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From: Ian Bell on 12 Jan 2010 05:22
Suppose you have a simple pot divider consisting of two equal value
resistors, say 10K each. Across the upper one you connect a series LC
circuit that resonates at 3KHz or thereabouts. If you drive this network
from a low impedance source and plot the response across the bottom
resistor, the Q of the resulting peak is not the Q of the series LC but
rather is determined by the pair of LC values. For example, choosing
l=150mH and C=18nF gives a Q of just over 2. Choosing L= 1.5H and
C=1.8nF gives a much higher Q.

What I need is a simple means of calculating L and C given the pot
divider resistor value and desired Q and f (assuming the Q of the LC
itself is much higher).

Cheers

Ian
From: Wimpie on 12 Jan 2010 08:51
On 12 ene, 11:22, Ian Bell <ruffreco...(a)yahoo.com> wrote:
> Suppose you have a simple pot divider consisting of two equal value
> resistors, say 10K each. Across the upper one you connect a series LC
> circuit that resonates at 3KHz or thereabouts. If you drive this network
> from a low impedance source and plot the response across the bottom
> resistor, the Q of the resulting peak is not the Q of the series LC but
> rather is determined by the pair of LC values. For example, choosing
> l=150mH and C=18nF gives a Q of just over 2. Choosing L= 1.5H and
> C=1.8nF gives a much higher Q.
>
> What I need is a simple means of calculating L and C given the pot
> divider resistor value and desired Q and f (assuming the Q of the LC
> itself is much higher).
>
> Cheers
>
> Ian

Hello Ian,

The LC series circuit "sees" the upper and lower resistor in parallel.
This means when Rupper or Rlower (or both) are zero, the Q factor will
be infinite (assuming lossless L and C).

So the Q factor of the loaded LCcircuit will be: Q = 2*pi*f*L/(Rp) Rp
= Rupp parallel to Rlow (that is Rp = 0.5*Rlow, when Rlow = Rupper).

However this Q factor does not directly relate to the frequency
response. So the -3dB BW of your circuit is not equal to BW = (center
frequency)/Q. This can be seen when Rupper = 0 and Rupper > 0 (all-
pass filter).

Given your circuit, when you double the value of L (and halve the
value of C to enable same resonant frequency), the width of the peak
halves.

Best regards,

Wim
PA3DJS
www.tetech.nl
the email address is OK, but without abc
From: Ian Bell on 12 Jan 2010 10:26
Wimpie wrote:
> On 12 ene, 11:22, Ian Bell <ruffreco...(a)yahoo.com> wrote:
>> Suppose you have a simple pot divider consisting of two equal value
>> resistors, say 10K each. Across the upper one you connect a series LC
>> circuit that resonates at 3KHz or thereabouts. If you drive this network
>> from a low impedance source and plot the response across the bottom
>> resistor, the Q of the resulting peak is not the Q of the series LC but
>> rather is determined by the pair of LC values. For example, choosing
>> l=150mH and C=18nF gives a Q of just over 2. Choosing L= 1.5H and
>> C=1.8nF gives a much higher Q.
>>
>> What I need is a simple means of calculating L and C given the pot
>> divider resistor value and desired Q and f (assuming the Q of the LC
>> itself is much higher).
>>
>> Cheers
>>
>> Ian
>
> Hello Ian,
>
> The LC series circuit "sees" the upper and lower resistor in parallel.
> This means when Rupper or Rlower (or both) are zero, the Q factor will
> be infinite (assuming lossless L and C).
>
> So the Q factor of the loaded LCcircuit will be: Q = 2*pi*f*L/(Rp) Rp
> = Rupp parallel to Rlow (that is Rp = 0.5*Rlow, when Rlow = Rupper).
>

Yes, that was my first thought but that still gives the wrong Q value.


> However this Q factor does not directly relate to the frequency
> response. So the -3dB BW of your circuit is not equal to BW = (center
> frequency)/Q. This can be seen when Rupper = 0 and Rupper > 0 (all-
> pass filter).
>

I think you meant the second to be Rlower, but I understand what you are
saying.

> Given your circuit, when you double the value of L (and halve the
> value of C to enable same resonant frequency), the width of the peak
> halves.
>

Yes, I already realised that. What I don't understand is how the Q of
the filter relates to the circuit values.

I have just simulated the first example again and I suspect the problem
arises because the 3dB points turn out not to be symmetrical about the
resonant frequency. The simulation yields a resonant frequency close to
3.1KHz and -3db points at about 1.1KHz and 8.5KHz which gives a Q of
about 0.42 (assuming Q = deltaf/f). Feeding this back in to find the
effective resistance of the RLC gives R = 6974 ohms which is rather
higher than the expected 5000 ohms.

I am still puzzled.


Cheers

Ian

> Best regards,
>
> Wim
> PA3DJS
> www.tetech.nl
> the email address is OK, but without abc
From: Ian Bell on 12 Jan 2010 10:36
Ian Bell wrote:
> Wimpie wrote:
>> On 12 ene, 11:22, Ian Bell <ruffreco...(a)yahoo.com> wrote:
>>> Suppose you have a simple pot divider consisting of two equal value
>>> resistors, say 10K each. Across the upper one you connect a series LC
>>> circuit that resonates at 3KHz or thereabouts. If you drive this network
>>> from a low impedance source and plot the response across the bottom
>>> resistor, the Q of the resulting peak is not the Q of the series LC but
>>> rather is determined by the pair of LC values. For example, choosing
>>> l=150mH and C=18nF gives a Q of just over 2. Choosing L= 1.5H and
>>> C=1.8nF gives a much higher Q.
>>>
>>> What I need is a simple means of calculating L and C given the pot
>>> divider resistor value and desired Q and f (assuming the Q of the LC
>>> itself is much higher).
>>>
>>> Cheers
>>>
>>> Ian
>>
>> Hello Ian,
>>
>> The LC series circuit "sees" the upper and lower resistor in parallel.
>> This means when Rupper or Rlower (or both) are zero, the Q factor will
>> be infinite (assuming lossless L and C).
>>
>> So the Q factor of the loaded LCcircuit will be: Q = 2*pi*f*L/(Rp) Rp
>> = Rupp parallel to Rlow (that is Rp = 0.5*Rlow, when Rlow = Rupper).
>>
>
> Yes, that was my first thought but that still gives the wrong Q value.
>
>
>> However this Q factor does not directly relate to the frequency
>> response. So the -3dB BW of your circuit is not equal to BW = (center
>> frequency)/Q. This can be seen when Rupper = 0 and Rupper > 0 (all-
>> pass filter).
>>
>
> I think you meant the second to be Rlower, but I understand what you are
> saying.
>
>> Given your circuit, when you double the value of L (and halve the
>> value of C to enable same resonant frequency), the width of the peak
>> halves.
>>
>
> Yes, I already realised that. What I don't understand is how the Q of
> the filter relates to the circuit values.
>
> I have just simulated the first example again and I suspect the problem
> arises because the 3dB points turn out not to be symmetrical about the
> resonant frequency. The simulation yields a resonant frequency close to
> 3.1KHz and -3db points at about 1.1KHz and 8.5KHz which gives a Q of
> about 0.42 (assuming Q = deltaf/f). Feeding this back in to find the
> effective resistance of the RLC gives R = 6974 ohms which is rather
> higher than the expected 5000 ohms.
>
> I am still puzzled.
>
>
> Cheers
>
> Ian
>
>> Best regards,
>>
>> Wim
>> PA3DJS
>> www.tetech.nl
>> the email address is OK, but without abc


Wim, I just changed Rlower to 1K which gives simulation results much
closer to what I expected i.e Q = 3, (resonance near 3KHz, -3dB points
at 2.5K and 3.5K respectively) and if you calculate Q from w*L/R where R
ie 10K//1K you get Q = 3.

Cheers

Ian
From: Tim Wescott on 12 Jan 2010 11:16
On Tue, 12 Jan 2010 10:22:17 +0000, Ian Bell wrote:

> Suppose you have a simple pot divider consisting of two equal value
> resistors, say 10K each. Across the upper one you connect a series LC
> circuit that resonates at 3KHz or thereabouts. If you drive this network
> from a low impedance source and plot the response across the bottom
> resistor, the Q of the resulting peak is not the Q of the series LC but
> rather is determined by the pair of LC values. For example, choosing
> l=150mH and C=18nF gives a Q of just over 2. Choosing L= 1.5H and
> C=1.8nF gives a much higher Q.
>
> What I need is a simple means of calculating L and C given the pot
> divider resistor value and desired Q and f (assuming the Q of the LC
> itself is much higher).
>
> Cheers
>
> Ian


.--||---UUU--. ____
Vin o--o ____ o----|____|----.
'---|____|---' |
===
gnd

Like this? How about simple circuit analysis?

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