# Capacitor + inductor filter - please check

Started by June 2, 2014
```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))

G  = 1/(1 + 1/(j&sup2;w&sup2;CL)) = 1/(1 - 1/(w&sup2;CL)

So for very high w (frequency),  G = 1/(1 - 0) = 1 which looks correct.

But at a certain frequency w&sup2;CL = 1,  G = 1/(1 - 1/1) = infinity. huh?

And for low frequency w&sup2;CL < 1,  G is negative.  180&deg; phase-shift?

Is this anything like correct or am I totally misunderstanding this?

```
```On 2.6.14 15:53, Dave Rove wrote:
> 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))
>
> G  = 1/(1 + 1/(j&sup2;w&sup2;CL)) = 1/(1 - 1/(w&sup2;CL)
>
> So for very high w (frequency),  G = 1/(1 - 0) = 1 which looks correct.
>
> But at a certain frequency w&sup2;CL = 1,  G = 1/(1 - 1/1) = infinity. huh?
>
> And for low frequency w&sup2;CL < 1,  G is negative.  180&deg; phase-shift?
>
> 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
where w&sup2;LC = 1, to get a Q of 100.

--

Tauno Voipio
```
```On Mon, 02 Jun 2014 16:37:13 +0300, Tauno Voipio wrote:

> On 2.6.14 15:53, Dave Rove wrote:
>> 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 where w&sup2;LC = 1, to
> get a Q of 100.

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.
```
```On Monday, June 2, 2014 10:26:05 AM UTC-4, Dave Rove wrote:
> On Mon, 02 Jun 2014 16:37:13 +0300, Tauno Voipio wrote:
>=20
>=20
>=20
> > On 2.6.14 15:53, Dave Rove wrote:
>=20
> >> Is this anything like correct or am I totally misunderstanding this?
>=20
> >=20
>=20
> > That's right. You have found the resonance of an unloaded filter.
>=20
> >=20
>=20
> > For more realistic scenario, put a resistor in parallel to the coil. A
> > suitable value would be 100 * w * L at the frequency where w=B2LC =3D 1=
, to
> > get a Q of 100.
I would have said to put the R in series with the coil. =20
>=20
>=20
> OK, thanks. Googling Q-factor now.
>=20
> So it is right, then. (Obviously only in the ideal case of a pure C and L=
=20
> and no load). I knew that you could form "tuned circuits" from Ls and Cs=
=20
> to act as narrowband filters, although I was hazy about the details, and=
=20
> hadn't realized that you could get an actual voltage gain like a=20
> transformer from two passive components.
Well see if you can calculate the the impedance of your filter when it is=
=20
at resonance (w^2 =3D 1/LC).
(How much current will it draw from the source?)

George H.

```
```On Mon, 02 Jun 2014 16:37:13 +0300, Tauno Voipio
<tauno.voipio@notused.fi.invalid> wrote:

>On 2.6.14 15:53, Dave Rove wrote:
>> 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))
>>
>> G  = 1/(1 + 1/(j&#2013266098;w&#2013266098;CL)) = 1/(1 - 1/(w&#2013266098;CL)
>>
>> So for very high w (frequency),  G = 1/(1 - 0) = 1 which looks correct.
>>
>> But at a certain frequency w&#2013266098;CL = 1,  G = 1/(1 - 1/1) = infinity. huh?
>>
>> And for low frequency w&#2013266098;CL < 1,  G is negative.  180&#2013266096; phase-shift?
>>
>> 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
>where w&#2013266098;LC = 1, to get a Q of 100.

More "realistic would be a resistor _in_series_ with the inductor...
or _in_parallel_ with the capacitor.

...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.
```
```On Mon, 02 Jun 2014 08:11:03 -0700, George Herold wrote:

> On Monday, June 2, 2014 10:26:05 AM UTC-4, Dave Rove wrote:

>> 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.

> 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?)

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.
```
```On Mon, 2 Jun 2014 12:53:13 +0000 (UTC), Dave Rove <daver@dr.invalid> wrote:

>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))
>
>G  = 1/(1 + 1/(j&#2013266098;w&#2013266098;CL)) = 1/(1 - 1/(w&#2013266098;CL)
>
>So for very high w (frequency),  G = 1/(1 - 0) = 1 which looks correct.
>
>But at a certain frequency w&#2013266098;CL = 1,  G = 1/(1 - 1/1) = infinity. huh?
>
>And for low frequency w&#2013266098;CL < 1,  G is negative.  180&#2013266096; phase-shift?
>
>Is this anything like correct or am I totally misunderstanding this?
>
>

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
```
```On Mon, 02 Jun 2014 14:26:05 +0000, Dave Rove wrote:

> On Mon, 02 Jun 2014 16:37:13 +0300, Tauno Voipio wrote:
>
>> On 2.6.14 15:53, Dave Rove wrote:
>>> 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 where w&sup2;LC = 1, to
>> get a Q of 100.
>
> 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.

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

```
```On Mon, 02 Jun 2014 11:10:41 -0500, Tim Wescott
<tim@seemywebsite.really> wrote:

>On Mon, 02 Jun 2014 14:26:05 +0000, Dave Rove wrote:
>
>> On Mon, 02 Jun 2014 16:37:13 +0300, Tauno Voipio wrote:
>>
>>> On 2.6.14 15:53, Dave Rove wrote:
>>>> 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 where w&#2013266098;LC = 1, to
>>> get a Q of 100.
>>
>> 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.
>
>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.

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.
```
```Tauno Voipio wrote:
> On 2.6.14 15:53, Dave Rove wrote:
>> 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))
>>
>> G = 1/(1 + 1/(j&sup2;w&sup2;CL)) = 1/(1 - 1/(w&sup2;CL)
>>
>> So for very high w (frequency), G = 1/(1 - 0) = 1 which looks correct.
>>
>> But at a certain frequency w&sup2;CL = 1, G = 1/(1 - 1/1) = infinity. huh?
>>
>> And for low frequency w&sup2;CL < 1, G is negative. 180&deg; phase-shift?
>>
>> 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
> where w&sup2;LC = 1, to get a Q of 100.
>
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..

```