DEBUG / HELP WITH - D*A*M FZ-673 (Tonebender ish)

Started by wildebelor, December 30, 2013, 08:23:18 PM

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wildebelor

Having some issues with this circuit!
Ordered the parts before christmas from LIC pedals - arrived after Christmas because of the damn post delays!
Eagerly put the board together exactly as the layout pertains and no sound!
I was hoping to give it to my Father for his belated christmas present, but not yet I'm afraid!  :icon_redface:


1.What does it do, not do, and sound like?
No sound when pedal is engaged. Bypasses effect when not engaged (like normal).

2.Name of the circuit:
D*A*M FZ-673

3.Source of the circuit (URL of schematic or project):


4.Any modifications to the circuit?
No

5.Any parts substitutions?
No

6.Positive ground to negative ground conversion?
No modification - just followed layout

7.What is the out of circuit battery voltage?
8.76

Now insert the battery into the clip. If your effect is wired so that a plug must be in the input or output jack to turn the battery power on, insert one end of a cord into that jack. Connect the negative/black meter lead to signal ground by clipping the negative/black lead to the outer sleeve of the input or output jack, whichever does not have a plug in it. With the negative lead on signal ground, measure the following:
Voltage at the circuit board end of the red battery lead =
Voltage at the circuit board end of the black battery lead =


I couldn't get this to work?


Here are my transistor values:
(All are OC75)

Q1
HFE 64
C = 8.76
B = 6.85
E = 8.74

Q2
HFE 122
C= 8.72
B= 8.63
E= 8.73

Q3
HFE 96
C = 8.72
B = 8.66
E = 8.73


Thank you in advance to anyone who can lend a hand - Just racking my brain as to why it's not running!

Nic
I can't think of anything funny just yet.

dwmorrin

I think it might be easier to consider the positive battery terminal, and all points connected, "0V", put your black lead there, and then take the voltage measurements again.

The emitters of Q2 and Q3 are both tied to 0V, so the measurement of 8.74V is confusing...

You said "I couldn't get this to work" after the basic power check instruction.
You couldn't get a voltage reading?  Row A is 0V and the right side of Row G is -9V.  (Doesn't the LED appear to be connected backwards...?)
You have to have a jack inserted into the input to turn on the battery.  Check that you wired the input jack so this happens.

Black lead should go to the red battery terminal, and red lead should check for 0V at Row A, and -9V at Row G... if that's good, then recheck/report those transistor voltages again.

wildebelor

Quote from: dwmorrin on December 30, 2013, 09:30:43 PM
I think it might be easier to consider the positive battery terminal, and all points connected, "0V", put your black lead there, and then take the voltage measurements again.

The emitters of Q2 and Q3 are both tied to 0V, so the measurement of 8.74V is confusing...

You said "I couldn't get this to work" after the basic power check instruction.
You couldn't get a voltage reading?  Row A is 0V and the right side of Row G is -9V.  (Doesn't the LED appear to be connected backwards...?)
You have to have a jack inserted into the input to turn on the battery.  Check that you wired the input jack so this happens.

Black lead should go to the red battery terminal, and red lead should check for 0V at Row A, and -9V at Row G... if that's good, then recheck/report those transistor voltages again.

Row A measures 0V and -9V at Row G

Recheck of transistor voltages are as follows:

Q1
C=8.82
B=7.36
E= 8.81

Q2
C= 8.79
B= 8.71
E= 8.79

Q3
C= 8.78
B= 8.73
E= 8.79
I can't think of anything funny just yet.

aron

Where are the transistors in this circuit? If your voltages are correct you do not have ground connected at some point and have the positive of the battery in 2 places (both on the collector and emitter side of the transistors). Is there a schematic??

wildebelor

Quote from: aron on December 31, 2013, 07:11:52 PM
Where are the transistors in this circuit? If your voltages are correct you do not have ground connected at some point and have the positive of the battery in 2 places (both on the collector and emitter side of the transistors). Is there a schematic??

I don't understand what you mean by "where are the transistors in this circuit"?  :icon_redface:
I literally just followed the layout, from my understanding this is a zonk machine?

I unfortunately don't have a schematic and can't seem to find one from a google search either  :icon_confused:
I can't think of anything funny just yet.

Kipper4

Ma throats as dry as an overcooked kipper.


Smoke me a Kipper. I'll be back for breakfast.

Grey Paper.
http://www.aronnelson.com/DIYFiles/up/

wildebelor

I can't think of anything funny just yet.

Quackzed

battery negative is going to that second to bottom row, with the battery connected, put the black lead of your voltometer on that row somewhere, so its connected also to ground(batt neg) then use the red lead to get a voltage reading on the 3 transistors. one of which has its collector on that same ground row so that ones collector should read 0v... because its connected to batt neg or 0 volts/...
nothing says forever like a solid block of liquid nails!!!

wildebelor

Quote from: Quackzed on December 31, 2013, 08:37:41 PM
battery negative is going to that second to bottom row, with the battery connected, put the black lead of your voltometer on that row somewhere, so its connected also to ground(batt neg) then use the red lead to get a voltage reading on the 3 transistors. one of which has its collector on that same ground row so that ones collector should read 0v... because its connected to batt neg or 0 volts/...


I ended up removing the whole circuit from the test box to double check all solder points then omitted the LED and the DC jack.
Retook voltages as follows:

Q1
C= 8.80
B= 1.55
E= 1.88

Q2
C= 0.07
B= 0.13
E= 0.00

Q3
C= 5.67
B= 0.06
E= 0.00

This is too strange! :-\
I can't think of anything funny just yet.

Quackzed

there we go, thats not bad , check q2's bias resistor... looks like q2's collector isnt going to ground via a resistor...  q1's emitter should be closer to 9v+ with that 4.7k resistor to 9v+, i'd check those things first...
nothing says forever like a solid block of liquid nails!!!

wildebelor

Quote from: Quackzed on December 31, 2013, 10:00:01 PM
there we go, thats not bad , check q2's bias resistor... looks like q2's collector isnt going to ground via a resistor...  q1's emitter should be closer to 9v+ with that 4.7k resistor to 9v+, i'd check those things first...

q2's bias resistor (8k2) is as follows:

0.07 row D column K

8.79 row G column K

q1's emitter is 2.49

p.s thank you so much for helping  :icon_redface:


I can't think of anything funny just yet.

Quackzed

ok,row g columb k is at 8.79, but if you look to the right on that row is - battery lead, so should be 0 volts?
thos points should all read the same as the - battery because they are all connected to the - battery lead. 0 volts... maybee the battery is backwards? or you may not be measuring with the black probe to ground.
nothing says forever like a solid block of liquid nails!!!

wildebelor

Quote from: Quackzed on December 31, 2013, 10:37:21 PM
ok,row g columb k is at 8.79, but if you look to the right on that row is - battery lead, so should be 0 volts?
thos points should all read the same as the - battery because they are all connected to the - battery lead. 0 volts... maybee the battery is backwards? or you may not be measuring with the black probe to ground.


I feel like punching myself in the face....  :icon_eek:
see the trace cut on - Row G Column F - well I was measuring the black probe from the opposite side of that cut, which was giving me readings that were wacky

Lets try that again:

Q1
C= 0.00
B= 5.51
E= 7.01

Q2
C= 8.71
B= 8.65
E= 8.78

Q3
C= 2.46
B= 8.68
E= 8.78

q2's bias resistor (8k2) is as ACTUALLY follows:

8.70 row D column K

0.00 row G column K

Sorry for the mixup!

I can't think of anything funny just yet.

Quackzed

#13
this looks like the schem, with a few value changes...

but were getting somewhere, still looks like q2's collector is high, try measuring resistance from q2's collector to ground... or reflowing solder around that 8k2 bias resistor... or double check those cuts on row b and c make sure they are not shorting...
BTW Happy New Year!!!! :)
  good luck , keep at it! i'll check back tomorrow.
nothing says forever like a solid block of liquid nails!!!

LucifersTrip

#14
Quote from: wildebelor on December 31, 2013, 10:51:32 PM

Q1
C= 0.00
B= 5.51
E= 7.01

Q2
C= 8.71
B= 8.65
E= 8.78

Q3
C= 2.46
B= 8.68
E= 8.78


first, link to the exact schematic you used. If using something similar to the schematic above. it looks like you have the DMM probes on the wrong rail. Q1C should be ~9V (your battery or wart V) and Q2, Q3 E  should be 0V (ground)

black lead on  ground, and red lead on the 3 lugs you're measuring
always think outside the box

wildebelor

Quote from: LucifersTrip on January 01, 2014, 02:54:57 AM

first, link to the exact schematic you used. If using something similar to the schematic above. it looks like you have the DMM probes on the wrong rail. Q1C should be ~9V (your battery or wart V) and Q2, Q3 E  should be 0V (ground)

black lead on  ground, and red lead on the 3 lugs you're measuring

hey LucifersTrip - I only used a layout - couldn't find a schematic.

I also got it working as per the layout, if I have anymore problems I'll report back but a big thanks to everyone!  :icon_biggrin:

Sounds HUGE!
I can't think of anything funny just yet.

Quackzed

cool. glad you got it going! it may be the first fuzz of 2014!  :icon_cool:
nothing says forever like a solid block of liquid nails!!!