Fender BXR300 Cutting Out

[bobloblaws]

The 1k resistor (R62, R67) are the sense resistors feeding the bases of the current limiter circuit, and are NOT part of the bias measurement test points.

But the bias test procedure described does measure across R62 and R71 in series. The R62 test point is connected to R63 and there is also continuity between the two opposite ends of both R62 and R71.

[nevetslab]

But the bias test procedure described does measure across R62 and R71 in series. The R62 test point is connected to R63 and there is also continuity between the two opposite ends of both R62 and R71.

While I don't have anything but the schematic you had posted on this circuit, the bias adjustment sets the voltage difference across the Q9 & Q13, those being the voltage gain stage of the circuit, with the bias circuit placed between those collectors. From there, you're feeding the bases of the two pairs of driver xstgrs, which feed the bases of the two pairs of output xstrs. Bias voltage of 10mV, which I recall reading earlier being close to 30mA of current flowing in Q20. I'm suggesting looking further so you're not taking the reading of that one xstr's collector current flowing thru R71. Looking across all four pairs of output xstrs may spot one pair that's conducting a lot more than 30mA. That's where the output current is flowing at idle. The total of all that current flowing times the collector voltage of the power xstrs is the bulk of the idle power being dissipated in the heat sink. All gets back to dealing with how to get the heat out of the system. There might be one or more of the power xstrs that are running higher idle current.

[bobloblaws]

While I don't have anything but the schematic you had posted on this circuit, the bias adjustment sets the voltage difference across the Q9 & Q13, those being the voltage gain stage of the circuit, with the bias circuit placed between those collectors. From there, you're feeding the bases of the two pairs of driver xstgrs, which feed the bases of the two pairs of output xstrs. Bias voltage of 10mV, which I recall reading earlier being close to 30mA of current flowing in Q20. I'm suggesting looking further so you're not taking the reading of that one xstr's collector current flowing thru R71. Looking across all four pairs of output xstrs may spot one pair that's conducting a lot more than 30mA. That's where the output current is flowing at idle. The total of all that current flowing times the collector voltage of the power xstrs is the bulk of the idle power being dissipated in the heat sink. All gets back to dealing with how to get the heat out of the system. There might be one or more of the power xstrs that are running higher idle current.

This is tricky as the emmiter ends of those resistors are hiding underneath the heat sink fins. Is there another way to measure whether one of the pairs is conducting more than the others?

Also, just to be clear, this is still measured at idle/no load?

[nevetslab]

Yes....still measured at idle, no load.

Ah......I hadn't thought about that. Both the emitter leads of the power xsts AND the emitter lead ends at the emitters are obstructed by the heat sink fins. Arghhhh! OK, I now see why they had you measure between the output buss and either of the current sense resistors R62 or R67. The 1k resistor is only adding to the input resistance of your DMM, which in DCV mode is probably 10M, so it doesn't change anything. But, due to the mechanical layout, you can't get at ANY of the emitter resistor leads where they connect to the emitters of the output xstr's. So, we can't find out if there's any substantial difference in collector current of the other output pairs. Only way to do it is by removing the single PCB assembly from the chassis, and flipping it over. Now, one could measure the case temperature of the power xstrs, to see if all are running the same temp, or if there are substantial differences. Granted, they're all bolted down to the heat sink to effectively average the temperature out, but, there would still be temperature differences that could be sensed. I don't suppose you have a Surface Probe. In my collection of Temperature meters/measuring gear, I have a Surface Probe for my Fluke Type 51 & 52 to do that sort of thing. I have one of their Infrared Temperature Probes 80TK-IR, which can make selective measurements.

[bobloblaws]

Yes....still measured at idle, no load.

Ah......I hadn't thought about that. Both the emitter leads of the power xsts AND the emitter lead ends at the emitters are obstructed by the heat sink fins. Arghhhh! OK, I now see why they had you measure between the output buss and either of the current sense resistors R62 or R67. The 1k resistor is only adding to the input resistance of your DMM, which in DCV mode is probably 10M, so it doesn't change anything. But, due to the mechanical layout, you can't get at ANY of the emitter resistor leads where they connect to the emitters of the output xstr's. So, we can't find out if there's any substantial difference in collector current of the other output pairs. Only way to do it is by removing the single PCB assembly from the chassis, and flipping it over. Now, one could measure the case temperature of the power xstrs, to see if all are running the same temp, or if there are substantial differences. Granted, they're all bolted down to the heat sink to effectively average the temperature out, but, there would still be temperature differences that could be sensed. I don't suppose you have a Surface Probe. In my collection of Temperature meters/measuring gear, I have a Surface Probe for my Fluke Type 51 & 52 to do that sort of thing. I have one of their Infrared Temperature Probes 80TK-IR, which can make selective measurements.

I've been measuring temperature with my Ryobi gun which I linked to in post #30. I did not see any temperature difference between any of the eight transistors.

I may be able to get measurements from the emmiter resistors despite the obstruction. I managed to measure across each one earlier. But to do both sides at the same time is doubly hard.

I may not get to this anyway. I'm heading down to their club for a field test. If there are no more issue I'll leave it there. If it still shuts down I'll bring it back and see what I can do.

[bobloblaws]

I've been measuring temperature with my Ryobi gun which I linked to in post #30. I did not see any temperature difference between any of the eight transistors.

I may be able to get measurements from the emmiter resistors despite the obstruction. I managed to measure across each one earlier. But to do both sides at the same time is doubly hard.

I may not get to this anyway. I'm heading down to their club for a field test. If there are no more issue I'll leave it there. If it still shuts down I'll bring it back and see what I can do.

Well I played it on 5/10 with their Jackson bass, active pickups through input jack 1. Apparently that is their bass player's typical setup. He wasn't there so I did the tryout myself. Played it steady for about 15 minutes and it really wasn't getting very hot at all. I felt around the back and it seemed like the fan was still on regular speed and the air coming out wasn't particularly warm. So crossing fingers that the bias adjustment did the trick. I'll know in a couple of days when they have a proper jam session.

[g1]

That's good news, and why I wanted to make the point earlier that just because you could still make it do thermal cycling, did not necessarily mean it was still faulty.
I guess you have verified that he does not also connect a second cab.?

[bobloblaws]

I guess you have verified that he does not also connect a second cab.?

I only know that he experimented with a second cab when I suggested trying it a couple of weeks ago when they first bought the amp for their club and started having shut down issues. I reasoned that dropping the load from 8 ohm to 4 ohm
would result in the amp not working as hard to get the desired volume. (I had already confirmed the internal speaker was 8 ohm rather than 4 ohm when they brought it to me initially to fix an input jack, although the actual reading I got on meter is about 5.7 ohm which makes me think the actual impedance is closer to 7 ohms than 8 ohms). In any case, when I tested it earlier tonight it was without a second speaker.

[Enzo]

At 4 ohms compared to 8 ohms, the amp works twice as hard.

The RESISTANCE was 5.7 ohms. That is not impedance. Look at any speaker response curve, you will see the impedance at any frequency varies, an 8 ohm speaker does not present a steady 8 ohms across the band.

[bobloblaws]

At 4 ohms compared to 8 ohms, the amp works twice as hard.

An amp with a specified 4 ohm output works twice as hard with the specified load than when the load is doubled?

The RESISTANCE was 5.7 ohms. That is not impedance.

That's why I said the "actual impedance" was probably closer to 7 ohms than 8 ohms. In my experience an 8 ohm speaker will usually read about 6.5 ohms on the ohmmeter.

[The Dude]

An amp with a specified 4 ohm output works twice as hard with the specified load than when the load is doubled?

Yes.

[bobloblaws]

Yes.

Obviously I defer to your expertise, but it seems counter-intuitive. In my notes I have this blurb under SS amp section: "...if output is 8 ohm and you hook up 16 ohm load, power will halve". So it seems to me if DOUBLING the load FROM specified output halves the power, it would follow that HALVING the load TO specified output would double the power. And I always thought increasing the power would increase efficiency. Where am I going wrong?

[The Dude]

You sort of answered your own question. "...if output is 8 ohm and you hook up 16 ohm load, power will halve", and "I always thought increasing the power would increase efficiency". So if power is half and efficiency less, it means the amp is not working as hard and less current is flowing at the output.

[Enzo]

Ohm's Law. The amp puts out a voltage, and the load determines the current draw. If you lower the load impedance, the current increases.

SAy 20v output. I=V/R. so with 8 ohms, I=20v/8ohms, or 2.5A Now change to 4 ohms, I=20v/4ohms, or 5A. Wanna figure power? P=V^2/I, so with 8 ohms, we get (20x20)/8. 400/8, or 50 watts. now 4 ohms, (20x20)/4, or 400/4 or 100 watts. So going from 4 to 8 cuts power in half, going from 8 to 4 doubles power. Same deal, different direction

If I have an 8 ohm speaker on an amp, lets say the example I made amp. 50 watts into that speaker. Now make it 16 ohms. 400/16 = 25 watts.

Remember you are not doubling the load going 8 to 16, you are doubling the load IMPEDANCE. 16 ohms loads the amp half as much as 8 ohms, only the impedance itself was doubled, not the load. This is maybe an area of confusion for you. Kinda like when we confuse "bias" versus"tube current" in some amp.


Efficiency is a whole separate thing. I just described textbook cases. In the real world, you will find amp specs not exactly two to one on load doubling.

[bobloblaws]

At 4 ohms compared to 8 ohms, the amp works twice as hard.

Efficiency is a whole separate thing.

So if we leave aside the theory for a moment, maybe it's only my terminology that is incorrect (but more likely I'm incorrect in general ) . When I talked about the amp "not working as hard" I wasn't thinking about how much current is flowing. I could put it another way and say that my understanding was that at the rated load the amp would run more efficiently (and cooler) at a given dB level than it would with twice the rated load. Is that still incorrect?