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On 29 Aug 2006 22:30:16 -0700, "Bill Bowden" <wrongaddress@xxxxxxx>
wrote:
>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. 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.
>
>So why does it work?
>
>-Bill
Emitter resistors are typically rather low - think 100 ohms or so.
The current flowing through the transistor drops voltage across the
resistor - a volt for the sake of discussion - that voltage moves the
emitter away from zero while the base resistors are maintaining a more
or less fixed voltage on the base. More current = greater drop on the
emitter resistor = less "on" bias = less current.
That is the DC theory - but it works with a signal imposed on the
emitter current as well - the bias, sans capacitor, will try to track
the signal - and will serve to decrease the amplification - bypass the
resistor with a sufficiently large cap and the DC bias is maintained
while the (higher frequency) signal is not affected.
You seem to be trying to think in terms of a static switch - this is a
transistor biased into its linear region to amplify a linear signal.
We aren't concerned with it turning on quickly - it is never fully on
or off (for class A operation). The cap isn't really charging and
discharging - it is just providing a low impedance path for AC.
(imposed on the DC)
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