On Sun, 03 Oct 2004 23:25:05 GMT, Doug Smith W9WI
wrote:

A capacitor has a very low impedance to high-frequency (i.e., 145MHz)
signals and a very high impedance to low-frequency (i.e., DC) signals.

An inductor is the other way around - very low impedance to low
frequency (DC) signals and very high impedance to high frequency (2m).

Sure, I'm aware of all this basic crap, but every reply I've had to
this question has thrown up a different answer and none of them make
much sense. Can't you guys come up with something in common that adds
up?
Thanks,

Steve
--

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

On Sat, 09 Oct 2004 22:40:20 GMT, Steve Evans
wrote:

On Sun, 03 Oct 2004 23:25:05 GMT, Doug Smith W9WI
wrote:

A capacitor has a very low impedance to high-frequency (i.e., 145MHz)
signals and a very high impedance to low-frequency (i.e., DC) signals.

An inductor is the other way around - very low impedance to low
frequency (DC) signals and very high impedance to high frequency (2m).

Sure, I'm aware of all this basic crap, but every reply I've had to
this question has thrown up a different answer and none of them make
much sense. Can't you guys come up with something in common that adds
up?

Perhaps if you posted the proper schematic it would help. What's the
transistor in question?
--

"What is now proved was once only imagin'd." - William Blake, 1793.

On 10 Oct 2004 00:26:35 GMT, (Michael Black)
wrote:

Then rephrase the question.

Are you wondering why the base of the transistor is at DC ground, or
wondering what the RF choke does to ensure that the base of the transistor is
not at ground for RF?

People have answered what they think you are asking, which isn't really clear.
Clarify that, and you might get an answer you want.

Thanks Michael,
I'm sorry I can't post the full diagram as my scanner's bust. I'll try
to clarify. I know why the base of the transistor (it's a 2n5771, in
answer to Paul's question) is at DC ground. I'm not so dumb as to
realise that a choke passes dc but not the high frequncy RF. I guess
it boils down to this: how did the designer arrive at the given value
of 0.4uH for this inductor? Why not 4nH? Why not 40uH? Why not 100mH??
What's the deal with this value and since I've only got a 1uH in my
junk box, will that be okay instead?

Thanks, all,

Steve.
--

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

On Sat, 09 Oct 2004 22:40:20 GMT, Steve Evans
wrote:

On Sun, 03 Oct 2004 23:25:05 GMT, Doug Smith W9WI
wrote:

A capacitor has a very low impedance to high-frequency (i.e., 145MHz)
signals and a very high impedance to low-frequency (i.e., DC) signals.

An inductor is the other way around - very low impedance to low
frequency (DC) signals and very high impedance to high frequency (2m).

Sure, I'm aware of all this basic crap, but every reply I've had to
this question has thrown up a different answer and none of them make
much sense. Can't you guys come up with something in common that adds
up?

it looks /IIRC schematics posted/ that that is a some kind of RF power
amp in "C" class ....

that inductor keeps shorted transistor base for DC to ground &
capacitor keeps away DC from previous stage.

Also that CR circuit is an impedance transformer (from previous higher
to transistor lower RF Z input & also a high pass filter @ the same
time.

If enough RF current would pass capacitor to rise the potential over
few hundred mV RF, every half-period of the signal will open
transistors E - C gate to conduct thru (parallell) LC tank that
should be connected there from C transistor terminal to power line.

LC tank will recover after one period also other (opposite)
semiperiod, so you will have a complete sinusoide (RF) to transmit (or
amplify it further).. usefull only for CW, modul.CW or FM (not AM or
SSB).

Hope I helped & understood the schematics & question.
--
Regards, SPAJKY ®
& visit my site @ http://www.spajky.vze.com
"Tualatin OC-ed / BX-Slot1 / inaudible setup!"
E-mail AntiSpam: remove ##

Steve Evans wrote:
. . .
I'm not so dumb as to
realise that a choke passes dc but not the high frequncy RF. I guess
it boils down to this: how did the designer arrive at the given value
of 0.4uH for this inductor? Why not 4nH? Why not 40uH? Why not 100mH??
What's the deal with this value and since I've only got a 1uH in my
junk box, will that be okay instead?

I'm guessing you're not experienced enough to realize that inductors are
far from ideal. Among their many imperfections, the most problematic in
an application like this is shunt capacitance. This resonates with the
inductance to create a parallel resonant circuit. At resonance, the
impedance is very high -- higher than that of the inductance alone. But
above the self-resonant frequency, the impedance drops, and at some
point becomes less, then very much less, than the impedance of just the
inductance. Well above self-resonance, all you see is the self
capacitance -- it looks like a capacitor, not an inductor.

That's why a designer doesn't just use 100 mH for everything. Just about
any 100 mH inductor looks like a capacitor at 145 MHz, with a very low
impedance, much lower than an inductor with smaller inductance value.(*)
The trick is to choose an inductor that's below or at its self resonant
point while still having enough impedance so it doesn't disturb the
circuit it's across. A good rule of thumb is an inductor whose reactance
is about 5 - 10 times the impedance it's across. (In your case, this
might not be easy to determine. You might be able to get it either from
knowing someting about the previous stage, or from the S parameter
specfication of the transistor.) 5 - 10 times is usually enough, and if
you try for too much, it won't work any better and you run the risk of
being above the self-resonant frequency. Of course, an individual
inductor can be measured to make sure its impedance is high enough at
the frequency of use, if you have the equipment to make the measurement.

So, now, is your 1 uH ok? It depends on its shunt C, which depends on
its construction. If you can't measure it, just try it. The worst that's
likely to happen is that it'll kill the signal (due to low impedance).
At only 2-1/2 times the value of the original, there's a good chance
it'll work ok. If it doesn't, and if your 1 uH inductor isn't potted,
you can get the value down to 0.4 uH by unwinding about 1/3 of the turns.

(*) Even this explanation is highly simplified. At higher frequencies,
the inductor will exhibit additional series and parallel resonances due
to various parasitic capacitances and inductances and transmission line
effects.

Roy Lewallen, W7EL

On Sun, 10 Oct 2004 11:57:58 GMT, Steve Evans
wrote:


Thanks Michael,
I'm sorry I can't post the full diagram as my scanner's bust. I'll try
to clarify. I know why the base of the transistor (it's a 2n5771, in
answer to Paul's question) is at DC ground. I'm not so dumb as to
realise that a choke passes dc but not the high frequncy RF. I guess
it boils down to this: how did the designer arrive at the given value
of 0.4uH for this inductor? Why not 4nH? Why not 40uH? Why not 100mH??
What's the deal with this value and since I've only got a 1uH in my
junk box, will that be okay instead?

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...
--

"What is now proved was once only imagin'd." - William Blake, 1793.

On Sat, 09 Oct 2004 22:40:20 GMT, Steve Evans
wrote:

On Sun, 03 Oct 2004 23:25:05 GMT, Doug Smith W9WI
wrote:

A capacitor has a very low impedance to high-frequency (i.e., 145MHz)
signals and a very high impedance to low-frequency (i.e., DC) signals.

An inductor is the other way around - very low impedance to low
frequency (DC) signals and very high impedance to high frequency (2m).

Sure, I'm aware of all this basic crap, but every reply I've had to
this question has thrown up a different answer and none of them make
much sense. Can't you guys come up with something in common that adds
up?


The problem is, there are quite a few possibilities for what the inductor
is doing. For example, it might be providing a DC path to ground for the
base of the transistor, or it might be supressing a parasitic oscillation in
the amplifier.

Or, for that matter, it might be part of an impedance transform from one
section to another. Without the complete schematic, there is no way to
tell.

Jim
N6BIU

On Sun, 10 Oct 2004 14:01:44 +0100, Paul Burridge
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

--

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

On Sun, 10 Oct 2004 05:41:23 -0700, Roy Lewallen
wrote:

[snip]

(*) Even this explanation is highly simplified. At higher frequencies,
the inductor will exhibit additional series and parallel resonances due
to various parasitic capacitances and inductances and transmission line
effects.

groan
Dur, nope. I still don't get it. Can anyone explain this stuff *in
simple terms*? I'm really struggling here with some of the
terminology. :-(
--

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

Steve Evans wrote:
On Sun, 10 Oct 2004 14:01:44 +0100, Paul Burridge
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


The transistor acts like a tiny capacitor from base to emitter, in
addition to its other jobs. Thus, an inductor in series with the base
will act like a bandpass filter at

f = 1/(2*pi*sqrt(L*C))

An inductor and capacitor with the same values in parallel will act like
a bandstop filter at that frequency.

Q is the "quality or merit factor" of the inductor, and is computed by
dividing the 'reactance' (or AC resistance) of the inductor by the DC
resistance *at the frequency in question*; it will generally be
different at different frequencies. Q is really the ratio of reactive
power in the inductance to the real power dissipated in the resistance,
and can be computed by using the formula:

Q = 2*PI*f*L/ri

where f is frequency, and ri is the resistance of the coil at f. You buy
inductors with a given Q range.

For an LC resonant circuit, the Q of the inductor will affect the 'Q' of
the resulting resonance, which simply means that with a bigger Q, the
passband or stopband will be wider. The Q of a resonant circuit is
defined as the resonant frequency divided by the width of the passband.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.