Dead Fender Stage 185-replacing thermisto

[kepeb]

i know the speakers are 4ohm.

if i can figure out exactly what you just said, lol. does that mean it doesn't matter how loud i have it? the maximum dissipation will stay the same?

[Phostenix]

I am too lazy. Use Ohm's Law. Calculate what the maximum power through the speaker load would be. At its impedance, solve for current. That current through the .22 ohms will tell you the maximum dissipation for that resistor. FOr safety, double that amount, and the result would be the minimal wattage for the part.

The rated power into the load is 160 Watts into 4 ohms. Power = current squared X resistance (i2 X R), so I = the square root of P/R. P/R is 160/4 = 40. The square root of 40 is 6.325a. Calculate the power of 6.325a through .22 ohms using the same P = i2 X R. i2 is that same 40a from above, but now R is .22 ohms. 40 * .22 = 8.8 watts, so double that gets you 17.6, hence the 20 watt resistor.

Power calculator:
Ohm's Law Equations Formulas Calculator - Resistance Given Power Current

[Phostenix]

Enzo, I'm still having a hard time understanding these resistors from speaker output ground to ground. First, how does it see the same current as the load? Is its purpose to isolate the ouput from ground to reduce hum? I understand it being used as NFB - is that the case here? I see the output coming back to the diff pair through 2 resistors (R148 & R149), and it looks like the ground leg of NFB is C64, R137, & that .22 ohm resistor. Is that right?

[kepeb]

The rated power into the load is 160 Watts into 4 ohms. Power = current squared X resistance (i2 X R), so I = the square root of P/R. P/R is 160/4 = 40. The square root of 40 is 6.325a. Calculate the power of 6.325a through .22 ohms using the same P = i2 X R. i2 is that same 40a from above, but now R is .22 ohms. 40 * .22 = 8.8 watts, so double that gets you 17.6, hence the 20 watt resistor.
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i think i'm getting it... and i thought i was doing alright up to now but i have a feeling i need to be corrected

not that my understanding is important to fix this problem but i would like to 'get it' so the next questions aren't really directly related to my fender.



so... just to see if I'm close. not for stress, like my amp...

1. to work out the maximum power through my speakers "Power = current squared X resistance" means i measure the current (in amps) at the speaker out?... get the square of it(times its self), then times that by the 4ohm of the speakers and i have the answer, but i need the current and resistance in order to work out the power?

2. i work out the current (amps) by working out the square route of power(160) divided by resistance (4) (= 40), so square route of 40 will equal the current (somewhere between 6a and 7a), but i cant work out the current without both power and resistance?

3.
then i calculate the power in the given current through 0.22ohms with p= I2 x r...
current squared (40) times 0.22ohm = 8.80watt?!

how do i even begin to remember all this?

[kepeb]

ok. i have the new opamp in all the rails still check out at 16v
the main out still seems dead tho.

[Phostenix]

There's a saying that's attributed to Einstein (but since it's the internet, it's probably not true) that goes something like, "Why would I remember something that I can look up in 5 minutes."

Now that we have the world at our fingertips, I'm thinking that pretty much means we don't have to remember anything....

We all learn Ohm's law & commit that to memory, but power calulation has squares and square roots and who can remember all that?

I remember Power as P = I * E (PIE, 'cause PIV doesn't have the same ring to it, and remembering P = I2 * R and P= E2 / R isn't so easy.).

So, the power output of an amp is the current through the speaker times the voltage across the speaker. Of course, the next thing you'll ask is - what's the voltage across the speaker?


The problem with amplifiers is that we only usually know 1 thing for sure - the load. We typically have a 4 or 8 ohm speaker load. We have to go by what the manufacturer states as the output power of the amp to try to work out the rest - especially if it's a non-working amp that can't be measured. I just went through all of this with my Crate amp. When it was completed, I did a bunch of measurements to understand it better.

You can make some guesses if you know the voltage rails of the amp, (but the power supply and other factors will limit what the amp can truly deliver). In your case, that's +50V & -50V. In an ideal world, that amp could generate an output voltage swing of 100v rail to rail (peak to peak) to the speakers. That's 50v peak times .7 to get RMS voltage, or 35v RMS. If you could deliver 35v across a 4 ohm load, that would be 8.75a of current through the speaker. You can calculate the power multiple ways from here, but I usually just multiply I * E - 8.75a * 35v = 306.25 watts. What, Fender says it's a 160 watt amp? That's because it's not capable of providing the full voltage to the load. The theory guys can explain why that is, but I think it's mostly about power supply capabilities.

I have all my notes from the Crate at home, but it was, in fact, not able to deliver as much voltage to a 4 ohm load as it could to an 8 ohm load. It was able to produce 145 watts into 8 ohms, but only 210 into 4. It "should" have been able to generate twice the power into 4 ohms as it could into 8 ohms (290 watts), but it didn't. It's rated at 350 watts into 2 ohms (instead of the theoretical 435 watts), but I don't have enough resistors to test that. Maybe down the road....

You'll probably never measure the current through the speakers directly. You can measure voltage across a speaker and divide by the resistance to get a close number (remember, speakers aren't purely resistive loads, blah, blah, blah). Trust me, you'll never hook up a speaker to an amp, put a 1kHz tone into it, turn it up all the way & happily measure the voltage across the speaker unless it's a really small amp or you have great earplugs & no neighbors. That's why everyone buys those giant 4 or 8 ohm resistors as dummy loads. It's not a true speaker load, but it's close enough.

If you know what the reistance of the load is & you know what the voltage is across that resistance, you can calculate the power output of the amp with P = I * E. If all you have is the stated power output of the amp and the value of the load, you can calculate the current through or the voltage across the load using the power formulas. It won't be exact, but close enough for most things.

What was the question again?

[Phostenix]

The next step is to go back to your audio source into the amp and use the audio probe to see where the signal stops. Start with U9, pin 7 - then pin1. If it's making it through there, go to R136 & check both sides of it.

If we get to that point, I'll be looking for some input from others about where you can safely probe in the power amp. What exactly is the probe you are using? I'm concerned about the high DC voltages that you may encounter if you start poking around in there.

Maybe the next thing to do there is get some DC voltage readings with your meter.

[kepeb]

ok. i got audio signal on pin 1 but not 7. but i do have audio on r136 now.

the probe i am using is a tool my pal made with a green capacitor and a plastic jack socket. using one leg of the cap to probe and the other is connected to the tip input on the jack theres a croc clip on the sleeve of the jack which you put to ground and i plug a mini amp into the jack socket to hear the results.

i'm a little confused now as to how i'm hearing audio on pin 7 but not on 1

[Phostenix]

Don't be confused, it's doing what it's supposed to be now! The switches on the Power Amp Input jack determine whether the signal goes from pin1 of U8 to pin 3 of U9 (normal - nothing plugged into the PA In) or from the PA In jack to pin 5 of U9. If you don't use the loop, the signal goes to U9B, if you do it goes to U9A.

Is the audio on both sides of R136?

Your probe has a cap on the tip, so that makes me feel a little better, but I have no experience using audio probes (only scopes).

[kepeb]

oh right ok, so 7 and 1 are both outs + and - for audio. so i'm missing half the signal right?

i went through the other opamps too, theres a huge variance in signal level but audio on all except u8 pin 7. this is different from u9 tho, u9 make no sound. in fact the general mini amp hum disappears when i touch that one like i've shorted it. on u8 it makes more of a buzzing sound than normal. it could just be that the signal is too much for the little amp tho.

[Phostenix]

If the audio is there on both sides of R136, then find C66. It's a 66pf cap, so it's probably a small ceramic. Be careful not to short the legs together as you probe it. Check for audio on both sides of it. Then, find R153 (I think we checked this one earlier) & check it on both sides.

[kepeb]

sorry missed your last one... ok then thats comforting

[Phostenix]

U8 pin 7 is the return from the reverb tank. It will be controlled by the Reverb knob and will be lower level than the others.

[kepeb]

yep.
audio on both of them

[Phostenix]

oh right ok, so 7 and 1 are both outs + and - for audio. so i'm missing half the signal right?

Pins 1 & 7 are both outputs. There are 2 opamps in that IC. The amp will use one or both depending on whether you are using the loop. U9B buffers the audio coming in from the Power Amp Input and gives you the Gain Offset control. If you aren't using the loop, it gets bypassed. U9A always gets used. U9B only gets used when you plug into the PA In.