Driving HD650 Balanced with DAC1

[Iron_Dreamer]

Quote:

Originally Posted by TrevorNetwork Would you mind explaining the concept of using the original signal, and inverted signal to drive the headphones? It would seem that if each driver is receiving an in-phase, and negative-phase signal that they would cancel eachother out, and result in no movement of the driver.

It is the same as how speakers are driven, AFAIK, look that up on the web and you should be able to get a good understanding of how a positive and negative singal together can equal sound.

 

[TrevorNetwork]

I hope somebody decides to break into the thread and explain this concept. It was my understanding that each driver had a signal, and ground input. The ground, is used for reference, and return. The signal's amplitiude, and frequency dictating the driver cone's amplitude (excursion), and frequency.

 

[TrevorNetwork]

I had thought balanced audio worked in such a manner, that both an in-phase, negative-phase, and ground are sent from the source. The receiver (amplifier, preamplifier) would then OR them together, and feed that difference with the in-phase. Effectively cancelling most of the noise incurred on the connection between the source, and amplifier.

 

[mulveling]

Quote:

Originally Posted by TrevorNetwork Would you mind explaining the concept of using the original signal, and inverted signal to drive the headphones? It would seem that if each driver is receiving an in-phase, and negative-phase signal that they would cancel eachother out, and result in no movement of the driver.

Hmm, let me take a stab at this (I have no education in electronics). I'll edit it out when someone tells me how wrong I am

You are thinking in terms of sound waves, which do cancel out when out-of-phase. In this case the electrical signal and its inverse have the opposite effect of cancelling out: ie a +2V signal and its -2V inverse makes a 4V difference, not a 0V one.

 

[Iron_Dreamer]

Quote:

Originally Posted by mulveling Hmm, let me take a stab at this (I have no education in electronics). I'll edit it out when someone tells me how wrong I am You are thinking in terms of sound waves, which do cancel out when out-of-phase. In this case the electrical signal and its inverse have the opposite effect of cancelling out: ie a +2V signal and its -2V inverse makes a 4V difference, not a 0V one.

That is the gist of it. Since the ground is used as a reference it is subtracted from the signal, not added, so 2-0=2 or if running balanced, 2-(-2)=4, so you have twice the power.

 

[mulveling]

Quote:

Originally Posted by Iron_Dreamer That is the gist of it. Since the ground is used as a reference it is subtracted from the signal, not added, so 2-0=2 or if running balanced, 2-(-2)=4, so you have twice the power.

Cool, thanks for confirming. I love threads like this; always learn something new.

 

[TrevorNetwork]

That does not make sense to me. Ultimately a driver is given a reference, and a signal. A variant reference is not possible afaik. If the coil receives +2v and -2v, with no reference, the result is 0v. You are think of electrical differential, not speakers. Of course, if I am wrong, then my apologies in advance...

 

[Iron_Dreamer]

An example of what typing "how speakers work" can find on google:

http://electronics.howstuffworks.com/speaker5.htm

 

[TrevorNetwork]

That seems to substantiate what I think:

Essentially, the amplifier is constantly switching the electrical signal, fluctuating between a positive charge and a negative charge on the red wire. Since electrons always flow in the same direction between positively charged particles and negatively charged particles, the current going through the speaker moves one way and then reverses and flows the other way. This alternating current causes the polar orientation of the electromagnet to reverse itself many times a second.

 

[Iron_Dreamer]

Quote:

Originally Posted by TrevorNetwork That seems to substantiate what I think: Essentially, the amplifier is constantly switching the electrical signal, fluctuating between a positive charge and a negative charge on the red wire. Since electrons always flow in the same direction between positively charged particles and negatively charged particles, the current going through the speaker moves one way and then reverses and flows the other way. This alternating current causes the polar orientation of the electromagnet to reverse itself many times a second.

The part you have highlighted is just a basic definition of AC, that it takes the form of a sine wave, fluxuating between positive and negative charge. If you put the same signal, out of phase to that on the red wire, on the black wire, you will have an outflow of electrons on the black wire at the same time you have an inflow on the red wire, or vice versa, which is known as push-pull, since you are controlling the diaphram with equal but opposite voltage on both wires. When the red wire pushes (positive flow on the positive terminal), the black wire pulls (negative voltage on the negative terminal), resulting in twice the amount of power delivered (assuming the same voltage on both inputs use as was used on the red wire for unbalanced).

Think of it as a tug of war. If the red team pulls, and the black team just stands there, the rope will move toward the red side. If the red team pulls and the black team pushs the rope forward (opposite action of what the red team is taking) the rope would move forward toward the red side much faster than in the first example. This is as clear an analogy as I can imagine to explain this concept.

 

[TrevorNetwork]

Iron_Dreamer:

I have read what you said in its entirety, and it does not seem any clearer. A signal of varying amplitude, and frequency comes in on the red wire. The speaker references the ground to decide which way to move (up or down, positive or negative respectfully). If the red wire passed a +1.5V (up), and the black wire passed at -1.5V the movements would effectively cancel eachother out. If the speaker moves (1.5)+(-1.5) it is not moving. To use a simply analogy (and I just did this) if I apply a positive voltage to a speaker (using a 9volt), it moves out, and stay there. If I apply +9v and -9v to a speaker it does not move at all. It stays neutral. Although, the speaker has had 18v of voltage differential applied to it, it has not had a frame of reference. If I had a ground, and a +18 I would have 18v worth of movement, without a reference, I have equal, and opposite movement, effectively, nothing. Using your analogy of a tug-of-war, if the right side moves with a certain force (+), and the left side moves in the opposite (-) direction with an equal force, there will be no movement.

 

[gaboo]

Quote:

Originally Posted by TrevorNetwork Iron_Dreamer: I have read what you said in its entirety, and it does not seem any clearer. A signal of varying amplitude, and frequency comes in on the red wire. The speaker references the ground to decide which way to move (up or down, positive or negative respectfully). If the red wire passed a +1.5V (up), and the black wire passed at -1.5V the movements would effectively cancel eachother out. If the speaker moves (1.5)+(-1.5) it is not moving. To use a simply analogy (and I just did this) if I apply a positive voltage to a speaker (using a 9volt), it moves out, and stay there. If I apply +9v and -9v to a speaker it does not move at all. It stays neutral. Although, the speaker has had 18v of voltage differential applied to it, it has not had a frame of reference. If I had a ground, and a +18 I would have 18v worth of movement, without a reference, I have equal, and opposite movement, effectively, nothing. Using your analogy of a tug-of-war, if the right side moves with a certain force (+), and the left side moves in the opposite (-) direction with an equal force, there will be no movement.

Forget about "absolute" voltages. There is no such thing with speakers and/or headphone transducers. All it matters is the instantaneous voltage drop/difference accross the transducer. If you have two wires a black and a red one going to the transducer, the transducer transforms the voltage difference V_red - V_black into sound. (remember that transducers are polarized).

If you have a grounded black wire (V_black = 0) and a "hot" red wire (V_red = V) you have get V - 0 = V across the transducer. If you have a differential signal, then at any moment in time you have V_red = V and V_black = -V. So the difference is always V_red - V_black = V - (-V) = 2V. q.e.d.

 

[gaboo]

Quote:

Originally Posted by TrevorNetwork Iron_Dreamer: If I apply +9v and -9v to a speaker it does not move at all. It stays neutral. Although, the speaker has had 18v of voltage differential applied to it, it has not had a frame of reference. If I had a ground, and a +18 I would have 18v worth of movement, without a reference, I have equal, and opposite movement, effectively, nothing. Using your analogy of a tug-of-war, if the right side moves with a certain force (+), and the left side moves in the opposite (-) direction with an equal force, there will be no movement.

Speakers are not grounded! The don't "know" what +9 or -9 is. All they care about is the difference.

 

[Iron_Dreamer]

That is all well and good, but your example uses DC, whereas speakers use AC. It invalidates your entire example.

Sure +9v DC + -9V DC = 0V

but AC is more complicated than that because of the sine wave function involved

imagine the function VAC = 9 * COS (T) when T = 0, it will be equal to +9V

however the inverted input is VAC = 9 * -1 * COS (T) when T = 0, it equals -9V

The former is at a peak, the latter a trough. There is a differential of 18V between the peak and the valley, at T = 0.

so subtract them: VAC = 9*COS(T)-(-9*COST(T)) = 18*COS(T)


That is what drives the speaker element. if it were two peaks (i.e. uninverted signal on both poles), or two valleys, the difference would be zero, as the two curves overlap each other exactly.

i.e. VAC = 9*COS(T)-(9*COST(T)) = 0*COS(T)

From a ground reference, of a flat line VAC = 0* T, the difference from the first is 9V, the distance from the line to the peak.

i.e. VAC = 9*COS(T)-(0*T) = 9*COS(T)

I wish I could make an easy graph or something to demonstrate this, as it really is quite simple.

Why do you doubt me so strongly? I mean I have the headphones set up this way, and they work, very well I might add. Others have heard it for themselves.

 

[TrevorNetwork]

I understand now!

Thanks to both of you for your time and patience. Basically if the red side is receiving +9v, and the black side is receiving -9v (peaks of sine waves), since the speaker is polarized, it is looking to the red side for direction, and the differential between the two for amplitude. it will move +18v, because the red side is receiving positive at the time. Sound right? *lol*