Re: amplifier
- From: default <default@xxxxxxxxxxxxx>
- Date: Wed, 30 Aug 2006 08:31:50 -0400
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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- References:
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- From: chrajesh911@xxxxxxxxx
- Re: amplifier
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