What went wrong?

[Stink Bear]

Recently I acquired a Fisher RS-2004A integrated receiver / amp. It had two burned fuses on the power board, both 6 amp if I remember correctly. I replaced them and powered it up on a dim bulb tester with a 60W bulb. It turned on and I got a music out of it for a short time and then a resistor on what I would guess is the main amp (I haven't looked at a schematic yet) board smoked and popped.
I was under the impression that the dim bulb tester was supposed to prevent damage.
I also read that a lower watt bulb gave more protection than a higher watt bulb, which I haven't made sense of yet but I accepted.
So what went wrong?
I thought the bulb was glowing more brightly than the one other time I used it on a unit that didn't have a power problem, but it still was not very bright. I do not remember a bright surge and then going dim either.
Maybe the tester is not supposed to directly prevent damage, but indicate to me if there is a problem so I can shut down the unit before damage occurs?
I'm busy and won't have time to work on the unit again for at least a few weeks but I would like to know where things went south so I can prevent problems like this in the future.
Thanks

 

[petehall347]

dbt is only really any good at saving output transistors . although it can and does save some other things . but not always everything .
at least it made things more graceful . best id burned resistor first to have an educated guess as to what is shorted out .

 

[petehall347]

yes lower wattage bulb allows less current .

 

[nj pheonix]

Dnt is a voltage divider.
Higher watt , lower resistance (impedance)
So you're dividing V though 2 series resistances . The bulb and your device.
The more v drop through bulb the less at amp.

 

[Stink Bear]

nj phoenix: So the light bulb provides a voltage drop, lowering the voltage available to the unit, which in turn lowers the amperage the unit draws? And the lower the wattage of the light bulb, the higher the voltage drop across it?

 

[Stink Bear]

Thanks for the replies.

 

[efzauner]

A higher watt bulb means a lower resistance since P= V^2/R lower R, = higher watts. However, recall that resistance increases with heat, so the way this light bulb in series works is that there is a very small voltage drop across the bulb when very low current is drawn by your amp. If there is a short somewhere, it will draw more current, thus increasing the power in the bulb fillament, increasing it's temp and resistance.. and therefore increased voltage drop and giving protection to your equipment.

 

[Stink Bear]

efzauner: Thanks for the reply. I understand your explanation of how the bulb works in relation to the current drawn by the amp. I'm lost on the equation. My electronics training was too little, too long ago I guess.
I dug out my old Electronic Communication (Schrader) book. Hopefully I will have time to start reading it soon. Maybe I can learn or re-learn some things.