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What is the significance of negative Q factor in a second-order low-pass filter?

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samiran_dam

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What is the significance of negative Q factor in a second-order low-pass filter? When negative Q factor arrives in the circuit?

Cheers
Sam
 

Re: Q factor

That's not possible! Q factor by definition is always is positive. Check for calculation error.
 
Re: Q factor

In one of my previous post you have replied that Qp=wp/2σ (with wp: pole frequency and σ: real part of the complex pole). So according to this equation, Qp can be negative if the real part is negative isn't it?. This is not going with the theory...I am confused!

Actually, what I am doing, in cadence I am simulating a second-order sallen-key lp filter with a single-pole roll-off verilog-A model of opamp with different values of opamp's ugb. For lower values of ugb, lets say for ugb=1MHz, with gain=60dB, dominant pole is = 184059 + i.0, with Qp = -0.5. It implies for positive real pole Qp < 0.

Could you please explain this phenomena?

Cheers
Sam
 

Re: Q factor

samiran_dam said:
In one of my previous post you have replied that Qp=wp/2σ (with wp: pole frequency and σ: real part of the complex pole). So according to this equation, Qp can be negative if the real part is negative isn't it?. This is not going with the theory...I am confused!
Click to expand...

Don't be confused. Since for a stable circuit σ ALWAYS is negative, of course the magnitude of σ must be inserted into the formula.
Thus:

Qp=wp/2|σ|.

Sorry for the confusion.
 
Re: Q factor

Okey, I got it; but what about the simulation result I got???? I mean do you have any idea about this? Is there any relation between sign of the Q and capacitive/inductive reactance behavior of the circuit?

Cheers
Sam
 

Re: Q factor

I do not understand the problem. What is the question? Where is a problem in simulation results?
 
Re: Q factor

In simulation, I got Qp = -0.5 < 0. Physically is it possible? and if it is possible, then what is the intuitive explanation for this ?
 

Re: Q factor

samiran_dam said:
In simulation, I got Qp = -0.5 < 0. Physically is it possible? and if it is possible, then what is the intuitive explanation for this ?
Click to expand...

Please explain in detail!
What means "I got..." ? Which kind of simulation?
Did the simulator show the value of Qp directly or did you use some other data to calculate the value of Qp? As mentioned, physically impossible for a stable circuit!
Show us the circuit, perhaps there is an error (unstable?).
 
Re: Q factor

2v2y62f.jpg
.

Above is the circuit diagram. Here the opamp used is modeled in verilog-A with a single (dominant) pole, a finite gain and non-zero output resistance, so I can control these three opamp parameters independently and for one ac simulation in cadence analog design environment (spectre simulator), I set ugf = 1MHz, gain = 60dB and output resistance = 1KΩ. Then I simulated and did a "pz analysis", so I directly got the location of poles and respective pole-Qfactor without doing any calculation.

Hence I got the above said results.

Has it explained my doubt? Please let me know if you need any more information regarding this.

Cheers
Sam
 

Re: Q factor

For an ideal opamp, your circuit is a 2nd order lowpass with a pole frequency at app. 500 kHz.
Thus, you need a real opamp which has a transit frequency of at least 20..50 MHz (if you really want a bandwidth of 500kHz).
May be, that´s the reason for your surprising (and unrealistic) result.
Therefore, try better opamp parameter.

Yes, probably I know the reason: The pole Q formula which is used by the simulator does not apply for Qp values below +0.5. With your (bad) opamp model the value of Qp will be lower than 0.5.
 
Re: Q factor

Yes, that's right the filter cutoff frequency is 500 KHz. Can you explain me how did you calculate the opamp transit frequency to be 20.5MHz? or refer some document/paper that I can go thru myself and understand?

And many many thanks for your patient replies....It is really really helping me alot.

Cheers
Sam

Added after 49 minutes:

Actually there is no problem with the opamp model I guess, because for higher values of ugf e.g. 100MHz, I an getting Qp = 0.715 as intended. Maybe what you suggested in the first place, that as ugf < minimum required opamp transit frequency I am getting the weired result. But it will really help if I get to know how this transit frequency is calculated for a particular filter cutoff frequency.
 

Re: Q factor

samiran_dam said:
Yes, that's right the filter cutoff frequency is 500 KHz. Can you explain me how did you calculate the opamp transit frequency to be 20.5MHz?
.........
Click to expand...

No, the UGF should be at least 20 MHz....50 MHz - which means: Much larger than the wanted pole frequency! Of course, there will be ALWAYS a deviation between real and ideal circuits. But as larger the UGF as smaller the error! That`s the whole secret.
 
Re: Q factor

This means maximizing (or if required minimizing) opamp parameters will drag filter characteristin solely dependent upon passive component in terms of cutoff frequency, but that will make the specification of the opamp tighter. So I was wondering if we can find out the minimum required ugf....just an inquisitiveness. But hey, really thanks ton for helping me.
 

Re: Q factor

samiran_dam said:
..............
So I was wondering if we can find out the minimum required ugf....just an inquisitiveness.
Click to expand...

Yes, there are some approaches (rough rules) to calculate the minimum required UGF for each particular filter circuit based on filter data and based on specified acceptable deviations from ideal values (Qp, Fp, gain) which, normally, are given as a percentage.
 
Re: Q factor

By deviation, did you mean to indicate the so-called sensitivities of Qp, Wc, gain of the filter on opamp parameters?? Can you redirect me to any reference where I can find the 'rough' rules you are talking about?
 

Re: Q factor

samiran_dam said:
By deviation, did you mean to indicate the so-called sensitivities of Qp, Wc, gain of the filter on opamp parameters?? Can you redirect me to any reference where I can find the 'rough' rules you are talking about?
Click to expand...

Yes, that's exactly what I mean.
Corresponding references (documents available via internet) are not at my hand, at the moment. But, several books on filter design deal with this subject. I will keep your desire in my mind.

Added after 7 minutes:

Here is a document, which gives some recommendations for the opamp choice (chapter 16.8.4)
 

Re: Q factor

By the way: Don't forget slew rate limitations at 500 kHz!
 

Re: Q factor

Hi LvW,

Could you please explain in brief, how opamp slew rate can limit LPF operation?

regards
Sam
 

Re: Q factor

When the opamp used is not able to produce the amplitudes as required by the operating frequency and the associated gain, there will be severe distortions of the output signal (triangle instead of sinus). This effect is connected with additional negative phase deviations. The reason is not the limited small signal bandwidth but the limited slew rate (large signal behaviour). Look at the "power bandwidth" graph in the opamp data sheet. This property often is overlooked in the design of high frequency filters.
 

Re: Q factor

some analytical explanation would immensely help me. So, please suggest me some papers/books which discuss this topic (slew rate limitation on high-frequency filters).

Regards
Sam
 

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