"Steve Evans" wrote in message
...
On Mon, 11 Oct 2004 14:59:11 -0500, "Steve Nosko"
wrote:

The Base-Emitter in a transistor is a semiconductor junction just like a
diode and in the (higher power) RF amplifiers behaves pretty much like a
diode. With RF applied to the base, there will be conduction on the
positive peaks only and this will constitute a DC current flow which must
have a DC path. The inductor provides such a path since the capacitor
can
not.

Okay, Steve, I'm gonna have to take your explanation in bite-size
chunks. Kindly indulge me...

I don't see that the inductor is necessary to provide such a path to
ground for the signal peaks, since they (the input signal pos. peaks)
turn on the transistor and complete the circuit to ground via the
base/emitter junction, which will be a low resistance path with
sufficient base drive level on the peaks. Can you tell me why this
path alone isn't good enough and there has to be an inductor across
B/E as well?
Thanks! Steve

Most certainly, Steve.... Indulge, I will.
This can get confusing... Steve Noskowicz here (K9DCI)

Joe Rocci tried, but I don't think he went through the proper step-by-step
explanation.

OK Here's what happens... Start with the capacitor completely discharged -
zero volts across it - right end same voltage as left side (whatever that
may be). On the first positive peak, some current flows through the base
emitter junction because the voltage on the right side of the cap is the
same as that on the left side. (this assumes there is at least 0.7 volts of
signal coming from the left. NPN transistors conduct current when the base
is about 0.7 volts positive.
The current is some quantity of electrons, right. Well these electrons
will start to "fill up" or charge the capacitor. Each time another positive
pulse happens, the right side of the capacitor collects more and more
electrons. This charges the cap more and more and makes the right side more
and more negative. There is no way to drain off this charge before the next
pulse comes along. The base-emitter junction will be reverse biased and not
conduct any current. I hope you see this because that's the key.
As the caps gets more charge from each pulse, the right side becomes more
negative. After each pulse, it will take more and more voltage on the caps
left side to get enough voltage on the right side (0.7 volts) to get to the
base-emitter conduction voltage and get any current to flow.
The end result is that the capacitor will charge to the peak input
voltage (less about 0.7 volts) and the base will never get to the 0.7 volt
level. At this point, the base voltage will be swinging (with the input
signal causing it) from about 0.7 volts plus to a value equal to
*negative* the peak-to-peak input voltage (less the 0.7 volts). That's
minus volts. The AC signal at the bacse will NOT be swinting equally around
zero volts.

This could be viewed as what we call a "clamper circuit" The AC voltage
AT THE BASE will have its positive peak "clamped" or moved to +0.7 volts and
the waveform will extend from there as negative as the waveform is tall.

IF we had 5 volts peak (10 volts peak-to-peak) the cap left side, the base
voltage will swing from +0.7 volts to -9.3 volts.

WHEW !
Did this work ???
--
Steve N, K,9;d, c. i My email has no u's.