How can I analyze the performance measure of a serial-parallel and serial-serial RLC circuits.

What are the performance measure you use for circuit analysis

  • Q-Factor

    Votes: 2 100.0%
  • Bandwidth

    Votes: 1 50.0%

  • Total voters
    2

Thread Starter

Harish G

Joined Oct 21, 2016
34
Hi everyone.

I am a beginner to circuit analysis.

I have got two circuits to study. I would like to measure their performance based on some quantity (such as output power or Q-factor).

So I have attached the circuits here please take a look.

I know how to calculate the Q-factor for a normal serial resonant circuit or a parallel resonant circuit. But what I dont know is how do you calculate the Q-factor of the whole circuit (loaded) when the circuit is not just a normal serial/parallel resonant circuit.

Should we make it to simple serial circuit/parallel circuit and then find the total resistance and the reactance of the circuit ? , I dont know. [And again I am not familiar with converting these circuits to serial equivalent or parallel equivalent, I know only if the circuit has only resistive elements but not when it has RLC components in it].

So if someone can explain me how the Q-Factor of the circuit can be calculated.
or
If some other performance measure can tell which circuit is better.

then that would be a lot of help. Thanks.
 

Attachments

drc_567

Joined Dec 29, 2008
703
Find the characteristic equation of the circuit. That will give you a resonant frequency.
 

Thread Starter

Harish G

Joined Oct 21, 2016
34
Find the characteristic equation of the circuit. That will give you a resonant frequency.
I made that simple. by making C1=C2 and L1=L2.
Now resonant frequency is simply, w0 = 1/2*pi*sqrt(L1*C1)

I guess I know what you want to say, but I dont have an idea on finding the B.W of this circuits.

PS: What happens if I want to operate at half power frequencies, say w1 and w2 and not the resonant frequency w0, how does the concept of Q changes then, sorry for question in a question, I am just curious to know.

Regards
 

drc_567

Joined Dec 29, 2008
703
Another idea, find the Thevenin Equivalent of the circuit. It will be a complex value, due to inductors and capacitors. That would be easier.
 

Thread Starter

Harish G

Joined Oct 21, 2016
34
Another idea, find the Thevenin Equivalent of the circuit. It will be a complex value, due to inductors and capacitors. That would be easier.
What is the use of finding equivalent voltage and resistance, when Q value of the circuit depends on equivalent resistance and reactance. like Q=Rp/Xp.

Sorry I dont know why you mentioned that.

Regards
 

drc_567

Joined Dec 29, 2008
703
Yes ... It would appear that you are able to find a Thevenin Equivalent impedance, that is looking in from the load resistance. This would lead to the series equivalent of a Thevenin impedance in series with a load resistance. You would need to find a resonant frequency to complete the solution. So what other problems do you see?
 

Thread Starter

Harish G

Joined Oct 21, 2016
34
Yes ... It would appear that you are able to find a Thevenin Equivalent impedance, that is looking in from the load resistance. This would lead to the series equivalent of a Thevenin impedance in series with a load resistance. You would need to find a resonant frequency to complete the solution. So what other problems do you see?
I understand that if I find thevenin equivalent circuit, then I can able to find the equivalent circuit reactance and equivalent resistance, so there by I can calculate the Q value of the circuit. Thanks soo much for clarifying that.

The only question now is. The resonance frequency drifts away as the load resistance increases. and in my case I am sure I wont be operating at resonance frequency but the half power frequencies. So all the above methods are still applicable ? even if I dont operate at resonant frequency ?, Not really I guess.

So now the quality factor calculation as per above methods is no more valid. Isn't it ?

So how can I calculate some performance measure for the circuits in this case ?

Regards
 

MrAl

Joined Jun 17, 2014
6,455
Hello again,

Assuming you know the shape of the response is a bandstop or bandpass, you could solve for the transfer function Hs as noted first, then solve for the center frequency Wc, then solve for the amplitude at the center frequency call that Ac, then equate Hs to Ac/sqrt(2) and solve for the two -3db points W1 and W2, then subtract Wb=W2-W1 then find the Q with Q=Wc/Wb. So the steps would be:
1. Find Hs.
2. Calculate Wc.
3. Find Ac.
4. Equate Hs=Ac/sqrt(2), solve for W1 and W2.
5. Subtract: Wb=W2-W1
6. Divide to get Q=Wc/Wb.
so there you have both the bandwidth Wb and the Q.

Note this system may be 4th or 5th order so it's not as easy to solve for these things as it is in a 2nd or 3rd order filter. More work involved and probably have to bring in a numerical solver if you dont want to do the calculations yourself.

Another modern measure of performance might be the settling time, the time from a step response to the time it takes for the output to settle to whatever specification you need such as 10 percent, 5 percent, 1 percent, 0.1 percent, etc.
 

drc_567

Joined Dec 29, 2008
703
One possibility would be to form an expression for the output current, that is the current actually going through the load resistor. This would be evaluated as the ratio of the Thevenin voltage divided by the load resistance. .. A numerical plot of the current should show a peak at resonance, as well as bandwidth at the half power frequencies.
... Do you have an example circuit that might be used to compare values?
 

Thread Starter

Harish G

Joined Oct 21, 2016
34
Hello again,

Assuming you know the shape of the response is a bandstop or bandpass, you could solve for the transfer function Hs as noted first, then solve for the center frequency Wc, then solve for the amplitude at the center frequency call that Ac, then equate Hs to Ac/sqrt(2) and solve for the two -3db points W1 and W2, then subtract Wb=W2-W1 then find the Q with Q=Wc/Wb. So the steps would be:
1. Find Hs.
2. Calculate Wc.
3. Find Ac.
4. Equate Hs=Ac/sqrt(2), solve for W1 and W2.
5. Subtract: Wb=W2-W1
6. Divide to get Q=Wc/Wb.
so there you have both the bandwidth Wb and the Q.

Note this system may be 4th or 5th order so it's not as easy to solve for these things as it is in a 2nd or 3rd order filter. More work involved and probably have to bring in a numerical solver if you dont want to do the calculations yourself.

Another modern measure of performance might be the settling time, the time from a step response to the time it takes for the output to settle to whatever specification you need such as 10 percent, 5 percent, 1 percent, 0.1 percent, etc.
--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Wow, thanks for the details.

I will post my step response, can you please take a look at that.

Thanks.
 

Thread Starter

Harish G

Joined Oct 21, 2016
34
One possibility would be to form an expression for the output current, that is the current actually going through the load resistor. This would be evaluated as the ratio of the Thevenin voltage divided by the load resistance. .. A numerical plot of the current should show a peak at resonance, as well as bandwidth at the half power frequencies.
... Do you have an example circuit that might be used to compare values?

Thanks. I am doing my analysis in LTspice, so I can able to get hold of those half power frequencies easily.

Here is my circuit in LTspice .asc format.

Thanks.
 

Attachments

MrAl

Joined Jun 17, 2014
6,455
Hi,

Yeah i was going to ask a similar question.

Are there any values given for this circuit yet?
That would be required for the most advanced cases.
As is, we might have to solve a 4th order equation to get the center frequency in the general case.
 

Thread Starter

Harish G

Joined Oct 21, 2016
34
Hi,

Yeah i was going to ask a similar question.

Are there any values given for this circuit yet?
That would be required for the most advanced cases.
As is, we might have to solve a 4th order equation to get the center frequency in the general case.

pardon the name of the file. but the ultimate aim is Q-factor calculation.

Thanks.
 

Attachments

MrAl

Joined Jun 17, 2014
6,455
Thanks. I am doing my analysis in LTspice, so I can able to get hold of those half power frequencies easily.

Here is my circuit in LTspice .asc format.

Thanks.

Hi,

Well if you do an AC response you can see the center frequency as well as the two -3db points if that's what you have. Are those the only values you need to use ? That makes it much easier.

I did the version with the output parallel cap and found two resonant peaks that are almost the same at about 2.79MHz.

For the version with the series cap i only did the simulation which shows peaks at about 1.6MHz and 3.6MHz roughly. For this version you may want to call the center frequency the frequency in the middle of those two, or else you have to solve for more -3db points (two peak frequencies).
 
Last edited:

Thread Starter

Harish G

Joined Oct 21, 2016
34
Hi,

Well if you do an AC response you can see the center frequency as well as the two -3db points if that's what you have. Are those the only values you need to use ? That makes it much easier.

I did the version with the output parallel cap and found two resonant peaks that are almost the same at about 2.79MHz.

For the version with the series cap i only did the simulation which shows peaks at about 1.6MHz and 3.6MHz roughly. For this version you may want to call the center frequency the frequency in the middle of those two, or else you have to solve for more -3db points (two peak frequencies).
I think I interpreted two frequency peaks as you mentioned in series cap case as -3dB points, apologies for that. It is not -3dB points, instead they are the two peak frequency components at which I want to operate my system.

So does anything change if I operate at these two peak frequency components or the calculation steps remain same? I guess I just have to take W1 as first peak frequency and W2 and second peak frequency, and Wc or Wo = W2-W1.

Regards.
 

Thread Starter

Harish G

Joined Oct 21, 2016
34
Hi,

Well if you do an AC response you can see the center frequency as well as the two -3db points if that's what you have. Are those the only values you need to use ? That makes it much easier.

I did the version with the output parallel cap and found two resonant peaks that are almost the same at about 2.79MHz.

For the version with the series cap i only did the simulation which shows peaks at about 1.6MHz and 3.6MHz roughly. For this version you may want to call the center frequency the frequency in the middle of those two, or else you have to solve for more -3db points (two peak frequencies).

If you want to select the one circuit among these two circuit, what will you consider ? based on what criteria's can you please comment a little on that.

Thanks.
 

MrAl

Joined Jun 17, 2014
6,455
I think I interpreted two frequency peaks as you mentioned in series cap case as -3dB points, apologies for that. It is not -3dB points, instead they are the two peak frequency components at which I want to operate my system.

So does anything change if I operate at these two peak frequency components or the calculation steps remain same? I guess I just have to take W1 as first peak frequency and W2 and second peak frequency, and Wc or Wo = W2-W1.

Regards.

Hi,

Well, no that is not the way to do it, sorry. The peaks are the maximum response and the -3db points define the bandwidth. As you go from the left -3db point to the right -3db point, that frequency difference is the bandwidth. So if you have two significant peaks you'd have to say you have two completely different resonant peaks unless your application can put up with that and use the single center frequency between those two peaks as the tuning point frequency, and then look for -3db points down from that. This will be very application specific though.

The parallel circuit exhibits two resonant peaks also, but they are so close together that they appear to be one peak, and that makes it easier to work with if you only need one resonant peak. There may be other resonant peaks also but are more insignificant because they dont change the output response by very much at all.

So if you only need one resonant peak then yes the parallel circuit seems to be better, at least given your constraint of C2=C1 and L2=L1. I did not look over your method of calculating the center frequency yet.

If you want to do this with the AC simulation, then look for a peak, then look to the left and right of that peak where the response is -3db down from the peak. This is sometimes hard to spot without using fine resolution though, but you can probably get some idea what you have. Once you have those two frequencies and the center frequency, you just divide:
BW=f2-f1
Q=fc/BW
 

Thread Starter

Harish G

Joined Oct 21, 2016
34
Hi,

Well, no that is not the way to do it, sorry. The peaks are the maximum response and the -3db points define the bandwidth. As you go from the left -3db point to the right -3db point, that frequency difference is the bandwidth. So if you have two significant peaks you'd have to say you have two completely different resonant peaks unless your application can put up with that and use the single center frequency between those two peaks as the tuning point frequency, and then look for -3db points down from that. This will be very application specific though.

The parallel circuit exhibits two resonant peaks also, but they are so close together that they appear to be one peak, and that makes it easier to work with if you only need one resonant peak. There may be other resonant peaks also but are more insignificant because they dont change the output response by very much at all.

So if you only need one resonant peak then yes the parallel circuit seems to be better, at least given your constraint of C2=C1 and L2=L1. I did not look over your method of calculating the center frequency yet.

Ok, I understand what you are saying.

So in case of serial case here, the Q-factor calculation impossible ? considering both peaks into the account (again considering the application in question needs both the peaks.)

Regards
 
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