As diyers most of us could give a hoot about commercial applicability. Current drive is -- at least in the modern age -- a little explored corner of audio and those who want to explore it benefit from sharing challenges and successes.
dave
Nope. In this case designing cinema systems, where Western Electric ruled in the US.
dave
I think you misunderstood my question. But assuming that these were not engineers coming from a transmission background can you explain why the matched impedance argument and matched at which point in the speakers impedance curve?
EDIT: Found the reference I was looking for. NP suggests 50 Ohms for an 8 Ohm speaker, not 8 for 8. So what was different in the WE case?
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Boy this discussion went to pot again. Please, voltage never get "converted" to current. And a current source is no different that a voltage source form the point of what causes the current to flow. It is the potential difference between the + and - terminals which forces electrons to move, i.e. produce a current. What is different is that the voltage source is designed so that the output voltage is linearly related to the input and the current floats based on the load. The current source is design so that the output voltage varies dynamically so that the current is linearly related to the input. And a loud speaker is not driven by current. It is driven by the voltage applied across the VC. Electrons have mass. They must be accelerate to move. Once in motion, unless we are talking about a super conductor, they collide with atoms and impurities in the conductor giving rise to the equivalent of friction which we call resistance. Thus to keep them moving a force must be applied to overcome this friction. That force is the voltage across the VC. It's not a chicken and egg thing. A voltage is applied across the VC. That voltage give rise to a current flow. That current flowing in the magnetic field of the gap gives rise to a force F = BL x I, which then accelerates the mass of the cone. So what drives the speaker? Maybe we should say BL drives the speaker since there would be no force generated if BL was zero. Obviously I'm being facetious. The point is that while the force that accelerates the moving components of the speaker is a result of a current flowing in a magnetic field, how are you going to get that current? What drives it? The voltage applied across the terminals.
And again, the so called advantages of current drive go out the window with anything but a fully active system where each driver is connected directly to the source.
Another thing that has to be realized is that even when looking at a single driver connected to a current source it is often stated that this eliminates the electrical damping of the driver. That is complete BS. It comes about because when you write the equation of motion for the driver for the voltage drive case and look at the forces you have
F = BL x( (V - BL x U)/ Re) where V is the voltage across the VC and U is the velocity of the motion of the cone (BL x U) is the back EMF. The current flowing is I = (V - BL x U)/Re
When the force is expressed for a current source you just write
F = BL x I because the current is prescribed to be fixed. But the reality is that I is still
I = (V - BL x U) / Re. All that is difference is that to maintain constant I, V applied to the VC is increased/decreased as necessary as the driver Z increases and decreases with frequency. Remember,
Z = Vapplied/I = Vapplied x Re /( Vapplied - BL x U) = Re /( 1 - BL x U/ Vapplied).
Now, anyone who know anything about the motion of a driver know that in the nominally linear range of performance the ratio of driver velocity, U, to Vapplied is a function of frequency only with peak at resonance, and BL is a constant. Thus,
Z = Re /(1 - G(f)), where G(f) will have a max at resonance less that 1.
This is a property of the driver and has noting to do with how it is driven.
Thus the highly undamped behavior of a driver when connected to a current source has nothing to do with eliminating electrical damping. It has to do with the increase in the applied voltage across the driver. Again, voltage is the driving force. That is, the force that is driving the current.
Now, you place a compensation network across this driver to flatten the resonance peak. What happens with a current source? Now the voltage does not rise around resonance. The reason is because the compensation network provides a low impedance path for the "excess" current. The driver now behaves exactly as it would if connected to a voltage source, drawing the same amount of current as with a voltage source, with all the excess current flowing through the compensation network. In reality, the compensation network is nothing but a "current" equalization network, so that the voltage across the driver is no longer a function of frequency and no longer rises at the driver's resonance peak.
With the compensation network it also negates the potential benefits of eliminating compression due to the effect of VC heating since as the VC resistance increases what will happen is that more current will be diverted through the comp network instead of through the driver.
And again, the so called advantages of current drive go out the window with anything but a fully active system where each driver is connected directly to the source.
Another thing that has to be realized is that even when looking at a single driver connected to a current source it is often stated that this eliminates the electrical damping of the driver. That is complete BS. It comes about because when you write the equation of motion for the driver for the voltage drive case and look at the forces you have
F = BL x( (V - BL x U)/ Re) where V is the voltage across the VC and U is the velocity of the motion of the cone (BL x U) is the back EMF. The current flowing is I = (V - BL x U)/Re
When the force is expressed for a current source you just write
F = BL x I because the current is prescribed to be fixed. But the reality is that I is still
I = (V - BL x U) / Re. All that is difference is that to maintain constant I, V applied to the VC is increased/decreased as necessary as the driver Z increases and decreases with frequency. Remember,
Z = Vapplied/I = Vapplied x Re /( Vapplied - BL x U) = Re /( 1 - BL x U/ Vapplied).
Now, anyone who know anything about the motion of a driver know that in the nominally linear range of performance the ratio of driver velocity, U, to Vapplied is a function of frequency only with peak at resonance, and BL is a constant. Thus,
Z = Re /(1 - G(f)), where G(f) will have a max at resonance less that 1.
This is a property of the driver and has noting to do with how it is driven.
Thus the highly undamped behavior of a driver when connected to a current source has nothing to do with eliminating electrical damping. It has to do with the increase in the applied voltage across the driver. Again, voltage is the driving force. That is, the force that is driving the current.
Now, you place a compensation network across this driver to flatten the resonance peak. What happens with a current source? Now the voltage does not rise around resonance. The reason is because the compensation network provides a low impedance path for the "excess" current. The driver now behaves exactly as it would if connected to a voltage source, drawing the same amount of current as with a voltage source, with all the excess current flowing through the compensation network. In reality, the compensation network is nothing but a "current" equalization network, so that the voltage across the driver is no longer a function of frequency and no longer rises at the driver's resonance peak.
With the compensation network it also negates the potential benefits of eliminating compression due to the effect of VC heating since as the VC resistance increases what will happen is that more current will be diverted through the comp network instead of through the driver.
John I fully agree with your post except for one thing. I think it is correct to state that it is the current in the VC that determines the force exerted on the moving assembly - the famous F = B*I*L, no?
Jan
Jan
Yes, I wrote that in the first paragraph. But how does the current come about? How to you get those electron to move along the wire? Where does the energy come form? Left to their own devices they just jump around randomly. There must be an applied voltage. Voltage drives the current, current flowing in the magnetic field generate the force to accelerate the moving mass.
Electrons aren't particularly motivated. They are like a bunch of drunks in a bar, meandering around banging into things. To get them to move in one direction take a driving force, like an attractive woman.
Reminds me of a joke told by a female colleague once. In semiconductor physics you have electrons and holes that carry negative and positive charge. When electrons are excited to high energy they are referred to as hot electrons. So there are about 5 men and this one women in an elevator at the conformance on semiconductor physics and as we are going up in the elevator she say, "Poor me. A single hole surrounded by a bunch of hot electrons." And yes, she was single and quite attractive. 🙂
It seems like maybe this confuses people into thinking voltage doesn't matter when it comes to driving a speaker.
I think a review of the fundamentals might be helpful for folks reading this and trying to grasp the detailed explanations John is providing.
And no, I'm not talking about Joe, Dave or Steve when I say that, though I do think the way they are phrasing some things can make this sound more complicated than it really should be.
The "conversion" of voltage to current is no different when using a voltage amp than when using a current amp.
Voltage, or potential energy, is where everything starts. You can have voltage without current, which is why it is called "potential" energy. But you can not have current without voltage.
When a voltage is applied to a conductor, current will flow, even with zero Ohm resistance added to the conductor. In fact, so much will flow that bad things will happen like melting the conductor or starting a fire.
Resistance does just what the name implies. It resists the flow of current. So the more resistance you have, the less current you will get, assuming voltage is constant. This is what Ohm's law is all about.
Capacitors and inductors can also be seen as resisting the flow of current, but both are dependent on frequency. (It is actually called reactance but that doesn't matter much here)
An inductor resists that flow more as frequency increases while a capacitor resists the flow more as frequency decreases, down to the point of completely blocking DC. An inductor just doesn't do anything when placed in a DC circuit, except maybe when the voltage is switched on or off.
Impedance is a combination of the total resistance to current flow from all the capacitors, inductors and resistors in a circuit, at a particular frequency.
What impedance really gives us is the ability to apply Ohm's law to an AC circuit. That's why impedance is measured in Ohms, just like pure resistance. The only real difference from DC is that any impedance value is for a specific frequency.
-Chris
Note 1: Of course audio amplifiers do not just handle one frequency but you can still apply the same basic math to figure out how much current should be flowing at any specific frequency.
Note 2: Moving a voice coil inside a magnetic field is where my head starts to spin as to what the overall circuit actually looks like at a given frequency, voice coil position, and power/heat level. But I do know that Ohm's law still applies if we know the reactance and resistance values.
@ Joe Rasmussen
Hi, always good to see alternative etc approaches & info/discussions etc. So Thanx for posting 🙂
@ john k...
On the "HOLE" i agree 😀
As for speakers being current devices, as you mention, without Voltage we would hear nothing ! But, the lower the impedance of drivers goes, then the more current they demand = more of a current device. For eg, many drivers have been available for some years with 4Ohm & 2Ohm coils & even 1Ohm. I know of one amplifier that can work into 0.5Ohm, so connecting multiple drivers as a 0.5Ohm load on that amp, would be an interesting experiment !
Hi, always good to see alternative etc approaches & info/discussions etc. So Thanx for posting 🙂
@ john k...
On the "HOLE" i agree 😀
As for speakers being current devices, as you mention, without Voltage we would hear nothing ! But, the lower the impedance of drivers goes, then the more current they demand = more of a current device. For eg, many drivers have been available for some years with 4Ohm & 2Ohm coils & even 1Ohm. I know of one amplifier that can work into 0.5Ohm, so connecting multiple drivers as a 0.5Ohm load on that amp, would be an interesting experiment !
You cannot 'convert' voltage to current. How would that work? One is a potential, the other is, well, a current.
What people *probably* mean is that current and voltage are related through impedance - the old tired law of Mr. Ohm.
Of course I understand that 'voltage conversion' sounds more sexy than 'ohms law', but that only adds to the confusion.
One reason that this thread is going nowhere is because of sloppy thinking, sloppy writing and unclear meaning.
Jan
I understand, and I actually typed out "there is really no conversion of voltage to current" then deleted it.
-Chris
There is people that have given their simulators, measurement
software, finished speaker designs also for free. It is usual for
any kind of hope for developing a business.
Free software like ARTA, Bagby spreadsheets, Boxsim, HolmImpulse,
REW, Unibox, XSim, etc... all Saints of diyAudio. 🙂
Maximum power transfer, the other i don't know.
Nelson is talking about a current amp. Different thing.
dave
Maximum power transfer makes sense, but that is just a function of valve amps with output transformers. No other way makes sense once you have an output transformer.
Now for curious theory of the day Valve Amps: Valve verses Solid-state amps makes the claim that ultralinear provides constant power to speakers with varying impedance unlike voltage or current drive. On inspection I don't buy it, but one to throw into the pot. 🙂
Now for curious theory of the day Valve Amps: Valve verses Solid-state amps makes the claim that ultralinear provides constant power to speakers with varying impedance unlike voltage or current drive. On inspection I don't buy it, but one to throw into the pot. 🙂
I don't know why one would design an audio amp for max power. In audio we are more concerned with efficiency and noise. Max efficiency pushes you towards low output impedance and high load impedance. Noise is a whole other issue. We've already discussed that current sources with high output Z lead to low efficiency when driving a loud speaker.
At risk of being pedantic, efficiency and max power transfer, whilst related are different variables to play with in toobland. But you could change your first sentence from 'for max power' to 'with tubes' and the rest would still hold true 🙂
And of course the true audiophile is more concerned with inefficiency 🙂
Also push-pull valve amps.
I disagree. Ultrasound HiFi has been making high efficiency speakers with their matching high Zout (up to 1 MOhm!) amps for more than 3 decades. You are not going to find any other true 93 dB/W/1m in the form of a compact 2 way stand mound speaker without compromising on other parameters (likely frequency response). These speakers are of course 3-4 ohm bass reflex speakers and of course no crossover because the drivers are perfectly complementary! Just a protection cap in series with the tweeter and no waste of power....
The theory of static damping factor is flawed and the intermediate and high Zout amps work at least as fine as low Zout amps. If not better when properly set-up....
So your sentence might be only true if you want to use a generic speaker (i.e. with drivers and design that are not suitable) driven by a high Zout amp.
Same thing when speaking of constant power sources (typical zero fb valve amps and similar).
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