Series to parallel conversion

*IF* your circuit is high Q you can use a simple approximation, get close. Really better to simply use LTspice.

Try this you have high Q where the series is L and C and R. High Q, XL and XC are large compared to R. For grins let XL=1000 ohms, XC=-1000 ohms and R=10 ohms Q is what? 100,

Instead of having the 10 ohm R in series if you MULTIPLY the R by Q^2 you have the equivalent Q but parallel loss, that would mean R=1MEG

Not 'accurate' but faster than a calculator.

and again USE LTspice!

Q with a paralleled resistor... value of resistor is Q*XL (or Q*XC, same difference, since XL=XC at resonance).

For an already degraded L with a series R, see...

...Jim Thompson

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| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
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| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  | 
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Thanks, As I calculate it, a 75,800 ohm parallel resistor will spoil the Q the same as inserting a series 30 ohm series resistor. Since R2series cancels out, then it would seem that this is good for any Q. Does the error get worse at higher Q or lower, and is the difference worth any concern. Say Q's from 100 to 1000. Thanks again, Mikek

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You would generally be OK. All I would worry about is the slight shift in the resonant peak. ...Jim Thompson

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| 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.

I get 76190, but I rounded to 105pF.

As long as the circuit is fairly high Q, if Xt/Rp = Rs/Xt, with Xt being the coil & cap impedance, Rp being parallel R, and Rs being series R, Rp and Rs will have pretty much the same effect at resonance.

Assuming that I have my math right, and didn't just hand it all to you upside down or something.

You must be starting out with something pretty high Q if you're knocking it down by that slight amount. If I were working with an RF circuit I'd be looking at ways that my loading resistance could be improving my circuit somehow, by increasing coupling to a following amplifier, or by being part of an attenuator that feeds the following amplifier or some such.

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Given the numbers he's citing, he's got to be tweaking a cap or coil in production anyway.

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No, just playing with AM broadcast band ferrite antennas. Yes, easy tweak of the variable cap. Wanted to make an easy experiment, easier clip in a parallel resistor then to cut and install a series resistor. Thanks, Mikek

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Not really a fairly low Q for a resonant AM ferrite rod antenna. Figuring Q = 150 @ 1MHz, Loss resistor is about 10 ohms. Adding a series 30 ohm or a parallel 75,800 ohm, the Q should drop to about 38. The crystal radio guys now have ferrite rod material to get Q's of

1000 to 1200 across the whole broadcast band. Mikek PS. If you want the whole story, send me an email. My address is good.

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I can't get this online calculator to output numbers that make sense with the formula R2p = L/R2s*C

240uH 105.54pf 30 ohms

It may be that it's a slightly different problem in that I have an existing 10 ohms and I'm adding 30 ohms.

Mikek

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Try calculating for 10 ohms and then for 40, and see if the difference in parallel equivalent resistance doesn't work out to the 10-ohm answer in parallel to your 75800-ohm resistor.

You know, you can just do this arithmetically with complex numbers -- just express the inductive and capacitive impedances as imaginary numbers, draw up your circuit, then (assuming that coil and cap are grounded) solve for the admittance at the junction of coil and cap -- the real part of the admittance will be the conductance of your parallel resistor.

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If I could just do this arithmetically with complex numbers, I wouldn't be here, asking for help from the adults. I appreciate that everyone pitches in and I usually get what I need to forge ahead. I'm algebra challenged, but then to make me think in admittance and conductance... I'll retire in a few years, if I get bored, I may take some algebra and trig classes, to make my electronic dabbling go a little smoother. I'll try the 10 ohm 30 ohm idea. I just went and tried the online calculator, I'm probably missing something, but it does not like my h=numbers that are very near resonance. 1MHz 240uH 105.54pf 30 ohms

Thanks, Mikek

Pursue the Heaviside approach to Laplace... jOmega = s...

Impedance of an inductor = Ls

Impedance of a capacitor = 1/(Cs)

R = R

That'll help you understand what I posted. ...Jim Thompson

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| 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.