Re: What's this inductor doin'?

From: Joe Rocci (joe_at_roccis.com)
Date: 10/11/04 

Date: Mon, 11 Oct 2004 21:01:46 GMT

Steve,
I'm not sure, but I think the original post said this stage was a frequency
multiplier with an OUTPUT frequency of about 145 MHz. If that's the case,
then the INPUT frequency would be 72 MHz or less. At that frequency, I don't
think the choke and the input capacitance of the transistor are anywhere
near resonance. Also, the coupling cap was stated as 1nF if I recall.

I think what we're looking at here is a DC -lock coupling cap and a
DC-return RF choke....nothing more.

Joe
W3JDR

Steve Nosko <suteuve.nosukowicuz@moutouroula.com> wrote in message
news:ckeork$sip$1@newshost.mot.com...
>
> "Steve Evans" <smevans@jif-lemon.co.mars> wrote in message
> news:r2ejm0t0lfstc6pa8edd2o3statd7a4kr7@4ax.com...
> > On Sun, 10 Oct 2004 14:01:44 +0100, Paul Burridge
> > <pb@notthisbit.osiris1.co.uk> wrote:
> >
> > >Well then it appears Reg's hunch was right. The transistor in question
> > >has an input capacitance of just under 3pF, so the 0.4uH inductor
> > >forms a parallel tuned circuit with it at 145Mhz. This prevents the
> > >input signal from partially shunted to ground via the input
> > >capacitance were the inductor not there. The idea is to allow as much
> > >of the input signal as possible to develop across the BE diode to
> > >maximise input impedance and gain. The input capacitance is in
> > >parallel with this diode and bypasses RF signals around it - which you
> > >*don't* want. Will a 1uH work instead? Do the maths and find out. But
> > >if you don't know the inductor's Q that probably won't help much...
> >
> > Sorry, but none of this makes sense to me. There's no diode involved
> > so I don't know where you get that from. And what's "input
> > capacitance" and "Q"?
> > Do try to speak in plain English!
> >
> > Steve
>
>
> Hi Steve (swell name, by the way),
> From this last comment, it appears that you have a lot to learn.
Paul's
> was a pretty good explanation of a first step at understanding what might
be
> going on in the circuit shown some time ago (an inductor in shunt with the
> base-emitter of the transistor).
>
> 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. If this makes no sense to you then you, indeed are in over your head
> in an attempt to understand because it is pretty basic and simple. You
will
> need to understand diodes and transistors first.
>
> you say you are "really struggling here with with some of the
terminology."
> Perhaps you can tell us which words are giving you heartburn?
>
>
>
> I will respond to Paul's content, however, with this. The BE
> capacitance of this device, in this aparent application, I am pretty sure
is
> not the dominant effect. The Rrverse biased capacitance is the wrong thing
> to focus on. While it is interesting that that it and the inductor are
near
> resonance, this probaly is not what is happening because this would make
the
> inpedance looking into the base very high and difficult to get power to
the
> base, contrary to Reg's hunch. The orignal ASCII cricuit simply had a
> coupling cap and a base-emitter shunt cap. It looks like class B or C. C
> more likely. Therefore the transistor is in conduction part of the time
and
> not for another part of the time. Therefore we have a nonlinear, large
> signal condition. The base impedance under this condition (pulsed
> conduction) will be quite low and dominate and therefore, it will need
some
> impedance matching to get enough into the base (from the preceeding
> collector). SO, I say that the inductor is :
> 1- Providing the obvious DC path and.
> 2- Impedance matching along with the "coupling" capacitor (did it have a
> value??)...BUT!
>
> The one monkey wrench I will throw in, is that the Miller effect will also
> have a very significant effect on the input impedance of the stage. The
Ccb
> is a path providing significant feedback and probably dominating the input
> impedance.
>
> If you don't recall, the Miller Effect describes the capacitance looking
> into the base which looks like Ccb times the voltage gain (call it A).
This
> is due to the fact that Ccb connects between the input / base and output /
> collector. Because the collector voltage is ~~ 180 degrees out of phase,
> with the base voltage, an input voltage change of, say one millivolt, on
the
> input side of Ccb results in a change in voltage on the output side of Ccb
> of one milivolt times the voltage gain, A. This results in a total change
> across Ccb of A+1 milivolts and therefore a current change A+1 times a
value
> that the 1 milivolt input change expected to see. This makes the
capacitor
> look A+1 times as big as it actually is.
>
> Finally, and possibly the most difficult to quantify (ok two monkey
> wrenches--nobody expects the Spanish inquisition), in RF circuits there is
> *very frequently* one other confounding factor and this is the circuit
board
> layout and/or the actual physical construction. All the previous talk
about
> how inductors and capacitors behave differently at high frequencies (I
> believe by Roy Lewallen) is nicely put, but the actual connection methods
> also can have a very significant effect on what value components are used.
> The "wiring" can add other capacitances and inductances which, very often,
> do not show up on the schematic. This can have profound effect on the
> components used, completely masking any hope of understanding of the
circuit
> from the schematic diagram. As the power level in the circuit goes up, the
> impedances go down and short wires or PC board runs can become significant
> impedances, either to help or hurt the desired matching circuit.
> --
> Steve N, K,9;d, c. i My email has no u's.
>
>

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