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mole42uk 10th Jun 2021 9:20 pm

Modifying an inductor?
2 Attachment(s)
I am trying to make a pi filter for 1MHz and a known 10H ferrite cored inductor resonates at about 850kHz. Two of the 10H inductors in parallel offer about 1.3MHz so I conclude that the impedance I'm after is in the region of maybe 12.7H.

I have these two inductors in my RF bits box. On one is the inscription 10.1619, the other reads 10.1691. They look similar but one has a slug and the other doesn't. Having tested these in my filter, they both offer a resonance at about 1.4MHz. My question is this: can anyone guide me how to modify the coils to gve me a resonance at 1MHz, and I'd prefer to use the one with the slug because I can tune that a bit. Each has ~40 turns, I wonder if I need to rewind at say 30 turns, or is that too simplistic?

Herald1360 10th Jun 2021 10:52 pm

Re: Modifying an inductor?
Isn't a pi filter two capacitors to ground with an inductor in the signal path across the top (or the converse for a high pass one)?

Won't having a self resonant inductor cause "interesting" problems?

How about a circuit of what's going on?

Terry_VK5TM 10th Jun 2021 10:58 pm

Re: Modifying an inductor?
Removing turns is the way to go, but, don't remove too many at a a time.

Take off a couple of turns at a time and turn the slug in and out to see the effective range

Bit slow and painful, otherwise, if you have more winding wire about that size, strip and rewind with new wire.

kalee20 10th Jun 2021 11:22 pm

Re: Modifying an inductor?
Hmm... 10uH resonates at 850kHz; 5uH (two 10uH in parallel) resonates at 1.3MHz, so to get 1MHz you're going to need something between 5uH and 10uH.

So, I'm not clear how you arrive at 12.7MHz, can you explain?

G0HZU_JMR 10th Jun 2021 11:48 pm

Re: Modifying an inductor?
The best advice I can offer is to screw the ferrite slug into the centre position of the former and then put the coil in series with a known good 3.3nF poly cap. Not a class 2 ceramic 3.3nF cap though.

Then put this series LC across the output of your sig gen on the Marconi 2955 (via a suitable connector) and use a x10 scope probe to probe the output pin of the connector fitted to the 2955. Then tune the 2955 sig gen in small steps across (say) 500kHz to 1.5MHz and look for the frequency where there is a deep null on the scope. From this frequency and the value of the 3.3nF cap you can work out the inductance of your coil with the slug in the middle of its range.

Then I'd suggest you stick with this inductance value and redesign the Q of your pi filter such that it uses this inductance.

This way you don't have to mess with taking turns off the coil. 10uH is already quite a low value for your application so I'd hesitate to remove any turns if I were you.

mole42uk 11th Jun 2021 5:18 am

Re: Modifying an inductor?

Originally Posted by kalee20 (Post 1381807)
Hmm... 10uH resonates at 850kHz; 5uH (two 10uH in parallel) resonates at 1.3MHz, so to get 1MHz you're going to need something between 5uH and 10uH.

So, I'm not clear how you arrive at 12.7MHz, can you explain?

Silly billy was adding, not dividing.

John_BS 11th Jun 2021 9:10 am

Re: Modifying an inductor?
If you're sure the new coils resonate at 1.4MHz, they must be less than 5uH. It seems strange that the one with a core resonates at the same frequency as I'd expect that to show at least 1.5 times the inductance with the same number of turns? Can you check the turns you have on both.

So, assuming the 1.4MHz figure is correct, that implies your new inductors are around 4uH, so you need to add turns, not remove them. You're aiming for around 7.5 to 8uH, so the number of turns should be increased by a factor of around 1.4

PS if you can post the details of those ceramic coils (diameter of actual coil, length of winding, no of turns), we can work out the inductance pretty accurately.

mole42uk 11th Jun 2021 11:35 am

Re: Modifying an inductor?
1.4MHz is an approximate figure, I didn't write it down at the time. The two coils resonate a different frequencies. I counted the turns at 40 on one of them, I'll check the other. They are the same diameter and both a close-wound single layer. I'll check the dimensions when I get home.

kalee20 11th Jun 2021 2:24 pm

Re: Modifying an inductor?
They may be both 10uH, but they could have different self-capacitance (which obviously affects resonant frequency in a circuit).

That being said, a winding of inductance 10uH is unlikely to have enough self-capacitance to affect things at 1MHz - particularly if single-layer jobbies.

So, looking forward to your report!

mole42uk 11th Jun 2021 4:25 pm

Re: Modifying an inductor?
Here's more information:

Both chokes are 6mm outside diameter, the winding is 40 turns, 9mm long, single layer, close wound. That seems to measure 34 swg?

Using a test setup with a 300Ω input, a 3.3n capacitor to ground on the input and an 8.2n capacitor to ground on the output terminated at 50Ω.

The air core resonates at 1.61MHz, the slugged core resonates at 1.32MHz with the core in the centre of the coil. Perhaps I'm using the wrong word here, the pi filter peaks at these frequencies - is that resonance or not?

Please don't be put off by my mention of the 10H choke earlier, that is one I was using until I found that its resonance was too low for this circuit.

G0HZU_JMR 11th Jun 2021 5:46 pm

Re: Modifying an inductor?
If I understand your latest info correctly it looks like the inductances are about 4uH without the ferrite and 6uH with the ferrite. A quick look on a coil calculator spreadsheet estimates the air core version to be just under 5uH.

I think these inductance values will be too low for your 1MHz pi match application. The Q of that coil won't be good either.

John_BS 11th Jun 2021 9:35 pm

Re: Modifying an inductor?
I agree with Jeremy. If the 6mm is measured over the wire, the coil without core is 4.6uH. To hit your 7.5uH target you'll need to add about 12 turns on top and use the core to trim.

G0HZU_JMR 11th Jun 2021 10:41 pm

Re: Modifying an inductor?
It might be worth revisiting what you want from the pi match filter. I think you obviously need to provide filtering of the unwanted harmonics from the square wave but you also need to produce 4Vpkpk at 1MHz at the output of your pi match if you ultimately want a 2Vpkpk sine wave at the output of the emitter follower.

I think a reasonable target for harmonic rejection at the output would be 35dB (prefer 40dB?) If you try for more than this from a basic pi match it means you end up with a very fussy/peaky circuit that requires accurate component values and a decent coil Q to prevent excessive insertion loss.

I would suggest there are two main approaches assuming you are happy with the 270R series resistor at the HC logic gate. This defines the source impedance as being about 300R.

If you want to convert the 5V square wave to a 4Vpkpk sine wave you can either use the pi match to step up in impedance and provide the correct termination resistance for the pi filter. Or, you can step down in impedance from 300R and choose to leave the pi match unterminated. Both methods are capable of delivering 4Vpkpk at 1MHz but I think that a typical pi match will probably need an inductance somewhere around 22uH for the step up case and maybe 12uH for the step down case assuming you are aiming for a sensible Q (as in a sensible/realistic filtering response) for the network.

It might also be worth exploring how the pi match works. This can be explained without needing equations. It is really quite an elegant network and it is very flexible.

mole42uk 12th Jun 2021 2:21 pm

Re: Modifying an inductor?
2 Attachment(s)
I feel as though I've missed a fundamental point. I made a choke on one of my 5mm formers, calculated to be 75 turns, two layers close wound, 10mm long for 15H. In the test circuit that gave me a very unacceptable result.

So, using one of my commercial 10H 2A chokes in the circuit that follows, I use the TTL output on my function genarator set to a 1MHz square wave connected to the open end of the 270Ω resistor. I then connect the junction of the choke and the 8.2n capacitor to my 'scope. The trace shows a nicely formed sine wave the level of which I can peak by altering the frequency control. The circuit as shown peaks at 735kHz.

If I then add another 10H choke, as in the second circuit, and change nothing else, I still get a lovely sine wave output, but now it peaks at 940kHz. Surely the total inductance is now ~5H but that doesn't make sense. It seems that, to get a peak at 1MHz, which appears to be the point of best harmonic rejection, I need a choke at less than 5H.

The numbers don't add up, so I must have missed the point. It seems quite straight-forward to use an LC pi filter to derive a sine wave from a square wave and, indeed, my breadboard tinkering confirms that the output is exactly what I expect. What isn't as I expect is that the output voltage isn't peaked at 1MHz and nothing I've tried seems to alter that.

G0HZU_JMR 12th Jun 2021 5:22 pm

Re: Modifying an inductor?
2 Attachment(s)
It looks like you are always using 3.3nF and 8.2nF caps in every version of pi filter and this is restricting what you can do.

Probably the best thing to do is study a basic L match circuit that attempts to match a higher resistance (say 300R) to a lower resistance (say 50R) at 1MHz. The lowpass version of the L match consists of a shunt C and a series L as in the image below.

In this case the shunt C always gets placed across the higher resistance (300R) and then the series L then feeds to the lower resistance (50R). This shunt C series L circuit only has one solution at 1MHz and I've given it in the first image below. The C is 1186.3pF and the L is 17.795uH.

For your application this simple L match doesn't give enough filtering rejection on its own as you can see there is only about 18dB rejection at 3MHz. Matching from 300R to 50R only has a ratio of 6 so the solution doen't produce enough selectivity. Also, in your case it won't produce enough output voltage at the 50R port for your application. You can see the blue trace shows the Vout is less than the red Vin trace.

You could swap the L match around and design for a step up from 300R to 500R and this would give enough voltage at the output. The circuit is given below but this only has a resistance ratio of 500/300 = 1.67. The harmonic filtering for this L match is therefore very poor with just 7.5dB rejection at 3MHz. However, at least it gave out a higher voltage on the blue trace. This is the second image below.

G0HZU_JMR 12th Jun 2021 5:25 pm

Re: Modifying an inductor?
3 Attachment(s)
We now know that to get frequency selectivity from an L match we need to match a 'big' ratio change in terms of resistance. But we also need to provide
adequate output voltage to drive your emitter follower with 4Vpkpk.

The solution to this (in your case) is to use two L matches back to back and this forms the familiar pi match network. Suddenly you now have oodles of design flexibility. You can now choose to match down from 300R to a very low resistance (say 13R) and this ticks the box for a high resistance ratio. But you can now choose to match back up from that 13R to a high resistance and get another bonus of even more filtering because you get two L matches back to back and they both have a high resistance ratio. If you match up to (say) 500R in the second L match you get a high output voltage.

This now ticks both boxes. Lots of selectivity and a high output voltage!

See the first image below. This is the first L match from 300R to 13R at 1MHz. Then look at the second image. This is the second L match from 13R up to 500R at 1MHz.

The third image shows these two L matches back to back. The first one matches 300R down to 13R at 1MHz and the second one matches the 13R back up to 500R at 1MHz. When you join them together to make a single 22uH inductor L1+L2 you get a classic pi match network.

When most people design this pi match network they don't think of it being two back to back L matches, they just use a spreadsheet or computer program to design it and they build it.

G0HZU_JMR 12th Jun 2021 5:46 pm

Re: Modifying an inductor?
1 Attachment(s)
Here's a link to a typical online Pi match calculator

I've included a screenshot showing a pi match that does the same as the two L matches back to back at 1MHz. If you study the L and C values they are the same. at 2493pF, 22.4uH and 1948pF

In the case of the online calculator there is a user prompt for circuit Q and I've entered 6.12 in order to get the same L and C network values above.

I think this number is derived from a simple equation based on the L match section with the highest resistance ratio. In this case it is the second section that matches back up from 13R to 500R. It's 30 years or more since I last looked at these equations but I think that is how the Q number is derived in the online calculator.

Q = sqrt((Rhigh/Rlow) -1) = sqrt((500R/13R) -1) = sqrt(38.46 -1) = sqrt(37.46) = 6.12

If you play with the online calculator you should spot one VERY useful feature of the Pi match. If you tinker with the Q number you get to alter the inductance. So it is possible to tweak the Q slightly if you want to end up with a regular (fixed) inductance value of (say) 22uH.

G0HZU_JMR 12th Jun 2021 6:49 pm

Re: Modifying an inductor?
3 Attachment(s)
If this is not making much sense yet then I can suggest a pi match network that should work well with your current breadboard setup with the 50R pulse gen and the series 270R resistor.

The excel spreadsheet below shows a pi match design that matches 300R to 301R at 1MHz with a Q factor of 6.22. This gives really nice circuit values of 3.3nF and 15uH and 3.3nF.

Try setting it up as in the circuit below and try increasing the value of the 300R resistor R2 until you see 4Vpkpk at 1MHz. This assumes your 50R pulse generator can produce a 5Vpkpk square wave at 1MHz.

If you study the excel spreadsheet below you can see that this pi match network is really a pair of L matches that match down from 300R to 7.58R and back up to 301R. This has a Q of 6.22 and this should give fairly good selectivity so you should see a nice sine wave at 1MHz and hopefully 4Vpkpk with R2 set to a suitable value (maybe 560R?).

G0HZU_JMR 12th Jun 2021 8:40 pm

Re: Modifying an inductor?
1 Attachment(s)
Here's a quick simulation using a square wave drive at 5Vpkpk. You can see the circuit outputs a 4Vpkpk sine wave at 1MHz with a 560R load.

A quick and dirty FFT analysis is included in the plot below and this shows that the worst case harmonic is better than 40dB down. It's probably close to 50dB down.

This amount of harmonic rejection is probably not necessary so if you can live with slightly higher harmonics on your sine wave then the network could be redesigned for a lower Q.

This lower Q is achieved by simply changing the 6.22 value down to something lower. This will mean the inductor value will go up.

A Q of 4.1 would deliver component values of 2.2nF and 22uH and 2.2nF. This circuit will have slightly less loss and will be less peaky when adjusting the inductor for max output at 1MHz. However, the price for this is slightly more distortion on the sine wave. I doubt you would be able to tell on a scope though.

mole42uk 13th Jun 2021 7:09 am

Re: Modifying an inductor?
Thank you, some light - I hadn't seen that the pi filter is two L match filters pushed together. That's rather neat and helps me understand the design.


Originally Posted by G0HZU_JMR (Post 1382210)
It looks like you are always using 3.3nF and 8.2nF caps in every version of pi filter and this is restricting what you can do.

I have been using those values because they are, with the 10H inductor, close E24 values to the numerically correct ones.

If I use the pi-match tool for 300Ω into 50Ω at 1MHz, and adjust the Q factor to 6.44 it gives me 10.011 for the coil, 3.4nF and 7.8nF. Unfortunately, one of the mathematical tricks I have never understood is how to turn a formula around so that I can calculate any particular value form the given ones. So what I need to do is calculate the Q from using E24 capacitors, which would be 3n3 and 8n2.

I have been heavily relying on LTspice, using the circuit that we've discussed, and looking at the output waveform as a rough indicator of Q. Now I have seen that, using 3n3, 15 and 3n3 there's a rather neat 4v p-p sine wave across the 560Ω output. So this morning I've learned that, to work properly, I need to understand the importance of the load resistor. Pi-match tells me that these values give me a Q of about 10.

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