maestro fz1-b info

[LucifersTrip]

2. Actually, I don't know what it does with the squelcher circuit, or what the squelcher circuit does, either.  

solidhex answered that one in another thread:

http://www.diystompboxes.com/smfforum/index.php?topic=97391.msg848825#msg848825


and just for the helluvit

[Mark Hammer]

2. Actually, I don't know what it does with the squelcher circuit, or what the squelcher circuit does, either. 

solidhex answered that one in another thread:

http://www.diystompboxes.com/smfforum/index.php?topic=97391.msg848825#msg848825

Hey, I guessed right!  I'm learning stuff!   

[pinkjimiphoton]

could be. 

[Gus]

The correct answer is the diode keeps a min current going to the base of the transistor Q4 with no signal if it was off how could the output ever start conducting?  The diode also keeps the .1uf only discharging by the two resistors and Q4

Note the path  +9 to the 470k to the diode to the resistors and .1uf cap and Q4

The .1uf can be thought of as part of the release time (RC time constant)

The more positive the output transistor collector goes the more current to the squelcher transistor base(raising the voltage at the .1uf) turning it on more and (connecting Q3 emitter more to ground raising the gain).  As the music stops the .1uf discharges to a lower level via the two resistors and the base of the squelcher transistor.

All the parts at the output interact for the charging and discharging of the .1uf,  resistor values, .1uf cap value, transistor hfe

[Gus]

pinkjimiphoton

I noted the picture of the board looks to have the resistors for the squelcher transistor.  I also noted the disk caps are bent to the right and might be covering the transistor hiding it from view.  I understand you don't want to harm the PCB or break the cap legs if you could look at the trace side of the PCB you could could figure out if there is a Q4.

Now looking at this circuit I see three parts I would adjust if I was to build this.  One I would rework the input emitter follower(EF) could be as simple as a MPSA18 and using a 470K  and adding a 1meg base to ground input divider bias resistors.  Two working out the bias resistors (C to B) for different transistor hfe for the two gain stages.  Three find a good time constant(s) for different  squelcher transistor hfe.

I would start with a MPSA18 input, 2N2222 gain stages and a 2N5088 Q4.

[pinkjimiphoton]

thanks guys...

i heard from glenn, i will get to the bottom of it probably wednesday...see him tuesday nite, and will get it then,

he gave me permission to disassemble it as necessary, so will do so.

i am really curious.....was i THAT baked, or is it really three trannies?


stay tuned!!

[LucifersTrip]

I noted the picture of the board looks to have the resistors for the squelcher transistor.  I also noted the disk caps are bent to the right and might be covering the transistor hiding it from view.  I understand you don't want to harm the PCB or break the cap legs if you could look at the trace side of the PCB you could could figure out if there is a Q4.

from his voltages (below), without looking at the trace side, you'd have to guess that if there is a 4th that he's not seeing, it would have to be Q3, since he's given two already with Qev = 0

q1
e 7,87
c 9.15
b 8.04

q2
e 0
c .84
b .53

q3
e 0
c .49
b .47

[pinkjimiphoton]

works for me. wouldn't surprise me that i missed it, my bad.

irregardless, i'll have it tuesday nite. 

[Gus]

Edit

[Solidhex]

The correct answer is the diode keeps a min current going to the base of the transistor Q4 with no signal if it was off how could the output ever start conducting?  The diode also keeps the .1uf only discharging by the two resistors and Q4

Note the path  +9 to the 470k to the diode to the resistors and .1uf cap and Q4

The .1uf can be thought of as part of the release time (RC time constant)

The more positive the output transistor collector goes the more current to the squelcher transistor base(raising the voltage at the .1uf) turning it on more and (connecting Q3 emitter more to ground raising the gain).  As the music stops the .1uf discharges to a lower level via the two resistors and the base of the squelcher transistor.

All the parts at the output interact for the charging and discharging of the .1uf,  resistor values, .1uf cap value, transistor hfe

Listen to Gus. He knows more about this stuff than I do.

[pinkjimiphoton]

break out the eggs, guys, i was wrong...it's a 4 transistor. no idea how the heck i coulda missed it looking at it, but i did.

so let loose the eggs of "told ya so"!!

more later!!! 

[Mark Hammer]

It's like that $20 bill you were sure was in those pants, and even after you went through all the pockets, you still couldn't find it...so how the hell did it end up in the dryer?   

[pinkjimiphoton]

ex-actly!!!!

near as i can figure, i must have been....well, i sure wasn't stoned!!!!

[joegagan]

ahh, mystery solved. thanks for taking the time to update us, jimi. 

[pinkjimiphoton]

 

all good. i don't know how i missed it!!