DIode tests both ways?
With the red wires off, there should be NO continuity between those two posts, and even ini diode test there should be no diode drop measured. If it checks between the two posts like there is a diode across them, that is not good, that is a shorted diode rectifier. if you meant it checks open even om diode test, then that is what we want.
There is a +HV rail and a -HV rail, and from each, the output power transistors connect to the speaker out. Their job is to condect EITHER positive or negative current to the speaker to make sound happen. But if both sides turned on at the same time then current would flow from the positive to the negative side directly. This will blow fuses.
In its simplest form, the bases of the + side transistors and the - side transistors are connected together. That way if the signal goes positive, it turns on th + side transistors, and when negative, it turns on the - side transistors. That is how it amplifies the current the speakers need.
In practice, we do not connect those bases together, instead we have a small circuit that allows the bases on each side to be slightly biased towards their ON voltage. SO instead of the opposing bases sitting at zero volts, our circuit biases the + side one at about +.5v, and the - side bases at about -.5v. Adding an extra half volt on eaither side, we have about +1v and -1v respectively at the pos and neg driver transistor bases.
Those TIP14x transistors - I am going to abbreviate transistor as xstr from here on - those TIP14x xstrs are darlington xstrs. That means there are two xstrs in one part, so they contain their own drivers essentially. SO I expect more or less +1v at the bases of Q3,4 and more or less -1v at the bases of Q5,6.
Look just to the right of th op amp and you see a string of four diodes CR7-10. They are between the + and - bases. At about half a volt each, that leaves the two volts (+1 to -1) total to hold those bases apart.
But what happens when one of those diodes opens? R73,74 will pull the bases of the + side xstrs towards +rail and that would turn them on fully. Likewise, on the negative side, R75,76 would pull the bases of the negative side xstrs towards -rail and that would turn them on hard. Pulling the base voltage of a bipolar towards its collector voltage is how to turn it on. Xstrs are really current devices and not voltage devices, but this description serves me better - put the voltage there and the current will flow.
SO as it sits, the current through R73,74,75,76 also flows through the four bias diodes. And their volatge drops add up to keep the xstr bases right on the edge of turn on. If that string opens, then BoTH sides turn on hard, and fuses blow.
I don't know that to be your case, but it merits exploration. A simple bypass should tell you. If we clip a shorting wire across the four diodes - basically short the base of Q3 to the base of Q5 - and power it up, and the fuses no longer blow, then we have found the area of trouble. If it still blows, fuses, then the bias string is not it.
With the red wires off, there should be NO continuity between those two posts, and even ini diode test there should be no diode drop measured. If it checks between the two posts like there is a diode across them, that is not good, that is a shorted diode rectifier. if you meant it checks open even om diode test, then that is what we want.
There is a +HV rail and a -HV rail, and from each, the output power transistors connect to the speaker out. Their job is to condect EITHER positive or negative current to the speaker to make sound happen. But if both sides turned on at the same time then current would flow from the positive to the negative side directly. This will blow fuses.
In its simplest form, the bases of the + side transistors and the - side transistors are connected together. That way if the signal goes positive, it turns on th + side transistors, and when negative, it turns on the - side transistors. That is how it amplifies the current the speakers need.
In practice, we do not connect those bases together, instead we have a small circuit that allows the bases on each side to be slightly biased towards their ON voltage. SO instead of the opposing bases sitting at zero volts, our circuit biases the + side one at about +.5v, and the - side bases at about -.5v. Adding an extra half volt on eaither side, we have about +1v and -1v respectively at the pos and neg driver transistor bases.
Those TIP14x transistors - I am going to abbreviate transistor as xstr from here on - those TIP14x xstrs are darlington xstrs. That means there are two xstrs in one part, so they contain their own drivers essentially. SO I expect more or less +1v at the bases of Q3,4 and more or less -1v at the bases of Q5,6.
Look just to the right of th op amp and you see a string of four diodes CR7-10. They are between the + and - bases. At about half a volt each, that leaves the two volts (+1 to -1) total to hold those bases apart.
But what happens when one of those diodes opens? R73,74 will pull the bases of the + side xstrs towards +rail and that would turn them on fully. Likewise, on the negative side, R75,76 would pull the bases of the negative side xstrs towards -rail and that would turn them on hard. Pulling the base voltage of a bipolar towards its collector voltage is how to turn it on. Xstrs are really current devices and not voltage devices, but this description serves me better - put the voltage there and the current will flow.
SO as it sits, the current through R73,74,75,76 also flows through the four bias diodes. And their volatge drops add up to keep the xstr bases right on the edge of turn on. If that string opens, then BoTH sides turn on hard, and fuses blow.
I don't know that to be your case, but it merits exploration. A simple bypass should tell you. If we clip a shorting wire across the four diodes - basically short the base of Q3 to the base of Q5 - and power it up, and the fuses no longer blow, then we have found the area of trouble. If it still blows, fuses, then the bias string is not it.
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