The bias supply doesn't make any sense to me. I was looking at the standard esl diagram and thinking; the bias supply is pointless. I MUST be missing something but i can't figure out what.
Say there is a 1kv bias, and 2kv across the transformer.
With the bias supply the first stator is 0v, the diaphram is 1000v, the second stator is 2000v.
Without the bias supply the first stator is -1000v, the diaphragm is 0v, the second stator is 1000v
The voltage difference between the elements are the same in both cases, therefor the force acting on the diaphragm is also the same. I am missing a key idea somewhere and it is causing me great woes.
An externally hosted image should be here but it was not working when we last tested it.
Say there is a 1kv bias, and 2kv across the transformer.
With the bias supply the first stator is 0v, the diaphram is 1000v, the second stator is 2000v.
Without the bias supply the first stator is -1000v, the diaphragm is 0v, the second stator is 1000v
The voltage difference between the elements are the same in both cases, therefor the force acting on the diaphragm is also the same. I am missing a key idea somewhere and it is causing me great woes.
ak_47_boy said:
Both stators are at the same voltage because they both connect to the same side of the bias supply through the transformer coil. So you have diaphragm at 0, stators at -1000V. Or diaphragm 1000V, stators (both) at 0V.
If the stators were spaced exactly the same distance from the diaphragm on both sides, and if the diaphragm weren't subjected to air currents, there would be no net force on the diaphragm. However, spacing isn't perfect and air currents are always around, so the diaphrgam will be sitting slightly closer to one stator than the other. The result is that the diaphragm will tend to pull toward one side or the other. The only thing that prevents it from hitting the stator is the mechanical tension on the diaphragm.
When you apply a signal from the amplifier, think of the diaphragm at 0 V relative to the AC signal and one stator at positive potential and the other at a negative potential (due only to the AC signal). Add the voltages at each point (bias + AC signal) and see what you get. For example, bias alone, diaphragm at +1000V, stators at 0V. AC signal: diaphragm at 0V, stator A at +300V, stator B at -300V.
Total voltage at stator A= 0+300=+300V. Total voltage at diaphragm= +1000+0=+1000V. Total voltage at Stator B= 0+-300V= -300V. The diaphragm will be attracted to stator B and repelled from stator A.
The reality is that the diaphragm is charged to a specific value, either positive or negative and the charge is supposed to be constant, NOT voltage, but it may be easier to wrap your mind around voltage than charge. The charge on the diaphragm responds to the changing E field between the stators and drags the diaphragm along with it.
I_F
But if stator A = -300v, stator B = 300v, and diaphragm = 1000v then the potential from A to the diaphragm is 700v and B to the diaphragm is 1300v.
Would that not be uneven force? I thought for example if one stator is 1000 volts over the diaphragm the other stator must be 1000 volts under the diaphragm.
Would that not be uneven force? I thought for example if one stator is 1000 volts over the diaphragm the other stator must be 1000 volts under the diaphragm.
SY said:
SY didn't say, perhaps he thought it was understood that the uneven stator to diaphragm voltage is what causes the diaphragm to move and that's how ESL panels convert AC signals to sound. Since the signal is applied through a transformer, only AC voltages cause the unbalance. The input transformer can not transmit DC. Getting the diaphragm to move is the whole point.
The sole purpose of the bias supply is to charge the diaphragm. Nothing else. Don't think in audio voltages and bias voltage acting together this will only make it confusing.
So can you see that the diaphragm is charged by the bias supply?
OK now when you do, we assume the charge on the diaphragm to be constant from now on (which with high resistance coating will be the case). Unplug the bias supply. Forget about voltages. Forget about capacitors. We just have a charged diaphragm between two metal plates.
Now we can look at the audio signal, which when applied to the stators creates an electric field between them (like I_F said). Physics says that a charge in an (uniform) electric field will have a force act upon it which is independant on it's position in the field. The magnitude of this force is linear with charge and with field strength according to the equation F = q * E, q being the charge and E being the field strenght. For simplicity we'll ignore direction.
Let's assume we have a drive voltage across the secondary of the transformer called Va (audio). The electric field will have a magnetude of E = (Va / d) where d is the spacing between the stators. So for a given geometry field strength only depends on this drive voltage.
Hence with constant charge, for a given geometry, the total force on the diaphragm is linear with drive voltage. This force is what moves the diaphragm. If we look at it this way there really is no 'push-pull' to talk about. We call this setup push-pull in distiction to the single ended electrostatic speaker (one stator).
If you want to look at it more closely the subject quickly becomes a lot more complicated as I made some assumptions and simplifications above. It's as simple as I could make it without becoming completely invalid.
So can you see that the diaphragm is charged by the bias supply?
OK now when you do, we assume the charge on the diaphragm to be constant from now on (which with high resistance coating will be the case). Unplug the bias supply. Forget about voltages. Forget about capacitors. We just have a charged diaphragm between two metal plates.
Now we can look at the audio signal, which when applied to the stators creates an electric field between them (like I_F said). Physics says that a charge in an (uniform) electric field will have a force act upon it which is independant on it's position in the field. The magnitude of this force is linear with charge and with field strength according to the equation F = q * E, q being the charge and E being the field strenght. For simplicity we'll ignore direction.
Let's assume we have a drive voltage across the secondary of the transformer called Va (audio). The electric field will have a magnetude of E = (Va / d) where d is the spacing between the stators. So for a given geometry field strength only depends on this drive voltage.
Hence with constant charge, for a given geometry, the total force on the diaphragm is linear with drive voltage. This force is what moves the diaphragm. If we look at it this way there really is no 'push-pull' to talk about. We call this setup push-pull in distiction to the single ended electrostatic speaker (one stator).
If you want to look at it more closely the subject quickly becomes a lot more complicated as I made some assumptions and simplifications above. It's as simple as I could make it without becoming completely invalid.
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