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#1 |
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diyAudio Member
Join Date: Sep 2006
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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. |
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#2 |
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diyAudio Member
Join Date: Sep 2006
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I made a mistake should be:
With the bias supply the first stator is -500v, the diaphragm is 1000v, the second stator is 2500v. But that shouldest matter because it is the same on both sides? Only a higher push-pull voltage change should increase output. |
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#3 |
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diyAudio Member
Join Date: May 2006
Location: Netherlands
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Forget about the audio signal for a minute and just look at the bias supply. Let's call the '-' of the EHT supply 0V and the other pole +Vb. Can you see how it pulls the diaphragm up to +Vb relative to the stators (which are both at 0V) and thus charges it?
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#4 |
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diyAudio Member
Join Date: Sep 2006
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Yes, but wouldn't the stators cancel each others effect?
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#5 |
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diyAudio Member
Join Date: May 2006
Location: Netherlands
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Look at it as two capacitors (stator - diaphragm), (diaphragm - stator) in parallel.
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#6 | |
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diyAudio Member
Join Date: Jan 2005
Location: Phoenix, Az.
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Quote:
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 |
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#7 |
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diyAudio Member
Join Date: Sep 2006
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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. |
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#8 |
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diyAudio Moderator
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They're only unbalanced like that in the presence of signal, i.e., yes, there's going to be a force displacing the diaphragm.
__________________
“Listening to records is like ****ing a picture of Brigitte Bardot.” - Sergiu Celibidache |
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#9 |
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diyAudio Member
Join Date: Sep 2006
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So its not really push/pull?
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#10 | |
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diyAudio Member
Join Date: Apr 2004
Location: Northern California
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Quote:
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