I'm learning how you use complex numbers to model reactance, and trying out a few examples.
So if I string a capacitor between the input and output, and put an inductor between the output and the ground like this:
Vin ---||--------Vout | ) ) ) |
------------------
Then the gain G = Vout/Vin = Xl/(Xl+Xc) = jwL/(jwL+1/(jwC))
So for very high w (frequency), G = 1/(1 - 0) = 1 which looks correct.
Is this anything like correct or am I totally misunderstanding this?
That's right. You have found the resonance of an unloaded filter.
For more realistic scenario, put a resistor in parallel to the coil. A suitable value would be 100 * w * L at the frequency
-- Tauno Voipio
OK, thanks. Googling Q-factor now.
So it is right, then. (Obviously only in the ideal case of a pure C and L and no load). I knew that you could form "tuned circuits" from Ls and Cs to act as narrowband filters, although I was hazy about the details, and hadn't realized that you could get an actual voltage gain like a transformer from two passive components.
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I would have said to put the R in series with the coil.
Well see if you can calculate the the impedance of your filter when it is at resonance (w^2 = 1/LC). (How much current will it draw from the source?)
George H.
More "realistic would be a resistor _in_series_ with the inductor... or _in_parallel_ with the capacitor. ...Jim Thompson
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| James E.Thompson | mens |
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I love to cook with wine. Sometimes I even put it in the food.
The series impedance being the denominator of that gain expression, zero, which I'd have guessed from analogy with how a high-gain transformer behaves, so infinite current. :)
I've actually impressed myself with this, figuring something about components that I didn't know before, just from trying out the maths.
That all sounds right. At series resonance, net LC impedance is zero, so any applied sine wave voltage makes infinite current hence infinite voltage across the L. It's one of those limiting-singularity situations.
In real life, you can get a practical 20 to maybe 100:1 voltage step-up, given a very light load.
-- John Larkin Highland Technology Inc www.highlandtechnology.com jlarkin at highlandtechnology dot com Precision electronic instrumentation
Yup. It's only good for a narrow band of frequencies, but within that band you can use it for impedance transformations up and down, and it's tunable to the components at hand.
If you look at vacuum-tube circuits from the 1920's through the 1960's, you'll see tunable resonant circuits between stages, that can be used for all sorts of things.
-- Tim Wescott Wescott Design Services http://www.wescottdesign.com
A lost art... I don't think they teach circuit design/analysis anymore in colleges... particularly passive circuits... just "IT" :-( ...Jim Thompson
--
| James E.Thompson | mens |
| Analog Innovations | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| San Tan Valley, AZ 85142 Skype: skypeanalog | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |
I love to cook with wine. Sometimes I even put it in the food.
NO; a realistic model would have a resistor in SERIES with the inductor to model its loss, and maybe a resistor in parallel with the capacitor to model its loss. Add in lead inductance and resistance as appropriate..
Transformers ARE passive..
CHECK!
Q from the peanut gallery: what is the practical unloaded voltage gain with a Q of 2,000?
-- Depends on the frequency.
Wait, wait, don't tell me...... 2000!
-- John Larkin Highland Technology Inc www.highlandtechnology.com jlarkin at highlandtechnology dot com Precision electronic instrumentation
"John Larkin" wrote in message news: snipped-for-privacy@4ax.com...
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