Please help debugging phase 90 (voltages included)

Started by ianmgull, June 04, 2008, 10:05:15 PM

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ianmgull

Any help would be very much appreciated.





1.What does it do, not do, and sound like?

No guitar signal is passing through. Just an very loud hum. Turning the speed pot sound like it throws the hum slightly out of phase, but I may just be imagining that.


2.Name of the circuit =

Tonepad Phase 90


3.Source of the circuit

Tonepad project file: (in pdf form)     http://tonepad.com/getFile.asp?id=42


4.Any modifications to the circuit?

No.


5.Any parts substitutions?

I didn't have a 250k trim pot so I used a 500k instead. All other parts are as listed on the project file.



Power supply voltage:   11.05v

Power supply voltage from board:  11.05v


Q1: D  0.62v
      S  0.62v
      G  0v

Q2: D  0.48v
      S  0.62v
      G  0v

Q3: D  0.62v
      S  0.62v
      G  0v

Q4: D  0.47v
      S  0.62v
      G  0v

Q5: C  0.25v
       B  0.1v
      E  0.62v

IC1: Pin 1  0v
       Pin 2  0v
       Pin 3  0.42v
       Pin 4  0v
       Pin 5  0.01v
       Pin 6  0.01v
       Pin 7  0.01v
       Pin 8  11.05v

IC2: Pin 1  0v
       Pin 2  0v
       Pin 3  0.62v
       Pin 4  0v
       Pin 5  0.48v
       Pin 6  0v
       Pin 7  0v
       Pin 8  11.05v

IC3: Pin 1  0v
       Pin 2  0v
       Pin 3  0.62v
       Pin 4  0v
       Pin 5  0.48v
       Pin 6  0.01v
       Pin 7  0.01v
       Pin 8  11.05v


Polerized Cap 1 (10uf)  Cathode  0v
                                  Anode     0.62v


Polerized Cap 2 (15uf)  Cathode  0v
                                  Anode     0v




I'm going to go take a walk so I don't pull my hair out. :-\   Thanks for looking this over, hopefully somebody can make sense of it.

Ian

Rob Strand

Check you have put the zener diode D2 and the 1uF cap in parallel with it in the correct way.  (I'm assuming you have actually used a zener for D2.)

Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

ianmgull

Hey thanks Rob. Both polarized caps (I assume you meant 10uf?) are oriented negative side up. The 5.1v zener is also oriented as on the schematic. The only substitution I made was the 500k trimpot. Any other ideas?

R.G.

Rob's right - check that the zener's in the correct way. You should see 0V on one end if the zener, and 3-6V on the other end, the voltage depending on which zener diode you bought.
R.G.

In response to the questions in the forum - PCB Layout for Musical Effects is available from The Book Patch. Search "PCB Layout" and it ought to appear.

petemoore

http://www.geofex.com/FX_images/p90ramp.pdf
It looks like all the voltage dividing is happening around the lower left corner/diodes/10k, and bias pot.
IC1: Pin 1  0v
       Pin 2  0v
       Pin 3  0.42v
       Pin 4  0v
       Pin 5  0.01v
       Pin 6  0.01v
       Pin 7  0.01v
       Pin 8  11.05v
   None of the biased pins are biased, pin4 reads Gnd. and V supply is there.
IC2: Pin 1  0v
       Pin 2  0v
       Pin 3  0.62v
       Pin 4  0v
       Pin 5  0.48v
       Pin 6  0v
       Pin 7  0v
       Pin 8  11.05v
  Same story for this one..I can't see much to check except the bias section connections or that there is something pulling the 'bias pins' of the OA down to 0v or slightly above...it looks like they all are sharing a common problem stemming from the vbias connections.
Convention creates following, following creates convention.

ianmgull

Hey thanks for the help everyone.


R.G.

I checked again, and the zener is oriented as indicated on the tonepad layout. One end (grounded side) is getting 0v as expected but the other side is getting 0.62v. I used a 5.1v zener. That means I should be reading 5.1v there doesn't it?


Pete

When you say "bias pin" are you referring to pin four of the IC's? I looked at the schematic but I'm not sure where the bias source is. I'm looking for the typical voltage divider around the DC jack but it doesn't look like what I've seen before. Should the bias pin be getting half of the main supply voltage? (4.5v or in this case around 5.5?).


Thanks again for the help.

Ian


petemoore

When you say "bias pin" are you referring to pin four of the IC's?
  Pin 4 is Gnd. = 0vdc.
  The input and output pins need to be biased somewhere between V+ and Gnd. to have room to swing the signal.
  I looked at the schematic but I'm not sure where the bias source is.

  It looks like the diodes keep a voltage potential for the bias, where the 10k and diode meet, there is marked Vref 3v, a voltage 'reference' which is connected through resistors to the opamp +inputs.
   I'm looking for the typical voltage divider around the DC jack but it doesn't look like what I've seen before.
  Yupp...no Vdivider here.
  Should the bias pin be getting half of the main supply voltage? (4.5v or in this case around 5.5?).
  I'm not the one in the real about ph90 bias, but that's what I recall my voltages were...
  basically something is off on all the pins that should be biased where the opamp can work, [inputs and output pins], something common to all of them probably, I guess enough about that when 1 pin gets biased the others will follow suit, and the ph 90 will phase [if everything else and bias pot is adjusted, jfets matched], and show bias voltages...like they 'should be.
   
Convention creates following, following creates convention.

ianmgull

Thanks a lot Pete, I'm going to check that out as soon as I get home from work.

Rob Strand

The zener voltage being not being correct is a show stopper (whatever the cause).

I suggest carefully lifting one leg of the zener out then temporarily placing a 10k resistor on the back of the board in it's place, then measure the voltage across that resistor.

If there is a problem with the zener the circuit will spring to life with the 10k and you will have to get a new zener.  If the zener is OK the voltages with the 10k present will possibly allow further diagnosis.

Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

ianmgull

So finally I get back to working on this one. I tried what you mentioned Rob, with a 10K in place of the zener. I'm still only getting about .6v on the positive end. Can someone explain how the bias network is set up in this? I realize that because everything's bias is off, this is the place to look.

mdh

Is it possible that you accidentally put a 100k resistor in for the 10k resistor from +9V to the cathode of the zener?  Because that would explain the voltage that you get at that junction with a 10k in place of the zener.  If the resistor is marked with 3 + 1 bands, a 10k would be brown-black-orange-(silver/gold/whatever); if it's a 4 + 1 band marking, it would be brown-black-black-red-(gold/whatever).  If you have a mixed stock, it would be easy to substitute a 4-band marked 100k (brown-black-black-orange...) for a 10k.

In general, the way that bias network is supposed to work is that the 10k resistor sets the reverse-biased current through the zener.  If the current and supply voltage are high enough, the zener will hold the voltage between its terminals at the rated voltage.  If not, then it won't drop the rated voltage.  So my 100k hypothesis could explain the behavior in both cases (with zener/with 10k in place of zener).  It's just a guess, though.

ianmgull

Hey thanks. I just pulled the 10k (the one from +9 to positive end of zener) and I'm still getting about .6v...   Does D1 have anything to do with the bias network? Or is it just the zener and the 10k resistor? Thanks.

ian

mdh

Quote from: ianmgull on June 30, 2008, 07:57:53 PM
Hey thanks. I just pulled the 10k (the one from +9 to positive end of zener) and I'm still getting about .6v...   Does D1 have anything to do with the bias network? Or is it just the zener and the 10k resistor?

No, D1 is just there for reverse-polarity protection.  If you have pulled both the 10k resistor and the zener, you no longer have a bias network.  What I would do is put in a new 10k (measure it first), and then put in either the zener diode or another 10k resistor to see whether you get something around 4.5-5V.  Ultimately you need the zener there, but this is just troubleshooting.  It's entirely possible that there's a short somewhere that's screwing up the bias even with good components of the right values in place.  One place I would check is between the leads of the trimmer.  Also, did you look at or check the resistance of the resistor you just removed?

ianmgull

Yeah, so far I've tried a couple different 10k resistors and zeners with the same result. I guess now I'm looking for solder bridges coming off of the +9v power trace. Not quite sure what exactly else to look at...

mdh

This is something of a shot in the dark, but have you tried replacing the 10uF electrolytic cap in parallel with the zener diode?  Also, does twiddling the trim pot change the voltage at the junction of the zener and the 10k resistor?  It shouldn't but if one of the legs of the trimmer is shorted to the wiper it will, and that would be a problem.

Edit: be sure to check the polarity of that electrolytic cap as well.

ianmgull

#15
Strange.... On one extreme of the trimpot I'm getting .61v while one the other extreme I'm getting .65v. You weren't reffering to that small of a change were you? I'm going to replace the cap you mentioned. thanks!