Bridging an amp... Circuit theory question???

[Sanitarium]

Hi Gurus... idiot here trying to make sense of things.

I always thought that a stereo amplifier running in bridged mono mode
inverts the mosfet output wave from one channel and couples that to the
other channel. This is how the circuit creates more power into one
channel.

How far off base am I?? :~)

Another thing. When a stereo amp is bridged mono pushing a 4 ohm load,
Is it true or false that the amps circuit sees this a a 2 ohm load? If
this is true, why?

Confused...
Garrett

[Scott Johnson]

Sanitarium > wrote in
:

> Hi Gurus... idiot here trying to make sense of things.

join the club!


>
> I always thought that a stereo amplifier running in bridged mono mode
> inverts the mosfet output wave from one channel and couples that to the
> other channel. This is how the circuit creates more power into one
> channel.

Yup


>
> How far off base am I?? :~)



>
> Another thing. When a stereo amp is bridged mono pushing a 4 ohm load,
> Is it true or false that the amps circuit sees this a a 2 ohm load? If
> this is true, why?
>
> Confused...
> Garrett
>

each channell sees 1/2 of the load. as in (2) 2ohm loads

why? i don't know

[Eddie Runner]

Sanitarium wrote:

> idiot here trying to make sense of things.
>
> I always thought that a stereo amplifier running in bridged mono mode
> inverts the mosfet output wave from one channel and couples that to the
> other channel. This is how the circuit creates more power into one
> channel.

thats pretty much how it works, one channel pills while the other channel
pushes
actually takes the + voltage from one channel and the - voltage from
the other channel and uses them together to have ahigher voltage.

> Another thing. When a stereo amp is bridged mono pushing a 4 ohm load,
> Is it true or false that the amps circuit sees this a a 2 ohm load? If
> this is true, why?

a 4 ohm load is still a 4 ohm load.
alot of folks say its 1/2 of the load but it IS NOT!
alot of folks say each channel sees half the load IT DOES NOT!
but alot of folks think so.

Eddie Runner
http://www.installer.com/tech/

[Kevin McMurtrie]

In article >,
Sanitarium > wrote:

>Hi Gurus... idiot here trying to make sense of things.
>
>I always thought that a stereo amplifier running in bridged mono mode
>inverts the mosfet output wave from one channel and couples that to the
>other channel. This is how the circuit creates more power into one
>channel.
>
>How far off base am I?? :~)
>
>Another thing. When a stereo amp is bridged mono pushing a 4 ohm load,
>Is it true or false that the amps circuit sees this a a 2 ohm load? If
>this is true, why?
>
>Confused...
>Garrett


Some amps always run with two channels inverted. That way a big bass
kick draws power from the positive power supply in one speaker and the
negative power supply in the other. It reduces the number of capacitors
that have to be packed into the amp. In such a case, the bridged mode
simply makes it mono.

Left + in --+---- non-inverting amp ----> Left + / Bridge +
|
| +---> Left -
Bridge switch (mono) |
| Ground --+---> Right +
|
Right + in --+---- inverting amp ----> Right - / Bridge -


Yes, bridging cuts the perceived output impedance in half. Think of it
this way. The two outputs are exactly opposite. Halfway inside the
coil they cancel to zero volts. It's like each amp is seeing a coil
half as long.


Cancels to zero volts
here in center
v
2 Ohms v 2 Ohms

^^^^^^^^^^^^^^^^^^^^
| |
| -- 4 Ohms -- |
| |
+ phase - phase

[Eddie Runner]

No it doesnt Kevin.
If the voltage was ZERO half way in the speaker coil then
there would be no current, and kurcheff tells us this is not
possible! In a circuit the current is the same everywhere!

Eddie Runner

Kevin McMurtrie wrote:

> Yes, bridging cuts the perceived output impedance in half. Think of it
> this way. The two outputs are exactly opposite. Halfway inside the
> coil they cancel to zero volts. It's like each amp is seeing a coil
> half as long.
>
>
> Cancels to zero volts
> here in center
> v
> 2 Ohms v 2 Ohms
>
> ^^^^^^^^^^^^^^^^^^^^
> | |
> | -- 4 Ohms -- |
> | |
> + phase - phase

[Lex]

ok eddie, then explain why bridged output @ 4 ohms is equalivlent to stereo
output @ 2 ohms. because each channel only sees half of the load from the
speakers

"Eddie Runner" > wrote in message
...
> Sanitarium wrote:
>
> > idiot here trying to make sense of things.
> >
> > I always thought that a stereo amplifier running in bridged mono mode
> > inverts the mosfet output wave from one channel and couples that to the
> > other channel. This is how the circuit creates more power into one
> > channel.
>
> thats pretty much how it works, one channel pills while the other channel
> pushes
> actually takes the + voltage from one channel and the - voltage from
> the other channel and uses them together to have ahigher voltage.
>
> > Another thing. When a stereo amp is bridged mono pushing a 4 ohm load,
> > Is it true or false that the amps circuit sees this a a 2 ohm load? If
> > this is true, why?
>
> a 4 ohm load is still a 4 ohm load.
> alot of folks say its 1/2 of the load but it IS NOT!
> alot of folks say each channel sees half the load IT DOES NOT!
> but alot of folks think so.
>
> Eddie Runner
> http://www.installer.com/tech/
>

[Kevin McMurtrie]

Don't confuse current and voltage. If one end is -10V and the other is
+10V, then a conductive path between them will have 0V at one point.
Voltage is relative, of course.



In article >,
Eddie Runner > wrote:

>No it doesnt Kevin.
>If the voltage was ZERO half way in the speaker coil then
>there would be no current, and kurcheff tells us this is not
>possible! In a circuit the current is the same everywhere!
>
>Eddie Runner
>
>Kevin McMurtrie wrote:
>
>> Yes, bridging cuts the perceived output impedance in half. Think of it
>> this way. The two outputs are exactly opposite. Halfway inside the
>> coil they cancel to zero volts. It's like each amp is seeing a coil
>> half as long.
>>
>>
>> Cancels to zero volts
>> here in center
>> v
>> 2 Ohms v 2 Ohms
>>
>> ^^^^^^^^^^^^^^^^^^^^
>> | |
>> | -- 4 Ohms -- |
>> | |
>> + phase - phase
>

[Gary Rodgers]

"Sanitarium" > wrote in message
...
| Hi Gurus... idiot here trying to make sense of things.
|
| I always thought that a stereo amplifier running in bridged mono mode
| inverts the mosfet output wave from one channel and couples that to the
| other channel. This is how the circuit creates more power into one
| channel.

Exactly.

|
| How far off base am I?? :~)
|
| Another thing. When a stereo amp is bridged mono pushing a 4 ohm load,
| Is it true or false that the amps circuit sees this a a 2 ohm load? If
| this is true, why?

Each half of the bridged amp will see a 2-ohm load, but the load itself
doesn't change. 4-ohms is 4-ohms, etc. The "bridged" amp sees the load as
it is. 4 ohms.

|
| Confused...
| Garrett

Don't be confused - just check my previous response for a detailed link -
scroll down the right hand side and select section 51 "Amplifier Bridging"
for all the information (and more than you'd EVER want) you'd ever need.

Gary

[Eric Desrochers]

Kevin McMurtrie > wrote:

> Don't confuse current and voltage. If one end is -10V and the other is
> +10V, then a conductive path between them will have 0V at one point.
> Voltage is relative, of course.

A voltage is a difference of potential between 2 points. If one point
is +10v and the other is -10v, the difference is obviously 20 volts.

10 - (-10) = 20

And since you have doubled the voltage, you will double the current in
the load as well, giving us (theorically) 4 time as much power than a
single channel.

The impedance of the load have not changed.
--
Eric (Dero) Desrochers

Hiroshima 45, Tchernobyl 86, Windows 95

[Mark Zarella]

> ok eddie, then explain why bridged output @ 4 ohms is equalivlent to
stereo
> output @ 2 ohms.

He already did. It's because the voltage is doubled. By the way, the
scenario you described above is often equivalent (or close), but sometimes
not.

> because each channel only sees half of the load from the
> speakers

Half of the load? What happens to the other half? The speaker doesn't say
to itself "gee, the amp must be bridged, so I better modify my impedance".
(hey, if amps can "see", then speakers can talk!)

[Kevin McMurtrie]

In article >,
(Eric Desrochers) wrote:

>Kevin McMurtrie > wrote:
>
>> Don't confuse current and voltage. If one end is -10V and the other is
>> +10V, then a conductive path between them will have 0V at one point.
>> Voltage is relative, of course.
>
>A voltage is a difference of potential between 2 points. If one point
>is +10v and the other is -10v, the difference is obviously 20 volts.
>
>10 - (-10) = 20
>
>And since you have doubled the voltage, you will double the current in
>the load as well, giving us (theorically) 4 time as much power than a
>single channel.
>
>The impedance of the load have not changed.

The virtual impedance that each channel of the amps sees does change.
Half of the impedance is cancelled by the other out-of-phase amp.
That's why a 2 Ohm stable amp can only drive 4 Ohms bridged.

[Eddie Runner]

Kevin McMurtrie wrote:

> The virtual impedance that each channel of the amps sees does change.

No, it does not change!

> Half of the impedance is cancelled by the other out-of-phase amp.

No it isnt! How can that happen??
When you bridge an amp you run the old 2 channels in SERIES!
On a series circuit every thing passes through everything in the series!
on the + side, the current rushed to the - side, on its journey it passes
through the ENTIRE speaker voice coil!

Then on the return trip from - to + the current again travels though the
ENTIRE speaker voice coil....

nothing is cancelled....

>
> That's why a 2 Ohm stable amp can only drive 4 Ohms bridged.

No, its because there is more current caused by the increased voltage!

Eddie Runner
http://www.installer.com/tech/

[Eddie Runner]

Gary Rodgers wrote:

> Each half of the bridged amp will see a 2-ohm load, but the load itself
> doesn't change. 4-ohms is 4-ohms, etc. The "bridged" amp sees the load as
> it is. 4 ohms.

Gary,
you recomend a decent web site to answer all the questions here.
but then you state exactly the opposite.... ha ha ha

the web site says this:

NOTE:
Some people say that when an amplifier is bridged onto a 4 ohm
load, it 'sees' a 2 ohm load. While it is true that the
same current flows whether the amp is bridged on a 4 ohm load or a 2 ohm
stereo load, the amplifier is driving a 4 ohm
load across its outputs. A single 4 ohm speaker can never be a 2 ohm
load.


and you say, Each half of the bridged amp will see a 2-ohm load

get with the program kiddo!


Eddie Runner
http://www.installer.com/tech/


Each channel does NOT see half!!!

[Gary Rodgers]

"Eddie Runner" > wrote in message
...
| Gary Rodgers wrote:
|
| > Each half of the bridged amp will see a 2-ohm load, but the load itself
| > doesn't change. 4-ohms is 4-ohms, etc. The "bridged" amp sees the load
as
| > it is. 4 ohms.
|
| Gary,
| you recomend a decent web site to answer all the questions here.
| but then you state exactly the opposite.... ha ha ha

Where? It says the bridged amp (complete circuit) sees the load just as it
is - 4 ohms? A load is a load...now, when you look at the circuit and the
way it is derived (bridged), each side of the bridge is acting as it would
with a 2 ohm load...


Gary

[Eddie Runner]

Gary Rodgers wrote:

> | Gary,
> | you recomend a decent web site to answer all the questions here.
> | but then you state exactly the opposite.... ha ha ha
>
> Where?

Where? In your last messages! Thats where... you said each channel sees half
the impedance, and it does not! And the website you recomended says if does not!

> It says the bridged amp (complete circuit) sees the load just as it
> is - 4 ohms?

Sure, how can it see anything else??

> A load is a load...

thats right!
So why make a reference to 1/2 of that load at all????

> now, when you look at the circuit and the
> way it is derived (bridged), each side of the bridge is acting as it would
> with a 2 ohm load...

No it isnt!!!!!!!!!!!!!!!!!!!!!!!!!
read it again, thats NOT what that web page says!

it says there is more current (like 1/2ing the load)
why even mention halfing the load at all though is still a moronic thing to do
cause
it makes folks that dont read very well think its the same as 1/2ing the load
and
it is NOT in anyway like halfing the load....

I am 6ft tall and out in the wood room I have a 2x4 (piece of wood) thats 6 ft
tall...
telling folks that the 2x4 is just like ME is kind of missleading .... ;-)

Yes, 1/2ing the impedance COULD make the current the same as when you
bridge and amp...
The current could also be the same if my pet dog ****es on the
amp circuitry!
Or if we dump semi wet concrete on the amp circuitry, or any
number of reasons it could be the same....

BUT WHY MENTION THEM IF ITS NOT THE REAL REASON we get more current???
WHY MENTION 1/2 impedance it ITS NOT REALLY 1/2 IMPEDANCE????

DUMBDUMBDUMB!!!

Eddie Runner
http://www.installer.com/tech/

[Gary Rodgers]

"Eddie Runner" > wrote in message
...
|
|
| Gary Rodgers wrote:
|
| > | Gary,
| > | you recomend a decent web site to answer all the questions here.
| > | but then you state exactly the opposite.... ha ha ha
| >
| > Where?
|
| Where? In your last messages! Thats where... you said each channel sees
half
| the impedance, and it does not! And the website you recomended says if
does not!
|
| > It says the bridged amp (complete circuit) sees the load just as it
| > is - 4 ohms?
|
| Sure, how can it see anything else??
|
| > A load is a load...
|
| thats right!
| So why make a reference to 1/2 of that load at all????
|
| > now, when you look at the circuit and the
| > way it is derived (bridged), each side of the bridge is acting as it
would
| > with a 2 ohm load...
|
| No it isnt!!!!!!!!!!!!!!!!!!!!!!!!!
| read it again, thats NOT what that web page says!
|
| it says there is more current (like 1/2ing the load)
| why even mention halfing the load at all though is still a moronic thing
to do
| cause
| it makes folks that dont read very well think its the same as 1/2ing the
load
| and
| it is NOT in anyway like halfing the load....
|
| I am 6ft tall and out in the wood room I have a 2x4 (piece of wood) thats
6 ft
| tall...
| telling folks that the 2x4 is just like ME is kind of missleading ....
;-)
|
| Yes, 1/2ing the impedance COULD make the current the same as when you
| bridge and amp...
| The current could also be the same if my pet dog ****es on the
| amp circuitry!
| Or if we dump semi wet concrete on the amp circuitry, or any
| number of reasons it could be the same....
|
| BUT WHY MENTION THEM IF ITS NOT THE REAL REASON we get more current???
| WHY MENTION 1/2 impedance it ITS NOT REALLY 1/2 IMPEDANCE????
|
| DUMBDUMBDUMB!!!

I stand corrected and I'm dumb dumb dumb.
But at least I'm not caught up in semantics.
Guess I better take that BSEE back to the place I got it.

I'll go back to my car and listen to music.

Gary

[Eddie Runner]

half of 4 is 2, half of 100 is 50....!
what does that have to do with bridging an amp?
NOTHING!!!!

Each channel (of the two before bridging) since they are in series
sees THE WHOLE LOAD!

NOT HALF OF IT!

So Tony, what are you rambling about???

Eddie Runner

Tony Hwang wrote:

> Hi,
> Rephrase it, bridged two amp circuits sharing the 4 Ohm load.
> Half of 4 is 2, right?
> Tony
>
> Gary Rodgers wrote:
>
> > "Eddie Runner" > wrote in message
> > ...
> > | Gary Rodgers wrote:
> > |
> > | > Each half of the bridged amp will see a 2-ohm load, but the load itself
> > | > doesn't change. 4-ohms is 4-ohms, etc. The "bridged" amp sees the load
> > as
> > | > it is. 4 ohms.
> > |
> > | Gary,
> > | you recomend a decent web site to answer all the questions here.
> > | but then you state exactly the opposite.... ha ha ha
> >
> > Where? It says the bridged amp (complete circuit) sees the load just as it
> > is - 4 ohms? A load is a load...now, when you look at the circuit and the
> > way it is derived (bridged), each side of the bridge is acting as it would
> > with a 2 ohm load...
> >
> >
> > Gary
> >
> >

[Eddie Runner]

Gary Rodgers wrote:

> I stand corrected and I'm dumb dumb dumb.
> But at least I'm not caught up in semantics.

Semantics are IMPORTANT!

>
> Guess I better take that BSEE back to the place I got it.

another college boy bites the dust!
WTF do the colleges teach the kiddos of today anyway???

Eddie

[Mark Zarella]

> > Guess I better take that BSEE back to the place I got it.
>
> another college boy bites the dust!
> WTF do the colleges teach the kiddos of today anyway???

If they're teaching him that a 4 ohm load becomes 2 ohms, then I think the
college professor should give his degrees back too...

[Eddie Runner]

good one Markie! ha ha

Mark Zarella wrote:

> > I am 6ft tall and out in the wood room I have a 2x4 (piece of wood) thats
> 6 ft
> > tall...
> > telling folks that the 2x4 is just like ME is kind of missleading ....
> ;-)
>
> Yeah, because a 2x4 is not also 6 feet wide...

[Lex]

alright, after rereading a few things I realized my memory of the situation
was a little skewed.. too much reading not enough soak in time lately and I
seem to be forgetting a few things. next question is, because I cannot find
this anywhere... are speakers voltage driven devices or current driven
devices
?

I want to say current, because if I remember my definitions right, current
is what does the work, not voltage, voltage was more of a potential to be
able to do work.. am I remembering this correctly?


"Eddie Runner" > wrote in message
...
> I cant explain it because it IS NOT TRUE!
>
> What is true is that your normal (unbridged) 2 channels each see any
> speaker as the impedance that it is .. IE: a 4 ohm speaker is 4 ohms
> a 2 ohms speaker is 2 ohms....
>
> And if you bridge the amp to one channel, then that channel sees any
> load you put on it just the same.. IE: a 4 ohm speaker is 4 ohms
> a 2 ohms speaker is 2 ohms....
>
> But, because when you bridge an amp there is more voltage to the speaker,
> the current will rise! so the amp may run harder...
>
> There are two ways to increase current.
>
> 1) lower impedance
> 2) raise voltage
>
> In the case of bridging an amp we raise the voltage NOT lower the
> impedance... But so many folks are confused by it and I dont know why...
>
> Eddie Runner
> http://www.installer.com/tech/
>
>
> Lex wrote:
>
> > ok eddie, then explain why bridged output @ 4 ohms is equalivlent to
stereo
> > output @ 2 ohms. because each channel only sees half of the load from
the
> > speakers
> >
>

[Mark Zarella]

> alright, after rereading a few things I realized my memory of the
situation
> was a little skewed.. too much reading not enough soak in time lately and
I
> seem to be forgetting a few things. next question is, because I cannot
find
> this anywhere... are speakers voltage driven devices or current driven
> devices
> ?

Not this again...

Question for you...why does it matter? How is it relevant? I'd like to
know, because this is the second time the question has been asked in the
past few weeks.

[Mark Zarella]

While what you say below is true, it does not directly address how the
speakers function. It is not voltage that sets up the force, but current.
Therefore, they're current driven and the magnitude of the current can
predict the electrical contribution to the cone motion. The same cannot be
said of voltage.

--
Mark Zarella
zarellam at upstate dot edu


"Kevin McMurtrie" > wrote in message
...
> In article <hyD%a.136534$Oz4.26769@rwcrnsc54>,
> "Lex" > wrote:
>
> >alright, after rereading a few things I realized my memory of the
situation
> >was a little skewed.. too much reading not enough soak in time lately and
I
> >seem to be forgetting a few things. next question is, because I cannot
find
> >this anywhere... are speakers voltage driven devices or current driven
> >devices
> >?
> >
> >I want to say current, because if I remember my definitions right,
current
> >is what does the work, not voltage, voltage was more of a potential to be
> >able to do work.. am I remembering this correctly?
> >
>
> They're voltage driven. The speaker's motion generates a back voltage
> so driving them by voltage provides some controlling negative feedback.
>
>
> Tuned enclosures depend on driving the cone by voltage too. Enclosures
> increase the sound levels by increasing the acoustic load on the
> speaker. That doesn't work the same way if the speaker is driven by a
> constant current (force). It might even work in the opposite way.
>
> Constant voltage:
> Watts= Volts^2 / Ohms
>
> Constant current:
> Watts= Amps^2 * Ohms
>
> With a constant voltage, an increased load (reduced Ohms) produces more
> power. With a constant current, it's the opposite.
>
>
> Of course both voltage and current are there. It's just that the
> amplifier regulates the signal voltage and lets the speaker draw any
> amount of current.

[Kevin McMurtrie]

In article >,
"Mark Zarella" <seesigfile> wrote:

>While what you say below is true, it does not directly address how the
>speakers function. It is not voltage that sets up the force, but current.
>Therefore, they're current driven and the magnitude of the current can
>predict the electrical contribution to the cone motion. The same cannot be
>said of voltage.
>
>--
>Mark Zarella
>zarellam at upstate dot edu

Of course it's the voltage that creates the current, and the current
that creates the force.

The current doesn't determine the cone motion, though; it determines the
cone force. The voltage is a bit better at determining the motion
because back EMF from motion causes some negative feedback. The
force/current of the cone when driven by voltage is (Vin - Vback) / Ohms.

There aren't any audio amps that regulate by current output. The audio
quality would be extremely unpredictable.

>
>"Kevin McMurtrie" > wrote in message
...
>> In article <hyD%a.136534$Oz4.26769@rwcrnsc54>,
>> "Lex" > wrote:
>>
>> >alright, after rereading a few things I realized my memory of the
>situation
>> >was a little skewed.. too much reading not enough soak in time lately and
>I
>> >seem to be forgetting a few things. next question is, because I cannot
>find
>> >this anywhere... are speakers voltage driven devices or current driven
>> >devices
>> >?
>> >
>> >I want to say current, because if I remember my definitions right,
>current
>> >is what does the work, not voltage, voltage was more of a potential to be
>> >able to do work.. am I remembering this correctly?
>> >
>>
>> They're voltage driven. The speaker's motion generates a back voltage
>> so driving them by voltage provides some controlling negative feedback.
>>
>>
>> Tuned enclosures depend on driving the cone by voltage too. Enclosures
>> increase the sound levels by increasing the acoustic load on the
>> speaker. That doesn't work the same way if the speaker is driven by a
>> constant current (force). It might even work in the opposite way.
>>
>> Constant voltage:
>> Watts= Volts^2 / Ohms
>>
>> Constant current:
>> Watts= Amps^2 * Ohms
>>
>> With a constant voltage, an increased load (reduced Ohms) produces more
>> power. With a constant current, it's the opposite.
>>
>>
>> Of course both voltage and current are there. It's just that the
>> amplifier regulates the signal voltage and lets the speaker draw any
>> amount of current.
>
>

[Mark Zarella]

> >While what you say below is true, it does not directly address how the
> >speakers function. It is not voltage that sets up the force, but
current.
> >Therefore, they're current driven and the magnitude of the current can
> >predict the electrical contribution to the cone motion. The same cannot
be
> >said of voltage.
> >
> >--
> >Mark Zarella
> >zarellam at upstate dot edu
>
> Of course it's the voltage that creates the current, and the current
> that creates the force.

Yes, but the voltage is not directly tied to the production of the force.
The current is.

>
> The current doesn't determine the cone motion, though; it determines the
> cone force.

I never said otherwise...

> The voltage is a bit better at determining the motion
> because back EMF from motion causes some negative feedback. The
> force/current of the cone when driven by voltage is (Vin - Vback) / Ohms.

That's correct, but regardless, the device is driven by current. Ampere's
law says so.

If you want to ultimately predict cone motion as you suggest above, it's no
easier to use the voltage-source method than it is to use the current-source
method. In either case, you still need to know all the nuances of the
impedance characteristics of the system. If given the voltage, you need
impedance to determine force and DCR (as a function of temperature). If
given the current, you need impedance to determine back emf. So in order to
accurately make a prediction, the electrical and mechanical properties of
the driver need to be known, as well as the "acoustical" properties of the
system. Doesn't matter whether you identify the waveform as i(t) or v(t).

[Mark Zarella]

> If
> given the current, you need impedance to determine back emf.

I realize this line is unclear. You don't need the impedance to determine
back emf. You need the impedance to determine what the contribution to the
current is because of the back emf.

[Tony Hwang]

Eddie Runner wrote:

> half of 4 is 2, half of 100 is 50....!
> what does that have to do with bridging an amp?
> NOTHING!!!!
>
> Each channel (of the two before bridging) since they are in series
> sees THE WHOLE LOAD!
>
> NOT HALF OF IT!
>
> So Tony, what are you rambling about???
Hi,
Both amps are looking into the load together, not individually.
Well?
Tony

> Eddie Runner
>
> Tony Hwang wrote:
>
>
>>Hi,
>>Rephrase it, bridged two amp circuits sharing the 4 Ohm load.
>>Half of 4 is 2, right?
>>Tony
>>
>>Gary Rodgers wrote:
>>
>>
>>>"Eddie Runner" > wrote in message
...
>>>| Gary Rodgers wrote:
>>>|
>>>| > Each half of the bridged amp will see a 2-ohm load, but the load itself
>>>| > doesn't change. 4-ohms is 4-ohms, etc. The "bridged" amp sees the load
>>>as
>>>| > it is. 4 ohms.
>>>|
>>>| Gary,
>>>| you recomend a decent web site to answer all the questions here.
>>>| but then you state exactly the opposite.... ha ha ha
>>>
>>>Where? It says the bridged amp (complete circuit) sees the load just as it
>>>is - 4 ohms? A load is a load...now, when you look at the circuit and the
>>>way it is derived (bridged), each side of the bridge is acting as it would
>>>with a 2 ohm load...
>>>
>>>
>>>Gary
>>>
>>>
>
>

[Eddie Runner]

BOTH!!
VOLTAGE AND CURRENT = WATTAGE
wattage is power!
one watt is the work done by one volt and one amp!

Speakers MUST have BOTH voltage AND current to produce
the WATTAGE it takes to do the work....

Eddie Runner

Lex wrote:

> alright, after rereading a few things I realized my memory of the situation
> was a little skewed.. too much reading not enough soak in time lately and I
> seem to be forgetting a few things. next question is, because I cannot find
> this anywhere... are speakers voltage driven devices or current driven
> devices
> ?
>
> I want to say current, because if I remember my definitions right, current
> is what does the work, not voltage, voltage was more of a potential to be
> able to do work.. am I remembering this correctly?
>
> "Eddie Runner" > wrote in message
> ...
> > I cant explain it because it IS NOT TRUE!
> >
> > What is true is that your normal (unbridged) 2 channels each see any
> > speaker as the impedance that it is .. IE: a 4 ohm speaker is 4 ohms
> > a 2 ohms speaker is 2 ohms....
> >
> > And if you bridge the amp to one channel, then that channel sees any
> > load you put on it just the same.. IE: a 4 ohm speaker is 4 ohms
> > a 2 ohms speaker is 2 ohms....
> >
> > But, because when you bridge an amp there is more voltage to the speaker,
> > the current will rise! so the amp may run harder...
> >
> > There are two ways to increase current.
> >
> > 1) lower impedance
> > 2) raise voltage
> >
> > In the case of bridging an amp we raise the voltage NOT lower the
> > impedance... But so many folks are confused by it and I dont know why...
> >
> > Eddie Runner
> > http://www.installer.com/tech/
> >
> >
> > Lex wrote:
> >
> > > ok eddie, then explain why bridged output @ 4 ohms is equalivlent to
> stereo
> > > output @ 2 ohms. because each channel only sees half of the load from
> the
> > > speakers
> > >
> >

[Eddie Runner]

its so funny to read all these posts argueing about VOLTAGE OR CURRENT!

And whats funny is there are a few smart guys in this thread and none of them
have said BOTH current and voltage or WATTAGE yet...

Its like the chicken and the egg almost to argue that because current cant
exist without voltage, so how can something even be current driven and not
voltage driven???

This useless arguement makes me laugh.

Eddie

[Kevin McMurtrie]

In article >,
Eddie Runner > wrote:

>its so funny to read all these posts argueing about VOLTAGE OR CURRENT!
>
>And whats funny is there are a few smart guys in this thread and none of them
>have said BOTH current and voltage or WATTAGE yet...
>
>Its like the chicken and the egg almost to argue that because current cant
>exist without voltage, so how can something even be current driven and not
>voltage driven???
>
>This useless arguement makes me laugh.
>
>Eddie
>
>

Don't worry, we're not attempting to redifine the universe.

I did mention several times that the voltage and current are directly
related. That's where the wattage, and the ability to do work, lies.

If you had a perfect, zero ohm, superconducting speaker, the current
would be the force, the voltage would be the speed, and the electrical
wattage would exactly match power formulas for mechanical motion.
Unfortunately most of the voltage, and therefore wattage, in a real
speaker goes into heating the thin wire that makes the voice coil.

[Eddie Runner]

yes, I did see where you said it was useless... I didnt respond to
one of your messages, I just responded in general not quoting anyone.

How can this CURRENT or VOLTAGE even be argued about??
(I guess Im asking the same questions as you Mark)

May I suggest to anyone who cares my world famous ohms law chart.
This copy of the chart is on installation bay walls all over the world.
http://www.installer.com/tech/ohmslaw.html

its fun to build these simple formulas into an EXCELL SPREADSHEET
so you can see what happens when you change any of the variables...

Eddie

Mark Zarella wrote:

> Eddie,
> If you notice earlier in the thread I already said it was useless. I said:
> "why does it matter? How is it relevant? I'd like to
> know, because this is the second time the question has been asked in the
> past few weeks."
>
> --
> Mark Zarella
> zarellam at upstate dot edu
>
> "Eddie Runner" > wrote in message
> ...
> > its so funny to read all these posts argueing about VOLTAGE OR CURRENT!
> >
> > And whats funny is there are a few smart guys in this thread and none of
> them
> > have said BOTH current and voltage or WATTAGE yet...
> >
> > Its like the chicken and the egg almost to argue that because current cant
> > exist without voltage, so how can something even be current driven and not
> > voltage driven???
> >
> > This useless arguement makes me laugh.
> >
> > Eddie
> >
> >

[Eddie Runner]

Kevin McMurtrie wrote:

> Don't worry, we're not attempting to redifine the universe.

ha ha

> I did mention several times that the voltage and current are directly
> related. That's where the wattage, and the ability to do work, lies.

sure, its all related!
Thats the secret of OHMS LAW that so many folks seem to miss!
And even those that do finally get it that they are all related, they seem
to still missunderstand exactly what the relationships are...

> If you had a perfect, zero ohm, superconducting speaker, the current
> would be the force,

you mean a DIRECT SHORT! ha ha
Alot of current for a short time till the amp or wires fry themselves...

> the voltage would be the speed,

Nope, Kevin!!
Slow down a little!
VOLTAGE IS NOT SPEED!

Electrons flow the same speed no matter if its 9 volts or 12 volts.


Eddie

[Kevin McMurtrie]

In article >,
Eddie Runner > wrote:

>Kevin McMurtrie wrote:
>
>> Don't worry, we're not attempting to redifine the universe.
>
>ha ha
>
>> I did mention several times that the voltage and current are directly
>> related. That's where the wattage, and the ability to do work, lies.
>
>sure, its all related!
>Thats the secret of OHMS LAW that so many folks seem to miss!
>And even those that do finally get it that they are all related, they seem
>to still missunderstand exactly what the relationships are...
>
>> If you had a perfect, zero ohm, superconducting speaker, the current
>> would be the force,
>
>you mean a DIRECT SHORT! ha ha
>Alot of current for a short time till the amp or wires fry themselves...
>
>> the voltage would be the speed,
>
>Nope, Kevin!!
>Slow down a little!
>VOLTAGE IS NOT SPEED!
>
>Electrons flow the same speed no matter if its 9 volts or 12 volts.
>
>
>Eddie
>

A superconducting speaker would be a dead short only if you stopped the
cone. Voltage does determine the speed, just as it does for a DC motor.
The current is (Vin - Vback)/Ohms. The higher the input voltage, the
faster the cone must move to create back EMF to balance the equation.
The amount of back EMF created by motion depends on the winding size.

Of course we're getting theoretical here. 4 or 8 ohms is a lot of
resistance so speakers are far from perfect.

Don't bring electrons into this. They move only a little faster than
grass grows.

[Eddie Runner]

Kevin McMurtrie wrote:

> >> If you had a perfect, zero ohm, superconducting speaker, the current
> >> would be the force,
> A superconducting speaker would be a dead short only if you stopped the
> cone.

YOU SAID zero ohms!
THATS A DEAD SHORT!

> Voltage does determine the speed, just as it does for a DC motor.

Voltage deteminines the speed for an electric motor because there
is MORE force to or from the permanent magnets...

NOT because higher voltage travels through a circuit faster!
High voltage or low voltage both have the current still have
the same speed through a circuit...

For instance, a far away radio signal thats STRONG doesnt arrive
ant faster than a far away rado signal thats WEAK....If that was
the case you would have something like the doppler effect when a
radio station fades.. And you dont...

There can be doppler effects in radio but not because a signal is
weaker or stronger...

> The current is (Vin - Vback)/Ohms. The higher the input voltage, the
> faster the cone must move to create back EMF to balance the equation.
> The amount of back EMF created by motion depends on the winding size.

But were not talking about back EMF.... And back EMF doesnt
affect the SPEED of voltage anyway, only the amount of the voltage.

If you apply voltagoe to a circuit (like a speaker) and the motion of the
voice coil through the magnet created ITS OWN VOLTAGE, then thats
called back EMF... The added voltage adds to the original voltage and
changes the potential.

To see the back EMF with out the original voltage simply hook your volt meter
to your woofer speaker trminals (with the woof disconnected) and then
push doen the woofer cone, your meter will tell you that voltage is being
produced by the voice coil moving..

But of course, nothing to do with speed...

NOW, you may be confused, thinking about your electric motor, because
when a back emf is dcreated by the motors movement, that could change
and even reduce the original voltage, so that reduced voltage might make
your electric motor run slower, but it still has nothing to do with the speed
of the juice through the wire...

> Of course we're getting theoretical here. 4 or 8 ohms is a lot of
> resistance so speakers are far from perfect.

Alot??
I guess that depends on how you look at it...

> Don't bring electrons into this. They move only a little faster than
> grass grows.

nearly the speed of light!
IMO, thats quite a bit faster than grass grows...

Eddie Runner

[Gary Rodgers]

"Eddie Runner" > wrote in message
...
| BOTH!!
| VOLTAGE AND CURRENT = WATTAGE
| wattage is power!
| one watt is the work done by one volt and one amp!
|
| Speakers MUST have BOTH voltage AND current to produce
| the WATTAGE it takes to do the work....
|
| Eddie Runner
|

DUMB DUMB DUMB
Speakers don't produce any wattage. Ever. Better get a refund on your
education as well.

BTW - Power (wattage) is defined in many ways - and it isn't just "Current &
Voltage" - ya got 3 variables...
Amongst MANY variations of Ohm's law are:
P=VI
P=I^2 * R
P=V^2 / R = (Vmax - Vmin)/R<==== BTW - this is the form of ohm's law that
applies to the bridging of amps, and the resultant non-linear function for
the performance of such - the confusion comes because the current needed to
drive the same stereo load increases non-linearly. While you are correct
that the circuit does not "see" a 2-ohm load with a 4-ohm stereo load
(since, as I have also stated, a 4 ohm load is a 4 ohm load), the net
result in the current required is such that the circuit ACTS as if you had a
single amp of twice the power with a 2-ohm load.

Also, btw, try looking at a little op-amp theory and you'll find out there
*are* uses for current sources, I'll leave the refresher up to you....and
try looking up Thevenin and Norton equivalents as well...

<snort>
Gary

[Eddie Runner]

Gary Rodgers wrote:

>
> DUMB DUMB DUMB
> Speakers don't produce any wattage.Ever.

Ever??
have you ever heard of an ACCOUSTIC WATT???

I believe that can be produced by a speaker! ;-)


> Better get a refund on your education as well.

My Education???
Who said I had one I had to pay for??
Wasnt it YOU that said you should rip on a person you
dont even know... ha ha ha

Im not one of the college boyees!
Hell I forgot which year of high school it was I dropped out of..! ha ha

All the STUPID college fellows just made me want to take a short cut
to the real world and miss what I thought at the time was a waste...

> BTW - Power (wattage) is defined in many ways - and it isn't just "Current &
> Voltage" -

I never said it was DID I??

> ya got 3 variables...
> Amongst MANY variations of Ohm's law are:
> P=VI
> P=I^2 * R
> P=V^2 / R = (Vmax - Vmin)/R<==== BTW - this is the form of ohm's law that
> applies to the bridging of amps, and the resultant non-linear function for
> the performance of such - the confusion comes because the current needed to
> drive the same stereo load increases non-linearly.

No the confusion comes because so many IDIOTS say that bridging
1/2s the impedance AND IT DOESNT! You can call around to stereo
shops and MOST of the PROS will tell you IT DOES 1/2 the impedance!

BUT IT DOESNT!
Its just an out of control piece of missinformation!

> While you are correct
> that the circuit does not "see" a 2-ohm load with a 4-ohm stereo load
> (since, as I have also stated, a 4 ohm load is a 4 ohm load), the net
> result in the current required is such that the circuit ACTS as if you had a
> single amp of twice the power with a 2-ohm load.

No, it ACTS like you increased the voltage!

IF IT WERE to act like 1/2 the impedance, WHY EVEN SAY SO??
If it was the same impedance as if I let my dog **** on the load, thats no
reason to run around telling folks ITS EXACTLY THE SAME AS IF
MY DOG ****ED ON IT....

See??
Why mention something that you dont need to...SPECIALLY when so many
folks are walking around saying it wrong because folks like you are saying
it is just like 1/2 ...


> Also, btw, try looking at a little op-amp theory and you'll find out there
> *are* uses for current sources,

Op amp current sources are NOT REALLY CURRENT SOURCES!
I have played a bit with op amps and well...


> I'll leave the refresher up to you....and
> try looking up Thevenin and Norton equivalents as well...

Why? I dont think they relate to anything we ar etalking about here ...

Eddie

[Kevin McMurtrie]

In article >,
Eddie Runner > wrote:

>Kevin McMurtrie wrote:
>
>> >> If you had a perfect, zero ohm, superconducting speaker, the current
>> >> would be the force,
>> A superconducting speaker would be a dead short only if you stopped the
>> cone.
>
>YOU SAID zero ohms!
>THATS A DEAD SHORT!
>
>> Voltage does determine the speed, just as it does for a DC motor.
>
>Voltage deteminines the speed for an electric motor because there
>is MORE force to or from the permanent magnets...
>
>NOT because higher voltage travels through a circuit faster!
>High voltage or low voltage both have the current still have
>the same speed through a circuit...
>
>For instance, a far away radio signal thats STRONG doesnt arrive
>ant faster than a far away rado signal thats WEAK....If that was
>the case you would have something like the doppler effect when a
>radio station fades.. And you dont...
>
>There can be doppler effects in radio but not because a signal is
>weaker or stronger...
>
>> The current is (Vin - Vback)/Ohms. The higher the input voltage, the
>> faster the cone must move to create back EMF to balance the equation.
>> The amount of back EMF created by motion depends on the winding size.
>
>But were not talking about back EMF.... And back EMF doesnt
>affect the SPEED of voltage anyway, only the amount of the voltage.
>
>If you apply voltagoe to a circuit (like a speaker) and the motion of the
>voice coil through the magnet created ITS OWN VOLTAGE, then thats
>called back EMF... The added voltage adds to the original voltage and
>changes the potential.
>
>To see the back EMF with out the original voltage simply hook your volt meter
>to your woofer speaker trminals (with the woof disconnected) and then
>push doen the woofer cone, your meter will tell you that voltage is being
>produced by the voice coil moving..
>
>But of course, nothing to do with speed...
>
>NOW, you may be confused, thinking about your electric motor, because
>when a back emf is dcreated by the motors movement, that could change
>and even reduce the original voltage, so that reduced voltage might make
>your electric motor run slower, but it still has nothing to do with the speed
>of the juice through the wire...
>
>> Of course we're getting theoretical here. 4 or 8 ohms is a lot of
>> resistance so speakers are far from perfect.
>
>Alot??
>I guess that depends on how you look at it...
>
>> Don't bring electrons into this. They move only a little faster than
>> grass grows.
>
>nearly the speed of light!
>IMO, thats quite a bit faster than grass grows...
>
>Eddie Runner
>
>

This thread is dead. Buy yourself an electronics book. My personal
favorite, and a favorite of many, is "The Art of Electronics" by
Horowitz and Hill. Winfield Hill even hangs out in
sci.electronics.design.

[Mark Zarella]

> > >> If you had a perfect, zero ohm, superconducting speaker, the current
> > >> would be the force,
> > A superconducting speaker would be a dead short only if you stopped the
> > cone.
>
> YOU SAID zero ohms!
> THATS A DEAD SHORT!

By "zero ohms" he was implying that the DCR is zero ohms (impossible,
superconducting or not). He said nothing of the reactive components. In
fact, even "theoretically" you can't reduce the reactive component to
nothing. You wouldn't be able to make the cone move! His initial point
still stands.

> > The current is (Vin - Vback)/Ohms. The higher the input voltage, the
> > faster the cone must move to create back EMF to balance the equation.
> > The amount of back EMF created by motion depends on the winding size.
>
> But were not talking about back EMF.... And back EMF doesnt
> affect the SPEED of voltage anyway, only the amount of the voltage.

No, it doesn't affect the amount of voltage coming from the amplifier. It
affects the current flow from the amplifier. Back EMF is an essential part
of the equation, like it or not. So it's perfectly valid to bring up in
this instance.

> If you apply voltagoe to a circuit (like a speaker) and the motion of the
> voice coil through the magnet created ITS OWN VOLTAGE, then thats
> called back EMF...

Well, actually it's not. Back EMF is something completely different. What
we've all been referring to as "back emf" is generally called "motional EMF"
(eg. J. King, J.AES, 1969?). Back emf involves the storage and release of
energy by an inductor.

> The added voltage adds to the original voltage and
> changes the potential.
>
> To see the back EMF with out the original voltage simply hook your volt
meter
> to your woofer speaker trminals (with the woof disconnected) and then
> push doen the woofer cone, your meter will tell you that voltage is being
> produced by the voice coil moving..
>
> But of course, nothing to do with speed...

The speed of coil motion through the field is one of the parameters that
determines the voltage amplitude.

[Mark Zarella]

> DUMB DUMB DUMB
> Speakers don't produce any wattage. Ever. Better get a refund on your
> education as well.
>
> BTW - Power (wattage) is defined in many ways - and it isn't just "Current
&
> Voltage" - ya got 3 variables...
> Amongst MANY variations of Ohm's law are:
> P=VI
> P=I^2 * R

Dumb dumb dumb. That's actually not ohm's law, but joule's law of heating.
:)

> P=V^2 / R = (Vmax - Vmin)/R

Forgot a squared.

> <==== BTW - this is the form of ohm's law that
> applies to the bridging of amps, and the resultant non-linear function for
> the performance of such

How is it nonlinear?

> - the confusion comes because the current needed to
> drive the same stereo load increases non-linearly.

Um...huh?

> While you are correct
> that the circuit does not "see" a 2-ohm load with a 4-ohm stereo load
> (since, as I have also stated, a 4 ohm load is a 4 ohm load), the net
> result in the current required is such that the circuit ACTS as if you had
a
> single amp of twice the power with a 2-ohm load.

Depends what you mean. The "current required" for what?

[Eddie Runner]

Mark Zarella wrote:

> > YOU SAID zero ohms!
> > THATS A DEAD SHORT!
>
> By "zero ohms" he was implying that the DCR is zero ohms (impossible,
> superconducting or not). He said nothing of the reactive components. In
> fact, even "theoretically" you can't reduce the reactive component to
> nothing. You wouldn't be able to make the cone move! His initial point
> still stands.

Oh, I see what he was getting at... But still ZERO OHMS is a short.

What he is saying still doesnt prove the impedance 1/2s when
bridging though... ;-)

> > But were not talking about back EMF.... And back EMF doesnt
> > affect the SPEED of voltage anyway, only the amount of the voltage.
>
> No, it doesn't affect the amount of voltage coming from the amplifier. It
> affects the current flow from the amplifier.

The inducted voltage (EMF) does affect the total potential of the circuit.

> Back EMF is an essential part
> of the equation, like it or not. So it's perfectly valid to bring up in
> this instance.

But it is still not proof that the impedance 1/2s
And this guy was trying to say that higher voltage travels faster than
lower voltage....

>
>
> > If you apply voltagoe to a circuit (like a speaker) and the motion of the
> > voice coil through the magnet created ITS OWN VOLTAGE, then thats
> > called back EMF...
>
> Well, actually it's not. Back EMF is something completely different. What
> we've all been referring to as "back emf" is generally called "motional EMF"
> (eg. J. King, J.AES, 1969?). Back emf involves the storage and release of
> energy by an inductor.

Thats exactly what I was saying Mark, I was refering to a Speaker motor.

>
> > But of course, nothing to do with speed...
>
> The speed of coil motion through the field is one of the parameters that
> determines the voltage amplitude.

He said SPEED is related to voltage.
your talking about something cvompletely different Mark
now your talking about frequency

Eddie

[Mark Zarella]

> > No, it doesn't affect the amount of voltage coming from the amplifier.
It
> > affects the current flow from the amplifier.
>
> The inducted voltage (EMF) does affect the total potential of the circuit.

Depends on the node you're referring to, I suppose. You didn't specify.

> > Back EMF is an essential part
> > of the equation, like it or not. So it's perfectly valid to bring up in
> > this instance.
>
> But it is still not proof that the impedance 1/2s
> And this guy was trying to say that higher voltage travels faster than
> lower voltage....

I agree with you on both points. He didn't make an argument about it. He
seemed to have gone off on a tangent.

> > > But of course, nothing to do with speed...
> >
> > The speed of coil motion through the field is one of the parameters that
> > determines the voltage amplitude.
>
> He said SPEED is related to voltage.
> your talking about something cvompletely different Mark
> now your talking about frequency

No, I was talking about speed, not frequency (even though cone speed is
proportional to frequency, but frequency is not the only parameter of
course). I'm not sure what speed you're talking about.

[Kevin McMurtrie]

In article >,
"Mark Zarella" <seesigfile> wrote:

>> > No, it doesn't affect the amount of voltage coming from the amplifier.
>It
>> > affects the current flow from the amplifier.
>>
>> The inducted voltage (EMF) does affect the total potential of the circuit.
>
>Depends on the node you're referring to, I suppose. You didn't specify.
>
>> > Back EMF is an essential part
>> > of the equation, like it or not. So it's perfectly valid to bring up in
>> > this instance.
>>
>> But it is still not proof that the impedance 1/2s
>> And this guy was trying to say that higher voltage travels faster than
>> lower voltage....
>
>I agree with you on both points. He didn't make an argument about it. He
>seemed to have gone off on a tangent.

The voltage matters in the cone's speed. This is easiest to see in
subwoofers with very large windings. They attenuate above a few hundred
Hz even through its no fault of the mechanics. The motion of the cone
generates a voltage that opposes the input voltage. A speaker is a
motor.

I said nothing of the relationship between electron speed and voltage.
There isn't one.


>> > > But of course, nothing to do with speed...
>> >
>> > The speed of coil motion through the field is one of the parameters that
>> > determines the voltage amplitude.
>>
>> He said SPEED is related to voltage.
>> your talking about something cvompletely different Mark
>> now your talking about frequency
>
>No, I was talking about speed, not frequency (even though cone speed is
>proportional to frequency, but frequency is not the only parameter of
>course). I'm not sure what speed you're talking about.
>
>

[Mark Zarella]

> The voltage matters in the cone's speed. This is easiest to see in
> subwoofers with very large windings. They attenuate above a few hundred
> Hz even through its no fault of the mechanics. The motion of the cone
> generates a voltage that opposes the input voltage. A speaker is a
> motor.

I thought you were talking about the driving voltage. Yes, the emf as well
as the coil inductance play a part in attenuating high frequencies.