Op-amp or TO-220 voltage regulator? - Page 1
Forum section: electronics / projects

author: FenderBender
author_rank: Super Contributor
timestamp: May 29, 2012, 07:23:42 pm
content: Hey guys,I've been dreaming of having a lab power supply for like 2 years now. Still don't have one. Kind of inhibits me working too much. What I have planned is a 2 rail supply, each from ~0-15V. Down to 0V is not an issue. 2V is probably fine.So I have a lot of the components already. Enclosure, Dual secondary toroidal transformer, tons of bulk capacitors and connectors etc...I don't actually have an idea as to how I actually want to do the voltage regulation part. A lot of power supplies that I see, like for example, the E3610 use op-amps with external pass transistors. A much more simple solution would just to use a 3-pin voltage regulator like an LM317.There's not going to be any official current limiting, but I will have some PPTCs and fuses and stuff.But in regards to an op-amp based regulator vs 3-pin TO-220. Any ideas?One other question, I see most op-amp voltage regulator schematics I have seen use BJT pass transistors, but I think a MOSFET would be better. Are these typically drop in replacements Not sure about biasing.Thanks.

author: free_electron
author_rank: Super Contributor
timestamp: May 29, 2012, 07:45:05 pm
content: MOSFET or BJT doesn't matter. you are burning up power anyway ...I'd go for the e3610 schematic. been in production for 25+ years , proven and bulletproof.

author: T4P
author_rank: Super Contributor
timestamp: May 29, 2012, 07:53:27 pm
content: Any proper lab power supply must have a proper current limitingusing something like a LM317 cannot get you levels down to 0vThere's a reason why every single lab PSU you see uses a opamp + series pass transistors ... there's no magic chip yet or else everyone's stepping onboard.

author: ejeffrey
author_rank: Super Contributor
timestamp: May 29, 2012, 08:48:44 pm
content: BJTs are a better choice for a simple lab power supply because they have a lower Vbe than a MOSFETs Vgs. A 0-15 volt supply might only need 17 volts input rail with an emitter follower output transistor, but a MOSFET source follower would require more like 20 V or an auxiliary supply. You can get around this with a common source output stage but this is harder to make stable and requires P channel FETs unless your supply is floating..For a more complicated design it is easy enough to add a charge pump for the aux. supply, and then MOSFETs can be a fine choice. However, MOSFETs and BJTs are rarely 'drop in' replacements. Their drive characteristics are sufficiently different that you generally need to make at least minor modifications.A lab power supply without an adjustable current limit is not a lab power supply it is just a regular power supply. Fuses and PTCs are there to protect the supply not the circuit under test.

author: Harvs
author_rank: Super Contributor
timestamp: May 29, 2012, 09:02:57 pm
content: Quote from: DaveXRQ on May 29, 2012, 07:53:27 pm+1Possibly the reason you don't see this as an essentail requirement is you haven't had it before? Until releatively recently I'd been making do with various high current fixed bench supplies for my designs as I only really worked at either 5V or 13.8V. That is until I picked up a nice Tek lab supply secondhand. The number of times since then the adjustable current limit has saved what I've been working on... I wouldn't consider a bench supply without it again.

author: FenderBender
author_rank: Super Contributor
timestamp: May 29, 2012, 10:56:16 pm
content: Well I guess I just have never used one as you say.Since I'm trying to build the supply as small as I can because my bench is cramped as it is, I can't bother to have much more circuitry. I like the L200 which would allow me to do voltage regulation and have a current limit on one chip. In order to do current limiting, I think I would use a 3-4 position rotary switch and have selectable resistors. Perhaps 100mA, 500mA, 1A and 2A would be decent.Simple. Not all that elegant, but I'm sure there are worse. Thoughts?

author: Unknown
author_rank: Unknown
timestamp: May 29, 2012, 11:27:17 pm
content: The volume taken up by a few opamps (eg. LM324 in SOIC) and some diodes and passives in 0805 is negligible compared to the transformer, smoothing caps, heatsink (capable of dissipating 60 W) and controls. Even smaller if you go down to something like QFN or TQFP. Feel free to go for a simpler design, but don't expect it to save much space.I don't think the L200 can do real constant current operation. The uA723 used to be common in low-end lab supplies, although it requires some tricks to have one continuous range from 0 V to > 7 V.

author: FenderBender
author_rank: Super Contributor
timestamp: May 29, 2012, 11:33:59 pm
content: What comprises "real" constant current operation?

author: FenderBender
author_rank: Super Contributor
timestamp: May 29, 2012, 11:57:43 pm
content: http://www.extremecircuits.net/2010/06/fully-adjustable-power-supply_17.htmlI am quite fond of this one. It's by Silicon Chip too! I'm trying to understand its operation. I used to never get how current limiting works, and I understand it better now. Kind of an extended application of ohms law and basic op amp/comparator theory.Even so, I'm still a newbie. Probably should have put this in the newbie section. Here's how I am understanding circuit operation.You have the 10k pot and the 270ohm resistor forming a voltage divider that's fed into the inverting input of the comparator. The voltage across the current sensing resistors is fed into the non-inverting input. If the voltage across the current sense resistors becomes significantly higher than the other input, the output goes low and voltage is limited thus output voltage is limited, which limits current.I don't really understand what exactly the other resistors are doing in there,which is why I am having problems. A while ago, I purchased an expensive 10-Turn 5k pot that I would like to use instead of the specified 10k pot.--1.) What values do I change in order to maintain a voltage range of roughly 1-15V?Also, I will not need a current limiting range of 0-5A. I would like it to be roughly 0-2A if possible.--2.) Do I reduce the value of the current sense resistors? Again, uncertain about how values were selected.I think I will keep a pass transistor since I don't think LM317 is speced to 2A or so. 1.5A I believe..--3.) But if I did remove the external transistor, I am assuming that all the components connected to the output of the MJ2955 would just be connected to the output of the LM317?MJ2955 only comes in TO-3 and it's pretty expensive. So I found, I think, a decent replacement, I hope.(D45H11) Transistor I would use:http://www.digikey.com/product-detail/en/D45H11/497-2578-5-ND/603603(MJ2955) Old transitor:http://www.digikey.com/product-detail/en/MJ2955/497-6712-ND/1038920Last question,--4.) Why were two 0.1R 5W resistors used in series? Why not just a 2.0R 5W resistor?Thank you so much. I think I am on the right track now.

author: T4P
author_rank: Super Contributor
timestamp: May 30, 2012, 03:26:49 am
content: You don't exactly see a 2Ohms current shunt in a power supply do you?2ohms is too big a VD most of the time

author: Harvs
author_rank: Super Contributor
timestamp: May 30, 2012, 07:42:31 am
content: That would be 0.2 ohm not 2 ohm. Typo I'd expect.Whilst I have no evidence of this, the follow is my conclusion. Being a Silicon Chip circuit, there main market is people in Oz buying it as kits or assembling themselves from parts available at Jaycar Electronics. So, from what I see, they tend to use parts available out of the Jaycar catalogue. Their 10W resistor range starts at 1.0 ohms, and for whatever reason they probably didn't think the 0.22R 5W resistor was adequate (either for temp or accuracy.) Therefore the next option is two 0.1R resistors in series.

author: Harvs
author_rank: Super Contributor
timestamp: May 30, 2012, 07:45:37 am
content: BTW being a SC circuit, if you want to make it easy on yourself you can probably just head over to siliconchip.com.au and buy their PCB.

author: FenderBender
author_rank: Super Contributor
timestamp: May 30, 2012, 09:23:51 pm
content: Ah yes late night typing. 0.2ohm. Sorry bout that. I suspected it might just be due to availability of parts.Does anyone have any other input for my other questions. I'd really appreciate it. I'll document the whole thing if I can get started.

author: Zero999
author_rank: Super Contributor
timestamp: May 31, 2012, 09:10:30 pm
content: Operating an constant current operation can be added to an LM317 power supply.It may be easier to design around a BJT or MOSFET but IC regulators, such as the LM317, offer other features such as over-temperature and safe area protection make them more robust than simple transistor regulators.

author: FenderBender
author_rank: Super Contributor
timestamp: June 02, 2012, 01:00:41 am
content: Decided to try something in Diptrace. I want a way for the PSU to be able to protect itself from overheating and especially the transformer. The transformer I plan to use is capable of 15V @ 1.7A + 15V @ 1.7A. It can probably be overdrawn up to 2A alright.I was trying to think of a simple way in which the PSU could monitor the transformer and be able to shut down the unit if being overdrawn. So I thought perhaps I would monitor the transformer via a thermistor? If it gets too hot, shut off the unit via a relay. Is this a good idea?So I drew a schematic but WARNING, but I'm looking for some general proof-reading. I'm 17 years old. Very little 'real' experience, but this is what I came up. I made up some of the resistor values, because I don't know exactly what parts I'm going to be using yet. I want to have a red light turn on when the protection is enabled, since it won't shut off the entire supply. Green light when everything is fine and the relay is closed. It's a normally open relay. The relay will be placed after the transformer. A separate light duty transformer will power the protection board and panel lighting etc.Here's my thinking here. I'm learning here, so my logic might be off. When the PTC is subject to "normal" temperatures, a small base current flows into the NPN 2N9304 and thus there is collector current, the relay closes, and a green LED shines. If the PTC is subject to "high" temperatures, it's resistance goes high which will prevent the transistor from turning on. Since there is no current flowing, the PNP 2N3906's base will be at 0V and no base current. Thus, current will flow through through the PNP lighting a red LED.On a scale of 1-10, how messed up is this circuit. I don't design too many circuits because I don't even have the time as a young gun. But here's something that was just on my mind.Would this work?

author: codeboy2k
author_rank: Super Contributor
timestamp: June 02, 2012, 05:33:24 am
content: Quote from: FenderBender on June 02, 2012, 01:00:41 amIt's a good first try, but there's some problems:1) the 100k at the base of the 3904 is too big. you will only get a base current of 44-45 microAmps. The 3904 has a DC gain (hfe) of between 100-300. Typically you will get something in the middle, but always design for the worst gain. That is 100. So your 44 microAmps base current enables 44uA x 100 = 4.4 mA of collector current. That's not enough for the LED, or the relay that you have in series with it.2) the Relay: some assumptions here: lets say you have a 5V relay, with a 300 ohm coil resistance, and it needs a minimum 10mA to pull in. So your 3904 has to sink at least 10mA for the relay, but then your LED might be on the DIM side. so choose 15mA. That's good for the LED and good for the relay. A 15mA collector current through a 3904, with hfe=100 is 15ma/100 = 150uA minimum base current needed. So double it to 300uA, just to be sure that the 3904 can really sink the needed 15mA for you. You will limit the actual collector current with a typical current limiting resistor in series with the LED and relay. So you want your base resistor on the 3904 to be about (5V - Vbe)/300uA = (5-0.6)/300uA = 14.6k, so choose 15k, 12k, 10k, it's not important. Lower is better. Now your 3904 can sink the needed current.3) the voltage drops in the collector leg of the 3904: write down your voltage drops from the top of the relay, across the resistor and the led. You'll see that 5V at the top of the relay is not going to be enough voltage, because you are dropping voltages all the way down to ground. The relay wants 5V across its coil, or it won't turn on, your 100 ohm resistor will drop 1.5V at 15mA and the LED will drop 1.9-2.0 V. the total voltage drop is now 5+1.5+2=8.5V; but you're only supplying it with 5V at the top of the relay. What's really going to happen is that the LED will turn on and drop 2.0V, the 100 ohm resistor in series with the 300 ohm relay coil is 400 ohm, and that total will drop the rest of the voltage, setting the current. I=V/R =>> (5V - 2V)/(100+300) = 7.5mA. You'll be lucky if the LED has a dim glow to it. And the relay will never turn on, because there's not enough voltage or current for it. So you have to supply the top of the relay with 15V and drop voltages so there is 5V across the coil and 2V across the LED, and the current is still 15mA. 15V-5V(relay) = 10V . 10V - 2V(LED) = 8V left. You have do drop 8V across the current limiting resistor, at 15mA. R=V/I = 8/15mA = 533 ohms. so 560 ohms would be fine.4) the 3906 on the RED side won't work at all in the configuration you've drawn. flip it around, so the emitter is at the top. But then it will turn on at the same time with the green LED because its base current will go down the collector of the 3904 too. So you really don't want a PNP there, you want an NPN too. The 3906 should be a 3904 like the existing one controlling the relay. Pull its base up to +5V with a 4.7k resistor or so, and connect that base to the collector of the existing 3904 through a diode ( Q1 collector ---|<)--- Q2 base ). When the relay is on, the diode is forward biased and takes all the base current away from the 3904 on the RED side. When the relay is off, the diode is reversed biased and the pullup resistor will turn on the red LED.

author: westfw
author_rank: Super Contributor
timestamp: June 02, 2012, 06:58:45 am
content: Any tips on debugging the "typical" op-amp based power supply? I have a Goldstar GP303 supply that died, and replacing the obviously damaged output transistor didn't fix it, so it'll take a little more effort. Since the major "expense" of one of these supplies in the the non-active components (transformer, meters, etc), I figure it would be worthwhile to poke around a bit more (perhaps completely replacing the electronics.)

author: FenderBender
author_rank: Super Contributor
timestamp: June 02, 2012, 05:43:25 pm
content: codeboy2k,Thank you, thank you, thank you. As I was in my bed yesterday, I was thinking of other ways to limit current or prevent the PSU from overheating, etc...and I might use a more tradition current limiter...BUT...Your explanation is really helpful, at least for me, since everything I know about transistors is still very new and I am still naive about the subject. Applause.My mistakes are very amateurish, so I feel a bit on the dumb side right now, but that's of course a good thing. I might use a transistor current limit on the output of the bridge rectifier/filter caps and have a set current. It would more accuratley shut off the unit when too much current is being drawn from the transformer. Any input on this?And regarding that original Silicon chip schematic I was looking at-http://www.extremecircuits.net/2010/06/fully-adjustable-power-supply_17.htmlI don't really understand what exactly the other resistors are doing in there,which is why I am having problems. A while ago, I purchased an expensive 10-Turn 5k pot that I would like to use instead of the specified 10k pot.--1.) What values do I change in order to maintain a voltage range of roughly 1-15V?Also, I will not need a current limiting range of 0-5A. I would like it to be roughly 0-2A if possible.--2.) Do I reduce the value of the current sense resistors? Again, uncertain about how values were selected.I'm trying!If anyone has any input, seriously, greatly appreciated.

author: codeboy2k
author_rank: Super Contributor
timestamp: June 03, 2012, 12:50:11 am
content: Quote from: FenderBender on June 02, 2012, 05:43:25 pmYou're welcome. Don't be discouraged. You're doing good, actually, and learning is all about making mistakes, and more importantly understanding your mistakes, what wrong thinking you had that caused those mistakes, and taking corrective action to undo that wrong thinking and thus learn from the mistakes.I started young like you, I started with electronics at age 10, and by age 12 I was pretty good, and could build most things I wanted to, but I still didn't quite understand these new transistor things; MOSFETS (FETs actually, MOSFETS came later) didn't really exist for me, when they first came out they were basically unavailable to the average kid at home.Transistor amplifiers (using BJTs) especially confounded me. I knew tubes pretty well by then because my father taught me all about those. Transistors were in common use by then, but I still didn't grasp them. Tubes are way easier to understand, conceptually. So I started reading up on the physical construction of transistors and the physics of the PN junctions. silicon, germanium, doping, electrons, holes, depletion regions, tunneling, zener effects, etc. This was when I was around 13 or 14 years old. It helped me to understand transistors better once I learned how they worked inside. But you don't need to know that to work with transistors, I am just saying it helped me because that's how my mind works.Just keep on studying, and continue to learn. I remember an amazing gratification I felt, when I finally grasped how it all worked, and I was finally able to design a BJT circuit all by myself, that actually worked.QuoteI'll take a look at the circuit. As you're aware already, you can't just drop in a 5k pot to replace the 10k pot, you'll need to change some things.

author: SeanB
author_rank: Super Contributor
timestamp: June 03, 2012, 06:30:54 am
content: To use the 5k pot change the 470R resistor to 240R. That will change the current through the pot to give the right output voltage.To change the current limit change the pot for current limit to a 47k pot, that will do it. Remember that the pot has to be wired so that at minimum resistance the pot is at zero current.

author: FenderBender
author_rank: Super Contributor
timestamp: June 03, 2012, 07:33:33 pm
content: Right thanks so much guys.

author: FenderBender
author_rank: Super Contributor
timestamp: June 03, 2012, 11:43:35 pm
content: Just trying to visualize the whole thing from an external design standpoint. This box is 8" x 6" x 3". It's quite a squeeze, but that's one of my challenges here. It's the standard box from Radio Shack. Cost me $7. came with a nice Aluminum cover like that.My main concern is to have a shallow power supply. My bench does not have much depth so like a skyscraper, I have to go UP and not out.Anyway, some NON-electronics concept work.