Go Back   UG Community @ Ultimate-Guitar.Com > Instruments > Guitar Gear & Accessories
User Name  
Password
Search:

Closed Thread
Old 11-22-2010, 01:35 AM   #41
AcousticMirror
loves cheesecake
 
AcousticMirror's Avatar
 
Join Date: Dec 2009
Quote:
Originally Posted by farmosh203
I'd like you to explain how running the supply voltage at 18V changes the signal.


input =! output
__________________
buffalo buffalo buffalo buffalo buffalo buffalo buffalo.
AcousticMirror is offline   Reply With Quote
Old 11-22-2010, 01:38 AM   #42
farmosh203
Registered User
 
Join Date: Aug 2010
Quote:
input =! output


So you're saying the 18V mod changes the bias current required for the amplifier on the input?
farmosh203 is offline   Reply With Quote
Old 11-22-2010, 01:42 AM   #43
AcousticMirror
loves cheesecake
 
AcousticMirror's Avatar
 
Join Date: Dec 2009
yes the 18volt mod goes into your amplifier. that's right.
__________________
buffalo buffalo buffalo buffalo buffalo buffalo buffalo.
AcousticMirror is offline   Reply With Quote
Old 11-22-2010, 01:45 AM   #44
mmolteratx
UG God
 
mmolteratx's Avatar
 
Join Date: Sep 2008
Location: Richardson, Texas
http://music-electronics-forum.com/t9095/

There's the supposed schem of the EMG preamp. Changing the supply voltage also affects Vin. So Vin increases=moar gain.
__________________
E-peen:
Rhodes Gemini
Fryette Ultra Lead
Peavey 6505
THD Flexi 50

Gibson R0 Prototype
EBMM JP13 Rosewood
Fender CS Mary Kaye

WTLT

(512) Audio Engineering - Custom Pedal Builds, Mods and Repairs
mmolteratx is offline   Reply With Quote
Old 11-22-2010, 01:46 AM   #45
farmosh203
Registered User
 
Join Date: Aug 2010
Quote:
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.

Quote:
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.
farmosh203 is offline   Reply With Quote
Old 11-22-2010, 01:49 AM   #46
farmosh203
Registered User
 
Join Date: Aug 2010
Quote:
http://music-electronics-forum.com/t9095/


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


The schematic does not match the resistor values on that circuit board.
farmosh203 is offline   Reply With Quote
Old 11-22-2010, 01:58 AM   #47
Roc8995
Moderator
 
Roc8995's Avatar
 
Join Date: Nov 2005
Location: Madison, Wisconsin
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.
__________________
rr, pe&a
Roc8995 is offline   Reply With Quote
Old 11-22-2010, 02:04 AM   #48
Zoot Allures
UG's Contrarian
 
Zoot Allures's Avatar
 
Join Date: Feb 2009
Quote:
Originally Posted by AcousticMirror
yes that's what the pa is. a single speaker.

wait actually yes you can drive headphones with active speakers.

problem solved.

Is this sarcasm? PA systems will always have a power amp in any live situation.
__________________
My fan club:
http://groups.ultimate-guitar.com/z...lures_fan_club/

〘◣_◢〙 Broken Clocks, click 'help i'm a rock'
HELP I'M A ROCK!

My Blog:
http://jackoswritings.wordpress.com//

A fan of whisk(e)y? Click here!
Zoot Allures is offline   Reply With Quote
Old 11-22-2010, 02:05 AM   #49
farmosh203
Registered User
 
Join Date: Aug 2010
Quote:
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?
farmosh203 is offline   Reply With Quote
Old 11-22-2010, 02:07 AM   #50
Roc8995
Moderator
 
Roc8995's Avatar
 
Join Date: Nov 2005
Location: Madison, Wisconsin
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.
__________________
rr, pe&a
Roc8995 is offline   Reply With Quote
Old 11-22-2010, 02:15 AM   #51
farmosh203
Registered User
 
Join Date: Aug 2010
Quote:
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.
farmosh203 is offline   Reply With Quote
Old 11-22-2010, 02:18 AM   #52
farmosh203
Registered User
 
Join Date: Aug 2010
Quote:
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.
farmosh203 is offline   Reply With Quote
Old 11-22-2010, 02:22 AM   #53
AcousticMirror
loves cheesecake
 
AcousticMirror's Avatar
 
Join Date: Dec 2009
Quote:
Originally Posted 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 ******.

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.
__________________
buffalo buffalo buffalo buffalo buffalo buffalo buffalo.
AcousticMirror is offline   Reply With Quote
Old 11-22-2010, 02:23 AM   #54
Roc8995
Moderator
 
Roc8995's Avatar
 
Join Date: Nov 2005
Location: Madison, Wisconsin
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?
__________________
rr, pe&a
Roc8995 is offline   Reply With Quote
Old 11-22-2010, 02:32 AM   #55
farmosh203
Registered User
 
Join Date: Aug 2010
Quote:
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.

Quote:
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 : 11-22-2010 at 02:33 AM.
farmosh203 is offline   Reply With Quote
Old 11-22-2010, 02:33 AM   #56
Roc8995
Moderator
 
Roc8995's Avatar
 
Join Date: Nov 2005
Location: Madison, Wisconsin
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.
__________________
rr, pe&a
Roc8995 is offline   Reply With Quote
Old 11-22-2010, 02:34 AM   #57
Cathbard
Grumpy Old Tech
 
Cathbard's Avatar
 
Join Date: Oct 2009
Location: Australia
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.
__________________
Gilchrist custom guitar
Yamaha SBG500
Telecaster
Randall RM100
Abbey Harmonic II
Marshall JTM45 clone
Marshall JCM900 4102 (modded)
Marshall 18W clone
Fender 5F1 Champ clone
Marshall 1960A
Boss GT-100


Cathbard Amplification

Last edited by Cathbard : 11-22-2010 at 02:37 AM.
Cathbard is online now   Reply With Quote
Old 11-22-2010, 02:36 AM   #58
Roc8995
Moderator
 
Roc8995's Avatar
 
Join Date: Nov 2005
Location: Madison, Wisconsin
Sure you can, Cath, but he's trying to cancel out the current, not calculate it.
__________________
rr, pe&a
Roc8995 is offline   Reply With Quote
Old 11-22-2010, 02:36 AM   #59
farmosh203
Registered User
 
Join Date: Aug 2010
Quote:
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.
farmosh203 is offline   Reply With Quote
Old 11-22-2010, 02:38 AM   #60
Roc8995
Moderator
 
Roc8995's Avatar
 
Join Date: Nov 2005
Location: Madison, Wisconsin
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.
__________________
rr, pe&a
Roc8995 is offline   Reply With Quote
Closed Thread


Thread Tools Rate This Thread
Rate This Thread:

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

vB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Forum Jump



All times are GMT -4. The time now is 04:22 AM.

Forum Archives / About / Terms of Use / Advertise / Contact / Ultimate-Guitar.Com © 2014
Powered by: vBulletin Version 3.0.9
Copyright ©2000 - 2014, Jelsoft Enterprises Ltd.