The whole thing has a circuit common, which we usually call ground. Your guitar cord plugs have the "sleeve" (the shaft that sticks into the jack), and the "tip", which is, well, the tip of the plug. The sleeve is the ground connection in the jacks. What I call the shell is the metal cover of the plug that screws down on it to cover the wiring. I find that shell a convenient place to clip by meter ground probe. I usually clip onto the edge of the hole where the wire enters the plug.
The metal box ought to be ground too. I just personally find clipping to the plug body shell more convenient.
Now the black wire from the battery snap is NOT ground. It is supposed to get CONNECTED to ground by the input jack, but only when a plug is in it. If you clip your meter to that wire, you can measure to the other end of the battery and get 9v, but that doesn't tell you if the ground is actually connected.
I suspect for some reason your input jack is not grounding that black wire. So by clipping your meter ground to the actual circuit ground, not the black wire from the battery, we now find if or not the 9v is actually getting into the system. You report that even the negative end of C5 has +9 on it. That is a dead giveaway that the negative end of the battery is not being grounded.
So do the test I describer earlier. NO battery or power adaptor. PLug a plug into the input jack. Connect your meter ground (black) probe to the sleeve/shell of the plug and measure resistance to that black wire from the battery snap connector. It is supposed to measure a connection - ie low resistance, continuity. But I am betting you will find it is not. Probably the jack's fault.
The metal box ought to be ground too. I just personally find clipping to the plug body shell more convenient.
Now the black wire from the battery snap is NOT ground. It is supposed to get CONNECTED to ground by the input jack, but only when a plug is in it. If you clip your meter to that wire, you can measure to the other end of the battery and get 9v, but that doesn't tell you if the ground is actually connected.
I suspect for some reason your input jack is not grounding that black wire. So by clipping your meter ground to the actual circuit ground, not the black wire from the battery, we now find if or not the 9v is actually getting into the system. You report that even the negative end of C5 has +9 on it. That is a dead giveaway that the negative end of the battery is not being grounded.
So do the test I describer earlier. NO battery or power adaptor. PLug a plug into the input jack. Connect your meter ground (black) probe to the sleeve/shell of the plug and measure resistance to that black wire from the battery snap connector. It is supposed to measure a connection - ie low resistance, continuity. But I am betting you will find it is not. Probably the jack's fault.
I'm still not exactly sure what you mean sorry. I understand the TS concept, that the sleeve is the ground, but where do i connect my leads when its plugged in. Hopefully you can point it out with the pictures.
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Also, I have all the jacks out of the casing so I could get to the components at the power supply. Does the input jack need to be reattached to the casing when testing it? Thanks
Also, I have all the jacks out of the casing so I could get to the components at the power supply. Does the input jack need to be reattached to the casing when testing it? Thanks
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Alright, nvm i figured it out. I am getting 9v At D2, C3,C4, and at (I believe) leads 2 & 3 on Q2. On what i think is lead 1 of Q2 (the lead right next to the positive lead on C5) I'm getting nothing, and I'm getting nothing at C5.
Then Q2 is probably bad. That assumes C5 is not shorted.
Once you get the 8v on C5, you must also make sure the op amp to the right is also working, making that 4v reference voltage.
Once you get the 8v on C5, you must also make sure the op amp to the right is also working, making that 4v reference voltage.
The 4v comes from IC1a, just to the right of C5 on the drawing. I see it foes appear to be the only IC on the board. The other half of the IC is used elsewhere in the circuit.
I am going to bet that the IC is OK, but until we see the 4v, we will not know.
I am going to bet that the IC is OK, but until we see the 4v, we will not know.
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