I need to make a "bullet proof" charge pump circuit that I could drop in to any circuit I throw at it. I know that's dangerous thinking since power is a finicky old lady. I also need these to be compatible with boss style supplies and daisy chain power supplies. I know I'm asking for the moon, but I want something that 90% of the time, the power circuit is over overkill, but for that other 10% the circuit is robust enough that there won't be any problems. This is what I've come up with so far. I'm still waiting on some LT1054s to arrive but I'm 99% sure this will work. Wanted to of course run it by all of you to take a look or possibly take something of value from it.
This is a mishmash of "road rage" (http://music.codydeschenes.com/wp-content/uploads/2013/09/Road-Rage-Madbean-Transfer.pdf) and things I've learned from here, btw R.G. is a national treasure.....
I imagine the whole idea originates from the klon centaur, electrosmash has an amazing writeup on the pedal if you haven't seen it already.
https://www.electrosmash.com/klon-centaur-analysis
(https://i.imgur.com/jputpyy.jpg)
Its LT1054 based for higher current demands than with a MAX1044 used previously. I included (I believe) RGs trick on switching/reverse polarity protection) power switching circuit with a 2n2907. From my understanding the 2n2907 is capable of providing 300mA and the LT1054 can't take more than 200mA so that should not be a problem.
1n4742A is for overvoltage protection (12v input max).
This entire thing needs a "true ground" and thats the grounding pin on the DC jack. Everything eventually hooks back up here via star ground.
Effect is powered on with the TS/TRS switching trick on the output jack. Inserting a jack shorts Vss on the ring to "true ground" on the sleeve thereby turning on the BJT and turning on the LT1054. Each Vss from any and all audio circuits (transistor boost stages or OP amps for example) connect to Vss on the audio jack ring forming a "star audio ground" to the ring. A single wire on the sleeve hooks up to true ground, so this "star audio ground" on the ring is shorted to true ground.
Each Vss pertaining to power (power conditioning caps for example) is connected to LT1054s Vss pin first (pin 3), and then THIS pin is connected to true ground via wire or thick 20mil trace.
Any and all LEDs in the audio circuit connect to true ground via star grounding.
A grounding spring (my personal perference) connects the metal enclosure to true ground (DC ground) vi thick 20mil trace or wire.
You may be wondering why I'm spending so much time on grounding. After reading so many of RGs posts on "sewer ground, I've learned proper grounding SEEMS simple enough, but it REALLY requires great attention to detail, maybe the most attention out of everything pertaining to guitar pedals. Google "site:diystompboxes.com "sewer ground" and just read everything you find.
The 15pF cap connecting pin 6 (Vref) to pin 7 (OSC ref) raises the internal clock outside the audio range to help mitigate "whining". All of this is from the datasheet (They recommend 5pF-25pF I think). In my research here, this whining is apparently a huge problem that is difficult to fix and trace/wire placement of ground with signal traces is critical. Cross talk is a problem with audio ins being next to audio outs and I'm guessing form some weird cross-talk loop. So keeping signal ins as far from possible from signal outs. Remember all this nonsense is just loops. Signal in forms a loop, signal out forms a loop and power is another loop.
1n5817 diodes are low forward voltage types thereby lowering the voltage drop at 18V output. Using 1n4004s or whathaveyou will cause 18v output to be lower, than with low forward voltage diodes.
LT1054s datasheet calls for tantalum caps for C29/C28. I normally go with a fat ass aluminum electrolytic cap and a smaller alumina electrolytic cap, "large" MLCC and a "small" MLCC all in parallel for power filtering like everyone else, however with the datasheet calling for low ESR tantalum caps, I'm not sure if you would then leave out or include the MLCC caps.
C8 in parallel with C33 and C6 in parallel with C34 are cost saving measures. Normally (per datasheet) these would just be large valued, low ESR tarantula caps, however they are balls expensive. So you can go with a fat electrolytic (C8 and C6) each in parallel with a low ESR, lower valued tantalum to save some moolah. All this is from the datasheet btw.
All the remaining 100uF aluminum electrolytic caps in parallel with 100nF MLCCs are decoupling caps for each power point, 18v, 4.5v and -9v. No need for these on 9v since the input decoupling already decouple 9v point.
Can't figure out the need for Q9 series pass transistor..
Quote from: antonis on July 09, 2020, 05:21:38 PM
Can't figure out the need for Q9 series pass transistor..
reverse polarity protection, current can only go one direction. I think also theres less voltage drop like if you used a diode.
I think R.G. came up with it.
Hello,
When I read your original post, I couldn't help but think about this (:icon_redface:) :
https://youtu.be/7YyBtMxZgQs?t=15
If it works as you expects (I'm not able to help), I wonder why you need, so much filtering, and such a drop-in solution (from a cost / effort / efficiency perspective, unless you decide to populate some sections only when you need the corresponding voltage).
But I'm probably missing something and it would be great if you could explain it :-)
Side note: the way you've annotated / oriented your diagram is quite difficult to read (at least for me).
Side question: is 20 mil really "thick" ? (all my PCBs' tracks use 20 mil as default, with 40 mils for power tracks)
Best regards,
M.
Quote from: Boner on July 09, 2020, 05:23:49 PM
reverse polarity protection, current can only go one direction. I think also theres less voltage drop like if you used a diode.
I think R.G. came up with it.
I also think R.G. will come and tell you that another BJT (n-p-n) is needed for shunting Q9 base current, in conjunction with a voltage set on the Base of the former BJT and a diode for preventing reverse B-E breakdown voltage..
Quote from: Boner on July 09, 2020, 04:26:07 PM
btw R.G. is a national treasure.....
Hey! You can't keep him all to yourself! :D
I see it as taking Fig 15 (the standard inverter) but with no regulation, merged with Fig 20 (the positive doubler),
https://www.ti.com/lit/ds/symlink/lt1054.pdf
FYI: As Antonis implied, you should follow RG's BJT version exactly. The BJT versions will do weird stuff if you just follow the MOSFET pattern. (You need to think of about what happens with the BE an BC junctions in with all fault cases. Remember the BE junction won't block like an Si diode it will break down at 7 to 10V like a Zener.)
If you want a low-drop reverse polarity protection then you might also look for a P-Channel MOSFET (with DS body diode) with a low Rds(on) like this:
(https://i.postimg.cc/xqJHQ8rq/2020-07-10-13-27-06-1-Schematic-C-Users-Niek-ten-Brinke-Dropbox-Eagle-Projects-Projects-Distortio.png) (https://postimg.cc/xqJHQ8rq)
How this works is that (in normal operation) the body diode will raise the voltage on the source to (f.e.) 8.3V. The gate is 0V so there is a negative Vgs voltage which will open the 'channel' (and which will now have a series resistance of Rds(on)). I've used the BS250. This one has a Rds(on) of 10ohms (which is pretty high actually, you can easily get around 1ohm or even lower), for a low-current overdrive pedal (say 25mA) this will mean a voltage drop of 25*10^-3 * 10 = 0.25V... :)
In reverse polarity mode, Vgs will be positive and the fet will remain closed :)
You just need to make sure that the Vgs threshold is low enough so it can actually turn on. Also the maximum ratings for the Vgs should not be exceeded
Quote from: Boner on July 09, 2020, 04:26:07 PM
.... low ESR tarantula caps, however they are balls expensive. ....
and, they will bite you if you don't pay attention to their polarity markings.
Wow that really is overkill :icon_lol:
I did this simple layout not too long ago so one could quickly make a drop-in board for any circuit that doesn't have the charge pump on-board already. I usually velcro them near the bypass switch or wherever it'll fit. I've had absolutely no problems with it.
Layouts here (https://drive.google.com/drive/folders/10qh8lq005E5i7RaFupnBPqFBzozQ_PG4)
-KM
Quote from: niektb on July 10, 2020, 07:33:44 AM
If you want a low-drop reverse polarity protection then you might also look for a P-Channel MOSFET (with DS body diode) with a low Rds(on) like this:
(https://i.postimg.cc/xqJHQ8rq/2020-07-10-13-27-06-1-Schematic-C-Users-Niek-ten-Brinke-Dropbox-Eagle-Projects-Projects-Distortio.png) (https://postimg.cc/xqJHQ8rq)
How this works is that (in normal operation) the body diode will raise the voltage on the source to (f.e.) 8.3V. The gate is 0V so there is a negative Vgs voltage which will open the 'channel' (and which will now have a series resistance of Rds(on)). I've used the BS250. This one has a Rds(on) of 10ohms (which is pretty high actually, you can easily get around 1ohm or even lower), for a low-current overdrive pedal (say 25mA) this will mean a voltage drop of 25*10^-3 * 10 = 0.25V... :)
In reverse polarity mode, Vgs will be positive and the fet will remain closed :)
You just need to make sure that the Vgs threshold is low enough so it can actually turn on. Also the maximum ratings for the Vgs should not be exceeded
BSP250 = rdsOn 0,25 Ohm or less
http://www.farnell.com/datasheets/454177.pdf
Datasheets are good..
Real word items aren't so good..
I Like the idea of the mosfet switch! How would you best incorporate it with the traditional "TS/TRS on switch" trick" to power on/off the effect?
If you're running at 18V, what's the need for the 4.5V "midpoint" supply, when you've already got a 9V actual midpoint?
Quote from: ElectricDruid on July 10, 2020, 05:30:11 PM
If you're running at 18V, what's the need for the 4.5V "midpoint" supply, when you've already got a 9V actual midpoint?
honestly none, put it there as a "If I need that, there it is" but will most likely not have it in any circuit
Quote from: ElectricDruid on July 10, 2020, 05:30:11 PM
If you're running at 18V, what's the need for the 4.5V "midpoint" supply, when you've already got a 9V actual midpoint?
There are a couple reasons I can think of:
1) You rarely get 18v once a load is on the charge pump circuit. It does work best if you use low ESR caps on the pump circuit but generally you are going to end up with something between 17-18v depending on the current demand.
2) Even in best case scenario (you have a steady 18v out of the pump circuit) even the "good" supplies aren't putting out exactly 9v. My One Spot is a steady 9.42v with or without any current draw on it.
Maybe there's a reason not to have virtual ground the same as the power supply input, as well? That's above my pay grade.
QuoteI Like the idea of the mosfet switch! How would you best incorporate it with the traditional "TS/TRS on switch" trick" to power on/off the effect?
Here's some of RG's pages showing the basic ideas.
1) Reverse Protection
http://www.geofex.com/Article_Folders/mosswitch/mosswitch.htm
This shows the correct orientation of a P-channel MOSFET for reverse polarity protection.
However, the switching is normal ie. it doesn't use the MOSFET as a switch.
2) Quiet Switching for Stereo Jack
http://www.radanpro.com/Radan2400/Napajanje/New%20Page%202.htm
Diagram has bug: swap ring and tip
It is important to wire the battery -V and the power supply -V directly to the circuit. Only the base/gate line goes to the ring.
I misunderstood the purpose of the transistor on your original ckt.
Quote from: antonis on July 09, 2020, 07:03:19 PM
Quote from: Boner on July 09, 2020, 05:23:49 PM
reverse polarity protection, current can only go one direction. I think also theres less voltage drop like if you used a diode.
I think R.G. came up with it.
I also think R.G. will come and tell you that another BJT (n-p-n) is needed for shunting Q9 base current, in conjunction with a voltage set on the Base of the former BJT and a diode for preventing reverse B-E breakdown voltage..
Quote from: Rob Strand on July 10, 2020, 07:41:20 PM
QuoteI Like the idea of the mosfet switch! How would you best incorporate it with the traditional "TS/TRS on switch" trick" to power on/off the effect?
2) Quiet Switching for Stereo Jack
http://www.radanpro.com/Radan2400/Napajanje/New%20Page%202.htm
So is a second BJT required in the Quiet Switching for Stereo Jack method?
QuoteSo is a second BJT required in the Quiet Switching for Stereo Jack method?
Good point. I was thinking to merge the two but unfortunately that won't work!
In order for the reverse protection to work, the direction of P-channel MOSFET passes current via the internal diode under normal polarity conditions. So that means the power won't turn off!
(With two MOSFETs you could probably add over voltage protection - perhaps a bit of feature creep.)
Quote from: bean on July 10, 2020, 07:19:09 PM....even the "good" supplies aren't putting out exactly 9v. My One Spot is a steady 9.42v......
We say "9V" so we don't get a 1.5V or a 67V battery. No battery is ever what it says on the pack. A "12V" car battery floats at 12.6V, is typically cycled 13.8V down below 12.0V.
The One Spot probably aims for "typical battery voltage", NOT "marketing number".
The fresh "1.5V" zinc cell was close to 1.56V (around 1960). Reliable enough to calibrate meters. (They keep dinking the chemistry and it may not be 1.56V now.) So a six-pack would be 9.36V.
Your Spot showing 9.42V is certainly possible for somebody's "9V batt", is well within 1% of my old-old reference.
Quote from: PRR on July 10, 2020, 09:02:51 PM
Quote from: bean on July 10, 2020, 07:19:09 PM....even the "good" supplies aren't putting out exactly 9v. My One Spot is a steady 9.42v......
We say "9V" so we don't get a 1.5V or a 67V battery. No battery is ever what it says on the pack. A "12V" car battery floats at 12.6V, is typically cycled 13.8V down below 12.0V.
The One Spot probably aims for "typical battery voltage", NOT "marketing number".
The fresh "1.5V" zinc cell was close to 1.56V (around 1960). Reliable enough to calibrate meters. (They keep dinking the chemistry and it may not be 1.56V now.) So a six-pack would be 9.36V.
Your Spot showing 9.42V is certainly possible for somebody's "9V batt", is well within 1% of my old-old reference.
Well, sure. But the point I was making is about why it might be a good idea to have a voltage divider in
this application instead of relying on the actual supply voltage. I think most people here realize 9v doesn't generally mean "or possibly 1.5v" :)
PS I'm being cheeky here
What I did in my first pedal was this, I did however omit it in the later pedals because I use them always in context of a pedalboard (and without option for battery) meaning that the input jack is always plugged in:
(https://i.postimg.cc/9rNjhXJp/2020-07-11-08-04-47-1-Schematic-C-Users-Niek-ten-Brinke-Dropbox-Eagle-Projects-Projects-Distortio.png) (https://postimg.cc/9rNjhXJp)
So basically I have 2 seperate grounds (one for the FET and a green LED that indicates power status) and the main ground. They are connected to the ring and sleeve of your input jack (meaning they will be shorted together as soon as you plug something in) and TADA! It works!
Quote from: bean on July 10, 2020, 07:19:09 PM
Quote from: ElectricDruid on July 10, 2020, 05:30:11 PM
If you're running at 18V, what's the need for the 4.5V "midpoint" supply, when you've already got a 9V actual midpoint?
There are a couple reasons I can think of:
1) You rarely get 18v once a load is on the charge pump circuit. It does work best if you use low ESR caps on the pump circuit but generally you are going to end up with something between 17-18v depending on the current demand.
2) Even in best case scenario (you have a steady 18v out of the pump circuit) even the "good" supplies aren't putting out exactly 9v. My One Spot is a steady 9.42v with or without any current draw on it.
I agree, Brian, but I don't think it's that crucial. Ok, maybe we've got a 17V rail, and maybe our incoming power is 9.5V. So things are a little bit asymmetrical, but we've still gained a ton of headroom, which is presumably the general idea. I'm not going to stress about a volt here or there.
Quote
Maybe there's a reason not to have virtual ground the same as the power supply input, as well? That's above my pay grade.
Yeah, this does sound like a risk, I agree, although its above my pay grade too. Still, worth a shot, right?!
So you guys are saying using 9v as a virtual ground in a 0/9/18 volt configuration might be asking for trouble, but a -9/0/9 and using actual groundy-ground as a virtual ground for audio signals would be more ideal?
That makes sense.
*edit*
but then if you're using JFET boosters, which I do all the time, wouldn't ground in this case now be -9 if you wanted 18 volts of headroom?
Could you use 18 volts referenced to zero (true zero) for jfet boosters and whathaveyou AND -9/9 with virtual ground being true zero for op amps and anything else requiring a virtual ground?
Wouldn't that solve any weird problems associated with using a power rail as a virtual ground?
*double edit*
came up with that after some spice tinkering. I'm embarrassed to admit that I can't get a BJT pair like you guys were talking about to work. However this seems to work....
(https://i.imgur.com/qu8xjzL.jpg)
Left out the 15pF timing cap that goes between pin 2 and 7 to make it more legible. This cap raises the clock freq to help with noise.
D1 protects from input voltages exceeding 12v. Q13 turns the circuit on when a jack is plugged into the output socket. Q9 turns on only for positive voltages (reverse protection). D16 protects Q9 from static discharge which could hurt it. All that nonsense ensures safe power coming into the rest of the circuit that is the charge pump..... Thoughts?
Quote from: Boner on July 11, 2020, 12:47:08 PM
So you guys are saying using 9v as a virtual ground in a 0/9/18 volt configuration might be asking for trouble, but a -9/0/9 and using actual groundy-ground as a virtual ground for audio signals would be more ideal?
That makes sense.
Yeah...be cautious of stuff that sounds like it makes sense but which no-one can actually justify. It might just sound plausible, but actually be nonsense.
I'm not offering any justification. ;)
Voltage is all relative, so there is no difference between +9, 0v, -9V and +18V, +9V, 0V. They are literally the same thing. The only question is how you derived those voltages, and how much noise that gives you on one wire over others and whether your circuit might be more sensitive to noise on one than another. The more I think about it, the less I find it makes sense to think it makes any odds.
One advantage to bi-polar circuits is they do reduce part count. Also (since I've taken a lot influence from Peter @ VFE) you can ditch the 100n decoupling caps on the +9 and -9 at the supply end and instead place them at critical points on the audio circuit. So, for example, a circuit utilizing a dual op-amp with a bi-polar supply: place the 100n decoupling caps right at pins 4 and 8 as close to the IC as you can get them. This is really good practice as far as circuit design.
Like I tried to explain in my last post, you can ditch the whole Q13 thingy if you connect the lower leg of R61 to the jack sleeve :) (I used that in my pedals and it absolutely works :))
QuoteLike I tried to explain in my last post, you can ditch the whole Q13 thingy if you connect the lower leg of R61 to the jack sleeve :) (I used that in my pedals and it absolutely works :))
Then the power doesn't turn off when you pull the jack - it powers through the MOSFET body diode.
If you don't use batteries you might not notice the problem because there's nothing to go flat. What you will notice with a power supply is the LED still goes off and on with footswitch when the Jack is out.
Quote from: Rob Strand on July 13, 2020, 04:07:20 AM
QuoteLike I tried to explain in my last post, you can ditch the whole Q13 thingy if you connect the lower leg of R61 to the jack sleeve :) (I used that in my pedals and it absolutely works :))
Then the power doesn't turn off when you pull the jack - it powers through the MOSFET body diode.
[...]
The ground is disconnected...?
(https://i.postimg.cc/9rNjhXJp/2020-07-11-08-04-47-1-Schematic-C-Users-Niek-ten-Brinke-Dropbox-Eagle-Projects-Projects-Distortio.png) (https://postimg.cc/9rNjhXJp)
QuoteThe ground is disconnected...?
The point of having the solid-state switch was so the power ground didn't pass through the input socket. The charge-pumps have current pulses on the input supply the solid-state switch stops those currents passing through the input socket altogether. A big input cap, like C29 on boner's schematic, certainly helps reduce those currents but if you want to be 100% sure those current are gone then it's best to add the solid-state switch.
It's a bit of a value judgement which way to go. The current pulses are likely to be more problematic on higher gain pedals than unity gain pedals.
I'm assuming boner wanted a solid-state switch as his original ckt has one (Antonis and I thought it was for reverse protection but that got cleared up in the thread).
To boner:
One thing bad about the input zener protection is if the supply is reversed polaritied the zener will forward conduct short circuit the power-rail just like a SI diode. That completely defeats the purpose of having the graceful disconnection of MOSFET reverse polarity ckt. One fix would be to add a silicon diode in series with the zener.
Quote from: ElectricDruid on July 12, 2020, 07:54:42 PM
Quote from: Boner on July 11, 2020, 12:47:08 PM
So you guys are saying using 9v as a virtual ground in a 0/9/18 volt configuration might be asking for trouble, but a -9/0/9 and using actual groundy-ground as a virtual ground for audio signals would be more ideal?
That makes sense.
Yeah...be cautious of stuff that sounds like it makes sense but which no-one can actually justify. It might just sound plausible, but actually be nonsense.
I'm not offering any justification. ;)
but man if you think about it like an electron it makes total sense man. (https://i.imgur.com/jHGuXNA.png)
Quote from: bean on July 12, 2020, 10:07:12 PM
ditch the 100n decoupling caps on the +9 and -9 at the supply end and instead place them at critical points on the audio circuit. So, for example, a circuit utilizing a dual op-amp with a bi-polar supply: place the 100n decoupling caps right at pins 4 and 8 as close to the IC as you can get them. This is really good practice as far as circuit design.
Thanks! 100n the norm?
Quote from: Rob Strand on July 13, 2020, 06:34:49 AM
add a silicon diode in series with the zener.
Good idea... like this? Wouldnt the Si diode short as well?
(https://i.imgur.com/R8LJh5Z.jpgf)
QuoteGood idea... like this? Wouldnt the Si diode short as well?
The diode needs to go the other way, so when the input voltage is reversed the diode and the zener are not conducting. When the input voltage is over voltage the zener kicks and the si conductors. That would increase the effective zener clamp voltage to 0.6+12 = 12.6V.
So with that mod, the circuit would have these two undesirable features,
- For the reverse-voltage case technically the circuit still has a problem since the BE junction of the switch transistor Q13 gets reverse biased when the input is more than say 8V (depending on the transistor).
- For the over-voltage case the zener is subject to large currents and is likely to blow-up. The zeners impedance isn't that low so under high faults the zener voltage could be raised well above the labelled zener voltage. Zeners aren't that tough. Some of the TVS devices are a little tougher.
Zener tolerances mean the zener could strongly conduct at a lower voltage than "expected". A 10% tolerance zener might have considerable conduction at 12V*0.9 =10.8V. Add the Si diode and that's 11.4V.
The voltage clamp boils down to choosing a voltage high enough not to kick-in undesirably and clamping the voltage to a low enough voltage to protect the circuit. Many circuits will handle much higher voltages than 10V but those charge pump devices have a tendency to fry quite close to their absolute max input voltage.
Suppose we allow the circuit to suffer under high input voltages. "Less angry" alternative would be to move the zener to the output size of the switch. In this position you probably don't need to add the silicon diode. What you would do is choose R37 to be large as possible so that it limit the fault current through the zener, Ifault = Icollector = hFe * Ibase. It's still not great since the switch requires R37 to be chosen based on 9V and the fault current will be determined based on Ibase at the maximum input voltage (say 40V?). It's still an "angry" method of protection just that it's not as angry as the straight zener and the "anger" is shared between the zener and the transistor Q13.
There are more graceful over-voltage protection schemes. They require more parts but not a lot more.
Image:
(https://www.electricaltechnology.org/wp-content/uploads/2019/11/Over-voltage-protection-using-Zener-Diode.png)
Link
https://www.electricaltechnology.org/wp-content/uploads/2019/11/Over-voltage-protection-using-Zener-Diode.png
From,
https://www.electricaltechnology.org/2019/11/simple-overvoltage-protection-circuit-using-zener-diode.html
There's no right answer with this stuff. It's a matter of how far you want to go. For larger more expensive products a fully protected supply input doen't make a big impact on the cost or the size of the product. But for a pedal, anything from no protection to fully protection is up for grabs. I guess it is a matter of knowing where you want to drawn the line.
Quote from: Rob Strand on July 13, 2020, 07:36:30 PMThere are more graceful over-voltage protection schemes. They require more parts but not a lot more.
Image: (https://www.electricaltechnology.org/wp-content/uploads/2019/11/Over-voltage-protection-using-Zener-Diode.png)
Is that really a good example for our purposes?
At a glance my so-called brain thought it looked like a cut-out. The Idiot gives the same trend. From small to 5.1V, out=in. Past 5.1V, out=zero. Since we can conveniently find regulators to hold a 5V output up to 35V input, I am not sure how this is better.
(https://i.postimg.cc/wtv3X96x/5-1-V-overvolt-42.gif) (https://postimg.cc/wtv3X96x)
QuoteIs that really a good example for our purposes?
At a glance my so-called brain thought it looked like a cut-out. The Idiot gives the same trend. From small to 5.1V, out=in. Past 5.1V, out=zero. Since we can conveniently find regulators to hold a 5V output up to 35V input, I am not sure how this is better.
The goal is to cut out, it disconnects the high voltage.
Sure a regulator will work, in fact that site I linked has such an example, but the regulator introduces a significant drop-out. Also with a high input voltage the regulator would have to handle the corresponding heat dissipation.
The cut-off method saves the day and we don't have to deal with any of the regulator issues.
We could design a universal input product but that's going to be a lot more work. Probably a universal input switchmode. Then we could create many output rails. Many products with universal main input products are like that but it's probably raising the bar too high for circuit complexity.