Big Muff buffer: advantages, inconvenients ?

[frank_p]

                                    Refs:
                                    http://www.basicaudio.net/Big-muff-tech.jpg
                                    http://www.diystompboxes.com/smfforum/index.php?topic=78111.0


1) It has a collector feedback resistor:
         
          Ref:
          http://www.allaboutcircuits.com/vol_3/chpt_4/10.html

 a)Supposedly good for minimising H_FE variations
 b)Preventing thermal rumaway of the BJT
 c)Can not go in saturation



2) It have an emitter feedback resistor.
             

  a) Like in the universal bias method: can it minimise the importance of H_FE



3) It have a base voltage divider.
  a) With the collector feedback, resistor, does the base voltage divider still acts as a stiff V_TH and R_TH
           



A) Are these the reasons why this bias setup is made like this?
B) Do these *features* interact negatively or they really can work together.
C) Is this a common biasing method for other applicatons than fuzzes ?
   a) If Yes, in what case and why ?
D) What are the disadvantages of using a buffer like this?
E) And other advantages ?

[frank_p]

It not that simple. It has few thing going on a the same time.  Some things to look up

Page 86 TAOE has something to look at.  The added voltage drop across 100 ohm in the emitter leg adds to the Vbe drop

http://www.nd.edu/~hscdlab/pages/courses/microwaves/labs/Agilent1293.pdf    Kind of like figure 6 but with an added emitter resistor.

[frank_p]

R2 Limits the signal in a bit as well as setting the impedance. Bigger lowers the impedance and signal.
R4 Is a feedback resistor which works along with R3, R5 and R6 to bias the transistor. Smaller = less gain
as it lets more feedback flow from collector to base.

R2, R3 and R4 set up the input and feedback for the transistor. All play multiple roles. R3 and R4 do indeed set biasing. But imagine that you replace Q1 with an inverting opamp stage, the + input being biased else where. That would make the gain of this stage be Rf/Ri = R4/R2, right? That is what is happening here, excepting for the transistor having a lower input impedance and less gain than a purpose-designed opamp. Because of the transistor's "imperfections" the gain is lower than you'd get from an opamp with the same resistors, but the operation is very similar. It's an inverting feedback stage.

[frank_p]

To add to R.G.'s post

The first stages gain is a combination of the open loop gain setting of the transistor (has to be lower than 15K /100) and the 470K/39K about X12 inverting feed from a 0 ohm source, a guitar is not a 0 ohm source.
so LESS than X12 inverting because of limited openloop gain and the source Z of the guitar.

If the transistor had a lot of gain like an opamp the gain would be very close to X12 inverting (ignoring the guitar Z).  The input Z would be the 39K and C1 because the node that the base and 470K and and C1 would be at a virtual ground.  The transistor is limited to a max of X150 by the 15k and 100 ohm (it is even less because you need to add re to the 100 ohm and re varies with current re=25/Ic, Ic is in ma).  Also where in the supply range closer to sat or cutoff

The 470K and 100K are a part of the biasing.  The higher the Hfe the less current thur the 470K to the collector node.  The Ic (collector current set by the designer selection of resistor value and collector operation point) "splits' at the collector node some current for the bias resistors and some for the emitter The higher the Hfe the less the loading of the resistor bias string.

There is even more. A lot going on.  IMO it does not lend itself to easy change this for that simple posts

What I suggest you do if you want to understand what does what.  Build the first stage use a 10K as a load after C3 (crude sim of the stages load). Use a regulated supply so the voltage does not change.  Measure the nodes voltages.  Then use a signal gen set to triangle build a sim of the guitar because that adds to the 39K input and set to a pickup you like to the  peak to peak voltage it generates.  Look at the signal with a scope then change out transistors that you measured the Hfe.  Then change the parts one by one maybe remove the 100 ohm emitter resistor.  Take notes maybe even listen.  If you change the bias resistor(s) and want the same gain the ratio needs to stay the same if they are lower value more current is shunted away From the emitter

I will build circuit or fragments to make sure I got the math right and test them with a gen and scope. 

Also If you start to get this you will understand the first stage of the Crybaby like inductor whas.

Here is an idea for the guitar sim think simple, take a pickup and volume control place them in a metal box can etc for shielding feed one end of the pickup from the gen the other end to the "top" of the volume pot, ground to gen and "bottom" of the volume pot and ground of the fragment effect, wiper to input of the effect.

[frank_p]

Code Select Expand


function bigmuff(form)
{r = parseFloat(form.x.value);
re = parseFloat(form.xe.value);
r1 = parseFloat(form.x1.value);
r2 = parseFloat(form.x2.value);
h = parseFloat(form.hfe.value);
d = parseFloat(form.vbe.value);
il = parseFloat(form.l.value);
vcc = parseFloat(form.v.value);

ib=(vcc-il*r-(r+r1+r2)/r2*(d+il*re))/(r1+(h+1)*r+(r+r1+r2)/r2*(h+1)*re);
ic=h*ib+il;
ie=(h+1)*ib+il;
i2=(d+ie*re)/r2;
i1=i2+ib;
i=i1+ic;
vc=vcc-i*r;
ve=ie*re;
vb=i2*r2;

form.vcres.value = vc;form.vbres.value = vb;form.veres.value = ve;form.icres.value = ic;form.ibres.value = ib;
form.ieres.value = ie;
form.i1res.value = i1;
form.i2res.value = i2;
form.ires.value = i;}


Ref:
http://www.aronnelson.com/gallery/main.php/v/mac/apps/

[mac]

Frank,
Those are the solutions to the Big Muff bias problem, or negative feedback bias, considering transistor's leakage, ie, germs.
In my model I assumed Vbe=constant, and ic = hfe*ib + iL, iL: leakage current to be determined using Rg method.
It works fine for silicons, and well for ge due to thermal runaways, etc., although the behaviour of the bias point are as expected when you simulate temperature changes by varying hfe and leakage.

I did not simplified things because it takes just half a page to solve the eqs. 

I did the same thing for voltage divider bias (Rangemaster) and the Fuzz Face. I'm thinking of adding Harmonic Percolator and Tonebender 3 Q3 bias.

mac

[Ben N]

Buffer? What buffer?

[frank_p]

Frank,
Those are the solutions to the Big Muff bias problem, or negative feedback bias, considering transistor's leakage, ie, germs.
In my model I assumed Vbe=constant, and ic = hfe*ib + iL, iL: leakage current to be determined using Rg method.
It works fine for silicons, and well for ge due to thermal runaways, etc., although the behaviour of the bias point are as expected when you simulate temperature changes by varying hfe and leakage.

I did not simplified things because it takes just half a page to solve the eqs.  

When plugging the numeric values I get the same value of Ib than you: but I think my eq. is not simpified enough.

The biasing circuit can be analysed with 3 loops of current.

Paths:
1)  IC
2) IR2
3) IB

Steps:
Isolate IR2 in the eq. of the IR2 current loop: because it is function of IC and IB only.
Plug current gain eq. and previous eq. : in eq. of the IB current path (for illimination of IC variable.

This eq will be function of only IB
Isolate IB -> Very long equations: Takes big sheets of paper...


         Vcc - IL*(RC+RE) - VBE - (Vcc - RC*IL)   *    (RC + R1)
                                                                      -----------------
                                                                      (RC + R1 + R2)
IB =  ---------------------------------------------------------------------
        (h+1)*(RC+RE) + R1 - (RC*(h+1)) - R1)  *     (R0+R1)
                                                                         -----------------
                                                                          (RC + R1 + R2)


IL: leakage
h: current gain
R1: feedback resistor
R2: the one to ground

The rest of the equations are the same as in your application.

This is working but the eqs. are a bit long: it does not tell you immediately what to do for redesigning one with different parameters.
Two feeback loops, a voltage divider and some non ideal transistors...

But I am continuing... ...

Buffer? What buffer?

The transistor amp at the front of the fuzz.
Amp stage, sorry.

[mac]

Frank,
I'll write down the differential eqs for AC... they are coupled, and I'll try to solve them assuming a voltage at the input of the form Vo*sin(wt)
Yeah, it is easier to use Spice 

mac

[frank_p]

I'll write down the differential eqs for AC... they are coupled, and I'll try to solve them assuming a voltage at the input of the form Vo*sin(wt)

The full Fourier spectrum of a A6th with trascients. Fun fun.

[mac]

This eq will be function of only IB
Isolate IB -> Very long equations: Takes big sheets of paper...

Naaa, just half a page,





mac

[frank_p]

Oh yes, thanks Mac ! That's much simpler than what I did.  I feel like I've been all around the block to go to my neighbour.

I passed one time over what you did and I think you have missed a R3. It's in your first eq. of Vcc(I2 path).  Second line it's no there anymore.  Would yield Ib(R1+R3(beta+2)... at the denominator in final Ib eq. (does not chage much).  But I will look again later to be sure.

[mac]

Oh yes, thanks Mac ! That's much simpler than what I did.  I feel like I've been all around the block to go to my neighbour.

I passed one time over what you did and I think you have missed a R3. It's in your first eq. of Vcc(I2 path).  Second line it's no there anymore.  Would yield Ib(R1+R3(beta+2)... at the denominator in final Ib eq. (does not chage much).  But I will look again later to be sure.

It's ic*, not ie*, what looks like written in bold... 

mac

[frank_p]

Ok, I've redone it a second time and everything is fine. Yes: I mixed ic* and ie*. Thanks Mac !