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Old October 12th 04, 12:54 AM
Joe Rocci
 
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Steve,

If the inductor was not there to hold the zero-crossings of the input sine
wave at zero volts, then the whole waveform would sink toward a lower DC
voltage because it is capacitively coupled. You can prove this to yourself
by taking a large capacitor and driving a diode that is connected to ground.
The test can easily be done with audio frequencies if you don't have RF
equipment. You could also simulate it on a program like SPICE.

If the choke were removed from the circuit, this input DC shift would
reverse bias the BE junction, preventing the abiltiy of the waveform to
drive current into the BE junction. No base current = no collector current =
no gain.

BTW, I was almost sure your original post said this was a multiplier
circuit. Did the word "multiplier" not appear in it anywhere? Hmmm...

Joe
W3JDR


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
--

Fat, sugar, salt, beer: the four essentials for a healthy diet.