H
HAYK
F=IBL
F in Newton I in Amper. L in meters. B in Tesla.
If you apply I in DC, the F will result a diaphragm displacement proportional to the current.
If you apply Iwft at resonant frequency Fwft will result speed of the diaphragm.
If you apply Iwft higher than the resonant, Fwft will result acceleration of the piston diaphragm proportional to the current.
We are talking of a full range speaker. With crossover filters, the current amp goes all wrong.
Moving cone and air means doing work and this requires power.
This is why voltage is needed as well as current.
Both are linked by speaker impedance.
Current drive increases voltage and thus SPL at high frequencies and especially around the bass resonance where the impedance is high.
That may or may not be desirable.
Typically the strong emphasis at the bass resonance is not desired with hifi.
Results with tube guitar amps using little or no NFB can be favorable.
Last edited:
If you now want to do V * I, what part of this is the reactive power? The current amplifier allows the current error to be reduced to a minimum and the voltage has to adapt. That's perhaps a simplified view, but it's close enough to reality.
I suspect that you mean the effect of electromotive force (EMF). My simulation shows that the voltage amplifier can offer even worse results. I have actually already presented the following post-oscillations in comparison:
The voltage amplifier delivers shorter oscillations, but with larger amplitude and distorted:
V * I gives apparent power, which is the vector sum of real power and reactive power.
Only the real power part produces work (and heat).
Exactly! This means that the output voltage cannot be used directly to determine the active power. Otherwise I don't see any other use for it.
Current or voltage square wave response doesn't mean much with a complex, frequency dependent speaker impedance.
The voice coil is driven by (real) power.
Speaker SPL response essentially follows power input at each frequency.
So we would have to compare power square wave response.
The voice coil is driven by (real) power.
Speaker SPL response essentially follows power input at each frequency.
So we would have to compare power square wave response.
The example above with the square wave signal shows the step response of the speaker model. FFT clearly shows that the voltage amplifier has more noise than the current amplifier. This just comes from driving the speaker unnaturally with the voltage.
FFT analyzer only evaluates what can be seen in the measurement window. I specially adjusted the simulation parameters so that only the ringing oscillations are available and not the square wave signal itself.
Maybe read up on FFT and associated artefacts with a limited window.
Anyway, only power square wave response matters.
Anyway, only power square wave response matters.
Then you would have had the same effect on both curves. Actually, even without FFT, it's good to see that I(R_v) has a steep fall first and then a slow rise and an even slower afterswing. This is not a clean sine wave! A lot of distortion was mixed in. I(R_i) only sounds at the natural frequency of the loudspeaker, but again much quieter.
How exactly do you want to calculate that? How much power is converted into sound? Simply sketched, without numbers.
It is kind of trivial that current drive gives a better current square wave response than voltage drive.
In fact a perfect current source must give a perfect current square wave, while a voltage source will give a perfect voltage square wave.
Question is how this translates to speaker SPL response.
To get the power signal P(t) = V(t) * I(t) you need to multiply voltage and current signals.
To answer your question:
Only about 5% of delivered power is converted to sound with a high efficiency speaker.
With SS guitar amps the ouput impedance is often increased by speaker current NFB in order to get a sound coloration similar to a tube amp.
In fact a perfect current source must give a perfect current square wave, while a voltage source will give a perfect voltage square wave.
Question is how this translates to speaker SPL response.
To get the power signal P(t) = V(t) * I(t) you need to multiply voltage and current signals.
To answer your question:
Only about 5% of delivered power is converted to sound with a high efficiency speaker.
With SS guitar amps the ouput impedance is often increased by speaker current NFB in order to get a sound coloration similar to a tube amp.
The Lorentz force F = I*B*L is the basis of all calculations. This is essentially the conversion element that converts electrical energy into sound energy. But there is a nuance. Mr. HAYK pointed this out. This concerns driving the speakers with the voltage.
The movement of the coil in the magnetic field induces the voltage Vind = v*B*L, which counteracts the output voltage of the amplifier Vout. v is the speed of the coil. Therefore the Lorentz force reduces to F = I*B*L = ((Vout - Vind)/R)B*L = ((Vout - v*B*L)/R)*B*L = (Vout/R )*B*L - v(B*L)²/R.
The first term (Vout/R)B*L corresponds to purely driving the loudspeaker with the current and the second term v(B*L)²/R is the electrical damping.
But because the speed of the coil can be completely arbitrary, i.e. depending on how many and which elementary waves are hitting each other at the moment, driving the loudspeakers with the voltage is fundamentally flawed.
The movement of the coil in the magnetic field induces the voltage Vind = v*B*L, which counteracts the output voltage of the amplifier Vout. v is the speed of the coil. Therefore the Lorentz force reduces to F = I*B*L = ((Vout - Vind)/R)B*L = ((Vout - v*B*L)/R)*B*L = (Vout/R )*B*L - v(B*L)²/R.
The first term (Vout/R)B*L corresponds to purely driving the loudspeaker with the current and the second term v(B*L)²/R is the electrical damping.
But because the speed of the coil can be completely arbitrary, i.e. depending on how many and which elementary waves are hitting each other at the moment, driving the loudspeakers with the voltage is fundamentally flawed.
Sorry, dropping obscure names doesn't impress me.
Being a physicist I know that a force by itself doesn't cause work or useable energy (as opposed to potential energy).
Not being a speaker expert I know that:
Being a physicist I know that a force by itself doesn't cause work or useable energy (as opposed to potential energy).
Not being a speaker expert I know that:
- Speaker SPL increases proportional to power.
- Speaker frequency response is typically flatter with voltage drive than with current drive. This might be due to the fact that electrodynamic speakers are optimized for voltage drive.
Last edited:
I'm even more surprised that you ignore the elementary laws of physics.
Let's assume the speaker diaphragm suspension has a spring constant k and a DC current flow through the coil. There must be a balance between Lorentz force and spring force. The diaphragm should therefore be shifted by the distance x: I*B*L = k*x, x = I*B*L/k. At the same time this means that the diaphragm has the potential energy: Epot = (1/2)*k*x^2. Now, when the current and accordingly the Lorentz force become zero, the spring force will pull the diaphragm back and so the diaphragm will start moving, dragging the air with it. In this way, the potential energy that was stored in the diaphragm with the help of the Lorentz force is converted into sound energy.
You don't seem to pay attention to what I'm saying.
A speaker does some work (and produces heat) like a motor, so you have to feed power.
Never ignore the law of conservation of energy.
I'm fed up with this thread.
Good luck.
A speaker does some work (and produces heat) like a motor, so you have to feed power.
Never ignore the law of conservation of energy.
I'm fed up with this thread.
Good luck.
Last edited:
Or not. According to the simulation, I was probably wrong. I adjusted the amplifiers with JFET and BJT for the same quiescent current.
The output power of the amplifier with JFET is slightly smaller.
But the distortions are less.
Last edited:
