If it's series LC doesn't the impedence drop to zero?

Cheers

Wally wrote:

Nav wrote:


Can you explain this. I though that for a parallel LC the impedence
went to infinite at resonance (f=1/2pi sqrt(LC)) so that Vin=Vout.

What am I missing?


Getting beyond my knowledge (or memory), but Gilly did say that the LC
circut is series - would that work better?


--
Wally
www.artbywally.com
www.wally.myby.co.uk/music

Nav wrote:

If it's series LC doesn't the impedence drop to zero?

Just had a play with a spreadsheet - there's a dip in Z in series, and a
rise with parallel.


--
Wally
www.artbywally.com
www.wally.myby.co.uk/music

That's what I thought. How can a low impedence increase voltage? Can you
explain Gilly's idea to me?

Cheers

Wally wrote:

Nav wrote:


If it's series LC doesn't the impedence drop to zero?


Just had a play with a spreadsheet - there's a dip in Z in series, and a
rise with parallel.


--
Wally
www.artbywally.com
www.wally.myby.co.uk/music

Nav wrote:

That's what I thought. How can a low impedence increase voltage? Can
you explain Gilly's idea to me?

'Fraid not, but I'll have a guess... :-)

With a parallel cct, Z either side of resonance is low, which would develop
a lower voltage across the two components. The series cct presents high Z
either side of resonance. IOW, the high voltage developed across each
component in the series cct is outside the tuned frequency - at low F on one
component, high F on the other.

Alternatively, and I really am guessing here - how about something to do
with phase effects at resonance? All that leading and lagging of current WRT
voltage.


--
Wally
www.artbywally.com
www.wally.myby.co.uk/music

Wally wrote:

Nav wrote:


That's what I thought. How can a low impedence increase voltage? Can
you explain Gilly's idea to me?


'Fraid not, but I'll have a guess... :-)

With a parallel cct, Z either side of resonance is low, which would develop
a lower voltage across the two components. The series cct presents high Z
either side of resonance. IOW, the high voltage developed across each
component in the series cct is outside the tuned frequency - at low F on one
component, high F on the other.

Alternatively, and I really am guessing here - how about something to do
with phase effects at resonance? All that leading and lagging of current WRT
voltage.


As far as I understand it, the voltage is just the current times the
impedence. Similarly, the current is only produced by the driver (Vin)
so how can Vout be bigger than Vin (there's no mention of a transformer)?

Cheers

The voltage across the inductor is 180 degrees out of phase with the voltage
across the capacitor.

A parallel LC resonator has a circulating current of the circuit Q times the
applied current. The components better be rated for high current.

These concepts are very basic radio theory, the type of questions asked on
the amateur radio exam.

In practice a pi or t type network may be better for tuning the ground
because it would have wider bandwidth than a series LC circuit.

It is beneficial to have the ground at a low impedance, that's why series LC
is used.


"Nav" wrote in message
...


Wally wrote:

Nav wrote:


That's what I thought. How can a low impedence increase voltage? Can
you explain Gilly's idea to me?


'Fraid not, but I'll have a guess... :-)

With a parallel cct, Z either side of resonance is low, which would
develop
a lower voltage across the two components. The series cct presents high
Z
either side of resonance. IOW, the high voltage developed across each
component in the series cct is outside the tuned frequency - at low F on
one
component, high F on the other.

Alternatively, and I really am guessing here - how about something to do
with phase effects at resonance? All that leading and lagging of current
WRT
voltage.


As far as I understand it, the voltage is just the current times the
impedence. Similarly, the current is only produced by the driver (Vin)
so how can Vout be bigger than Vin (there's no mention of a transformer)?

Cheers

Gay Sailor wrote:

The voltage across the inductor is 180 degrees out of phase with the voltage
across the capacitor.

A parallel LC resonator has a circulating current of the circuit Q times the
applied current. The components better be rated for high current.

So what's the voltage across the resonator?

Cheers

Depends on the circuit driving it.



"Nav" wrote in message
...


Gay Sailor wrote:

The voltage across the inductor is 180 degrees out of phase with the
voltage
across the capacitor.

A parallel LC resonator has a circulating current of the circuit Q times
the
applied current. The components better be rated for high current.

So what's the voltage across the resonator?

Cheers