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"Bill Bowden" <wrongaddress@xxxxxxx> wrote in message
news:1156915815.971190.91800@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> I never fully understood the idea of increasing the gain with a bypass
> cap across the emitter resistor. The problem you have is discharging
> the cap after it has charged through the emitter circuit. Suppose the
> emitter resistor is large at maybe 1 Megohm and the cap is 100uF. The
> transistor can charge the cap very quickly as it turns on and therefore
> the gain is increased at the collector due to the extra current.
Sorry, but you're thinking of this in the wrong terms. You're really
thinking about how the capacitor behaves in DC terms, when the
important consideration here is the impedance seen at various
frequencies. At sufficiently high frequencies (depending on the
capacitor value and that of the emitter resistor, etc.), the cap
is "bypassing" the emitter resistor - it is essentially a short to
ground for AC at those frequencies. Work through the gain
calculations (this is assuming that you're taking the output
from the collector, of course - a typical CE amp configuration)
and you'll see that impedance in the emitter-to-ground leg actually
REDUCES the gain.
> But
> now the capacitor must discharge through the 1 Meg parallel emitter
> resistor which takes RC time to fall 63% or about 100 seconds in this
> case. If the frequency is high, the cap will not have time to discharge
> much and cannot charge much on the next cycle.
That's the whole point - a large enough cap CANNOT change
its charge state rapidly enough to respond to rapidly-changing
AC currents in this manner. That is really what makes it a
low impedance to AC - the fact that you CAN'T change the
voltage across the cap very quickly.
Bob M.
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