vintage AD9 output very quiet

[thegnu]

So I got these two non-working Ibanez AD9s, and I'm trying to get them fixed. There are a couple issues, and I'm focusing on one of them right now. The schematic is here:
http://www.dirk-hendrik.com/Ibanez_ad9_analog_delay.pdf

1. DIODE
In B7/B8, there's a diode, D4. it was cracked on one of the AD9s, and the other pedal was grounding out my DC Brick, so I put the "good" diode in the pedal that didn't short the power supply, and lo and behold, the power supply shorted on that one. And now the other one turns on. So I'm functioning without that diode there, and that's weird. What does this diode do? Is this just a bad diode?

2. SIGNAL
I bridged a broken trace, and now it outputs signal, but it's very very attenuated on the Main Out. Dry Out works fine. So I looped signal on my DD5, and grounded the sleeve of a cable to the chassis of the AD9, and started touching signal path to test out where the signal goes bad.

3. RESISTOR AND OP-AMP
So I'm tracing back through the circuit with the tip of my guitar cable. Signal is good @ the place the output signal is soldered to the board. On the schematic in A1-3, the signal attenuates once over R65, and again at pin 6 of the JRC4558D. Is this normal behavior? The resistor will lower the signal, right? But the opamp?

Since my signal is fine from the Dry Output jack, the input and dry signal stage is fine. Since both the dry and wet signal coming out the Main Output jack are weak, then my problem has to be after wet/dry rejoin (in A1-3,B2-4 area), right?

[es44]

I am not experienced enough to read the complete schematic, but at least i know, that D4 is sitting there to protect from reverse polarity of the PSU.
Ideally, it should blow the fuse in the PSU connected, but if that fuse is a lot more than 1A, maybe even slow blow, then the diode protects the fuse by commiting suicide

Best regards
Ebbe

[Nigel Goodwin]

As es44 says, D4 is a reverse protection diode, it's blown because the power had been connected the wrong way round.

This could have damaged other components, you need to work through the circuit, either injecting, or tracing, a signal to find out where it's getting lost.

[thegnu]

OK, cool. I figured the diode was either a reverse polarity thing, or a voltage bleed in case you hooked up too high a voltage. I've got the right polarity, but it IS pretty troublesome that it works WITHOUT the diode.

I think I mis-identified the pin on the opamp, but if I'm on the input side of the opamp, and the signal coming out the output is quieter, then it's either not getting the right voltage or bias, or the opamp is bad, right? Because the opamp is suppose to amplify stuff, right?

[EDIT: just found the 4558 pinout, and I was right... (EDIT: that pin 6 is input) http://gaussmarkov.net/wordpress/par...1-description/ -- 3/5 down the page, the second diagram]

I've been learning quite a lot about electronics with my current ventures, lately, but I'm still like a child in the woods.

[Nigel Goodwin]

Originally posted by thegnu
OK, cool. I figured the diode was either a reverse polarity thing, or a voltage bleed in case you hooked up too high a voltage. I've got the right polarity, but it IS pretty troublesome that it works WITHOUT the diode.

Not at all, that's a nice bonus

I think I mis-identified the pin on the opamp, but if I'm on the input side of the opamp, and the signal coming out the output is quieter, then it's either not getting the right voltage or bias, or the opamp is bad, right? Because the opamp is suppose to amplify stuff, right?

If it's an inverting mode opamp, then the input pin will have essentially nothing on it, because it's a virtual earth point, so you can't signal trace (or inject) on the non-inverting input.

[thegnu]

Originally posted by Nigel Goodwin
If it's an inverting mode opamp, then the input pin will have essentially nothing on it, because it's a virtual earth point, so you can't signal trace (or inject) on the non-inverting input. [/B]

So does that mean that the signal is supposed to be coming in on pin 5 (the positive input)? So I would inject the signal there? Or did I misunderstand?

I appreciate you sharing your knowledge.

[Nigel Goodwin]

Originally posted by thegnu


So does that mean that the signal is supposed to be coming in on pin 5 (the positive input)? So I would inject the signal there? Or did I misunderstand?

No, you misunderstood, the input is at pin 6 (the non-inverting input), BUT you can't see it, or inject it, at that point - because it's a virtual earth mixer - you could inject it via a suitable resistor though.

You need to check at the input resistors feeding it, R44 and R47.

[thegnu]

Originally posted by Nigel Goodwin



No, you misunderstood, the input is at pin 6 (the non-inverting input), BUT you can't see it, or inject it, at that point - because it's a virtual earth mixer - you could inject it via a suitable resistor though.

You need to check at the input resistors feeding it, R44 and R47.

Someday I will understand. For now, I will follow your instructions.

Thanks.

[thegnu]

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