som den skal. :-)
"A theoretical efficiency of 50% is obtainable with transformer output coupling and only 25% with capacitive coupling,"
dualsupply.
hvileeffekt, 8 ohms load. Det giver +-16V udsving, og +-2A i load.
transistor er nul, mens den leverer 2A til load.
Overser jeg noget, eller er der fejl i Wiki?
-- Ulrik Smed Aarhus
Hej Ulrik
Med formler:
/Glenn
Den 15-03-2014 18:42, Ulrik Smed skrev:
med kunstigt nulpunkt ved 16Volt.
transistor.
rammer 50%.
Karl Erik.
grader af udgangs-signalet.
100% ampere i den positive forsyning.Da man beregner effektivitet'en over 360 grader af musik-signalel, skal man huske at det er hele den positive periode af sinus'en tilsammen vi
transistorer i varme. )
lav.
kommer til realitet'erne.
/ Lars
Den 15-03-2014 18:42, Ulrik Smed skrev:
= 8W drift / 32W Tomgang = 25%. ...
Den 15-03-2014 20:16, Lars / Viborg skrev:
-- Ulrik Smed Aarhus
Den 15-03-2014 19:36, Karl Erik Christensen skrev:
-- Ulrik Smed Aarhus
Den 15-03-2014 19:30, Glenn skrev:
Hah, der var forklaringen! De 25% er single-ended.
Another type of Class-A amplifier uses the same circuit as Figure 1, but
signal, the transistors can draw up to double the quiescent current at waveform peaks, or zero at the opposite peak (one transistor will be double and the other zero as the waveform alternates). Dynamic analysis of this arrangement is harder than either of the other topologies described, but if we examine the quiescent state and the maximum output, a reasonable estimate is possible. Using the same values as before, we have 20V across each transistor, at a quiescent current of 1.25A. Each transistor will therefore dissipate 25W with no signal, and maximum power is also 25W. Efficiency is therefore ...
Eff = 25 / 50 * 100 = 50%
-- Ulrik Smed Aarhus
Den 15-03-2014 19:30, Glenn skrev:
ensretteren. ;-)
-- Ulrik Smed Aarhus
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