current-voltage conversion factor which is known (usually a NFB resistor). When you need more than a small boost (more than most stompboxes need), feedback pairs can give very much better performance than two separate stages.

> gain correspond to increase in Db with a 100mv guitar signal ?

AS R.G. says, the whole premise of "linear amplifiers" is that the level does not matter. An ideal amp with gain of 4.4 will output

1mV in gives 4.4mV out
100mV in gives 440mV out
1V in gives 4.4V out
100V in gives 440V out

Obviously our amps are not "ideal": you won't get 440V from a 9V stomp. And if you put dead-zero in you always get some hiss out. Over the range from under 10 microVolts to over 100mV you should get very-very nearly the same gain at any level. In a simple practical amp the gain may appear to drop just before it goes into clipping, but you need sharp tools to prove it. And with a sharper amplifier (even a 741 run at gain of 4.4 at low audio frequencies) you will be very hard-pressed to prove that gain changes at all above noise and below clipping.

[R.G.]

How does one measure gain when bypassing the emiter with a cap ?

Paul's told you a lot of the details.

Another way to look at it is that when you bypass the emitter with a cap, you have made the emitter resistor ineffective at producing the feedback that made the gain easily predictable. With that gone, you're down to analyzing the details of both the circuit AND the transistor, including the operating current and that Shockley resistance I mentioned as well as the (variable!) current gain.

So I am thinking that with a beta of >300 thats a huge difference from 4.4,which doesnt seem right logically to me
...
I have done this and am not getting THAT big of a difference as in 4.4 vs 300
Is that because the Hfe is current gain and we are calculating the voltage gain?

The last. You need not to confuse current gain with voltage gain. A unity-gain buffer has a voltage gain of one, but potentially a huge current gain. A common-base stage with the signal injected into the emitter has a current gain of about one but a quite large voltage gain. And a common-emitter stage can have both voltage and current gain at the same time. They're different.

And once you get past thinking about the transistor stage in isolation, Paul's right, you do have to consider both the input loading and output loading. They both affect the operation of a bipolar gain stage a lot, and must be included in the design work if you have to get a specific gain. If you only have to get enough, then rougher approximations are useful.

[Brymus]

Thank You Paul
Thank You RG,
I didnt get time to work on,read any more on this today.
Hopefully before I go to bed I will get some time,I was pretty close to having some values(my own) to actually try on my breadboard
but was too tired after correcting the units in the calculations to rework them again.
Lesson learned from the EQ...
I think I am pretty close to having a pretty linear gain stage with out copying an already known one (the goal)
Might seem a small accomplishment to some,but for me its an important first step.
I am already envisioning ideas to try in regards to setting the Q for clipping the top or bottom of the signal.
using the cut off and saturation points.

I do have a few questions again right now.
How does one determine the current draw that should be used for each stage when designing from scratch ?
Obviously the data sheets will have the max and minimum values.
Do you just base it on battery or no battery? The ma and divide that by the number of active devices ?
I was going to use 5ma for my first attempt,is that a good current draw to aim for ,for a single 2N3904 ?
How about two cascaded ?

[PRR]

> How does one determine the current draw that should be used

That's an endless question.

As a quick sanity check, figure what the LOAD needs.

Say 0.5V in 50K, a 10-foot guitar cord at 10KHz. (In audio, capacitance matters.) 0.5V/500K is 10 microAmp.

Say 10V in 600 ohms, a hot old-skool broadcast-gear input (interface designed to swamp small capacitance). 10V/600 is 17 milliAmps.

For resistance-coupled stages (the only kind you should be considering now), run the stage 5 or 10 times hotter than the load current needs.

10V into 600 ohms suggests like 100mA, and also 2*1.414*10V= 28V, or three Watts, far beyond what you can do with jellybean transistors. And that's not today's lesson.

0.5V in 50K, 10 uA of signal, suggests 100uA or 0.1mA of device current.

Taking a 9V battery, assuming "half" across the resistor, say 5V across the resistor. 5V at 0.1mA is a 50K resistor.

Base-emitter voltage is not known within 0.1V, so we want an emitter resistor 10X larger to swamp the uncertainty. 1V emitter resistor at 0.1mA is 10K.

You have NOT asked how big the base-bias resistors should be. They should be small enough that the VERY uncertain base current causes "small" error. Or IOW the through-current should be maybe 10 times the base current.

Modern 23-skidoo small silicon transistors have hFE of 100 to 400. So if we aim for 0.1mA or 100uA emitter current, the base current is 1uA to 0.25uA. If error is "small" for 1uA it will be smaller for 0.25uA, so we use the low-hFE large-current numbers.

If 1uA is not to cause much error, then the base resistors must flow at least 10uA.

The base voltage will be the 1V Re voltage plus the ~~0.6V Vbe, 1.6V.

The lower resistor drops 1.6V and flows 10uA. Its value is 160K.

The upper resistor drops 9V-1.6V= 7.4V and flows 10uA plus base current, as much as 11uA. 7.4V/11uA is 672K.

We have our four main DC resistors.

The input impedance is the two base-bias resistors, plus the transistor base impedance, all in parallel.

160K||672K= 129K.

The base impedance is about hFE times Re. For Re=10K and hFE 100-400, 1Meg to 4Meg.

129K||1Meg= 114K.
129K||4Meg= 125K.

We have reduced a 400% spread in hFE to a 9% spread of in-circuit impedance.

The output impedance is, in this case, for practical audio purpose, the collector resistor 50K. That's comparable to a guitar, or some 1950s hi-fis, though in any other modern audio application this would be considered a "high" impedance. In particular a 50K impedance driving 10 feet or 300pFd of cable will roll-off at 10.6KHz. That's acceptable for guitar but most audio-heads would panic.

The un-loaded gain is about Rc/Re or 5. (Note that this is also the ratio of what we could get from the available battery leaving some room to swing to what we had to drop to swamp the Vbe uncertainty.)

The 114K-125K input impedance will cause some loading on typical guitar and some stompboxes.

The 50K output impedance is slightly loaded by typical 500K guitar-amp inputs, heavily loaded by 68K effects inputs.

It's a functional stage for many uses. For lower output impedance (without added parts), you could run it at higher current. Indeed 0.1mA is on the low side. For higher input impedance (without more parts) you could run it at lower current.

That's all with un-bypassed Re. If bypassed then voltage gain "could be" over 150, but input impedance could be as low as 24K... use the back of this quiz-paper to confirm this.

> I was going to use 5ma for my first attempt

It's a good value. Do be aware that the input impedance will tend to be low, and suck-down the output of some typical sources. If you had asked me, I'd say 1mA. Not only is the impedance likely to be less of an issue, most of your math gets simpler (5V/1mA= 5K). And being able to quick-estimate values and results without resort to a calculator speeds understanding.

> How about two cascaded ?

Don't get ahead of yourself. Yet.

Don't over-complicate.

As a rough-shot, it's just building blocks. If you have two 1:5 levers, and cascade them, you have a 1:25 lever. Although in levers as well as amplifiers, stray friction/loading issues leave you short of simple estimates.

And two 1:5 stages taking a 0.2V guitar signal should be 5V output. But the maximum Sine RMS possible on a 9V battery is 9V/2.828= 3V. So such a cascade will distort strong guitar. Cascading directly is not very useful in most audio. We have to stop every stage or two and do other stuff. That "other stuff" often dominates practical design.

[Brymus]

Can someone check if I did this right ?
Using PRR's suggestion of a 1 ma current draw,and trying to create an ultralinear stage (without rounding to standard resistor values)
This is what I get,it has a gain of 10.4
Providing I did this right
How can I keep it ultra linear and bring the gain down to 3 ?
And what would be my input/output impedences ? about 10k/4k7 ?
I realize this would load a guitar down horribly,so it needs to be changed or have a buffer in front of it.
But I am just trying to grasp how to design a stage from scratch with out copying someone elses work.

[R.G.]

Can someone check if I did this right ?

Lessee here: 1ma collector current as a starting criteria (we'll check this later) and gain of nominally 10.
1. Are the collector and emitter voltages possible?
1ma in the collector resistor is 4.7K*0.001A = 4.7V, about half of 9V supply OK.
1ma in the emitter resistor is 450*0.001 = 0.45V. OK.
2. Is the base bias point correct for making the collector/emitter side predictions come true?
There is a mildly touchy point here. That point is that base-emitter voltage is approximately constant, but only approximately. It's a good guess to set it for 0.6V, but it actually can be (and I have measured) values from 0.45V to 0.75V in working circuits. 0.45-0.5V is what happens with a very high gain transistor (300 and up in my experience) in low-current operation. 0.7 is at the other end of things, low gain transistors at higher currents.

The way to handle things when you can't get at the exact data is to make the most likely assumption, then see if that leads to a plausible and checkable result. We'll stay with 0.6, then see what happens if it's different.

The base voltage then needs to be 0.45V +0.6V, or 1.05V. The divider chain you show puts it at 1.15V in the absence of any loading. The base current will drag it down some. The base will need Ic/hfe, and we'll guess at a gain of at least 100, which is easy to get. the base current is then 0.001/100 or 10uA.

The easy way to handle how much this drags down the base divider is to convert the divider to its Thevenin Equivalent. A resistive divider can be modeled as a voltage source equal to its unloaded voltage, and a resistance in series with it of the parallel combination of the two resistors. This one reduces to 1.15V in series with 11.5k||78.5K, or 10.03K. The base current then causes a voltage drop of 10uA*10.03K = 0.1V. The base then sees a voltage of 1.05V. Bang on the design voltage. This one seems right for the desired bias point.

So let's check out the real gain. To do that we have to check the shockley resistance too. The Shockley resistance is 26mV/Ie. This can be simplified by multiplying by 1000/1000 to get 26/Ie when Ie is in units of milliamps. For example, 1.2ma of Ie is calculated as 1.2. In this case, the Shockley resistance is 26/1.0 or 26 ohms. The emitter resistor is then in effect 450+26 = 476. This makes the gain 4700/476 = 9.87. Very close to the desired gain.

Before we get all balled up in 9.87 not being 10, let's think about sources of error. In a real amplifier, the resistors will have a 1% tolerance at best, and probably 5%. So the real gain you'd get from breadboarding this would be anything between 11.5 (Rc=5%, Re=-5%) down to 9.45 (Rc=-5%, Re=+5%) and the bias point would vary similarly. And there will be some variance from Vbe not being 0.6 and hfe not being 100.

Still - good job. I would not expect to get it any closer than you have. If the gain has to be 10.00000, other methods are needed.

Note that the emitter voltage is subtracted from your possible voltage swing, so you lose 0.45V from the possible 9.0V swing before distortion.

How can I keep it ultra linear and bring the gain down to 3 ?

Smaller ratio of Rc/Re, or make Rc be two resistors in series with output taken from the middle.
Note that Re being 1/3 of Rc eats up more of your undistorted swing.

I'm a little uncomfortable with the term "ultra linear". Single transistor amplifiers simply aren't ultra anything. A single transistor stage will have a distortion of between 0.1% and 2% distortion depending on swing. This goes down as the gain goes down as a result of the emitter resistor providing more AC feedback at the same time it reduces gain.

And what would be my input/output impedences ? about 10k/4k7 ?

Input impedance is the parallel combination of the two resistances in the base bias divider and the base impedance, which is the hfe times the Shockley resistance plus the emitter resistor. The combination of the two base resistors is 10.03K, and the shockley plus emitter resistors is 476 ohms (about!). The hfe is what it is. If w say it's 100, then the base resistance is 47.6K, and the input impedance is 10K||47k, or 8.25k. Yep, big load for a guitar.

But I am just trying to grasp how to design a stage from scratch with out copying someone elses work.

Looks like you're grasping it as well as I did when I got an A in EE3213, Transistor Circuit Design.

[Brymus]

Thank you ! RG that means alot coming from you 
So my next questions are how does adding another resistor in series at the collector affect the calculations ?
And how do you select one ?
And again how would adding a feedback resistor from collector to base,affect the input impedence and the calculations ?
And how do you select one ?
I think ideally you want an input impedence of around 1M for the guitar,I just dont see how to get that without adding a buffer in front of this.
I did notice that the emiter resitor seems to be 1/10th the collector resistor despite whatever current draw I select.
When trying to keep it out of saturation and cut off.

[R.G.]

So my next questions are how does adding another resistor in series at the collector affect the calculations ?
And how do you select one ?

The simple way is to split the 4.7K you have there now into two, one 1/3 of 4.7K and another 2/3 of 4.7K. That's comes out to 15.6666yK and 31.3K - not standard values.

So you pick the nearest standard values and recalculate the rest of it slightly. 15K is a standard, and so is 30K, which are neatly 1/3 and 2/3 of 45K. It is rare for it to come out that neatly. Now all you have to do is change the emitter resistor to get the (Rc1+Rc2)/Re to be 9, and you have a gain of three. That becomes 5K, which is not a standard value, so you can either (1) use 4.7K and live with a final gain of (1/3)*(45K/4.7K) = 3.191... or (2) use 5.1K and live with a gain of 2.941...  or go to 1% resistors to get closer if that's really, really, really necessary or (3) split Re into two resistors and bypass part of it to ground with a cap to make it "vanish" for AC signals to trim in the AC gain. Note that a trimpot for Re does much the same thing. This is what's done, for instance, on changing gain in the fuzz face.

And again how would adding a feedback resistor from collector to base,affect the input impedence and the calculations ?

If you connect it DC (i.e. directly) it feeds DC from the collector to the base and increases the base voltage/current, so it changes the bias point. It also decreases the input impedance. But it gives both DC and AC negative feedback, so it stabilizes the bias point of the transistor. This is how the early transistor circuits were biased. It's voltage mode feedback. The emitter resistor is current mode feedback. These are different kinds of negative feedback, and have different effects. Sorry - you're getting into some advanced concepts.

And how do you select one ?

Much the same as with a resistor divider. You envision the bias point you want, then calculate the resistor from collector to base to make it so, then check to see that what you did really does make it so, within a reasonable tolerance.

I think ideally you want an input impedence of around 1M for the guitar,I just dont see how to get that without adding a buffer in front of this.

It is really quite difficult to get a single NPN common emitter gain stage to have an impedance of near 1M input impedance. Darlingtons, maybe. In most cases you'll go to an emitter follower with no gain, or use a JFET with a naturally higher input impedance rather than do the extreme measures for input impedance that may prevent you from getting the other things you want. Which is a fancy way of saying you're correct when you say "I don't see how." I don't either.

I did notice that the emiter resitor seems to be 1/10th the collector resistor despite whatever current draw I select.

That's because you set a gain of ten, I think. The actual current draw doesn't matter much if you select for gain. It only changes the specific value of the resistors, not their ratio.

When trying to keep it out of saturation and cut off.

Aye, there's the rub. 

[PRR]

> The Shockley resistance is 26mV/Ie.

For clarity: this is the same formula as my "The intrinsic emitter impedance is... 30/Ie ohms, where Ie is in milliAmps."

Shockley derived 26mV from hard physics and an assumed temperature. Real Silicon tends a little higher, and in my world Silicon often runs warmer (not so true in 9V work). And "30" is easier to compute on thumbs. It may be pessimistic, your circuit may work a hair better than predicted with "30", though rarely as good as "26" predicts. And if it matters, you swamp it away so it DON'T matter. As with your large emitter resistor.

> a good guess to set it for 0.6V, but it actually can be ...from 0.45V to 0.75V in working circuits.

I'm sure R.G. attended this lecture, but for others: For a specific die-size (actually a specific current density), Vbe rises 60mV for each decade of current.

So pretend that a specific transistor shows 0.600V at 1mA. At other currents we should predict these values:

10uA = 0.480V
100uA = 0.540V
1mA = 0.600V
10mA = 0.660V
100mA = 0.720V

Note that this is essentially R.G.'s "0.45V to 0.75V".

Transistors come in sizes. We use a lot of 50mA-max transistors (2N5088?). We also like the 500mA parts (2N4401). And not here but in other fields, TIP41(?) is a dandy 5A part. Assuming the medium 500mA part has Vbe=0.6V at 1mA, we can predict:

small 50mA device at 1mA = 0.660V
medium 500mA device at 1mA = 0.660V
big 5000mA device at 1mA = 0.540V

Vbe also decreases 2mV per degree C of temperature rise. This is moot for well-designed stages in shirt-sleeve environment. It does predict disaster if you attempt a FIXED bias voltage, and makes trouble down hot oil-wells or in power amplifiers.

The output impedance of this plan will always be essentially your R1. Technically this is in shunt with the transistor "plate impedance", but in any low-voltage work this will be far-far larger than your 4.7K resistor, and the "error" from neglecting it is far less than your resistor tolerance, or the precision you need when working with audio impedances.

> How can I keep it ultra linear

How ultra?

With no feedback (your cathode degeneration), a naked BJT stage doesn't really clip, but gets really badly bent as the output swings near the supply. You can assume the THD at this point is 25%. In 9V work, this would be approximately 2.5V output.

As level decreases, THD goes down, and a linear approximation is good enough. 0.25V out gives around 2.5% THD, mostly 2nd, which is not offensive nor really audible.

With emitter degeneration, THD goes down roughly by the ratio of emitter impedance to emitter resistor. You have 26-30 ohms in the emitter and 450 ohms resistor, THD is about 0.06 times the naked transistor case.  Perhaps 0.2% at 0.25V out, perhaps 2% at 2.5V out. That's very clean.

> the emiter resitor seems to be 1/10th the collector resistor despite whatever current draw I select.

If you are shooting for gain-of-10, with this plan, that will always be true.

Also: you are (wisely) shooting for a part-Volt on the emitter resistor and 1/2 of supply on the collector resistor. At 9V supply this will tend to be around 5 or 10 every time. But pencil a 300V supply. You could probably still aim at 1V on the emitter resistor. But you might take 300/2 = 150V on the collector resistor. At 1mA this gives 150K:1K, a 150:1 ratio. (And yes I have built a 300V amp with TO92 trannies and 100K collector resistors. It drove a CRT. Even with the driver board right under the neck of the CRT, the high output impedance against stray capacitances gave barely 19KHz bandwidth. But that was good enough for audio class: you could see what you could hear.)

To reduce gain, I would take a larger emitter resistor. 4K7 up top, pencil 1K5 in the emitter. At 1mA that wants 1.5V on the resistor and say 2.1V at the base. You lose around 1V of total swing due to the emitter sitting higher than your 0.45V design, but this is often acceptable. And if you run the numbers you will see the input impedance can be higher.

> want an input impedence of around 1M

Very possible, but you have to bend everything else toward this goal. If you study old transistor guitar amps you will find that most only managed 50K-200K at the pickup jack. If you study old IRE or IEEE papers you find well-designed ~~1Meg input circuits.... the fact that you could get a paper published for this shows that it isn't a beginner thing. And sometimes the speed or swing of those old hi-Z inputs was quite modest.

The JFET is the quick way to a high-Z input, kinda makes it "silly" to piddle around with BJTs for hi-Z inputs.

[Brymus]

Thank You RG
And
Thank You Paul, I am beginning to understand why op-amps are so desirable design wise outside the parts count.
And how different parts can make a circuit easier to complete to a desired result,as in using a FET for the input stage.(an op amp for the gain)
But for the sake of learning I spent the last couple of hours trying to find how to bias a BJT emitter follower.
But after reading PRR's post I am wondering about the gain stage above.
Since the gain is about 10 wouldnt it try to swing to 10vpp on a 1vpp signal ?
Which would easily put it into clipping aside from the increase in THD,2% THD I can live with for guitar use.Most time that little coloration is what gives an effect character.

So a gain of 3 would be max for a BJT stage without inducing clipping and excess THD. (with 9v supply ,1vpp input signal) Is this correct ?

Back to the emitter follower ,please check if you would.Seems easy enough...
I believe this to have an input impedence of 500k an output of 3k8 and a gain of just under 1
The idea was to put this in front of the gain stage to match the impedence of the guitar to the circuit,obviously a FET would be better as it could do both in one stage.

[Brymus]

Two things dawned on me this morning 
An LPB-1 is just a common emitter stage. (I am sure most of you know this)
And I forgot to figure in the shockly resistance into my emitter follower.

So wouldnt an LPB-1 be better with a buffer on the front end ? and a gain less than 27 ?

[R.G.]

But for the sake of learning I spent the last couple of hours trying to find how to bias a BJT emitter follower.

Couldn't be simpler. There are three "magic concepts" in bipolar transistors:
1. The base-emitter junction will always be in the linear conduction region (0.5-0.7V as a rule of thumb).
2. The emitter current is equal to the collector current plus the base current - and the base current can usually be ignored for hfe>100 or so.
3. The collector voltage goes wherever the emitter current and any voltage/resistance it's connected to tell it to be. The collector is a CURRENT output (well, input, as an NPN collector sucks current in)

So you decide what output voltage you want on the emitter. Let's say you have a 9V supply and want it at 4.0V. Then you decide how much current you can afford to run through there. Let's pick 2ma out of the air. The emitter resistor has to have 4V across it and 0.002A through it, so Georg Ohm tells us that the emitter resistor is R = 4V/0.002A = 2K. The base HAS TO BE 0.5v to 0.7V above that. Call it 0.6 and you won't be very wrong. So the base voltage is 4.6V. Now you need a divider to make that happen.

The emitter current is 2ma. The base current is less than 1/100 of that, or less than 20uA if you have a transistor with a gain over 100. So design a two resistor divider to make the base voltage be 4.6V when the base is pulling 20uA.

(note - there are some highly recommended further refinements here which I have not listed so you can get the concepts down first)

Since the gain is about 10 wouldnt it try to swing to 10vpp on a 1vpp signal ? Which would easily put it into clipping aside from the increase in THD,2% THD I can live with for guitar use.Most time that little coloration is what gives an effect character.

You are correct. A gain of ten is different from being able to swing 10V. The available peak to peak swing on the collector of the gain-of-ten stage is equal to the difference between the transistor being turned off (that is, Ic =0 and Vc =9) and the transistor being saturated, with perhaps Vce = 0.7V. When it's saturated, 9V = Vsat + Vrc +Vre. Then Vrc + Vre = 8.3V, and Ic= 8.3V/(Rc+Re) = 8.3/(4.7K +450) = 1.6ma, and that makes the voltage on the collector be 9V-(4700*0.0016) = 1.425V. And the total swing this thing can ever do is 9V-1.425V = 7.57V peak to peak. The power supply and resistors won't let it do any more.

So the biggest input signal it can take, for a gain of 10, is 0.757Vpeak to peak, 0.379V peak, 0.268Vrms.

So a gain of 3 would be max for a BJT stage without inducing clipping and excess THD. (with 9v supply ,1vpp input signal) Is this correct ?

It's actually less than that, because you lose the voltage across the emitter resistor and collector-emitter when the transistor is as "on" as it can get. There are losses which keep the transistor from swinging fully from ground to +9V. This is generally true of all amplifiers except certain special cases.

I believe this to have an input impedence of 500k an output of 3k8 and a gain of just under 1

Unfortunately, no. The base is set at 4.5V, minus the bias current through the parallel combination of the two 1Ms. (that is, the base bias current makes the voltage divider sag a bit) 

The emitter voltage will be roughly 4.5V (we'lll figure the sag later) minus 0.6V Vbe, or 3.9V. The current is then 3.9V/3.8K = 1.026ma. Assume hfe=100, so base bias current is 10.26uA. 10.26uA through the parallel combination of 1M/1M is a voltage of V = 10.26uA*500K = 5.1V. Gulp. It won't work. Can't work. The bias network can't supply enough current to pull the base up.

As a rule of thumb (do you detect my design approach? Get it close with rules of thumb, then do the "perfect" design as a refinement.) you need to make the current through a resistor divider be 10X the load to get the voltage not to sag too much. Your base is pulling 10uA and your divider is only running 4.5uA through it even unloaded.

The divider needs a standing current of at least 100uA as a starting point. That makes the resistors have to be 9V/100uA = 90K in two equal resistors, so they need to be - shoot, call it 47K each. Now the base voltage sags from 4.5V by the base current times 23.5K, or 0.2411V. This puts the base at 4.5V - 0.2411V = 4.26V, the emitter at about 3.66V, and the emitter current at 0.962ma. All these are first approximations.

The second appproximation takes the calculated emitter current, divides by 100 to get 9.62uA, and we note that this is only different from 10.26uA from the last calculation by about 6%. The resistor values are variable by 5%. If I wanted to spend my life getting designs perfect, I'd run through the calculations again a few times refining, but for me, I'd say it's close enough, especially since there are many transistors with gains way over 100.

This leaves you with an input impedance of the two bias resistors paralleled with the base impedance. The base impedance of an emitter follower is approximately the current gain times the emitter resistor, or or 380K in this example. The two bias resistors are 23.5K in parallel with this, so the input impedance is the parallel combination. Shoot, I'm in a hurry - it's about 20K, shooting from the hip.

An emitter follower can't get you to 1M unless you can bias it with an impedance over 1M. In fact, it cant' get you there with a 1M bias because the base input impedance can't get you to 1M because of the low hfe.

Isn't Mother Nature ugly? 

[Brymus]

Thanks again RG
I thought that was too easy to be right.
But hey I am enjoying learning all this,so hopefully I can use what I learn to understand more complicated ideas as I progress.
All kinds of things are starting to click with what little I have learned the past couple of weeks.
I really appreciate all the guidance you and Paul are giving me 
So back to reading.

[PRR]

> believe this to have an input impedence of 500k an output of 3k8....

R.G. shot most of this down.

"IF" it worked with several volts across several K emitter resistor, the small-signal output impedance would be 26-30 ohms. What can you do with a "too low" output impedance?

Go back to your common-emitter stage in post 7.

It has gain of like 10, a maximum output of almost 4V peak, so maximum input of 0.4V peak. R.G. did the math for the input impedance, got 8K. Therefore that stage needs 0.4V/8K= 0.05mA peak input current.

Put your emitter follower in front. It has to deliver 0.05mA peak to the CE stage's input. The EF could run 0.1mA and manage that, especially at this low voltage-swing.

So pencil the EF as 40K emitter resistor, 0.1mA emitter current. The base current is 0.001mA or less. You could flow 0.01mA in the base-bias network. A pair of 470K resistors.

The bias-net impedance is now 235K. The emitter drives 40K||8K or say 6K. At the base this looks like 600K or more. It's up to 168K and will bias-up fine.

At 0.1mA emitter current the output is near 300 ohms, negligible for driving the 8K of the next stage. Although this is higher than before it is still plenty-low. Concept: when a parameter is "too good", see if by making it less-good you can gain something else.

We are getting there.

There is another simplification. R.G. mentioned Darlington. This is a 2-transistor pair, the first working as an emitter-follower into a second transistor which may be used several ways. The most obvious is a common-emitter stage but assume the "transistor" (a Darlington pair) has hFE of like 100*100 or 10,000. Then your emitter current may be 1mA and the base current will be near 0.000,01mA, 0.01uA. Or maybe more: the first device in a Darlington may be starved and not near nominal current gain. Still you can try 1uA flowing in the bias network.

Where to bias? You have two Vbe or 1.2V, a bit less because the first device runs at very low (but poorly known) current. The larger effective Vbe means you want good drop in the emitter resistor. But you already know that you can't take a ton of voltage gain with hot guitar and small battery. So try the 1.5K emitter resistor, 1.5V here, say 2.7V at base. The lower bias resistor may be OTOO 2.7Meg and the upper resistor is left as an exercise.

Counting on my thumbs I am seeing >1Meg total input impedance and gain of 3. I know a "negligible" which now matters a little, but I think in practice something like this will be plenty hi-Z for any guitar use.

However a stage with this much leverage and the poorly-set current in the first device needs very careful figuring or a little tweakery. If you will build a million for $39, you want to get a lot of data on the transistors and figure how far-off a worst-case part on worst-case stage will be. For one-off a few 2.7Meg 2.2Meg 4.7Meg 6.8Meg resistors will zone-in a working bias for that part.

There's a cleverer way for this specific requirement if you use a PNP-NPN "quasi-Darlington". But I'm reaching deep in a bag-o-tricks and you want to work-through the most common implementations first.... they are most-common because they are usually best.

[Gus]

Next work out what is going on in the beginner project.  Set to min gain what is it's input resistance?

look at this
http://freestompboxes.org/viewtopic.php?f=7&t=9255&p=99794&hilit=bootstrap#p99794

[Brymus]

OK Paul I was thrown off by you saying the input impedence was 1M  
By my calculations the base impedence is 15M (beta*beta*Re) the top bias resistor needs to be 6M7 to bias at 2.7V
Which makes the total input impedence 1.6M , did I do that right ?
But I dont understand how to figure the gain for the darlington pair follower as there is no collector resistor,for a gain of 3 it would need a 4K5 Rc ?
So is it the Darlington has to be done as a common emitter and not a follower to have a gain of 3 ?
So the 4k5 has to be there,or else the gain would be unity or less than 1 ?
As you can see I am still confused but getting closer ( I hope)

As I was getting confused from the stuff I had read about biasing ,so I figured the Tube Screamer input impedence to see how that works,with the 510K bias resistor coming from Vref and the 10K emitter resistor the total input impedence would be approx 330K,correct ?
I did notice that the TS follower has a very low current draw,which I think is the only wa to get a decent input impedence without using a Darlington (super alpha) pair

So my question now is, is the emitter resistor in an emitter follower chosen simply to set the I of Re ?
And the voltage swing is determined by the two bias resistors ?
Because with the TS emitter follower, it has a 10K Re wich would make 9V  but I measured mine and its actually the Vref minus the .6  base emiter voltage,meaning I am getting 3.9v at the emitter not 9v, which I am finding confusing.

Gus I will figure out the beginer project too,(next) thanks for the suggestion.

[Brymus]

Gus ,using this schematic> http://www.diystompboxes.com/smfforum/index.php?topic=763.0
To the best of my understanding,using an Hfe of 100 the base impedance is then 500K,and as far as I understand bootstrapping that would make the input impedance 108K.
Is that correct ?
I did read where the bootstrapping can have a mutliplying effect on the input impedance but couldn't find a better explanation of how.
The best I found was that the bootstrap R is in series with the input impedance without it.
The output would be close to 5k and the gain ,I am unsure of because of it being bypased with the cap.unbypassed it would be 2

I have been searching for hours and cant find anything outside a basic explanation on Darlington pairs.
What I have read suggests the output is always taken from the emitter and has a gain of 1 or less,which is confusing as Paul said the example he gave has a gain of 3,also for Pauls example to have an input Z of 1M the top bias resistor would need to be 1M5.