Power Supplies Part 1 and 2 - Page 1
Forum section: electronics / general

author: 45Overload
author_rank: Contributor
timestamp: November 28, 2011, 11:20:01 pm
content: Thanks, a fun design project not requiring critical board layout. I have found that the only way to get a supply to put out 0 volts exactly is to supply the op amps with a small negative rail. Of course you have to watch your ground reference. The current demand on the minus supply is very low.45O

author: amspire
author_rank: Super Contributor
timestamp: November 29, 2011, 12:00:04 am
content: Instead of the LM334 current sink IC, you can probably do better with a NPN transistor with a constant base current.But if the supply is going to be powered by a transformer, then you can always organize a negative supply with a few capacitors and diodes. Then even a simple resistor from the negative supply to the + output would probably be good enough. if you don't want the current to change with output voltage, then have the resistor from the negative supply going to the emitter of a NPN transistor. Base to 0V and collector to + output.With the negative supply, you can genuinely get down to 0V.Richard.

author: Bored@Work
author_rank: Super Contributor
timestamp: November 29, 2011, 08:05:29 pm
content: Quote from: amspire on November 29, 2011, 12:00:04 amIf you have such a transistor, and if you control the PSU with an MCU you can control the transistor with the MCU, too, and perform active down programming.

author: Short Circuit
author_rank: Frequent Contributor
timestamp: November 29, 2011, 08:48:59 pm
content: Quote from: amspire on November 29, 2011, 12:00:04 amActually, you'd want to use a current mirror for that to get rid of trainsistor gain variantion (temperature dependence and tolerance):http://en.wikipedia.org/wiki/Current_mirror

author: amspire
author_rank: Super Contributor
timestamp: November 29, 2011, 10:48:49 pm
content: Quote from: Short Circuit on November 29, 2011, 08:48:59 pmI did think of a current mirror, but without a matched transistor pair, you still will have to adjust the current. Is it worth the complication?Definitely the beta will change with temperature, so if you set my circuit for 1mA, then if the power supply got really hot, it might get close to 2mA. Typically, if you set my circuit for 1mA at 20 degC, then at 0 degC it might sink 0.7mA and at 40 degC, it might be about 1.3mA. Good enough I think.Anyway, as I said, I would only use this if the supply was not powered by a transformer. If it is powered by a transformer, I would have a negative supply as I described so the voltage can go down to a true zero volts.Richard.

author: 45Overload
author_rank: Contributor
timestamp: November 29, 2011, 11:55:01 pm
content: A constant current to the base of the regulator transistor is fine for a minimal parts count constant voltage. This arrangement may give you constant base current, but the transistor's Beta will change with load and temperature, making for a loosely regulated supply. Also you would need a minimum load on the supply to source the E-B current.Since constant voltage and constant current power supply functions are closely related and do not require a big parts investment, it makes sense to use an inexpensive quad op amp and a small current sense resistor in the output. That way, you are actually monitoring output current. That would be especially important near 0 volts output, because of the non-linear turn-on characteristic of the BPT. Plus you will have plenty of gain blocks to make your feedback networks. Even op amps that are designed to work with a single supply rail will not go all the way to 0 volts output. It is just irritating, but not essential, if your supply won't actually zero.45O

author: amspire
author_rank: Super Contributor
timestamp: November 30, 2011, 12:28:43 am
content: Quote from: 45Overload on November 29, 2011, 11:55:01 pmNot quite sure what you mean. The need to have 1mA load on the supply is a compromise, and I am replacing an accurate 1mA sink (the LM334) that allows the supply to go to 0.9V with a transistor that will typically vary between about 0.7mA and 1.3mA, but will allow the supply to go down much lower - maybe 0.3V or something. It will have no effect on voltage regulation whatsoever.The E-B current would come from the source supply to the Dave's circuit, not from the load, so the base current will be there all the time. It will be unaffected by what the load is doing.QuoteIf you put a 1 ohm current resistor after the regulator IC, then it will add 1 ohm between the voltage regulator and the load. So if the supply is set to 5V and you are drawing 1A, the load voltage will be 4V. Not a good idea.Dave is trying to make use of the IC's stabilized voltage regulator, so he needs the current limit resistor before the voltage regulator. That regulator has no voltage feedback path external to the IC.If you are going to change that, then you are really going back to a standard regulated power supply design which will have full 0V to Max voltage regulation and full 0-1A current control, but you will have to try and do your own stabilizing of the feedback circuit for all possible loads. That is the hardest bit of the design, and Dave was trying to avoid that.Richard

author: 45Overload
author_rank: Contributor
timestamp: November 30, 2011, 06:20:45 pm
content: What a great way to have fun, make mistakes and produce a necessary piece of bench equipment. And, you don't need a multi-domain digital scope to do it! Actually, you don't need a scope at all. (Ok, the supply might oscillate and throw you a curve ball, but you can spot the symptoms with a good multimeter.)I wasn't trying to re-design Dave's circuit, but instead I was speaking in general about getting a single linear pass transistor voltage regulator to operate near 0 volts output. The LT3080 does all this for you, but you haven't really learned much about power supply designs. Yes, you can use your voltage reference or current reference to tighten up the voltage supplied to the adjustment pin on your LT3080. But just for the sake of ground-up design experience, lets assume you are working with all discreet parts.Any transistor BP or FET will have a very non-linear turn-on characteristic which will require a feedback circuit with enough loop gain to control your device at those very low levels of base current (or gate voltage). Without closed loop operation, you are depending on the transistor's Beta which will swing widely around the turn-on point and with temperature. And what better way to provide the necessary voltage AND current feedback than a current sensing resistor in series with the output? The "improved" LT3080-1" has the sense resistor built right into the package (by popular demand according to the manufacturer).The usual configuration for the regulator is to have the NPN emitter on the output side of the supply. A power MOSFET would be an even better choice for low voltage output because of the smaller D-S leakage current when the device is near or at turn-off. Just another idea for fun experimentation, not set in stone.Finally, the use of a current sense resistor in the output is not the horrible idea that most people believe it to be. Taking maximum current output into consideration, the sensing resistor can be (and often is) well under an ohm. 1 ohm was chosen in the video to make calculations simple. A precision length of circuit board trace is just what the doctor ordered.Dave mentioned an important point to keep in mind - don't get carried away with the size of your output filter capacitors or any capacitors used in the feedback circuitry. That way, the supply's ability to respond to dynamic load variations will be enhanced.45O

author: Short Circuit
author_rank: Frequent Contributor
timestamp: November 30, 2011, 06:48:16 pm
content: Quote from: amspire on November 29, 2011, 10:48:49 pmIt's easy to take a matched pair like BCV61, but even with two separate transistors, you get the benefit of more stable current over temperature. (if they are reasonably thermally coupled of course).However, the real reason to mention this is to indicate that sometimes there are problems that can be solved with simple & elegant discrete solutions, instead of picking yet another specific purpose ic. A bit like the PIC vs 555 tradeoffAnd smaller one-off projects like a DIY power supply are ideal to play around with 'new' techniques. Otherwise Dave would probably have thrown a L200 in the mix and been done in a single blog, actual build includedQuote from: 45Overload on November 30, 2011, 06:20:45 pmDoctor? Better consider that a quackPrecision length is no problem, but you have limited control over copper thickness (electroplating) tolarance and exact trace width (underetching)

author: BravoV
author_rank: Super Contributor
timestamp: December 01, 2011, 05:02:10 am
content: Quote from: 45Overload on November 30, 2011, 06:20:45 pm45O. do you have any link or reference where I can find such good linear lab psu design that utilizing discrete components like transistors, op-amp and precision voltage reference ?Really like to learn (or even build it) especially at the one that has an excellent design for the voltage and current control loop mechanism.

author: Unknown
author_rank: Unknown
timestamp: December 01, 2011, 05:21:04 pm
content: The Agilent E361xA series is one example and has a service manual with full schematics. Same for most of the other E36x series, but the newer models don't have schematics. Most of the commercial lab supplies will have this design, except for the compact switchers or the high power ones and other specialty stuff. I believe Agilent's latest design is different.Some of the cheaper ones use an LM723, but its constant current mode is not great.

author: BravoV
author_rank: Super Contributor
timestamp: December 02, 2011, 03:01:54 am
content: Quote from: alm on December 01, 2011, 05:21:04 pmAlm, which part number actually ? Appreciate if you have link to download just one example.Quote from: alm on December 01, 2011, 05:21:04 pmDo you mean 723 used in those Agilent/HP power supply ?

author: sonicj
author_rank: Frequent Contributor
timestamp: December 02, 2011, 04:09:27 am
content: Quote from: BravoV on December 02, 2011, 03:01:54 amE3610AE3611AE3612Athose are the 30W range i believe

author: BravoV
author_rank: Super Contributor
timestamp: December 02, 2011, 06:30:57 am
content: Thanks a lot Sonicj !Looks like its not that difficult to build ourself isn't it ? Ok, apart from the transformer that has multiple windings for different power sources other than the main power, for me this is no big deal since I got lots of those small current xformers from wallwart type that should enough to cover the reference voltage source and for the meter.The rest of the discrete components are not that special aren't they ? Ok, I noted few that could be tad expensive like the multturns quality pot, thats it right ? Even my unused voltage reference in my components junk bin is better than that LM336. Or did I miss other ?Is this considered a good design ? or just mediocre ?

author: FreeThinker
author_rank: Frequent Contributor
timestamp: December 02, 2011, 02:53:57 pm
content: I'm a little confused (again) when would you actually need a supply voltage of 0 volts? Or am I confusing Zero Volts with 0 Volt WRT Ground?

author: McMonster
author_rank: Frequent Contributor
timestamp: December 02, 2011, 04:07:35 pm
content: That's an interesting question, what're the applications of variable single supply that can go to 0 V? I know that some microcontrollers that can operate as low as 0.7 V, but can someone give an example of a possible benefit of such supply?

author: Unknown
author_rank: Unknown
timestamp: December 02, 2011, 09:12:30 pm
content: One example that comes to mind for close to 0V: measuring very low resistances, like current shunts. Assume you want to measure a 20A shunt with a resistance of about 3 mOhm. There's no way that you can measure this accurately with a normal DMM due to the extremely low voltages involved if you put 1mA through it (3 uV voltage drop). So you connect it to a lab supply and put a few A through it. The voltage drop is now in the order of 10mV, which you can measure with a normal setup. Way below 1.25V or 0.7V, however. A similar application would be finding shorts in PCB traces by measuring voltage drop along the traces. This trace is not going to drop 1V, more likely in the millivolt range.I'm not aware of any Agilent supplies using the 723, but some of the cheap Asian supplies do. It's a relatively easy way to design a power supply with a variable voltage and current, but it's much better as a voltage source than as a current source. I believe the manual sonicj posts is the one I mean, it covers multiple models. There's also an older HP power supply handbook covering some of the basics of linear power supplies, including constant voltage and constant current. It only contains simplified schematics, however.

author: IanB
author_rank: Super Contributor
timestamp: December 03, 2011, 07:36:49 am
content: Quote from: FreeThinker on December 02, 2011, 02:53:57 pmZero volts as a lower limit means there is (theoretically) no lowest output voltage that can be selected. So, for example, if the lowest available voltage was 0.5 V and you happened to need 0.2 V for some particular purpose you would be out of luck. But if the lowest available voltage was 0 V you would be OK.

author: FreeThinker
author_rank: Frequent Contributor
timestamp: December 03, 2011, 09:09:00 pm
content: Quote from: IanB on December 03, 2011, 07:36:49 amThanks for the reply, now I understand. Just couldn't get my head around usefulness of Zero Volts output but now see the reasoning.

author: RJSC
author_rank: Regular Contributor
timestamp: December 05, 2011, 03:26:08 pm
content: Since the LT3080 is hard to come by in most countries (like mine) and you have to order it anyway, I have found the LT3083.It's a 3 A capable version of the LT3080, for the people interested in higher current without the voltage drop of the resistors needed when paralleling LT3080's.I am also thinking of using one of those universal laptop switching power supplies which are adjustable and control it with a digital pot in place of the feedback resistor and have it allways being 2 volts above the output to minimize power dissipation, allowing for better energy efficiency and a smaller heatsink.

author: shebu18
author_rank: Frequent Contributor
timestamp: December 08, 2011, 09:51:36 pm
content: What would be a replacement for the LT3083, i can't find it in my country,

author: McMonster
author_rank: Frequent Contributor
timestamp: December 08, 2011, 09:58:47 pm
content: I think it would be a good idea if future "lets design a..." blog videos included mentioning possible alternatives for featured parts (if any worth mentioning exist of course). Apparently LT3080 is either hard to obtain or expensive in different parts of the world.

author: Unknown
author_rank: Unknown
timestamp: December 08, 2011, 09:59:46 pm
content: That's an issue with Dave's design in my opinion. The traditional design used in most commercial linear supplies consists of some low-end opamps (LM324 is OK), diodes, power and small signal transistors and some three terminal regulators to supply the opamps. They can be build from 1970s technology, which means you can get (slightly lower performance) replacements almost anywhere in the world. The LT308x is a fairly unique chip (anyone aware of any alternatives?), and Linear is not the most widely distributed semi manufacturer since their parts tend to be expensive and exotic. I don't believe Mouser carries them, for example.

author: RJSC
author_rank: Regular Contributor
timestamp: December 08, 2011, 11:52:12 pm
content: The commercial power supplies circuit using a discrete pass transistor and opamp require compensation (filters, usually low-pass RC) to eliminate the output oscillation with capacitive or inductive loads. Almost any circuit you connect to it will have capacitors, so, it is critical, and the math involved in calculating the compensation network is intimidating (systems and control theory).Dave chose an all-in-one regulator, because the engineers who designed it did all the hard math to find how to stabilize the output.Anyway, a discrete output transistor power supply is far more flexible allowing you to chose over a wide range of output currents. As for the output voltage, it is very simple, the LT3080 supports up to 30 volts, you just need to provide it on the input stage, and if you're doing the digital control, increase the non inverting amplifier gain suitably.Dave has faced that problem before on eevblog #95

author: shebu18
author_rank: Frequent Contributor
timestamp: December 09, 2011, 12:24:41 pm
content: What about the LT1963AEQPBF?http://www.alldatasheet.net/view.jsp?Searchword=LT1963AEQPBFIt is about the same price, easier to find and has almost the same caracteristics.

author: EEVblog
author_rank: Unknown
timestamp: December 09, 2011, 12:30:18 pm
content: Quote from: shebu18 on December 09, 2011, 12:24:41 pmMight as well use the LM317, unless you need low dropout, as the LT1963 only goes down to 1.2V.Dave.

author: shebu18
author_rank: Frequent Contributor
timestamp: December 09, 2011, 12:38:16 pm
content: I want a regulated power supply for my projects, the one thing is that i will need around 5 amps at 7,4V for my rc-car tests. I have made a power supply with a lm371 and a power transistor 9as shown in the schematics). The thing is that i can't control the amps.I like the design you made because i can also control the supply via a uC and the current protection.Offtopic, what was the forumula for the lm371 for calculating the output voltage? I found one in the schematics but i don't know what Iadj is and where to find it.Thanks. You do a great job!!@ejeffrey: Thank You.

author: ejeffrey
author_rank: Super Contributor
timestamp: December 09, 2011, 01:25:19 pm
content: Iadj is about 50 microamp, so it should provide a negligible offset in most circumstances. It is listed in the data sheet. For the standard configuration where you use a voltage divider with a fixed 240 ohm feedback resistor, it will give approximately a 1% error term. If you make it programmable by driving the adjust pin from an op-amp, the adjustment current has no effect. If you use larger feedback resistors, the adjustment pin current can cause a more significant offset.

author: RJSC
author_rank: Regular Contributor
timestamp: December 09, 2011, 10:39:13 pm
content: If you feed a negative voltage to the opamp's negative rail instead of ground, you can force the LM317 to Zero, by driving the ADJ pin to -1.25 V. With a pot, you can connect its low side to -1.25V.Given that the ADJ pin current is so low, you can generate the negative supply with a dedicated charge pump IC or with an oscillator (555 for example) to avoid the need of a second small transformer or a big transformer with 2 windings.You can read something about it here:http://www.stephenhobley.com/blog/2011/03/28/creating-a-negative-voltage-from-a-positive-one/Maxim has a lot of charge pump IC's that require only external capacitors, but I don't know of any that should be available at every electronics store like the LM317.

author: McMonster
author_rank: Frequent Contributor
timestamp: December 09, 2011, 11:07:01 pm
content: Quote from: RJSC on December 09, 2011, 10:39:13 pmI've seen a few circuits where forward biased diodes are used in series from some point in the circuit to ground to get a small reference voltage equal to voltage drop across diodes below ground, for example in LM35DZ's datasheet. Is this applicable here? Where else it would work and where not?

author: RJSC
author_rank: Regular Contributor
timestamp: December 09, 2011, 11:28:49 pm
content: Quote from: McMonster on December 09, 2011, 11:07:01 pmIt is applicable but is has some problems.The voltage you are dropping on the diodes times the current will be dissipated as heat, and you have to provide an higher voltage transformer to account for that drop or have a lower output with the same transformer.Also, a diode does not switch on cleanly at 0.7V, meaning that at a light load they will drop less than 0.7V and with high loads (1 A +) they can drop as much as 1V. Diodes also exhibit something called a "dynamic resistance" which you can understand as a small value resistor in series with the diode.As you can see on this graph, the voltage drop on the diode varies with the current passing trough it (although by a small value):If you want some precision on your supply you shouldn't mess with the ground path, thats why Dave put the current sense resitor on the high side with a differential amplifier instead of putting it on the ground return.

author: ejeffrey
author_rank: Super Contributor
timestamp: December 09, 2011, 11:31:08 pm
content: Quote from: RJSC on December 09, 2011, 10:39:13 pmIf you are going to do this, see if you can make the negative supply able to sink 5 milliamps. That will let you meet the load current requirement of the LM317.

author: RJSC
author_rank: Regular Contributor
timestamp: December 09, 2011, 11:38:52 pm
content: Quote from: ejeffrey on December 09, 2011, 11:31:08 pmGood point! We need to sink 5 mA from the LM317 for it to be stable or we will be on the same situation of Dave's circuit that needs a load connected to be able to drop to 0V (it's not able to be a reference for an high impedance circuit).I hope Dave doesn't get mad that we are messing his whole design...Anyway, even using the LT3080(3) a negative supply capable of sinking the minimum load current would be neat, as it would allow the current source to keep drawing current down to 0v output into the negative supply.

author: Unknown
author_rank: Unknown
timestamp: December 09, 2011, 11:44:51 pm
content: Quote from: RJSC on December 08, 2011, 11:52:12 pmNo argument here, but that doesn't make the availability of the LT3080 (in some parts of the world) any better. The single-sourced part with no close replacements would also be risky in a commercial design, although I guess Linear is better than some of their competitors in delivery schedules and discontinuing parts.

author: shebu18
author_rank: Frequent Contributor
timestamp: December 10, 2011, 03:36:29 pm
content: Search for LT1963.

author: Unknown
author_rank: Unknown
timestamp: December 10, 2011, 04:55:27 pm
content: To repeat Dave, how us that better than any random adjustable three terminal regulator like the LM317? The special feature of the LT3080 is that it supplies a constant current to be used as reference voltage, which allows the voltage to go down to 0V.

author: shebu18
author_rank: Frequent Contributor
timestamp: December 10, 2011, 07:27:08 pm
content: i found it as a replacement with near specs to lt3080. what can we use instead of the lt?

author: Bored@Work
author_rank: Super Contributor
timestamp: December 11, 2011, 11:30:08 am
content: Quote from: shebu18 on December 10, 2011, 07:27:08 pmAlmost everything or nothing.What you haven't got is that you can use whatever fulfills your needs and adapt the circuit if necessary. Dave's schematic is rather generic, isn't rocket science and can be adapted to many adjustable regulators. Including, if necessary, a negative offset to drive a non-zero Volt regulator down to 0V, or a transistor to sink more than your OpAmp can sink.Of course, if you don't want to adapt the circuit then you may end up with no suitable replacement at all.All this depends on your needs. Do you need output adjustment down to 0V? Down to 1.2V, to 2.5V?

author: FreeThinker
author_rank: Frequent Contributor
timestamp: December 11, 2011, 11:54:48 am
content: As Dave has said prior in the thread this is not the finished article, he still has a few tricks up his sleeveand his is promising to release it as a kit so supply problems are moot. Just hang in there and wait for the final release and then decide if it is for you or needs to be modded..... Remember Santa is coming SOON!