I've been playing with oscillators a bit here, and I wanted to ask if the way
I'm intuitively thinking about the operation of a tapped capacitor tank is the
way other people do:

Say we have an inductor L in parallel with tapped capacitors, C1 and C2. If
the loaded Q of the tank is reasonably high (say, 10 -- which you'd want for
any decent oscillator anyway), we can pretty much ignore the "input" and
"output" of the tank, as their effect on what happens within the tank isn't
significant. As such, we can just look at the current circulating around the
tank, and from the "top" of the tank to the "bottom" you get 90 degrees phase
shift going from C1 to ground and then another 90 degrees from ground through
C2. Hence, from the top to the bottom you have 180 degrees of phase shift,
with the values (ratios) of C1/C2 just setting the magnitude of the output
voltage vs. the input. Sound good?

I started going through the math involved and determined that you can readily
confuse yourself if you don't make the "high Q" assumption. For instance,
something like a 1 ohm source and 1 ohm load feeding a tank of 159.2mH (j1
ohms) and 2 318.3mF caps (-j/2 ohms) has such a horribly low Q that there
isn't a 180 degree phase shift top to bottom, nor is the resonant frequency
1Hz.

Thanks,
---Joel Kolstad

Joel Kolstad wrote:
I've been playing with oscillators a bit here, and I wanted to ask if the way
I'm intuitively thinking about the operation of a tapped capacitor tank is the
way other people do:

Say we have an inductor L in parallel with tapped capacitors, C1 and C2. If
the loaded Q of the tank is reasonably high (say, 10 -- which you'd want for
any decent oscillator anyway), we can pretty much ignore the "input" and
"output" of the tank, as their effect on what happens within the tank isn't
significant. As such, we can just look at the current circulating around the
tank, and from the "top" of the tank to the "bottom" you get 90 degrees phase
shift going from C1 to ground and then another 90 degrees from ground through
C2. Hence, from the top to the bottom you have 180 degrees of phase shift,
with the values (ratios) of C1/C2 just setting the magnitude of the output
voltage vs. the input. Sound good?

Sorry, no. EH antenna inventors and apologists notwithstanding, the
phase of current doesn't change as it goes through a capacitor.

. . .

Roy Lewallen, W7EL

Hi Roy,

"Roy Lewallen" wrote in message
...
Sorry, no. EH antenna inventors and apologists notwithstanding, the phase of
current doesn't change as it goes through a capacitor.

Agreed. I'm being rather loose in my terminology here, what I mean by "you
get 90 degrees of phase shift going through the capacitor" is that the voltage
across the capacitor ends up (must be) 90 degrees out of phase with respect to
the reference current as you consider the capacitor's effect.

You don't disagree that the voltage at the top of a tapped capacitor L-C tank
is 180 degrees out of phase what that at the bottom at resonance, do you?

---Joel

On Nov 20, 4:36 pm, "Joel Kolstad"
wrote:
Hi Roy,

"Roy Lewallen" wrote in ...

Sorry, no. EH antenna inventors and apologists notwithstanding, the phase of
current doesn't change as it goes through a capacitor.Agreed. I'm being rather loose in my terminology here, what I mean by "you
get 90 degrees of phase shift going through the capacitor" is that the voltage
across the capacitor ends up (must be) 90 degrees out of phase with respect to
the reference current as you consider the capacitor's effect.

You don't disagree that the voltage at the top of a tapped capacitor L-C tank
is 180 degrees out of phase what that at the bottom at resonance, do you?

---Joel

Joel, what do you mean by "at the top" and "at the bottom"? Voltage
measured with respect to what point? Clearly, if there is the same
current through two capacitors in series, the voltages across each of
them must be in phase, so the total voltage across the two is the
simple arithmetic sum of the voltage across each. That assumes that
they are ideal capacitors, but practical ones will come close. On the
other hand, the currents in the two may not be exactly in phase, as you
perhaps discovered with your low-Q tank. If the current at the node
between the capacitors is contributed to by some external load or
source, and if that additional current is not small compared with the
tank's circulating current, then the capacitor currents may not be in
phase, and the voltages then would also not be in phase, and you'd need
to add them vectorially to get the right result.

Of course, Spice will simulate all this for you very nicely, using
either a transient analysis, or a frequency-domain analysis if all the
parts can be assumed to behave linearly.

Cheers,
Tom

K7ITM wrote:
Of course, Spice will simulate all this for you very nicely, using
either a transient analysis, or a frequency-domain analysis if all the
parts can be assumed to behave linearly.

Cheers,
Tom

Andy writes
I'll second that. J , try your ideas out on SPICE. Try to figure
out
what it will do before you simulate it, and then use the simulation to
check your conclusions..... I've found it to be invaluable in
straightening
me out on some of the ways I "thought" nature worked, and in getting
a lot of things straight.....

Same with EZNEC, if you have it....... I learned a lot more about
how
antennae really work from EZNEC than in 30 years of cutting wires and
hooking up SWR meters......

Simulators are wonderful things. And no one has to know about the
times one turns out to be wrr...wrrro....wr.r...........
imprecise....

Andy W4OAH

"Joel Kolstad" wrote in message
...
[...]

I started going through the math involved and determined that you can
readily
confuse yourself if you don't make the "high Q" assumption. For instance,
something like a 1 ohm source and 1 ohm load feeding a tank of 159.2mH (j1
ohms) and 2 318.3mF caps (-j/2 ohms) has such a horribly low Q that there
isn't a 180 degree phase shift top to bottom, nor is the resonant
frequency
1Hz.

Thanks,
---Joel Kolstad



Made an inductance meter based on a resonant LC circuit and used a PIC to
back calculate the inductance and Q values based on known F and C and
currents. Circuit was able to work down to Q values of 1 at frequencies
down to 5Hz. Worked well.
Did the programming and found it just wouldn't calculate the correct L
values when working at low Q. Spent hours doing tests until it dawned on me
that I'd stupidly programmed in the "standard" resonance formula.
Suddenly became sweetness and light when I reprogrammed for ...
Frequency= 1/2*pi * sqrt(1/LC -R^2/L^2).
I continue to live and learn
john



--
Posted via a free Usenet account from http://www.teranews.com

Quite a stutter you have there... However my wife uses that word
frequently while, like you, I can't quite agree with her...

denny / k8do
AndyS wrote:

Simulators are wonderful things. And no one has to know about the
times one turns out to be wrr...wrrro....wr.r...........
imprecise....

Andy W4OAH

Denny wrote:
Quite a stutter you have there... However my wife uses that word
frequently while, like you, I can't quite agree with her...

denny / k8do
AndyS wrote:

Simulators are wonderful things. And no one has to know about the
times one turns out to be wrr...wrrro....wr.r...........
imprecise....

Andy W4OAH

Denny wrote:
Quite a stutter you have there... However my wife uses that word
frequently while, like you, I can't quite agree with her...

denny / k8do


Andy comments:

My wife has no problem with saying "YOU ARE WRONG !!""
Her speech has always been flawless...

Personally, I think that overuse of a particular description shows
a character flaw.... But I love her.... and keep my mouth shut....

Andy in Eureka

Hi Tom,

Joel, what do you mean by "at the top" and "at the bottom"? Voltage
measured with respect to what point?

Here's a pictu
http://pg.photos.yahoo.com/ph/jkolst...cd.jpg&.src=ph

I don't believe I was very clear about the capacitor's tap pointed being
grounded. The idea is that the voltage at the top of that tank is 180 degrees
out of phase from that at the bottom using the explanation I posted (as you
can see, that picture comes from SPICE, so I know that actually is the case, I
just wanted to insure my explanation was correct).

---Joel