Is there a middle ground between 'Charge Pump' and 'SMPS'?

[Scruffie]

Charge pumps such as the MAX/TC1044, ICL7660 & LT1054 are great for simplicity and ease of use in circuits to get an inverted voltage or for a few stages of doubling but after about 4 stages they start to take up quite a bit of board space.

SMPS based around the 555 timer and MAX1771 are nice for getting a high voltage in the 150-300V range and don't take up too much board space but are a bit finicky and require a few slightly specialised component choices and some very careful use to get the best out of them.

Is there a middle ground between the two that can put out from a 12V supply say... 60-120V and a decent (at least 10) amount of mA and can use some fairly standardised components with a circuit about the same sort of size as an SMPS or is that just being a bit too picky about the already available options and I should count my self lucky 

I know there has been SMPS style supplies built with the Charge Pump ICs in a hybrid fashion so perhaps something in that direction.

[samhay]

There are various options, but I guess you are probably not interested in incorporating e.g. a flywheel into your stompbox?

There are plenty of DC:DC converters like the Murata NKA series, but these are often best for generating bipolar supplies for e.g. op-amps.
http://www.murata-ps.com/data/power/ncl/kdc_nka.pdf

I found this one, which looks promising if you are willing to pay for it:
http://uk.rs-online.com/web/p/isolated-dc-dc-converters/7331102/

[duck_arse]

hah! I was wondering how to go about asking about the inductors for the mc33063 dc-dc converter, now someone else can do it for me.

I'm just started looking at a 12V in, 50V out for a xcaster, but I find winding theraputic. and that's the IC's I've got. I can post wot I find if yr interested, but somebody has probably already done it years ago.

[edit:] should say mc34063.

[AdamM]

this is still a SMPS, but looks relatively simple if you can find your way through the multilingual description - and cheap, if you have a dead atx psu hanging around

should be good for a few mA out...

http://320volt.com/en/atx-stantby-trafosu-ile-dc-yuksek-voltaj/

[Scruffie]

this is still a SMPS, but looks relatively simple if you can find your way through the multilingual description - and cheap, if you have a dead atx psu hanging around

should be good for a few mA out...

http://320volt.com/en/atx-stantby-trafosu-ile-dc-yuksek-voltaj/

Looks good but I should have stipulated non-transformer based.

There are various options, but I guess you are probably not interested in incorporating e.g. a flywheel into your stompbox?

There are plenty of DC:DC converters like the Murata NKA series, but these are often best for generating bipolar supplies for e.g. op-amps.
http://www.murata-ps.com/data/power/ncl/kdc_nka.pdf

I found this one, which looks promising if you are willing to pay for it:
http://uk.rs-online.com/web/p/isolated-dc-dc-converters/7331102/

Umm... no I hadn't planned on adding a flywheel.

Wow that's pricey, it'd be good if it was on a £3 8-Pin DIP... but I think for the £30 saving i'd just put up with the board space taken by a charge pump.

Perhaps just diddling with the 555 based SMPS will be the best way.

[R.G.]

Let's play power supply. 

Capacitive up-verters suffer from high impedance. The stock capacitive voltage multiplier is something like a C@ckcroft-Walton, and it can produce incredibly high voltages, but is really disabled for any significant currents. To get any significant current, you have to go with transformers or inductive up-verters.

So no, there isn't any easy halfway between charge pump and SMPS without transformers. However, it *is* incredibly easy to design an SMPS today. The control chips are largely condensed down to one eight pin dip for all the controls, and one MOSFET for the switch. They take care of all the ugly gotchas that used to bedevil me when I messed with this stuff.

With either capacitive or inductive (well, I guess transformer too!) you have to know your load - that is, how may volts you want out and how much maximum current it will supply. You're likely wanting to power tubes from low voltage (if not, correct me) so you're looking for one to six plate loads, possibly 12AX7 plates at less than 1ma per. So call it two to six ma. Let's guess at a voltage. Will 200V be enough?

If so, you need 200V * 0.006 = 1.2W. Since transformers are out, you will need an inductor. Inductors store energy as E = 1/2 * L*I^2. It can store, then dump this amount. I has to be chosen such that the L and the number of turns and the core and all that stuff do not saturate the core.

There are several ways to go, depending on what you pick first. If you pick frequency, (for example, 50kHz), then your inductor has to produce 1.2W, and it does this by dumping its bucket of energy 50,000 times per second, so each bucket of energy has to be at least 1.2W/50,000 = 24uJ. Setting that equal to the energy in the inductor, you have L*I^2=2*24uJ, or 48uJ. Alls you gots to do is to pick an inductor that your input voltage can charge to the necessary current in less than half of 1/50,000, and you're in.

This is because the current in an inductor is V = L di/dt. "dt" is half the period of a 50,000 wave, which is what you picked to start this, or 10uS, and V is your supply voltage, less your power switch saturation voltage at peak current. "di" is the current in the inductor from zero (*note, non-return-to-zero versions exist!) to a peak current less than its maximum current before saturation.

Actually, this is pessimistic. An inductor charging from one voltage and dumping to a higher voltage dumps faster than it charges. If you use 12V to charge, and dump at 200V, then the dump time is 12/(200-12) as long as the charge, or about 0.07, so you can spend 92% of your cycle charging. But you have to be sure that you always give it time to fully dump (in return-to-zero upverters). But if you use 50%, you're safe; you just buy a bigger inductor than you would otherwise.

The classical solution is two equations in two unknowns.
L*I^2 = 48E-6 and  L*I /(10uS) = 12V

It's probably better solved for the L*I product that works. Any inductor with a bigger L*I product can be made to work.

I can *hear* all the eyes glazing over this. 

Seriously, it's not that bad.

[duck_arse]

well, I some specifics question, and I'm here now, so I may as well still arks it.



with regard to the attached (modified) page from the spec sheet for the mc34063, I'm having trouble understanding the voltages encountered in the circuits shown.

in the fig 6 configuration, does my chosen output voltage appear at the collector of Q1, or only at the K of the
1N5819? if at the collector, then the internal 40V Vce shown in the maximum ratings limits the (minimum parts count) output voltage possible.

so, we put an external transistor, maybe a 2N6517, (modern times means a mosfet?) as shown in the current boosted config of fig 7a. the internal
transistor is now only seeing 0V6 less (?) than before, and is no solution.

but in fig 7a, the inductor is shown with a tapped winding to the internal collector. if the tap is at 50%, is there now only half the Vout voltage
appearing on the internal? in this config, with a Vin of 12V, can I set the Vout to, say 65V without exceeding the max ratings? or do I really
need to add a secondary to the inductor?

would I be better off leaving these and the low ESR caps in the back of the drawer, and finding a different method?

[R.G.]

in the fig 6 configuration, does my chosen output voltage appear at the collector of Q1, or only at the K of the
1N5819? if at the collector, then the internal 40V Vce shown in the maximum ratings limits the (minimum parts count) output voltage possible.

The inductor voltage rises so the sum of the supply voltage and the inductor voltage forward biases the 1N5819. This is the only way the diode can let current from the inductor through to the output. So the voltage at the collector has to be the maximum output voltage plus the instantaneous forward voltage of the diode. Yes - the internal 40Vce limits the maximum output voltage in this configuration.

so, we put an external transistor, maybe a 2N6517, (modern times means a mosfet?) as shown in the current boosted config of fig 7a. the internal
transistor is now only seeing 0V6 less (?) than before, and is no solution.

You're correct. However, I would connect the collector of the internal chip transistor to the supply voltage, not to the collector of the external boost transistor. This would limit the voltage on the internal switch transistor. Using a MOSFET will require more tinkering to be sure you have the correct gate drive. It's probably better to use a chip with a more modern internal setup.

but in fig 7a, the inductor is shown with a tapped winding to the internal collector. if the tap is at 50%, is there now only half the Vout voltage
appearing on the internal? in this config, with a Vin of 12V, can I set the Vout to, say 65V without exceeding the max ratings? or do I really
need to add a secondary to the inductor?

Yes, the tapped inductor offers a lower-voltage connection for the internal collector. However, this also lowers the collector voltage during the drive portion of the cycle. Perhaps this diminishes the internal dissipation  on the chip - a good thing - and this was why it was noted. But again, I would use a vastly superior MOSFET for switching, now that these are cheap and easily available.

would I be better off leaving these and the low ESR caps in the back of the drawer, and finding a different method?

I don't think so. I'd go to a similar chip specialized for MOSFET output devices, as is most common today. There are a number of them available.

[duck_arse]

thanks rg. when I was looking for samples from ti, I missed seeing the LT range. I've only got 2 of these, so it won't be long before I catch up with the new-fangled methods. and I do like winding.

[Scruffie]

Let's play power supply.  

Capacitive up-verters suffer from high impedance. The stock capacitive voltage multiplier is something like a C@ckcroft-Walton, and it can produce incredibly high voltages, but is really disabled for any significant currents. To get any significant current, you have to go with transformers or inductive up-verters.

So no, there isn't any easy halfway between charge pump and SMPS without transformers. However, it *is* incredibly easy to design an SMPS today. The control chips are largely condensed down to one eight pin dip for all the controls, and one MOSFET for the switch. They take care of all the ugly gotchas that used to bedevil me when I messed with this stuff.

With either capacitive or inductive (well, I guess transformer too!) you have to know your load - that is, how may volts you want out and how much maximum current it will supply. You're likely wanting to power tubes from low voltage (if not, correct me) so you're looking for one to six plate loads, possibly 12AX7 plates at less than 1ma per. So call it two to six ma. Let's guess at a voltage. Will 200V be enough?

If so, you need 200V * 0.006 = 1.2W. Since transformers are out, you will need an inductor. Inductors store energy as E = 1/2 * L*I^2. It can store, then dump this amount. I has to be chosen such that the L and the number of turns and the core and all that stuff do not saturate the core.

There are several ways to go, depending on what you pick first. If you pick frequency, (for example, 50kHz), then your inductor has to produce 1.2W, and it does this by dumping its bucket of energy 50,000 times per second, so each bucket of energy has to be at least 1.2W/50,000 = 24uJ. Setting that equal to the energy in the inductor, you have L*I^2=2*24uJ, or 48uJ. Alls you gots to do is to pick an inductor that your input voltage can charge to the necessary current in less than half of 1/50,000, and you're in.

This is because the current in an inductor is V = L di/dt. "dt" is half the period of a 50,000 wave, which is what you picked to start this, or 10uS, and V is your supply voltage, less your power switch saturation voltage at peak current. "di" is the current in the inductor from zero (*note, non-return-to-zero versions exist!) to a peak current less than its maximum current before saturation.

Actually, this is pessimistic. An inductor charging from one voltage and dumping to a higher voltage dumps faster than it charges. If you use 12V to charge, and dump at 200V, then the dump time is 12/(200-12) as long as the charge, or about 0.07, so you can spend 92% of your cycle charging. But you have to be sure that you always give it time to fully dump (in return-to-zero upverters). But if you use 50%, you're safe; you just buy a bigger inductor than you would otherwise.

The classical solution is two equations in two unknowns.
L*I^2 = 48E-6 and  L*I /(10uS) = 12V

It's probably better solved for the L*I product that works. Any inductor with a bigger L*I product can be made to work.

I can *hear* all the eyes glazing over this.  

Seriously, it's not that bad.

Perfect answer although it's going to take a couple of reads and some putting it in practice to grasp it but up shot, just tweak an SMPS to suit.

One more question though (and yes you were right about this being to power tubes) what about an SMPS to provide say 50mA at 120V? for low powered output pentodes? Or should I just get taking in what you've already said.

[R.G.]

One more question though (and yes you were right about this being to power tubes) what about an SMPS to provide say 50mA at 120V? for low powered output pentodes? Or should I just get taking in what you've already said.

You're on the right track, only the numbers have changed.

The first step is to know your load. 50ma at 120V(dc?) is  6 watts. Before I get too deeply into this, that has to be the maximum power your load needs. If it's a class A single ended or class A push-pull, you're probably OK calling it that. If its the idle current for a Class AB, there will be peaks of power bigger than the idle current; that's what Class AB is all about, getting the idle power down so the actual used power is more like the  actual output power. But let's say you have a setup that will make 3W of audio out of that 6W of DC by running class A push-pull.

So you have the same questions to answer: what frequency? If it's 50kHz, you need to push in 6W/50,000  = 120uJ at 50,000 times per second. This needs an inductor that can store and dump 120uJ 50,000 times per second. It needs an inductance that will let it ramp up to 120uJ when driven from your incoming voltage source fast enough to get fully charged up and then dump it back out in 1/50,000 second.

By the way, the numbers are much the same for any flyback converter with an isolating transformer, with the exception that the dump time gets longer if you use the turns ratio to make the voltage transformation easier and the spike on the collector/drain of the power switch lower, and that you can get isolation from the primary side ground.

There are issues of return-to-zero versus non-return-to-zero in the inductor current and how that gets controlled, but those are fancier issues on top of the power design.

Practically every switching power supply under 100W that's not a charge pump is either a stepdown (buck) inductor regulator or a flyback step up (boost) design.

[zambo]

1363/4 smps will do 150v up to 200v dc at 50 ma. ive run a 6v6 ,6k6 and an el84 power tube off of them. small and easy to use and for 13 bucks hard to beat.