Mic Pre-Amp

[MrAl]

MrAl,

I am already putting part of your suggestion into the schematic, leaving
the 10 uF caps in the signal path while changing C3 to 100 uF. This gives
a better bottom end response than with the 1 uF caps you suggested. I
will leave out the resistor after C3, because it kills the gain too much. I
could put a 100 ohm R paralleled with a 500 ohm VR at that point to vary
the output from ~140mV (max resistance) to ~480mV (min resistance).
My friend suggested maybe putting more than 1 value cap in for C3 on a
switch to vary the bass contour. I'll see what I'm up for when I get there.

CeaSaR

Oh so 10uf worked better than 1uf? That's good to know too.
The lowest frequency on the guitar is around 80 Hz if i rem right
but if you are saying 10uf sounds better than that's just as well.

Yes, i'd like to know how this turns out once you get it all built
up and used it a few times.

[Bob Scott]

Hi all and Merry Christmas.

I have one schematic solution for a 50Db preamp. I took Bigglez basically worthy circuit and massaged it to add higher AC gain and added a few missing doodads. I have not tested it, but I did triple check the part values. The part names have been changed and there are more parts than before.

I added R4 and C2 to decouple the input circuit from power supply noise and feedback.
R2 and R3 set the bias and the input impedance to 50K so it's compatible with all mics that I know of.
R1 and R11 are added just to bleed away any DC while the inputs are not connected. That way there are no annoying large pops when devices are plugged into the input or output.
R10 sets the preamp's output impedance to 10K. This preamp's circuit gain is 56Db, but the output drops 6DB going through R10 if the preamp is driving a 10K load, leaving a net gain of 50Db. If you feel that you don't need this 10K output impedance and short out R10, the output will become 6DB higher. You will then have to raise the value of R7 from 75 to 150 Ohms to reduce the net gain back down to 50 Db.
R9 is a little "low output swing helper". The preamp should now have about 14Db headroom on the output.
DC gain is still the same as Bigglez original but the values of the determining resistors has been scaled by 10X in order to keep the value of the new R7 high.
AC gain is determined by the ratio of R6 + R7 / R7. This happens when the impedance of C3 is equal to the value of R7 at about 10 Hz. So, the low frequency roll off point is 10 Hz (3 Db down). If R7 is changed to 150, then it rolls off at 5 Hz. If I hadn't scaled the values, C3 would be humungous!

Well that was a nice change in pace to design a discrete transistor audio device. Maybe someone would like a design for a tube model next? I'll get out my old tube manual and look up 12AT7's and 12AX7's just in case!

Ah, op-amps are sometimes just too simple.

[MrAl]

Hi Bob,


What is R10 for?

Also, i think maybe some Thev and Norton analysis can help to reduce the
parts count a little. Care to take a look?

Has this circuit actually been built and tested or just simulated?


Thanks...

[Robert Reed]

Nice circuit Bob. I presume this is a stand alone device in its own chassis? If I am correct in assuming this, I always add a follower to drive any cabling to another device - nice low output impedance and so simple. If my assumption is wrong and it connects directly to additional in chassis circuitry then it is just fine as it stands.

[Bob Scott]

What is R10 for?

Hi Mr. Al.

It sets the output impedance to 10K, like I said in the post. The signal at the collector of Q2 has almost zero impedance due to feedback control, so I added 10K in series with the output, and I boosted the gain to 56Db to make up for the 6Db loss through R10, which happens when the output is loaded; ie: connected to a power amp with a 10K input impedance.

In that post I also mentioned that if you don't want this output impedance, you can short out R10 to zero ohms, but then you should lower the gain back to 50Db by changing R7 to 150 Ohms.

Also, i think maybe some Thev and Norton analysis can help to reduce the parts count a little. Care to take a look?

I don't believe that the parts count can be reduced by any sort of analysis. Every part has a purpose. But sure, I would be interested and amazed if you succeed. Can you do it?

Has this circuit actually been built and tested or just simulated?

No, I hope someone else does. I didn't build it. The simulation was done the old fashioned way, in my head, not a simulation program. I'm pretty confident it works fine because it is so simple, but sometimes there is something unforseen that can cause problems..... like part tolerances causing a drift in bias levels (but I "simulated" those), amplifiers that oscillate, oscillators that amplify, inductance in electrolytic caps, etc.

Nice circuit Bob. I presume this is a stand alone device in its own chassis? If I am correct in assuming this, I always add a follower to drive any cabling to another device - nice low output impedance and so simple. If my assumption is wrong and it connects directly to additional in chassis circuitry then it is just fine as it stands.

Hi Robert. I agree that changes would improve this design. First thing I'd do is use an op-amp or two or three, but CeaSar (the OP) specified limitations. Only those particular transistor part numbers will do, and I assume he wanted to keep it as simple as possible.

I could add a transistorized buffer output. I'll add it, what the heck! Class A or push-pull? A pre-amp with its own DA built in! Haha. (distribution amplifier)

Merry Christmas all,
Bob

[MrAl]

Hi again,

Ok maybe im nit picking a little here

I'll take another look at the circuit and see what i can find.
Thanks.

Merry Christmas to you and yours also.

[MrAl]

Hi again,


With respect to R6,R7,R8, and R9 of the first schematic (two transistors
circuit) the following changes seem to eliminate the need for R9:

R6=7007 ohms
R7=11 ohms
R8=1491 ohms
(and possibly increase C3 accordingly)

This eliminates R9.

Also, it looks like if R5 is also changed to 7007 ohms, another
increase in gain results.

I quoted exact values rather than approximations because these
represent the solutions in case anyone wants to go over the math.

Another possible elimination is R4, but i would first ask what
that is for in the first place. Is that to eliminate ripple from
getting into the input stage or just for general filtering?
Not that it's a bad idea though.

Of course the circuit would have to be breadboarded next.

BTW it is a nice circuit too, in my haste to get the possible improvement
ideas out i almost forgot to mention that. The three transistor
circuit even better. Not sure what the "Z=0 ohms" means though.

Take care,
Al

[Bob Scott]

With respect to R6,R7,R8, and R9 of the first schematic (two transistors circuit) the following changes seem to eliminate the need for R9:

R6=7007 ohms
R7=11 ohms
R8=1491 ohms
(and possibly increase C3 accordingly)

This eliminates R9.

Yes, C3 would have to be increased by your scale factor of 6.8 also. It would increase C3 to a humungous 6,800 or 13,600uF (depending on which schematic you look at). I put R9 in there so that I could reduce the value of C3 to where it is now.

Also, it looks like if R5 is also changed to 7007 ohms, another increase in gain results.

????? I don't see R5 as a gain detemining component. R5 is just there to help turn off Q2. The voltage across R5 is always at ~0.65V, the forward voltage of the B-E juction of Q2.

The only DC and AC gain determining parts are R6, R7, R8, C3.

Another possible elimination is R4, but i would first ask what
that is for in the first place. Is that to eliminate ripple from
getting into the input stage or just for general filtering?
Not that it's a bad idea though.

"not a bad Idea", hehehe. Yes it is. R4 is part of the R/C filter for keeping power supply noise out of the input, which is sensitive to microvolts! Take a look at the schematic posted by sghioto on the first page of this thread. Also look at the DC path to the preamplifiers in the schematics of most hifi amps ever made. You'll see the same type RC circuit.

You seldom see suttle details like that filter in beginner hobbyist or toy grade circuitry.

Of course the circuit would have to be breadboarded next.

Anyone? I have too much other stuff to do from now until forever.

BTW it is a nice circuit too, in my haste to get the possible improvement ideas out i almost forgot to mention that. The three transistor circuit even better. Not sure what the "Z=0 ohms" means though.

Thanks. It was based on the low gain circuit posted by Bigglez. I just changed it slightly.
Z=0 ohms means the output impedance is zero or close to it. If you load that output with, say, 10K to ground, you should not see any attenuation of the output voltage.

[MrAl]

With respect to R6,R7,R8, and R9 of the first schematic (two transistors circuit) the following changes seem to eliminate the need for R9:

R6=7007 ohms
R7=11 ohms
R8=1491 ohms
(and possibly increase C3 accordingly)

This eliminates R9.

Yes, C3 would have to be increased by your scale factor of 6.8 also. It would increase C3 to a humungous 6,800 or 13,600uF (depending on which schematic you look at). I put R9 in there so that I could reduce the value of C3 to where it is now.

Also, it looks like if R5 is also changed to 7007 ohms, another increase in gain results.

????? I don't see R5 as a gain detemining component. R5 is just there to help turn off Q2. The voltage across R5 is always at ~0.65V, the forward voltage of the B-E juction of Q2.

The only DC and AC gain determining parts are R6, R7, R8, C3.

Another possible elimination is R4, but i would first ask what
that is for in the first place. Is that to eliminate ripple from
getting into the input stage or just for general filtering?
Not that it's a bad idea though.

"not a bad Idea", hehehe. Yes it is. R4 is part of the R/C filter for keeping power supply noise out of the input, which is sensitive to microvolts! Take a look at the schematic posted by sghioto on the first page of this thread. Also look at the DC path to the preamplifiers in the schematics of most hifi amps ever made. You'll see the same type RC circuit.

You seldom see suttle details like that filter in beginner hobbyist or toy grade circuitry.

Of course the circuit would have to be breadboarded next.

Anyone? I have too much other stuff to do from now until forever.

BTW it is a nice circuit too, in my haste to get the possible improvement ideas out i almost forgot to mention that. The three transistor circuit even better. Not sure what the "Z=0 ohms" means though.

Thanks. It was based on the low gain circuit posted by Bigglez. I just changed it slightly.
Z=0 ohms means the output impedance is zero or close to it. If you load that output with, say, 10K to ground, you should not see any attenuation of the output voltage.

Hi again,


Well, the schematic i was looking at the 220uf would have to change
to roughly 1000uf, that's about it, but that's provided we want to
keep the same low frequency end response.

The change in R5 (to 7007 or some value lower than 47k) helps
increase the overall gain believe it or not, but that's after the other
resistor changes are in effect.

I asked about the Z=0 ohms output impedance only because i
estimated the output Z to be about 500 ohms, but you did say
that you were more or less comparing the output Z to 10k ohms
load, so 500 ohms is still very low compared to 10k.

I had a feeling that that 10k 'filter' resistor was there for the purpose
of keeping noise out of the front end, that's why i didnt remove that
one before making sure that was the original intent.

Unfortunately i dont have these particular transistors around.
I know they are common types, but i end up using other types
more frequently so i end up buying 10 or 20 other types usually.

Very interesting circuit, and yeah it's interesting to look at these
older style circuits again. Op amps are too simple and boring

[Robert Reed]

Hi all
As to emitter follower out put impedance, there is a very simple test for this at audio frequencys. Merely shunt the emitter resistor with a capacitnce value that will have very lo reactance at the test frequency and run that in series to a decade resistance box to ground. Open the decade's path and inject an audio signal into that stage, measure the output, then connect the box and dial in until the audio signal drops 6 db (50%) at the emitter. Now the parrallel combination equals the output 'Z'. But like Op-Amps, there will be a difference between small signal and large signal measurments. In looking at that stage and without doing the math, I would guess some where in the 200-300 ohm output "Z'. This is based on the Hfe of the transistor (can vary 3:1) and the idle current flowing through it (4.5ma?). Want lower 'Z' - higher Hfe and lower Re (i.e, emitter current) but there are limits.

[MrAl]

Hi Robert,

Yeah you're right, and i just looked quickly and here's how i got my
"approximate" output impedance:

The lower resistor is 1k, and in order to bias the output stage's output
to exactly 1/2 of the supply voltage the transistor collector emitter
has to be such that it's dynamic resistance would have to equal exactly
1k ohms. Now with that 1k to Vcc and the other 1k to ground, the
output impedance is the parallel combination of those two 1k resistors,
which equals 500 ohms.

This is only true when the output is biased to 1/2 Vcc, but that's
usually close enough just to get a rough estimate.
If for example the output was biased to 1/3 Vcc that would require
the dynamic resistance of the transistor to be 2k, and that combined
with the lower 1k would come out to 667 ohms.
Of course if biased to 2/3 Vcc then that would make the transistor
dyno R around 500 ohms and so the combined would be 333 ohms.
So a very good estimate i think would be 330 ohms to 670 ohms.

I realize you know this already but just thought i would mention it
for the benefit of others perhaps.

[rshayes]

The output impedance will be quite low. The dynamic resistance of the transistor output in the emiter follower connection is about 26 ohms divided by the emittter current in milliamps. Since the emitter current is about 4 milliamps, the transistor output impedance will ideally be about 6.5 ohms. There will be some bulk resistance (prbably a few ohms) inside the transistor in series with this impedance, probably raising it to the 10 ohm level. There is another series term due to the source impedance feeding the emitter follower divided by the beta of the transistor. Due to the feedback in the previous stage, this impedance will be low, and the division by beta will make it even lower. This might raise the output impedance by an ohm or two. This will be in parallel with R10, which is 1000 ohms. This will lower the output impedance by about 1 percent. As a rough estimate, the output impedance will be in the 10 to 15 ohm range.

A resistor in the 100 ohm range is series with the output might be a good idea. This will raise the output impedance, but it will also make the emitter follower more stable if a capacitive load is applied (3 or 4 feet of audio cable with a capacitance of about 30 picofarad per foot).

Q1 and Q2 form a feedback amplifier, and may need some additional parts to allow operation without oscillation. One possibility is a fes picofarads between the collector and base of Q2.

The most effective bias filtering would be obtained by making R2 and R4 equal (possibly 150K).

The input impedance can be raised and some temperature compensation obtained by adding a PNP emitter follower in front of Q1. This will cancel the effect of temperature on the base-emitter voltage of Q1.

[MrAl]

The output impedance will be quite low. The dynamic resistance of the transistor output in the emiter follower connection is about 26 ohms divided by the emittter current in milliamps. Since the emitter current is about 4 milliamps, the transistor output impedance will ideally be about 6.5 ohms. There will be some bulk resistance (prbably a few ohms) inside the transistor in series with this impedance, probably raising it to the 10 ohm level. There is another series term due to the source impedance feeding the emitter follower divided by the beta of the transistor. Due to the feedback in the previous stage, this impedance will be low, and the division by beta will make it even lower. This might raise the output impedance by an ohm or two. This will be in parallel with R10, which is 1000 ohms. This will lower the output impedance by about 1 percent. As a rough estimate, the output impedance will be in the 10 to 15 ohm range.

A resistor in the 100 ohm range is series with the output might be a good idea. This will raise the output impedance, but it will also make the emitter follower more stable if a capacitive load is applied (3 or 4 feet of audio cable with a capacitance of about 30 picofarad per foot).

Q1 and Q2 form a feedback amplifier, and may need some additional parts to allow operation without oscillation. One possibility is a fes picofarads between the collector and base of Q2.

The most effective bias filtering would be obtained by making R2 and R4 equal (possibly 150K).

The input impedance can be raised and some temperature compensation obtained by adding a PNP emitter follower in front of Q1. This will cancel the effect of temperature on the base-emitter voltage of Q1.

Hi again rshayes,

Im not sure if you read mine and Roberts posts, but the output impedance
cant be that low because we usually dont want to bias too close to the
Vcc supply line.
With the output bias point set at 4.5v (one half Vcc) the transistor
impedance has to be 1k or else we couldnt get that dc output voltage,
and since the Vcc line is reqarded as 0 ohms that puts the transistor
CE in parallel with that 1k resistor, which works out simply to 500 ohms.
Of course that's just an estimate, and as the transistor pulls the output
up by 1v (2v peak to peak output) that puts the new output voltage
at 5.5v and the transistor dyno R at 636 ohms, which in parallel with
1k comes out to 389 ohms. As the transistor allows the output to
dip down to -1v from the bias point the new output is 3.5v, which
makes the transistor dyno R equal 1571 ohms, and in parallel with
1k that comes out to 611 ohms. Thus, the average output impedance
is (611+389)/2, which is 500 ohms, but since we care more about
the lowest output impedance we should go with 389 ohms, which means
that anything too much lower than that will cause bad distortion.

We also know that when we have a voltage source that has an impedance
of Rx ohms, which loaded by an equal resistor of Rx ohms, gives us
an output voltage of one half the open circuit source voltage, or V/2.
Thus, if we load a source with a resistor equal to its source impedance
we will measure one half the open circuit voltage.
If we were to load this circuit with 36 ohms, we would cause bad
distortion and although the output voltage would drop to possibly 1/2 on
one peak, it would be so badly distorted on the other peak we couldnt use
the amplifer very well.
Why does this happen?
It happens because although the output impedance of the transistor
itself can go down to 10 ohms (perhaps) that doesnt mean that is
a symmetrical impedance. Sure, it can make the output appear
to 'source' an output voltage with very low impedance, but it can
not help to 'sink' the output current very well on the other half of
the cycle. What makes up the other half of the
impedance is the lower resistor (1k in this circuit), and that is partly
responsible for the entire makeup of the output waveshape.
To approximate this, i take the total output resistance and divide by
two, because if we approximate the transistor Z by itself as very
low like 10 ohms and add this to the output resistor which is 1k
we get 1010 ohms, and the average of that is close to 1/2 of the output
resistor value, or 500 ohms. It is an approximation of course, and
we can get a little lower than this in real life.
If we were to bias closer to the Vcc supply rail we could also lower
the output impedance, but at the cost of an increase in quiescent
power dissipation.

Another way of looking at this is that although the transistor impedance
is very low when it is turned 'on', it is also very very high when it is
turned 'off', and the linear operation is somewhere in between, which
makes its overall impedance work more in accordance with the output
resistor than according to its own impedance which is usually much
lower.

Still yet another way to think about it is as in a switching circuit, where
the transistor is switched from 'on' to 'off', with no linear mode.
In such a circuit, the impedance is close to 1/2 of the resistance of
the resistor in series with the transistor (C or E) for a 50 percent
duty cycle switch cycle.

I think if you take another look you will agree.

[Robert Reed]

Hello Sreven
Good to see you back with your excellent postings. My assumption (and a quick one) was based on the fact that the emitter current was limited by the 1K resistor only since it is capacitor coupled to the out put. In that regard, the available current Re and bias current would be somewhat limited. I dont really see a lot of drive coming from the previous stage. So in keeping the whole shebang within tolerable limits of distortion one could only load it to a certain point (my reference to small signal- large signal). Of course, it is late and I am sleepy, so I may be overlooking some things. Wish I had time to throw a quick breadboard together for proof of perfomance, but my bench is so littered with current projects, I cannot even find my breadboard.

[rshayes]

Consider an emitter follower with its base connected to a 5 volt source, the collector connected to a 12 volt source, and the emitter connected to a 4 milliamp current sink. Further assume that the base-emittter voltage is .65 volts under these conditions. The output voltage is then 4.35 volts.

Now apply a load such that the output voltage drops by .026 volts. This increases the base-emitter voltage by .026 volts, which increases the emitter current by a foctor of 2.718. The increase in emitter current is thus 1.718 times 4 milliamps, or 6.872 milliamps. The small signal output impedance is thus .026 volts divided by 6.872 milliamps, or 3.78 ohms.

If the transistor has a beta of 30, the base current will initially be 129 microamps, which will increase to 351 microamps when the load is applied. If there is a 100 ohm resistor between the voltage source and the base, the initial conditions will have the base 12.9 millivolts below 5 volts. There will be a similar decrease in output voltage. Increasing the load current will decrease the output voltage still further, since the base current is now 351 microamps and the drop across the 100 ohm resistor is now 35.1 millivolts. The 100 ohm resistor has resulted in an additional 22.2 millivolts voltage drop under load. The output impedance is now (.026 + .0222 volts) /(6.872 milliamps), or slightly over 7 ohms.

The output impedance is lowered slightly if the emitter follower uses a resistor rather than a current sink. In this case, the difference will be less than 1 percent with a 1000 ohm resistor.