Using Nixie tube DC high voltage power supply for tube based pedal builds?

[Marcos - Munky]

Rob, do you have a layout for that smps? I'm drawing the schematic to make a layout, but if you have one, it would help a lot.

[caspercody]

This SMPS is part of a pedal PCB, so I do not have a layout. I could take a picture of the top and bottom to show you how they laid it out?

[Marcos - Munky]

I ended up modding the schematic and layout of a 555 smps I already had here on eagle, so I got a layout ready to test. It ended up pretty small (1.65 x 1.30 inches, iirc). If all goes right, I'll go to the parts store tomorrow or on the day after, then I'll try to etch the board on the weekend. Just need to find a high current tube circuit to test it and see how much current I can get out of it.

Here's a preview on how it went. The pads with no parts drawn around them are for voltage out (top left), inductor (bottom left), voltage in (bottom right) and trimpot to set the voltage (top-ish right).


Edited: I was checking if I had the high voltage 22uF cap and found a 450V one. It's bigger than what I recall, the biggest one I'm used to is 250V. So I'll need to modify the layout a bit to make it fit. Not a really big deal.

[Rambozo96]

Some things I noticed:
1- you're lacking a input cap.
2- I'm not sure if S1 will do anything to the circuit.
3- those 2 220K resistors near output1 can be swapepd for a single 470K resistor (or even a single 220K resistor). The same goes for the 2 ones near output2.
4- you probably can get rid of those 22K resistors. You already have a pot to set resistence.

I was just going by how Maestro designed the dry signal side's preamp. I did find it very odd there was no input cap on it. Probably wouldn't hurt to add one.

[Rob Strand]

I was gonna try a similar design using a MC34063, providing I can get one. It says the output power is 20W, from 40V to 400V. That makes 50mA @ 400V, so high current. This is the design: http://imajeenyus.com/electronics/20111010_40-400V_supply/

I can give a try on this one, but using a 100mH inductor (that's all I have). Do you know how much current this one outputs?

I'm pretty sure you guys mean 100uH and 150uH.

FYI, the output power strongly depends on the inductance the saturation current for the inductor.   If you change the circuit but keep the inductor the same you won't see much improvement.   You could reduce some losses and gain a bit of power that way.  A significant difference between a 20W and a 2W converter is the choice of inductor.

Based on the size of the inductor in the original photo I'm fairly convinced the 100uH inductor has a saturation current of around 1A.   You might be able to push 3A though it but it will not contribute to output power like a 100uH 3A rated inductor.   Once you go past the saturation current the inductor doesn't store much more energy.

Imagine you are pouring buckets of water into a stream.  If the bucket is certain size the only way you can get more flow is increase the rate you are filling and pouring the buckets, or, by getting a bigger bucket.

The "bucket" size is the energy in the inductor,

E = (1/2) L Isat ^2    [J]    ;

Which in you case, L=100uH and saturation current is Isat ~ 1A.
So, E =  50uJ.

Each cycle the inductor energy dumps in the output capacitor.     This is like dumping buckets into the stream.
The average flow is the output power,

P = E * frequency

If the frequency is 50kHz, then

P = 50e-6 * 50kHz = 2.5 W

So that is the *absolute* maximum output power that inductor can produce.   If this assumes no losses and the fact the circuit is fully charging the inductor.    Reality gives a much lower figure, which could be down to 50% or 60%, so only 1.25W to 1.5W.  That equates to about 6mA to 7.5mA at 200V.

If you forced the inductor current to 3A you might get a bit more but the losses will no doubt increase and the efficiency will drop.

The only way to get more power output with that inductor is to increase the frequency.   However, when you increase the frequency the losses go up.   What you could do is play with the timing cap to get the best efficiency for a given load.   You really need to measure the inductor current to know when you have hit the limit.

This guy gives a good account of some circuit variations and some efficiencies.  Keep in mind efficiencies are generally higher at high-ish current, as shown in the graph I posted earlier.    He notes the active pull-down transistor doesn't do much for the 555 since the 555 already has a relatively strong pull-down.  The MC34063 has a different output stage and that generally does benefit from the active pull-down.

https://leap.tardate.com/electronics101/555timer/nixiepowersupply/

[Marcos - Munky]

Yep, 100uH. My bad, since I don't use inductors I'm not used to their correct values.

This is the one I'm using for the other smps I've built:
https://postimg.cc/QKTKS3jP
Unfortunally, I barely have info on the inductor and don't know how to do measures on inductors. All I know is the seller told me it is a 100uH inductor, rated for 3A. For circuits such as preamps, it works nice, and it's small size helps to keep the builds small sized. While I don't really mind trying to get another inductor (or even making my own), they're very hard to find down here. This one is the only one I was able to get at the time, had to buy it online and now they're out of stock.

[Rob Strand]

My best guess is the inductor is about 8mm diameter and 10mm high.

From here it's the S0810 size
https://etwinternational.com/1-1-1-radial-fixed-inductor-127944.html

Unfortunately no data on the S0810, but we can see a smaller S0608 is 710mA(0.44 ohm)  for 100uH and a larger one S1012 is 1.2A (0.18ohm) for 100uH.    So 1A  (and maybe 0.3 ohms) seems like a very good estimate for the S0810 size.

Typical cheap drum from some other data was:
Size:   8.7mm diam 12mm high (max)
FDR0810-101K     1A,  RDC=0.28 ohm

Which is pretty close.

[Marcos - Munky]

I measured it, and it's closer to S0608 than the S0810. So I'll have to find another one or even make my own. Hope I can find another one to buy or materials to make my own.

[duck_arse]

Marcos-M - if you can find a dead power supply from a big box computer - ?ATX? whatever - they will have an assortment of ferrites inside, and some winding wire, and you can wind your own. easy.

[Rob Strand]

I measured it, and it's closer to S0608 than the S0810. So I'll have to find another one or even make my own. Hope I can find another one to buy or materials to make my own.

If you want to use those inductors, putting two in series would give you nearly double the output power/current.  The  series combination is like a 200uH inductor at the same current rating.  The down side is the inductor resistance doubles.   I think there's enough slack in the design for 200uH to work.

[lietuvis]

I have built this one some time ago, it worked well and is small footprint layout.


I've built it as Hi-Fi SRPP preamp and used 0805 smd resistors as most of them are standing so 2.5mm gap+pad was ok to solder them as per layout:





Her is the web for smps :
https://drugscore.blog.fc2.com/blog-entry-90.html#comment215

[marcelomd]

Anyone here tried to design a real boost converter for tubes? With, for example, a LM3478, or something?

I was looking into it today, and I think it's the same difficulty level of a 555 SMPS, since the datasheets come with layout examples.

I may do it in the (near) future. A module with two supplies, Adjustable HV (150-300V) and filament. Should we open a new thread to discuss this?

[Marcos - Munky]

Marcos-M - if you can find a dead power supply from a big box computer - ?ATX? whatever - they will have an assortment of ferrites inside, and some winding wire, and you can wind your own. easy.

I have one at work. Found two ferrite rods, both with 140mm length and 5mm diameter. For winding wires I have some at home (AWG 19, 20, 22, 24, 25, 30, 32 and one I think it's 16). I found this calculator: https://coil32.net/online-calculators/ferrite-rod-calculator.html, but I don't know the value for "initial magnetic permeability" and didn't got to understand what "shift from the rod center" means. Any clue?

If you want to use those inductors, putting two in series would give you nearly double the output power/current.  The  series combination is like a 200uH inductor at the same current rating.  The down side is the inductor resistance doubles.   I think there's enough slack in the design for 200uH to work.

Based on your maths and assuming the efficiency is 50%, the output will be 12.5mA @ 200V. That's a good amount of power, enough to power a few 6CG7 for example. That's something that can be done for testing purposes if the homemade inductor won't work.

I have a smps board based on the UC2842. This one was simulated by Transmogrifox and it'll be very interesting to try a better inductor on this one.

[Marcos - Munky]

Anyone here tried to design a real boost converter for tubes?

The 555 smps can be considered a real boost converter for tubes, since it can power tube designs. It's just a matter of adding a regulated 6.3V from the input voltage to power the heaters. The main difficulty is not the circuit itself, since they all are very similar, but to get a good inductor.

[marcelomd]

Anyone here tried to design a real boost converter for tubes?

The 555 smps can be considered a real boost converter for tubes, since it can power tube designs. It's just a matter of adding a regulated 6.3V from the input voltage to power the heaters. The main difficulty is not the circuit itself, since they all are very similar, but to get a good inductor.

Of course they are. I'm not bashing this design. It works and is widely used for a reason =)

But I meant something designed with application specific controllers, which, and I may be wrong here, "feel" more efficient and reliable.

[Rob Strand]

I don't know the value for "initial magnetic permeability" and didn't got to understand what "shift from the rod center" means. Any clue?

The initial permeability is the permeability of the material.  Common rod materials have permeabilities of 150 to 1000, so the default 600 is a good starting point.  It normally doesn't affect the results much.
[Your length to diameter is very large, so not knowing the permeability accurately starts to affect the results.]

The shift is the position of the centre of the *coil* relative to the end of the *rod*.    You should place the coil centrally so the offset is just half the rod length.

The rod inductor will have a maximum current just like a commercial inductor.    I posted some formulas how to estimate the maximum current in reply #14 of this thread,

https://www.diystompboxes.com/smfforum/index.php?topic=113329.msg1049269#msg1049269

If you choose the turns to get 100uH (or whatever), that automatically determines the maximum current.  The way to get more current is with more ferrite, so you would have to bunch up a few rods in parallel.

Ferrite rod inductor have a strong external magnetic field which can get into other circuits.    The drum/bobbin inductors are kind of like rods but they reduce the external field by keeping a lot of the field between the end-caps of the drum.    You will see ferrite rods used for DC filters since they don't produce large AC fields in that case and they also have relatively high current ratings.
-----------------------
EDIT:   I checked the web calculator against some formulas I worked out a long time ago.  If you set s = 0, implying the coil is centered on the rod, the inductance was within 10% to 15%.  So it looks like s is the offset of the center of the coil to the center of the rod.

For 100uH, I got about 29 to 30 turns with a saturation current of 1.5 to 1.6 amps.   I used 1mm wire for no reason other than to plug in some numbers.

[Marcos - Munky]

Thanks for the explanation on the shift. For the external magnectic field, while it scares a bit when thinking on using the smps for audio applications, I think a Faraday's cage around the smps may solve the problem. Or not

140mm length and 5mm diameter

Crap. I mean 14mm.

The calculator says coil won't fit the ferrite rod. Can I overlay the windings?

[Rob Strand]

The calculator says coil won't fit the ferrite rod. Can I overlay the windings?

Ah, yes I saw that when I was playing

There's no problems making an inductor with two layers.   The problem is the calculator doesn't handle it.

One way to trick the calculator is to set the inductance to (1/4) the final inductance.
The calculator will calculate the number of turns.  When you wind the inductor you wind two layers with that number of turns.

The idea here is inductance depends on the turns squared. By using two layers you get four times the inductance.

If you calculate the saturation current you use the real inductance and real number of turns.

[Marcos - Munky]

So, I calculated a 100uH inductor made by 30AWG wire on a ferrite rod with 14mm length and 5mm diameter, and got N = 62 turns.

From Imax = N * A * Bsat / L and supposing Bsat is 0.3 T (I used the same value you used),
Imax = 3.65 A

(then for the next step I noticed I made an error on the math for two of that inductors in series earlier, but anyway)

E = (1/2) L Isat ^2 = 666.125uJ
P = E * frequency (supposing 50kHz again) = 33 W -> wow!

So, for let's say 250V, I'd get 132 mA assuming no losses, and 66 mA assuming an efficience of 50%. That's a big current! Is the math correct?

[Rob Strand]

So, for let's say 250V, I'd get 132 mA assuming no losses, and 66 mA assuming an efficience of 50%. That's a big current! Is the math correct?

The math is correct *but* the conclusion isn't.   Confused?  Unfortunately there's more to it.

The NE555 is set-up so the off-time is fixed at around 15us.  The on-time varies to regulate the voltage.   In the original circuit with a light load the on-time shrinks and the frequency ends up at 45kHz to 50kHz.  When the output voltage is low, the on-time increases which charges-up the inductor more and then dumps more energy into the load.   If the output voltage is very low NE555 free oscillates at about 31kHz (period about 32us) and the largest on-time is 17us.

The 17us on-time (Ton) limit  prevents the inductor charging up indefinitely.    If the input voltage is Vin=12V and the total voltage drop for the inductor and MOSFET  is Vdrop =1V, there is 11V charging up the inductor.     You can work out the inductor current at the end of  the on-time period as,

Ipk  = (Vin - Vdrop) * Ton / L

So when you do that,  you get ipk = 1.87A ; nowhere near 3.65A.    Also notice when Ton = 17us  the frequency has dropped to 31kHz.  That means less power.   So using our rough calculations,

E  = (1/2) ip^2 * 100uH  = 175uJ
Pout = efficiency * f * E   = 0.5 * 31e3 * 175e-6 = 2.7W;
or 13.5mA @ 200V.

If you change the NE555 circuit to make Ton larger the inductor has more time to charge up.   That further drops the frequency but you will still win power.  That would need Ton=33us and with Toff=15us, the frequency ends up at 21kHz and Pout = 7W.

If the inductance is smaller it can charge up to the high current within the 17us Turn-on time.  A smaller inductance on the same
core will have a higher current rating.  Maybe somewhere around 33uH to 47uH.

BTW, the higher inductor current means more ripple on the output.

If you aren't interested in too much voltage adjustment the frequency can be increased by reducing the NE555 Toff time a bit.