Balanced High Pass Filter

[earthsled]

That's probably because the inductor is not shielded. You can make a shield out of mild steel (cheap) or Mumetal (expensive), or you can use two inductors of half-value arranged in a humbucking configuration - that's probably the most efficient solution.

I like the half-value humbucking idea. Can I use an inexpensive 1:1 transformer to accomplish this?

Now you understand why inductors became out of fashion and were replaced with active RC circuits!

Yes. An exploration of Sallen-Key designs may be coming soon for me.

Thanks!

 

[abbey road d enfer]

I like the half-value humbucking idea. Can I use an inexpensive 1:1 transformer to accomplish this?

Yes, no problem.

 

[earthsled]

Here's a new set of graphs. The test circuit is the same except for the HP inductor is made from two 2.7H coils in series (and humbucking). This results in a value of 5.4H for the value of L1 rather than the ideal 4.5H.

Once again, green is HPF+XFMR and red is HPF alone.

I'm happy to see the frequency response graphs are more consistent here, although it looks like the cutoff frequency is closer to 1K as opposed to 100Hz. Could this be due to the increased inductance?

Thanks!

 

[PRR]

> Here's a new set of graphs.

Look at your slope. 5.5dB/octave. This is a 1-pole filter.

Look at the bottom. There is NO reason it should be shelving flat below 40Hz.

The green curve suggests "flat to DC" and that's just not possible with the caps. If this is jury-rigged, I bet there is some sneak-path around the filter.

Sumthin is wrong.

> HPF_test_setup.gif

If you "know" your transformer is 11.8H, why add a choke?

0.2348uFd will ring 11.8H at 100Hz.

To stay balanced, you need two 0.47uFd caps.

Reactive impedance of 11.8H or 0.234uFd at 100Hz is about 6K.

For Q=1 (Butterworth) the source can be much-much less than 6K (it is) and the reflected load at the primary needs to be up near 6K.

If that is a 1:1 feeding a 10K load, it's close enough to run with. Q is 1.6+ which is a 4dB bump, which tends to ear-mask the fact there's noting below 100Hz. Or shave the loading down to 6K.

If reflected loading is a more typical 2K, rather 2K||11.8H, then Q is 0.3 which for the first octave down is same-as a single-pole filter. To get Q back up near unity, total inductance should be nearer 11.8/3 or 3.6H (yes, 10H||5H does that) and caps can be 0.47uFd*3 or 1.5uFd(!!).

> Maybe I should explore a HPF that doesn't use an inductor?

After two years?

 

[earthsled]

Okay. Some followup here.

Attached is my current circuit using PRR's suggested components...

If reflected loading is a more typical 2K, rather 2K||11.8H, then Q is 0.3 which for the first octave down is same-as a single-pole filter. To get Q back up near unity, total inductance should be nearer 11.8/3 or 3.6H (yes, 10H||5H does that) and caps can be 0.47uFd*3 or 1.5uFd(!!).

 

[earthsled]

...and here is the corresponding sweep I measured.

 

[earthsled]

This graph shows the results of including the inductor (blue) versus without an inductor (orange). I thought it was interesting that the low-end curve reversal seems to improve somewhat when the inductor is omitted. The cutoff frequency appears to be somewhere between 200Hz and 300Hz and the slope looks the same at about 5.5 dB per octave.

What am I doing wrong here?

 

[abbey road d enfer]

...and here is the corresponding sweep I measured.

The 10k load at the secondary reflects as 200r at the primary; the filter is overdamped.
For a correct Butterworth alignment, the secondary should be loaded by ca. 70k.
You may want to insert an output pad made with two 33k (series) and one 22k (shunt), that would give about 20dB attenuation when connected to the 10k input of the RME.

 

[earthsled]

Here is the circuit with Abbey's suggested output pad...

I assume the output pad serves the purpose of 70k loading; or do I need to add an additional resistance for that?

 

[earthsled]

...and here is the resultant frequency graph.

This looks a bit more flat, but the cutoff frequency is still too high.

 

[abbey road d enfer]

Here is the circuit with Abbey's suggested output pad...

I assume the output pad serves the purpose of 70k loading; or do I need to add an additional resistance for that?

You don't need any additional resistance.

 

[abbey road d enfer]

This looks a bit more flat, but the cutoff frequency is still too high.

and the slope is still 6dB/octave...have you tried reducing the input level?

 

[earthsled]

have you tried reducing the input level?

I tried a test with the input level reduced by 20dB, but the response graph appears identical.

Is there something else I can try? Thanks!

 

[abbey road d enfer]

have you tried reducing the input level?

I tried a test with the input level reduced by 20dB, but the response graph appears identical.

Is there something else I can try? Thanks!

I must admit it's a real brain teaser. The only stone left unturned is the DCr of the xfmr and inductor.

 

[earthsled]

The inductor measures 780r total. The input transformer's primary is about 24r, and the secondary is about 1.4k.

FWIW, I am still using a small signal transformer to serve as the inductor (L1). The primary and secondary of L1 both measure about 2.6H. When I measure the inductance of the coils connected in series, my meter only shows about 12mH (rather than the expected 5.2H). Would I be correct in thinking that transfer from one coil to the other is throwing off my meter?

Thanks!

 

[abbey road d enfer]

The inductor measures 780r total. The input transformer's primary is about 24r, and the secondary is about 1.4k.

FWIW, I am still using a small signal transformer to serve as the inductor (L1). The primary and secondary of L1 both measure about 2.6H. When I measure the inductance of the coils connected in series, my meter only shows about 12mH (rather than the expected 5.2H). Would I be correct in thinking that transfer from one coil to the other is throwing off my meter?

Thanks!

Wow! had you mentioned that earlier, we could have solved the mistery before I lost my hair!
When connecting in series two identical windings wound on the same core, the inductance is quadrupled. That's because the inductance is proportional to the sqare of the number of turns. In your case, the inductance should mesure 10.2H.
BUT, if the windings are out of phase, the main inductance is nulled, and what remains is only the leakage inductance AND the DCr.
So basically, you have a 780 resistor instead of a 5.2H inductor.
If you wire the windings correctly, you will have a cut-off frequency of about 60Hz, but the DCr will overdamp the response.

I've done a quick sim:
You could either wire only one winding and increase the caps to 2uF, or wire the two windings in proper series and the 1.5uF, the former being a tad sharper with a slight overshoot (ca. 0.5dB), the latter being less efficient IMO.
But the best option is to wire both windings in parallels: this won't change the inductance (the number of turns is unchanged, right), but it will halve the DCr, making the filter more efficient at VLF, by about 6dB.

 

[earthsled]

Wow! had you mentioned that earlier, we could have solved the mistery before I lost my hair!
When connecting in series two identical windings wound on the same core, the inductance is quadrupled. That's because the inductance is proportional to the sqare of the number of turns. In your case, the inductance should mesure 10.2H.
BUT, if the windings are out of phase, the main inductance is nulled, and what remains is only the leakage inductance AND the DCr.
So basically, you have a 780 resistor instead of a 5.2H inductor.

Sorry about that. I have definitely learned a valuable lesson today! 

I've done a quick sim:
You could either wire only one winding and increase the caps to 2uF, or wire the two windings in proper series and the 1.5uF, the former being a tad sharper with a slight overshoot (ca. 0.5dB), the latter being less efficient IMO.
But the best option is to wire both windings in parallels: this won't change the inductance (the number of turns is unchanged, right), but it will halve the DCr, making the filter more efficient at VLF, by about 6dB.

The model that I have been using recently is nice because it is potted in an electrostatic shield. The ratio is 7k:10k, so the turns don't quite match even though the inductance and resistance seem to be nearly equal in both coils. I'm worried about hum and noise in this stage, so the shielded / humbucking configuration remains appealing. If the ratio isn't 1:1, can I still use the transformer wired in proper series?

I have been accruing quite a collection of these tiny transformers in my attempts to find the right inductance values. For the test below, I have a 10k:10k with an inductance of about 6.5H in parallel shown in orange and the shielded 7k:10k wired in proper series shown in blue. The 10k:10k is a bit larger and unshielded.

Can we assume the odd LF response of the orange graph is hum and noise?

Thanks!

 

[earthsled]

Here is the 10k:10k with a comparison of -20dB input signal level. The change in level seems to have more of an effect now.

 

[abbey road d enfer]

The ratio is 7k:10k, so the turns don't quite match even though the inductance and resistance seem to be nearly equal in both coils.

That is odd. If the number of turns is off, that creates losses when they are parallels-connected. These losses appear as a resistor in parallels. Only experimentation can tell you whether it is acceptable or not.

I'm worried about hum and noise in this stage, so the shielded / humbucking configuration remains appealing. If the ratio isn't 1:1, can I still use the transformer wired in proper series?

Yes.

Can we assume the odd LF response of the orange graph is hum and noise?

Yes.

 

[abbey road d enfer]

Here is the 10k:10k with a comparison of -20dB input signal level. The change in level seems to have more of an effect now.

Lack of info there. What other changes have you done? You must have changed the caps...