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A Basic question about outputs stage driven
A Basic question about outputs stage driven
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Old 18th October 2020, 01:29 AM   #31
K-amps is offline K-amps  United States
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Quote:
Originally Posted by john_ellis View Post
H
If you plan to halve the load impedance (double the output current) then yes, the driver will see a doubled everything,and has to be able to handle it (or parallel the drivers too).
Very helpful thanks.
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Old 18th October 2020, 01:47 AM   #32
BesPav is offline BesPav  Russian Federation
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Originally Posted by MarsBravo View Post
So, after some 40+ years in electronics, it proves on diyAudio (of all ahum places) all used formulas and models I have used sofar are wrong
Ok.
Choose the correct statement, please, from your formulas and models:
1. Base current are prime while Vbe voltage drop are parasitic.
2. Vbe voltage are prime while base current are parasitic.
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Old 18th October 2020, 09:36 PM   #33
N101N is offline N101N
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In any PN junction, the presence of a Potential Barrier is indispensable. In forward bias condition, with a voltage of opposite polarity applied to the junction diode, an increment of ionic charge (current) takes place at the negative terminal of the diode.

Also under reverse bias, a Potential Barrier that opposes the applied voltage (force) will necessarily develop. In order for that to happen, strong covalent bonds need to be broken. They do not break voluntarily nor easily, must be forced to do so.

In place of the lousy semantics like "driven", "electronic and radiative excitation" would be a more suggestive nomenclature. The important acceleration of electric charge is accomplished by electromagnetic force fields.
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Old 18th October 2020, 11:59 PM   #34
traderbam is offline traderbam  Canada
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I guess whether you call something a voltage-driven device or a current-driven doesn't really matter since the two are related. I find it helpful to think of driving power transistors in terms of charge transfer.

Switching MOSFETs are often described in terms of Id vs gate (to source) charge; a charge-driven approach. A bipolar is also charge-driven except that it continually leaks charge rather than only integrating it. Because transistors store charge they have a capacitive characterstic. And a bipolar will turn itself off via its own charge leakage so can be "push" driven as in a darlington, but a MOSFET has no such leakage so needs to be "push-pull" driven.
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Old 19th October 2020, 02:39 AM   #35
PRR is offline PRR  United States
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A Basic question about outputs stage driven
> a charge-driven approach

SHHHH!!!!

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