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#41
Quote by farmosh203
I'd like you to explain how running the supply voltage at 18V changes the signal.


input =! output
Prs se Holcomb is the answer
#42

input =! output


So you're saying the 18V mod changes the bias current required for the amplifier on the input?
#45
yes the 18volt mod goes into your amplifier. that's right.


Can you be clear in your responses? You aren't making any sense.

There is an amplifier on EMG pickups (also known as a pre-amp), that's what I am talking about.

There's the supposed schem of the EMG preamp. Changing the supply voltage also affects Vin. So Vin increases=moar gain.


How does changing the supply voltage amplify the input signal? If you wanted to gain the signal, you just adjust the feedback resistor.
#47
The amount of amplification is the result of more air being moved.
Air being moved can be measured in the output power of the amp, in watts.
Electrical watts can be found by multiplying the voltage by the current.
You have not measured the current.


Instead of just measuring the voltage, which is only half of the (very simplified, in this case) equation, why not measure the current as well? I think you'll find that it is very, very small. Therein lies your answer - this is like saying, the tires on this toy car spin at the same rate as a real one, why isn't it going as fast?

You are missing important parts of the equation.
#49
The amount of amplification is the result of more air being moved.
Air being moved can be measured in the output power of the amp, in watts.
Electrical watts can be found by multiplying the voltage by the current.
You have not measured the current


Instead of just measuring the voltage, which is only half of the (very simplified, in this case) equation, why not measure the current as well? I think you'll find that it is very, very small. Therein lies your answer - this is like saying, the tires on this toy car spin at the same rate as a real one, why isn't it going as fast?

You are missing important parts of the equation.


You have confused me beyond belief with what you said...

Power = V*I = V^2/R = I^2*R

You only need two of the variables to solve for Power (Voltage and the impedance of the speaker = 8 ohms), but then again you have to make sure that your amplifier can drive the speaker with enough current (hence the need for an output transformer on a tube amplifier).

Air being moved is not proportional to the amount of power of an amp. What's the amount of air moved in a 1000 Watt short circuit?
#50
The current changes as well. You're adding power to the system from the electrical grid. Your numbers are wrong because you assumed no additional power was being added to the system.

Air being moved is proportional to the power, when all other variables are held the same. Turn the volume knob up on your amp sometime while measuring the power draw from the wall.

Your short circuit analogy is of course totally irrelevant. A more powerful car won't go faster than a less powerful one if it's up on blocks, but that proves nothing.

If you want further proof that you're incorrect, do your same calculations using the P=IV formula for a 50 watt amp and a 100 watt amp (which, by the way, will output up to 80 and 200 watts fully driven). The input doesn't change, and the voltage doesn't change, and neither does the impedance, but the power does.
#51
The current changes as well. You're adding power to the system from the electrical grid. Your numbers are wrong because you assumed no additional power was being added to the system.

Air being moved is proportional to the power, when all other variables are held the same. Turn the volume knob up on your amp sometime while measuring the power draw from the wall.

Your short circuit analogy is of course totally irrelevant. A more powerful car won't go faster than a less powerful one if it's up on blocks, but that proves nothing.


Current changes when the voltage changes. V = IR. My numbers are not wrong. Power = V*I = V^2/R = I^2*R. The electrical grid power is converted to DC power supplies inside the amplifier...

I'm just saying, not every speaker is the same. It's a big assumption to assume that everyone's speaker variable should be held constant.

Measuring power at the wall won't give you 100% correct readings, the power supplies have an efficiency, so it's bettery to measure the power at the output of the power supply feeding the amplifier.
#52
The input doesn't change, and the voltage doesn't change, and neither does the impedance, but the power does.


A 50W amp and a 100W amp will sound exactly the same loudness if you drive them at the same voltage and drive the same speaker. That's like saying 1 lb. of feathers is lighter than 1 lb. of brick.
#53
Quote by farmosh203
A 50W amp and a 100W amp will sound exactly the same loudness if you drive them at the same voltage and drive the same speaker. That's like saying 1 lb. of feathers is lighter than 1 lb. of brick.


no but a pound of brick is denser.

do you understand that trained electrical engineers are trying to educate you very very simply without coming right out and calling you a retard.

current is important dude.

again input=!output the 18v mod doesn't do shit to your input. because your input signal is your guitar.

omfg please just make this stop.
Prs se Holcomb is the answer
#54
But they will not. When you play louder, you're getting more current.
The plates on the tubes of both amps will be at the same voltage. Let's say it's 300. The plates of the tubes will see 300V the whole time, and the amp will see 120V from the wall the whole time.

What the tubes do is to turn a small change in current (like in the pickups) into a large current swing on the other side. That's how they turn the 1/10 of a watt of your pickups into the 100 watts of output. The voltages are all pretty much fixed.

The thing you are not taking into account, and I don't get why, is the current. You cannot simply cancel it out in this case - this is not a high school R-C circuit diagram with a lightbulb and a battery. This is complicated stuff, and there's a reason that the current variable is in that equation. It's necessary.


Let's use another car analogy.
Long version:
Your car can put out 100 horsepower in third gear and in fifth gear. In third, you're getting lots of torque but less speed (at the wheels - in this analogy, the gearbox is the output transformer, so the tach would read the same but the wheels would be moving different speeds). In fifth, the speed is higher but the torque is lower. In both cases, the power is the same.

If you said that since the power is the same and since the RPMs on the tach are the same, the car would be going the same speed, you would be wrong. You are missing an important variable because you are trying to simplify an equation in a way in which it cannot be simplified.

Short version:
Is your car always going the same speed when the tachometer reads 3000 rpm?
#55
current is important dude.


I know, but you solve for current with V = IR, it's a simple linear equation, you don't need to do a Laplace transform or solve all 4 Maxwell's equations to solve for current. You just need the impedance of the speaker and the voltage going into the speaker.

again input=!output the 18v mod doesn't do shit to your input. because your input signal is your guitar.


So please tell me how the 18V mod affects the output when the signal doesn't get close to 9V? The power rail for an op amp does not affect the gain.
Last edited by farmosh203 at Nov 22, 2010,
#56
It is not a simple linear equation. If you don't get that, we can't help you.

Try measuring and solving for all three variables. Measure input voltage, current, and impedance, then output current, voltage, and impedance. Then try your equation. It will not work.
#57
Well actually ....... if you know the voltage and the impedance you can calculate the current. The impedance changes with frequency but as long as your test tone is fairly low you will be pretty close to the rated impedance, higher frequencies will actually have higher impedance and therefore lower current than calculating with the rated impedance. The real point is that there is a lot more going on than simple amplification, all the devices colour the sound, not just the output transformer.
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Last edited by Cathbard at Nov 22, 2010,
#59
But they will not. When you play louder, you're getting more current.
The plates on the tubes of both amps will be at the same voltage. Let's say it's 300. The plates of the tubes will see 300V the whole time, and the amp will see 120V from the wall the whole time.

What the tubes do is to turn a small change in current (like in the pickups) into a large current swing on the other side. That's how they turn the 1/10 of a watt of your pickups into the 100 watts of output. The voltages are all pretty much fixed.

The thing you are not taking into account, and I don't get why, is the current. You cannot simply cancel it out in this case - this is not a high school R-C circuit diagram with a lightbulb and a battery. This is complicated stuff, and there's a reason that the current variable is in that equation. It's necessary.


You are telling me that V doesn't equal I*R. I'm sorry but no matter what you say, V = IR.

If the voltage is constant, then the impedance of the speaker would have to drop lower for the current to change. The current going into a fixed impedance isn't equivalent to an active circuit like an FPGA where the current can vary. The only way for the current to vary on a speaker is for the voltage to vary.
#60
I'm not telling you that it doesn't equal that. I know it equals that.

What I am telling you is that you cannot cancel out I because it is not equal on both sides of the equation. Power is being added to the system.
#61
R (Z actually) changes with frequency. It's a complex load. However the lowest impedance presented to the lowest frequency will be pretty close to the rated impedance.
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Marshall 1960A
Boss GT-100


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Last edited by Cathbard at Nov 22, 2010,
#62
It is not a simple linear equation. If you don't get that, we can't help you.


It is. What equation are you using? The equation is V = IR, it's that simple.

You're also getting confused because the voltage coming into the amplifier is 120 Vrms whereas the 300V for the tube is DC. You need an AC / DC converter for this. Sure the voltage on the amplifier is 300V, but that's the positive rail, not a fixed voltage, the amplifier varies from 0-300V.

I can only tell you so many times, you can't break the rule V = IR.
#63
I'm not sure what the problem is, here.

Are you trying to measure the voltage of your power amp, or something? It seemed you were having trouble understanding just how much tube amps amplified at first, but now I'm not sure where the misunderstanding is.

I think another thing you'll find interesting that you aren't considering Impedance. You're treating the speakers like resistors, but they're not! If you measure resistance of an 8 ohm speaker, you won't get 8 ohms. The fact that it's called an "impedance" implies there's an imaginary component.

EDIT: This post was WAY late. I need to refresh after reading threads.
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#64
What I am telling you is that you cannot cancel out I because it is not equal on both sides of the equation. Power is being added to the system.


I'm sorry, I honestly don't mean to be rude, but I don't think you understand what you are talking about.

You can ignore the 120Vrms coming from the wall, that's the input voltage to the AC/DC converter. You have a tube where the Anode (sorry if I'm using the wrong terminology, I'm used to FETs, which would be the Drain of an NFET) is connected to 300V. This 300V is your maximum voltage, it doesn't hold the voltage at a constant 300V.
#65
Ok... so I agree that there is an imaginary component, but you aren't going to go from 50W to 100W just by changing the frequency of the signal and keeping the voltage constant (well within 20-20kHz).

You can ignore the 120Vrms as the input voltage. This just goes to a power supply, you only look at the output voltage of the power supply and then just do an efficiency calculation of the power supply.
Last edited by farmosh203 at Nov 22, 2010,
#66
As I know nothing of electronics, this thread is blowing my mind...
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#67
I'm real sorry, I think I read the original question wrong. I've been up too late.
I still don't think you're doing the power equations properly, but we can put that aside. Someone else can set us both straight on that if need be.

I really do think the tubes have a lot to do with the warmth of a tube amp. There are a few tube amps that do not have output transformers and they are plenty warm. However, the OT does add something to the tone as well. The people I talked to with the OT-less design had to do quite a bit of electrical gymnastics to get the sound right, though I don't remember 'warmth' being one of the things they had to re-inject. It seemed more like compression and 'fatness' were the things lost with the OT.
#68
Colin knows full well what he's talking about, and everything he's said that I've read is true.

Tubes take the power from the outlet and use it to amplify the signal from your guitar by a HUGE factor. The voltage is amplified, then, as Craig explained earlier, the voltage is stepped down so that the current can be stepped up which drives speakers. Where's the misunderstanding?
Feel free to call me Kyle.

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#69
Tubes take the power form the outlet and use it to amplify the signal from your guitar by a HUGE factor. The voltage is amplified, then, as Craig explained earlier, the voltage is stepped down so that the current can be stepped up which drives speakers. Where's the misunderstanding?


If you have a static load (ok so it's not 100% static but you aren't going to get variation of impedance from 1 ohm to 8 ohms in a 20-20kHz bandwidth), and decrease the voltage, it's impossible for the current to increase. If you have an ACTIVE load, then sure, you can hold voltage constant and your current is going to change because your load is changing.

Ok, the 120Vrms comes from the wall and goes to an AC/DC power supply. This power supply supplies 300V. In this process, you're either going to have less or more current than the input to your power supply (I haven't done an AC/DC power supply design and I don't want to look up for the formula for rectification...).

So let's just say for very simplistic sake, that the wall voltage is 120V DC, which it isn't, but let's just say it is. You would feed this to a boost converter, which would boost your voltage to 300V. So if you're driving an 8 ohm speaker at 30Vrms, your output power is 112.5Watts.

Assuming 100% efficiency in the power supply, your current coming from the wall would be 120*current = 112.5Watts, which would mean your input current is 112.5W/120V = 0.9375Amps.

What was the current at the output? Well 112.5W / 300V = 0.375Amps.

Can we all agree on this?
Last edited by farmosh203 at Nov 22, 2010,
#70
I really do think the tubes have a lot to do with the warmth of a tube amp. There are a few tube amps that do not have output transformers and they are plenty warm. However, the OT does add something to the tone as well. The people I talked to with the OT-less design had to do quite a bit of electrical gymnastics to get the sound right, though I don't remember 'warmth' being one of the things they had to re-inject. It seemed more like compression and 'fatness' were the things lost with the OT.


I hate to argue with you more, but how did they drive the speaker without a transformer? The tube has way too high of an output impedance.
#71
They used some sort of induction system - I looked up the patent, and it wasn't a whole lot more specific than that. I'm trying to dig up the conversation now so I can get more specifics. The amp was new at NAMM 2009, I think. It definitely didn't have an OT.
#72
omfg. can we agree that you have no idea wtf is going on. there's so much ish going on here that your not accounting for.

1. your guitar pickups pick some shit up and send it to the output jack. It doesn't matter if you have actives or passives. The output will be at a certain range.

2. plug it into a pedal or whatever..pedal boost signal by some amount.

3. plug into the guitar amp. - The first stage after the input is an impedance buffer. You lose a lot of signal strength due to impedance coupling.

4. first volume pot soaks up more of the signal.

5. any additional stages will distort at the input and be stepped down at the output until the

6. phase inverter which splits the signal and sends it to the power tubes.

the signal prior to the pi needs to be impedance matched, that's why effects loops require buffers that supply unity gain when switching from lo-z to hi-z and back.

think about it logically and then you can work it out when you get to the other equations in your text book...what's coming out of your guitar output jack is very quiet. what's coming out of your pedal...still very quite. What's coming out of your loudspeakers...very loud.

the science is going to back up the reality.

dude must be homeschooled

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otl amps have been around forever.
Prs se Holcomb is the answer
Last edited by AcousticMirror at Nov 22, 2010,
#74
apparently otl's are dangerously dangerously unsafe since there is no isolation tranny.
Prs se Holcomb is the answer
#75
the signal prior to the pi needs to be impedance matched, that's why effects loops require buffers that supply unity gain when switching from lo-z to hi-z and back.

think about it logically and then you can work it out when you get to the other equations in your text book...what's coming out of your guitar output jack is very quiet. what's coming out of your pedal...still very quite. What's coming out of your loudspeakers...very loud.

the science is going to back up the reality.

dude must be homeschooled


All I'm saying is V=IR, why are you trying to argue with me? I must be homeschooled because V = IR?

Look, I measured about 5V coming out of my pedal peak to peak. 5V goes into the amplifier. Want 30W of power to an 8 ohm load? Amplify 5V by about 80. What is there to argue? V = IR, V=IR, do I need to repeat myself more?
Last edited by farmosh203 at Nov 22, 2010,
#76
That's the first page of the patent, more for demonstration purposes than as a full explanation. Read the rest, or one of the other explanations on the site. It's a matching device, but since it's so different it really doesn't apply to the OT discussion we were having. From what I understand it has almost nothing in common tonewise with a usual OT.
#77
I may be dumb but that schematic just confuses me. I'm trying to figure out how that's transformerless when I see magnetics all over that schematic.
#78
Quote by farmosh203
I may be dumb but that schematic just confuses me. I'm trying to figure out how that's transformerless when I see magnetics all over that schematic.


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what piece is the transformer.
Prs se Holcomb is the answer
#79
This thread is epic. Makes me feel much better about myself, thanks guys.
Quote by farmosh203
I may be dumb but that schematic just confuses me. I'm trying to figure out how that's transformerless when I see magnetics all over that schematic.
We're talking output transformerless, that schematic doesn't have an OT. On the schematic, find the output tubes, trace the wire out of the plate through the coupling cap towards the resistor called "load". That's representing the speaker.
Quote by farmosh203
All I'm saying is V=IR, why are you trying to argue with me? I must be homeschooled because V = IR?

Look, I measured about 5V coming out of my pedal peak to peak. 5V goes into the amplifier. Want 30W of power to an 8 ohm load? Amplify 5V by about 80. What is there to argue? V = IR, V=IR, do I need to repeat myself more?
You're funny!
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Last edited by Kanthras at Nov 22, 2010,
#80
Optimus prime is a transformer. The thing in the patent is a transformer. The idea is that it's so unlike a traditional transformer that you can call it not-a-transformer. They're similar in name only, and in very very basic function.

Look at that schematic against a traditional tube amp schematic, you'll see the difference right away.