Series to parallel conversion

amdx · Mar 2, 2015

I want to spoil the Q of an LC circuit with a parallel resistor.
I don't know what information is required.
Can a I just say I want the equivalent of a series 30 ohm resistor
in parallel to the circuit?
Or is the circuit info required?
Circuit; 240uh parallel 105.54pf (resonance) at 1 MHz with 10 ohms loss.
I want to add 30 ohms in series, but I want to use the equivalent of a
parallel resistance.
I tried a calculator, but things got strange at resonance.

Thanks, Mikek

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RobertMacy · Mar 3, 2015

On Mon, 02 Mar 2015 09:47:50 -0700, amdx <nojunk@knology.net> wrote:

I want to spoil the Q of an LC circuit with a parallel resistor.
I don't know what information is required.
Can a I just say I want the equivalent of a series 30 ohm resistor
in parallel to the circuit?
Or is the circuit info required?
Circuit; 240uh parallel 105.54pf (resonance) at 1 MHz with 10 ohms loss.
I want to add 30 ohms in series, but I want to use the equivalent of a
parallel resistance.
I tried a calculator, but things got strange at resonance.

Thanks, Mikek

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*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!

Jim Thompson · Mar 3, 2015

On Mon, 02 Mar 2015 10:47:50 -0600, amdx <nojunk@knology.net> wrote:

I want to spoil the Q of an LC circuit with a parallel resistor.
I don't know what information is required.
Can a I just say I want the equivalent of a series 30 ohm resistor
in parallel to the circuit?
Or is the circuit info required?
Circuit; 240uh parallel 105.54pf (resonance) at 1 MHz with 10 ohms loss.
I want to add 30 ohms in series, but I want to use the equivalent of a
parallel resistance.
I tried a calculator, but things got strange at resonance.

Thanks, Mikek

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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...

<http://www.analog-innovations.com/SED/Q-Equivalent_Series_Parallel.pdf>

...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

amdx · Mar 3, 2015

On 3/2/2015 12:53 PM, Jim Thompson wrote:

On Mon, 02 Mar 2015 10:47:50 -0600, amdx <nojunk@knology.net> wrote:

I want to spoil the Q of an LC circuit with a parallel resistor.
I don't know what information is required.
Can a I just say I want the equivalent of a series 30 ohm resistor
in parallel to the circuit?
Or is the circuit info required?
Circuit; 240uh parallel 105.54pf (resonance) at 1 MHz with 10 ohms loss.
I want to add 30 ohms in series, but I want to use the equivalent of a
parallel resistance.
I tried a calculator, but things got strange at resonance.

Thanks, Mikek

---
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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...

http://www.analog-innovations.com/SED/Q-Equivalent_Series_Parallel.pdf

...Jim Thompson

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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Jim Thompson · Mar 3, 2015

On Mon, 02 Mar 2015 13:33:43 -0600, amdx <nojunk@knology.net> wrote:

On 3/2/2015 12:53 PM, Jim Thompson wrote:
On Mon, 02 Mar 2015 10:47:50 -0600, amdx <nojunk@knology.net> wrote:

I want to spoil the Q of an LC circuit with a parallel resistor.
I don't know what information is required.
Can a I just say I want the equivalent of a series 30 ohm resistor
in parallel to the circuit?
Or is the circuit info required?
Circuit; 240uh parallel 105.54pf (resonance) at 1 MHz with 10 ohms loss.
I want to add 30 ohms in series, but I want to use the equivalent of a
parallel resistance.
I tried a calculator, but things got strange at resonance.

Thanks, Mikek

---
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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...

http://www.analog-innovations.com/SED/Q-Equivalent_Series_Parallel.pdf

...Jim Thompson

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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Tim Wescott · Mar 3, 2015

On Mon, 02 Mar 2015 13:33:43 -0600, amdx wrote:

On 3/2/2015 12:53 PM, Jim Thompson wrote:
On Mon, 02 Mar 2015 10:47:50 -0600, amdx <nojunk@knology.net> wrote:

I want to spoil the Q of an LC circuit with a parallel resistor.
I don't know what information is required.
Can a I just say I want the equivalent of a series 30 ohm resistor in
parallel to the circuit?
Or is the circuit info required?
Circuit; 240uh parallel 105.54pf (resonance) at 1 MHz with 10 ohms
loss.
I want to add 30 ohms in series, but I want to use the equivalent of a
parallel resistance.
I tried a calculator, but things got strange at resonance.

Thanks, Mikek

---
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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...

http://www.analog-innovations.com/SED/Q-
Equivalent_Series_Parallel.pdf

...Jim Thompson

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

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.

--
www.wescottdesign.com

Tim Wescott · Mar 3, 2015

On Mon, 02 Mar 2015 12:49:07 -0700, Jim Thompson wrote:

On Mon, 02 Mar 2015 13:33:43 -0600, amdx <nojunk@knology.net> wrote:

On 3/2/2015 12:53 PM, Jim Thompson wrote:
On Mon, 02 Mar 2015 10:47:50 -0600, amdx <nojunk@knology.net> wrote:

I want to spoil the Q of an LC circuit with a parallel resistor.
I don't know what information is required.
Can a I just say I want the equivalent of a series 30 ohm resistor in
parallel to the circuit?
Or is the circuit info required?
Circuit; 240uh parallel 105.54pf (resonance) at 1 MHz with 10 ohms
loss.
I want to add 30 ohms in series, but I want to use the equivalent of
a parallel resistance.
I tried a calculator, but things got strange at resonance.

Thanks, Mikek

---
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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...

http://www.analog-innovations.com/SED/Q-
Equivalent_Series_Parallel.pdf

...Jim Thompson

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.

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

--
www.wescottdesign.com

amdx · Mar 3, 2015

On 3/2/2015 3:07 PM, Tim Wescott wrote:

On Mon, 02 Mar 2015 12:49:07 -0700, Jim Thompson wrote:

On Mon, 02 Mar 2015 13:33:43 -0600, amdx <nojunk@knology.net> wrote:

On 3/2/2015 12:53 PM, Jim Thompson wrote:
On Mon, 02 Mar 2015 10:47:50 -0600, amdx <nojunk@knology.net> wrote:

I want to spoil the Q of an LC circuit with a parallel resistor.
I don't know what information is required.
Can a I just say I want the equivalent of a series 30 ohm resistor in
parallel to the circuit?
Or is the circuit info required?
Circuit; 240uh parallel 105.54pf (resonance) at 1 MHz with 10 ohms
loss.
I want to add 30 ohms in series, but I want to use the equivalent of
a parallel resistance.
I tried a calculator, but things got strange at resonance.

Thanks, Mikek

---
This email has been checked for viruses by Avast antivirus software.
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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...

http://www.analog-innovations.com/SED/Q-
Equivalent_Series_Parallel.pdf

...Jim Thompson

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.

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

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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amdx · Mar 3, 2015

On 3/2/2015 3:03 PM, Tim Wescott wrote:

On Mon, 02 Mar 2015 13:33:43 -0600, amdx wrote:

On 3/2/2015 12:53 PM, Jim Thompson wrote:
On Mon, 02 Mar 2015 10:47:50 -0600, amdx <nojunk@knology.net> wrote:

I want to spoil the Q of an LC circuit with a parallel resistor.
I don't know what information is required.
Can a I just say I want the equivalent of a series 30 ohm resistor in
parallel to the circuit?
Or is the circuit info required?
Circuit; 240uh parallel 105.54pf (resonance) at 1 MHz with 10 ohms
loss.
I want to add 30 ohms in series, but I want to use the equivalent of a
parallel resistance.
I tried a calculator, but things got strange at resonance.

Thanks, Mikek

---
This email has been checked for viruses by Avast antivirus software.
http://www.avast.com

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...

http://www.analog-innovations.com/SED/Q-
Equivalent_Series_Parallel.pdf

...Jim Thompson

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

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.

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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amdx · Mar 3, 2015

On 3/2/2015 4:13 PM, amdx wrote:

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

http://www.daycounter.com/Calculators/Parallel-Series-Imedance-Conversion-Calculator.phtml

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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Tim Wescott · Mar 3, 2015

On Mon, 02 Mar 2015 16:22:27 -0600, amdx wrote:

On 3/2/2015 4:13 PM, amdx wrote:

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

http://www.daycounter.com/Calculators/Parallel-Series-Imedance-
Conversion-Calculator.phtml

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.

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.

--
www.wescottdesign.com

amdx · Mar 3, 2015

On 3/3/2015 1:18 AM, Tim Wescott wrote:

On Mon, 02 Mar 2015 16:22:27 -0600, amdx wrote:

On 3/2/2015 4:13 PM, amdx wrote:

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

http://www.daycounter.com/Calculators/Parallel-Series-Imedance-
Conversion-Calculator.phtml

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.

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.

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

Jim Thompson · Mar 3, 2015

On Tue, 03 Mar 2015 08:03:11 -0600, amdx <nojunk@knology.net> wrote:

On 3/3/2015 1:18 AM, Tim Wescott wrote:
On Mon, 02 Mar 2015 16:22:27 -0600, amdx wrote:

On 3/2/2015 4:13 PM, amdx wrote:

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

http://www.daycounter.com/Calculators/Parallel-Series-Imedance-
Conversion-Calculator.phtml

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.

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.

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

Series to parallel conversion

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