Step down or step up voltage regulators with input (output) in series - Page 1
Forum section: electronics / beginners

author: Unknown
author_rank: Unknown
timestamp: May 07, 2014, 02:01:16 pm
content: POST REMOVED AT USERS REQUEST

author: benjamin545
author_rank: Contributor
timestamp: May 08, 2014, 12:59:11 am
content: one of the big limitations of connecting PS in series or parallel is are the supply "floating". take your ohms meter and measure the input negative to the output negative and input positive to output positive. do this in both directions. if you see a complete open in all forms, its a good indication that it could be floating (or internal circuitry is preventing you from seeing continuity in the supplies off state). assuming you have one of those low volt dc-dc converters you see all over ebay. im going to assume you do not have a floating supply.if the supply was floating, it means it has a complete galvanic disconnect between the input and output, means no metal physically touching. this is accomplished by using some sort of transformer or isolated inductor. if you did not have a floating supply, imagine what would happen if you tried connecting 2 supplies output in series, you would take one outputs positive output, and tie it to the others negative output, just like a battery, except the 2nd supplies negative output, technically is also its negative input, its the same all the way though, you would have both supplies inputs tied in parallel to the input power, so the second supplies negative rail is the also the same as the first supplies negative rail, so you basically made a short circuit. in the parallel setup, you basically have each supply providing a voltage offset from the negative rail, if you tied both positive outputs together, any slight variation in voltage would cause the higher one to back feed the lower one. this isn't awful, and could be made to work, linear regulator circuits like the lm317, lm350, lm338 often show a parallel configuration to get higher output current. they rely on each output having their own low resistance to the common tie together point so that the load doesn't try and source all its current from one supply.

author: rs20
author_rank: Super Contributor
timestamp: May 08, 2014, 10:30:18 am
content: Can you send a link to exactly what kind of part you're using? I can think of only two types:-- Standard linear regulators, which have only 3 terminals and obviously no galvanic isolation. These can't be stacked as far as I can imagine, unless you have both negative and positive regulators, which would be a rather strange arrangement.-- Isolated DC-DC converters. The outputs of these can be stacked freely. Stacking the inputs would, again, be pretty strange.

author: benjamin545
author_rank: Contributor
timestamp: May 08, 2014, 01:08:42 pm
content: Ok. See when you said cheap ebay regulators. A lot of us would think completed circuit board designs that are ready to use power supplies. You are talking about the individual regulator chips. That's a whole different bag of potatoes. The one you linked in there is a switching regulator. So the plain and simple answer is. No you could not hook the chips up in parallel or series to get greater voltage or amps. While you could hook up linear regulators in parallel to get greater current, a switching regulator is constantly switching the switch output from completely off to completely on. There is no way to get two chips to sync that output. Nor would you ever want to. What you would do is have that switch output drive a large power transistor that can handle the conditions you need it to. The mc34063 has diagrams in its data sheets showing such a configuration.

author: rs20
author_rank: Super Contributor
timestamp: May 08, 2014, 01:49:35 pm
content: Quote from: pereczes on May 08, 2014, 10:43:04 amAlmost all non-isolated DC to DC converters have their negative input shorted to their negative output, so they should be represented as a three pin device in your schematic (just like a plain linear regulator). Look at the example schematic in the data sheet you sent, there's no negative in or negative out, just ground and ground. If you try to redo your schematic with three terminal parts, you'll see it can't be done.What voltages are you actually trying to convert between, and how much current?Quote from: benjamin545 on May 08, 2014, 01:08:42 pmThis isn't quite right. You can, at least in theory, connect switching regulators in parallel to increase output current capacity, just connect *after* the inductors. The OP is right, there's little difference between a linear reg and a complete switching reg circuit in this regard. No need to sync them, either. In both switching and linear case, in practice, getting the two parts to share current equally is a challenge though.Your final point is valid, though. Use an external transistor.

author: benjamin545
author_rank: Contributor
timestamp: May 08, 2014, 03:13:06 pm
content: There is no v galvanic isolation regulator chips. You have to do galvanic isolation somewhere else. If you look at mains voltage lm317 circuits. They always have a isolation transformer. You would have to have an isolation transformer for each chip to get individual isolated circuits. In commercial switching designs you see the isolation being done at the inductor since a isolation transformer as mains ac frequencies gets really huge for the current most of these supplies are designed to handle. By isolating at the inductor you can Avoid having a large isolation transformer.You seem to be pretty stuck on the idea of putting multiple regulators in parallel or series to get the power ratings you want. The only time this is done is with simple linear regulators. I'm not even sure if you would do so with a LDO one. you would only even bother doing so as a quick and simple way to double or triple your current. Usually it would be better to just find a higher current chip or to use the regulator to Control a power transistor.

author: Jeroen3
author_rank: Super Contributor
timestamp: May 08, 2014, 07:38:09 pm
content: Quote from: benjamin545 on May 08, 2014, 03:13:06 pmThere are DC-DC isolated power supply blocks pincompatible to the LM78xx and LM317 families. However your "decent price" is likely to cause problems here.The isolated versions are typically low voltage in, low voltage out. Just for separating a piece of circuitry.Putting switch mode regulators in parallel must be done with proper thoughts, maybe find reference in the datasheet or application note this is possible.Putting them in series isn't such a big deal, as long as their minimum specs are met and maximum ratings are not exceeded.Connecting them like you did is asking for trouble.However, creating 150 Volt DC is more likely to be done using a small high frequency (or pulse?) transformer with a low voltage transformer driver.The secondary (high voltage) side then uses rectifier diodes and possibly an ldo or high voltage regulator, a zener or nothing to limit/regulate your output.Random googled example(flash charger circuit):http://www.linear.com/product/LT3484Please note that 1 Watt in is less then 1 Watt out, following the laws of physics this translates to low U with big I -> high U with low I.

author: benjamin545
author_rank: Contributor
timestamp: May 08, 2014, 08:23:07 pm
content: if you need to go from low volt to high volt you have some options, if you have an ac input voltage like 120v ac, you could use a regular old laminated steel transformer that is rated to handle whatever voltage you need it to support. then you could try and transform it up to a higher voltage and work with it from their.but it sounds like you have a DC input voltage from a regular power supply. so there are 2 ways you can boost that up. if you don't need very much current, like maybe a few millivolts, you can use a charge pump. they make charge pump chips, or you can just build your own out of a simple multi-vibrator circuit or 55 timer and stack a bunch of diodes and capacitors up to make a multi stage charge pump. this apparently was done frequently in old picture tube TV's to make the several thousand volts required to do some of the electron gun stuff. i don't know much about it, but in one of the videos dave did about the Sinclair pocket tv he talked briefly about a multi stage charge pump in there. i would definitely recommend this if all you need is a few millivolts, its pretty easy and doesn't really require a lot of concern about the parts being rated for your high output voltage except the decoupling capacitor you are going to charge up to the high voltage.the other option is to make a boost smps converter. i think you already have started to look at this option, but i don't think you really understand what all is involved with this method. this would method would really only be necessary if you need a decent amount of current output, like over 100mv where the charge pump would just require too big of capacitors to work efficiently. a full smps step up design is not exactly easy to do, and can be dangerous at those kinds of voltages. the hard part about it is not really choosing the regulator chip, its choosing the right inductor and also understanding what that inductor is going to do. how instance, you dont just grab any inductor, for any decent efficiency you need an inductor that will work well at your operating frequency. depending on your design, such as a flyback converter, you may have to worry about back emf emitted by your inductor during switch off times.

author: Jeroen3
author_rank: Super Contributor
timestamp: May 08, 2014, 08:55:47 pm
content: I assume you've mixed up the volts and amps here?Or you save yourself all design and validation effort by using an off-the-shelf module.http://www.recom-international.com/pdf/Innoline/Rxx-B.pdfBut that kinda takes the challenges out of the game.

author: benjamin545
author_rank: Contributor
timestamp: May 10, 2014, 03:11:40 am
content: yes. i did mean milliamps, not millivolts, sorry if that was confusing.good to hear that a charge pump is a suitable concept. of your two links only the first would actually load. and honestly it looks a bit more complicated than you really need to worry bout, i think thats what you might find inside of a ready made charge pump chip, but for a quick and easy build it yourself approach, id just watch daves video about dixon voltage doublers. for the clock signal i think he uses a waveform generator but you can use any decent square wave generator like a multi-vibrator circuit built with 2 transistor and 2 capacitors, or you can do a ne555 timer. you can build a pretty decent mock up of what you need in "small scale" by boosting 10volts to 50volts. if you can do that easy enough, then you just gotta feed in the right voltage to get up to what you want as your output, providing that your output decoupling capacitor can handle the output voltage. at low volts, its pretty easy to run into efficiency issues as the diods will knock good fractions of volts off the output voltage of each stage, but at higher volts it becomes relatively less and less of an issue.

author: benjamin545
author_rank: Contributor
timestamp: May 10, 2014, 07:11:53 pm
content: http://en.wikipedia.org/wiki/Common_collectorthat is a pretty basic and very common transistor configuration. basically when you drive a positive voltage into the Vin position, the transistor will turn on, current will flow through the transistor, and then through the resistor to ground. you would of course make the resistor be a suitable size so that you aren't dumping a bunch of power through the circuit. if you measure the voltage at vout at this time it will be (assuming the transistor is fully on, aka saturation mode) equal to the positive voltage rail minus whatever voltage drop the transistor would induce, probably a little under 1 volt or something like that. if you turn off the transistor the current is interrupted. now if you measure the vout point, it will be at ground because the resistor in that circuit is acting as a pull-down resistor. now, you wouldn't want to make that resistor value be too high since that will impede its action as ground, but a good balance between not beiing too much of a short when transistor is on, and not being too much of an impedance wen transistor is off, is what you want.that will act as your clock switch. i notice your design features no diodes, thats probably not right, the diodes are there to act as switches without needing a full transistor switch.

author: benjamin545
author_rank: Contributor
timestamp: May 10, 2014, 07:22:47 pm
content: http://en.wikipedia.org/wiki/Common_emitteractually i think it would be better to use that as a model for a switch. the problem with the one i posted before is it would be difficult to get the transistor to go into saturation because you would need the input voltage to be higher than the positive rail, its a good circuit when you want to keep the transistor out of saturation, like for a signal amplifier, but for a switch thats easy to go into saturation, use the common emitter. its the same principal except its not a pull down resistor, its a pull up resistor.

author: benjamin545
author_rank: Contributor
timestamp: May 11, 2014, 11:20:24 pm
content: Not exactly. If you watch Daves video about Dixon voltage doublers you will see he doesn't use any transistors at all in his circuit. The only time you need transistors is for your clock signal. Once you generate an oscillation between ground and your supply voltage, you can rely on diodes for controlling everything else. Do however watch out for capacitor ratings. They will need to meet the voltage of each stage for it to be safe and not burn them out.When I was talking about the emitter follower voltage circuit. I was using that to show how to drive that clock signal as a ground when off, rail voltage when on, signal. But once you have that, you don't need to do it for each stage.Watch Daves video on Dixon voltage doublers. He makes a 2 stage charge pump that basically show exactly why you are trying to do, only you would need to extend it 2 more stages and secure it for 200 or so volts.

author: rs20
author_rank: Super Contributor
timestamp: May 12, 2014, 11:07:02 am
content: Quote from: pereczes on May 12, 2014, 10:36:55 amBeware. You seem to be suggesting using mains voltage in a non-isolated kind of way. Mains 220V is more dangerous than isolated 220V, because almost everything slightly conductive around you is a potential return path to complete the circuit that includes your heart. You need to buy an isolation transformer at the very least, if not put this project on hold until you gain more experience and understanding. Designing your circuits to be testable is a skill gained through experience, please sacrifice some op-amps in a bunch of low-voltage projects instead of sacrificing yourself.Also, danger of death aside, when you say you have 220V. wikipedia says Belgium has 230V. Moreover, that means 230V ACRMS± 5%. That's 325 V peak (± 5%) or 651 V peak-to-peak (± 5%). So a simple doubler on its own is going to pop your cap.I'm working on a flash circuit myself; I'm using 30V isolated DC input into a flyback converter. This is over-the-top, and I'm doing it this way for interest's sake, but it's somewhat safer than working on mains directly. You could use a multi-stage charge pump to achieve the same thing more simply*. The high voltage will be isolated from earth, and I'll be able to probe the voltage via a voltage divider that doubles as a bleeder resistor (you are going to have bleeder resistors, right?) so that I can insulate and hide away the HV side of the circuit entirely while still monitoring its voltage indirectly. I don't know how you'd propose to achieve the same thing in something tied to mains safely.* See this video for a circuit that generates many thousands of volts from some batteries -- there's no need to start from such a dangerous voltage if you're only trying to reach half a kilovolt.