Kenwood KR-3130

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So you have 0V on Q1 base and approx 1.5V on the base of Q613.

Well the 'good' news is that also shows a problem. 0V on the base of Q1 means you should also see 0V on the emitter leg of Q613 as they are connected together.

If you have approx 1.5V on the base leg of Q613 then that means you should be seeing 1.5V directly across the B/E junction of Q613 and if so then that means it is faulty. The B/E voltage even at very high currents limits out at no more than around 0.8 volts.

So check carefully as before but this time the first thing to check is just to see if you have 1.5 volts across B and E by measuring directly on the legs (and be careful not short anything). That smaller type of transistor would be more likely to have this kind of fault... something you don't see in the big output type devices.
 
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OK, lets work with that. As you will see, the devil is in the detail unfortunately.

If Q613 emitter is at 0.8 volts then you now have to reconcile why:

Q1 is reading 0V
.

The base of Q1 and the emitter of Q613 are the same point. I think this is what you are describing... so is there a break between the base of Q1 and Q613 emitter?

(The bulb tester being dim is OK, that's normal)
 

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