Wavebourn said:
Indeed nobody, and to add insult to injury they are watching us.
http://people.csail.mit.edu/rahimi/helmet/
Lumba Ogir said:
I'm not sure what you want me to elaborate on. I've stated that in order to change ic, the ac collector current, a change in E field is needed. E fields have energy and changing said energy requires time. Thus dw/dt = power, P, during the change of energy. But P = i*v. To change E, it takes non-zero i AND v. Both are needed.
Let's simplify to a simple parallel plate capacitor. In the static state it has a charge Q0, a capacitance C, and a voltage V0. If we wish to change V0 to V1, then it takes time to transport charges. During this time, there has to be a non-zero current I. Or as we all know, i = C*dv/vt. The E field is V/x, if x is the separation of the plates. If E is changing, then V is changing. But V cannot change if I is zero. Thus it takes I and V both in unison to change E.
My refutation is to those who insist that vbe is preeminant in changing the E field. The currents ib, ie, and the voltage vbe, all must change in order for the E field value to change. Thus at the micro level, it is NOT 1 quantity that accounts for bjt action, but all of them. This is why every semiconductor physics text uses charge control at the atomic level. To change E we must transport charge. Inevitably ib, vbe, & ie, MUST be non-zero to achieve this.
Too many think that E is solely due to V, and H is solely due to I. Static conditions seem to suggest the same. But when fields change, both I and V must be non-zero. A similar analogy holds for changing H fields. Without V, I cannot change, as v = L*di/dt.
So is it ib, or vbe, or ie that is responsible for E changing? Answer: all of them, in NO specific order.
My 1964 General Electric transistor manual from 1964 spelled it out. Tonight I'll quote from it. Basically it stated that the change in collector current can be regarded as due to the change in base current, OR the change in base emitter voltage. They happen together and neither one is dominant.
The e/m field laws support no other view. When fields, E or H, change with time, both I and V are non-zero. It's a fact. Thanks for your interest and by all means don't hesitate to ask for clarification. BR.
Please keep the discussion on topic
Lumba Ogir said:
At one time, quite a lot. But over the last 2 decades, not much. But, I am a drummer and use to record myself live with bands. Also, I've recorded in the studio. I played in a very large church orchestra which recorded, gave concerts, and was featured on local tv.
Audio is something I find fascinating as I play music myself. I just began taking voice lessons. Right now I just do scales and such, as I've only had 2 lessons.
Have I answered your question? Best regards.
Claude
Iain,
thank you very much for correcting me, I`m certainly not a professor of English, I did mean "extensive knowledge". An "excessive knowledge" can be interfering, as in audio rather simple designs through unsophisticated thinking, backed up by a special sense (Wavebourn has that) often is more advantageous.
thank you very much for correcting me, I`m certainly not a professor of English, I did mean "extensive knowledge". An "excessive knowledge" can be interfering, as in audio rather simple designs through unsophisticated thinking, backed up by a special sense (Wavebourn has that) often is more advantageous.
Claude my apologies for being so harsh. I thought we were talking practical engineering not the pedantic details of the physics. There are huge classes of circuits, VGA's, multipliers, log domain filters, log amps, bandgaps etc., where beta is basicly just a nuisance factor. I don't think conserving energy at quasi-DC/audio frequencies by accounting for the energy stored in junction capacitances or the power delivered by the base current into the Vbe yields enormous additional insight into circuit operation. Your mileage obviously varies.
For instance the analysis of the Brokaw bandgap using only the Vbe equation yields correct first order results.
For instance the analysis of the Brokaw bandgap using only the Vbe equation yields correct first order results.
Re: Re: NFB clamp
Weeeeelll I had to adapt it because the VAS is unipolar and (depite how it's drawn in that simplified schematic) I require a clamp that tracks the supply rails for the power output devices.
Cheers,
Glen
Edmond Stuart said:
Hi Glen,
I feel flattered that you've adopted my idea of the nfb clamp. However, you have only used the feedback part. So, I would have been even more flattered if you also had adopted the other part (which monitors Vce of the drivers), instead of adapted it.....................
For further reading, please look here: http://www.diyaudio.com/forums/showthread.php?postid=1825871#post1825871
Cheers,
Edmond.
Weeeeelll I had to adapt it because the VAS is unipolar and (depite how it's drawn in that simplified schematic) I require a clamp that tracks the supply rails for the power output devices.
Cheers,
Glen
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