Network Filter Challenge, Can you find this value?

[MrAl]

Hi there folks,


I've run into this circuit a number of times, most recently right here on ET Forums. It's interesting enough where it should be studied at least a little bit.

See the attachment for the circuit details, and i'll explain what this is all about.




In Fig. 1 of the attachment we see the parallel LC network, the Equations for that network, and the impedance vs frequency F. F0 is the center frequency and as shown the network has highest impedance at that frequency.

With the equations we can calculate either the center frequency F0, or L or C.

Because of the impedance characteristic, an LC network in series with the output will block a particular frequency, while an LC network in parallel with the output will allow a particular frequency to pass and shunt other frequencies.

Now it is known that with proper filtering a square wave can be filtered into a pure sine wave (or nearly pure). It is also known that the third harmonic is the strongest unwanted frequency and the fifth is fairly strong too, so we want to filter those two out with two LC networks. Fig. 2 shows two networks in series and they are in series because they are intended to block the third and fifth harmonics. These networks are shown as L1 and C1, and L2 and C2.
The third LC network shown is in parallel with the output because we want to tune that for high impedance so that it will pass 100kHz but shunt other frequencies like the third and fifth and higher.

Using the Equations shown, we start with C1=1nf and C2=1nf, and we tune L1 so that L1 and C1 block 300kHz, and we tune L2 and C2 to block 500kHz. These are the third and fifth harmonics of the fundamental which is 100kHz. With C1 being 1nf we calculate L1=0.0002814477 Henries. With C2=1nf we calculate L2=0.00010132118 Henries.
With the third parallel LC network made of L3 and C3, we start with
L3=0.00253302959 Henries, a known inductance rather than a known capacitance.

The fixed value of R1 is 5000 ohms.

The question is, what is the best value for C3 with this chosen value of L3?

Remember we are trying to pass 100kHz and block all others, with two of the networks made to specifically block the 3rd and 5th harmonics.

 

[Ratchit]

MrAl,

The question is, what is the best value for C3 with this chosen value of L3?

By best value, I think you mean the highest peak value. I figure C3 should be 7 nf for a output of 24 volts peak, when the square wave is 1 volt peak. Evidently there is some amplification. It is tempting to say 1 nf, which will resonate with L3 at 100khz, but it appears the components interact with each because the 1st, 3nd and 5th harmonics are too close to each other to calculate them separately. I believe the recomendation is that the frequencies should be 10 times apart to consider the components isolated.

Ratch

 

[MrAl]

Hi Ratch,


Ok, so how did you arrive at that value and what was the motivation for looking for a cap other than 1nf?

 

[Warpspeed]

I agree with Ratch.

At 100 Khz, L1/C1 and L2/C2 will have some residual reactive (inductive) impedance that will not be negligible.
And if the source impedance of the driver is low, that is going to interfere with the tuning of L3/C3.

Any additional reactive components on either side of L3/C3 are going to detune it.

Think about what happens if you place two parallel tuned circuits in parallel.
You don't achieve two separate resonant frequency peaks, but one new resonant frequency that is unrelated to either of the originals.

You would be far better off with a single very high Q tuned circuit, very loosely coupled to the driver, followed by a very high impedance buffer.

 

[Ratchit]

MrAl,

Ok, so how did you arrive at that value and what was the motivation for looking for a cap other than 1nf?

I calculated the transfer function for the fundamental and the two harmonics using V*Z3/(Z1+Z2+Z3). I quickly saw that the harmonics were insignificant compared with the fundamental. The transfer function for the fundamental was (8.600409314E8j*cos(6.283185308E5*t))/(1.125790828E17j*C3+3.5835035E7-7.88053651E8j) . So to maximize the function, we have to minimize the denominator. Calculating C3 = 7 nf cancels out the j terms in the denominator and gives the maximum value for the transfer function. The cos term is the numerator is set to 1. Dividing the remaining terms in the numerator by the denominator gives 8.600409314E8j/3.5835035E7 = 24j .

I figured that since the frequencies were so close together, there would be some interaction between each filter element.

Ratch

 

[MrAl]

Hi,


Yes, you are both pretty much right on. I was hoping some others would look at this too to see how unusual it is intuitively. It appears at first that C3 could be 1nf or near that, but it turns out to be 7 times higher than that. And yes, interaction is the key here.

I used a similar technique, working from the transfer function can calculating the value of C3, and that's how i found the higher value which was a little surprising at first.

Also on a related note, the value of C3 does indeed approach 1nf when the generator series resistance increases to near 5000 ohms. Below that though it is always more than 1nf.

Here's a formula for calulating C3 outright (Rs is the generator series resistance, and w is 2*pi*100000):

C3=-((((w^8*C1*C2^2+w^8*C1^2*C2)*L1^2+(-w^6*C2^2-2*w^6*C1*C2)*L1+w^4*C2)*L2^2+((-2*w^6*C1*C2-w^6*C1^2)*L1^2+(2*w^4*C2+2*w^4*C1)*L1-w^2)*L2+w^4*C1*L1^2-w^2*L1)*L3+(((-Rs^2*w^8*C1^2-w^6)*C2^2-2*w^6*C1*C2-w^6*C1^2)*L1^2+(2*Rs^2*w^6*C1*C2^2+2*w^4*C2+2*w^4*C1)*L1-Rs^2*w^4*C2^2-w^2)*L2^2+(((2*Rs^2*w^6*C1^2+2*w^4)*C2+2*w^4*C1)*L1^2+(-4*Rs^2*w^4*C1*C2-2*w^2)*L1+2*Rs^2*w^2*C2)*L2+(-Rs^2*w^4*C1^2-w^2)*L1^2+2*Rs^2*w^2*C1*L1-Rs^2)/(((((Rs^2*w^10*C1^2+w^8)*C2^2+2*w^8*C1*C2+w^8*C1^2)*L1^2+(-2*Rs^2*w^8*C1*C2^2-2*w^6*C2-2*w^6*C1)*L1+Rs^2*w^6*C2^2+w^4)*L2^2+(((-2*Rs^2*w^8*C1^2-2*w^6)*C2-2*w^6*C1)*L1^2+(4*Rs^2*w^6*C1*C2+2*w^4)*L1-2*Rs^2*w^4*C2)*L2+(Rs^2*w^6*C1^2+w^4)*L1^2-2*Rs^2*w^4*C1*L1+Rs^2*w^2)*L3)

This may be simplified.

 

[The Electrician]

Plotting a family of frequency response curves for the source resistance varying from 0 to 5000 ohms shows that L3 is not needed.

Just make C3 6 nanofarads and eliminate L3; there is essentially no difference in the performance of the filter above 300 kHz when this is done.

 

[MrAl]

Hi there Electrician,

Ok i'll check that myself too.


LATER:

Ok, i find the same results. C3 comes out to very close to 6nf as you said when the generator resistance is zero.

Here's the new formula (R is the generator series resistance and w=2*pi*100000):

C3=-(((w^6*C1*C2^2+w^6*C1^2*C2)*L1^2+(-w^4*C2^2-2*w^4*C1*C2)*L1+w^2*C2)*L2^2+((-2*w^4*C1*C2-w^4*C1^2)*L1^2+(2*w^2*C2+2*w^2*C1)*L1-1)*L2+w^2*C1*L1^2-L1)/(((w^8*C1^2*C2^2*L1^2-2*w^6*C1*C2^2*L1+w^4*C2^2)*L2^2+(-2*w^6*C1^2*C2*L1^2+4*w^4*C1*C2*L1-2*w^2*C2)*L2+w^4*C1^2*L1^2-2*w^2*C1*L1+1)*R^2+((w^6*C2^2+2*w^6*C1*C2+w^6*C1^2)*L1^2+(-2*w^4*C2-2*w^4*C1)*L1+w^2)*L2^2+((-2*w^4*C2-2*w^4*C1)*L1^2+2*w^2*L1)*L2+w^2*L1^2)

and if we assume that L1 and L2 will be tuned to 300kHz and 500kHz with C1 and C2 respectively the formula simplifies to:
C3=(72*C1*C2^2+24*C1^2*C2)/(576*w^2*C1^2*C2^2*R^2+9*C2^2+6*C1*C2+C1^2)

and of course if C2=C1 we end up with a neat little formula:
C3=6*C1/(36*w^2*C1^2*R^2+1)

which with R=0 gives us a value for C3 of:
C3=6e-9

and with R=50 we get a value for C3 of:
C3=5.79413118256813e-009

and with R=500 we get:
C3=1.31779576444803e-009

 

[The Electrician]

It's not necessary to go to so much trouble if all one wants to do is to filter a square wave into a sine wave. A single series resonant tank will do the job if the impedance level is made appropriate for the load.

A .253302959 henry inductor in series with a .01 nF capacitor between the source and load will do a good job. The component values are not practical, but in theory it would work well. The harmonics are sufficiently attenuated to give a quite good looking sine wave output. There is no resonant rise in output voltage if the circuit is lightly loaded as with the other circuit.

I've attached two images. The first shows the original circuit in this thread with the source resistance swept from 0 to 5000 ohms. The second image shows the response of the single series resonant tank.

There is a related situation that is often discussed. How can we use a passive filter to convert the output of a grid frequency inverter that outputs a square wave into a good sine wave?

There are several problems associated with this. The filter must output a sine wave with any load from open circuit on the output, to a heavy load. There must not be any resonant rise at the fundamental changing with load; the output should be proportional to the input. There should not be significant loss through the filter.

Such filters are not easy to design, and when a good design is found, typically the component values are not practical. In particular, the inductor typically ends up being about as large as the inverter transformer.

 

[Warpspeed]

It is not that difficult.
Radio transmitters convert class C pulsed power to sine waves of excellent purity at very high power levels, and so do high power induction heaters. It is all in the design and tuning of the resonant tank circuit.

If you really want a high Q circuit, some positive feedback will do wonders.
Phase locking a high quality sine wave oscillator to an incoming square wave would be another approach.

 

[MrAl]

Hi again Electrician,

Well the reason for looking at this circuit in particular is because it comes up now and then in questions on forums. Not that it's the best way to do a filter.

Also, with L=0.2533H and C=0.01nf, the Q is so high that even a tiny change in C throws the center frequency off by enough to cause a major output amplitude change. So although it looks nice and wonderful with C=0.01nf, a small change like C=0.0101nf (a +1 percent change in C) the output drops to less than 1/3 of it's previous value with the perfect cap. That makes it fairly hard to tune and keep it tuned. Perhaps lowing the Q would help.

BTW, your circuit is a generator in series with the inductor in series with the capacitor in series with the load, and you're looking at the voltage across the load right or no? And the load resistor is how many ohms?

 

[The Electrician]

Hi again Electrician,

BTW, your circuit is a generator in series with the inductor in series with the capacitor in series with the load, and you're looking at the voltage across the load right or no? And the load resistor is how many ohms?

All in series, and the load was 5000 ohms, just like the filter that started this thread.

 

[MrAl]

Hi again,

Ok thanks. I was seeing a difference in the response with the single LC circuit so i wanted to find out why. I was using 500 ohms so i was seeing a very high Q.
With 5000 ohms the Q is about 10 times less which makes it less sensitive to small component variations.
It's kind of interesting that with a variation of only 1 percent in the capacitance i saw a change in output amplitude of 66 percent, with 500 ohms.
It's also kind of interesting that with a output resistance of 50 ohms and small capacitance variation the filter actually starts to cut at 100kHz. I guess the Q then is something over 3000 though, which makes it very hard to tune and keep tuned

 

[The Electrician]

The two notch filters for the 3rd and 5th harmonic in your original circuit have essentially infinite Q. How hard would it be to keep them tuned?

 

[MrAl]

Hi again Electrician,

I wasnt really talking about the original circuit, but the same rule of thumb would apply...

The 'valley' in the response is centered directly at the center frequency so we get maximum cut. Because of the high Q (limited also by inductor series resistance) the slope just to the right or left of the center frequency is quite steep (sharp response). With small component variation, the notch tunes to a slightly different frequency, which means now at the center frequency the cut isnt as deep. It's quite dramatic too. With a Q of 500 and component C variation of only 1 percent we can see a +20db increase in the amplitude at the original center frequency. For example, with that 300kHz parallel tuned part tuned to exactly 300kHz and some small inductor series resistance we might see a cut of -70db, but with a change in capacitance of only 1 percent the cut at 300kHz is only about -50db because the notch moves to the left or right of the 300kHz center frequency. This of course means we dont get rid of as much of the unwanted signal as we wanted to.

It's pretty well known that high Q tuned circuits are less stable frequency wise and im sure a quick search would tell more about this. In radio it is considered harder to tune but more selective. It's harder to tune because even a small variation in a component shifts the notch or peak, and if the peak is sharp (as in high Q) we quickly loose the characteristic we were after in the first place by using a high Q. This would mean for example that we would be turning the variable cap in one direction and with just a tiny twist we'd jump past the required center frequency, then turning back the other way we'd jump past it in the other direction, etc., so we'd be turning it back and forth trying to twist it less and less so we get it right on. This also means more temperature dependence too so the notch (or peak) would drift.

With a low Q the notch still drifts, but it doesnt matter as much because the response is more flat so we dont loose as much cut (or gain).

 

[The Electrician]

Hi again Electrician,

I wasnt really talking about the original circuit, but the same rule of thumb would apply...

I knew you weren't talking about the original circuit; I was just wondering why you hadn't discussed the problem of high Q with respect to it.

 

[MrAl]

Hi again,


Oh ok
I guess i just assumed that anything with high Q might present a practical problem.
I have a feeling this is a homework assignment that comes up now and then.