L200 CC/CV bench power supply - Page 1
Forum section: electronics / projects

author: microbug
author_rank: Frequent Contributor
timestamp: May 16, 2013, 05:35:55 pm
content: Hi,I'm fairly new to electronics and one things that seems to be recommended by everyone to do is to make a bench power supply, so I'm designing one (see attachments).It's a bit of a mish-mash of different designs, but the base is fromhere.I am posting this because I am pretty much certain there will be problems with the design that I haven't spotted, and would like input on what I should change.Thanks in advance for any help!

author: mariush
author_rank: Super Contributor
timestamp: May 16, 2013, 06:07:24 pm
content: I sincerely hope you're not connecting this directly to 220-240v, without an isolation transformer. Fuse and capacitor on input certainly seem to hint that.What's the specs of this psu?The L200 is an interesting chip, but it seems it can only go as low as 2.85v (the vref is aprox. 2.77v) so if you want to do 1.8v, 2.1v, 2.5v, 2.8v it doesn't seems you'd be able to... and has about 2-2.5v voltage drop which is kinda big.

author: microbug
author_rank: Frequent Contributor
timestamp: May 16, 2013, 06:10:59 pm
content: Sorry, I should have clarified! The input will be from a 30V 50 - 60VA transformer hooked up to the mains. I don't mind the minimum voltage of 2.85V; at some point I will probably make/buy a low power supply that is more accurate. The circuit diagram that I based this on says that if you connect a negative voltage to point A, you can achieve 0V output.http://obrazki.elektroda.pl/7582279200_1309598107.jpg

author: mariush
author_rank: Super Contributor
timestamp: May 16, 2013, 07:36:46 pm
content: Ok, then you don't really need a 0.1 250v capacitor, a 0.1uF 100v cap will work and is probably cheaper.if the transformer is 30v rms, then the peak voltage will be about 42v. At that VA rating, you'll have about +10-15% at low loads, so don't be surprised if the peak reaches about 45v... before the bridge rectifier. The rectifier will drop the voltage by 0.8-1v.The 50v rating of the capacitor would be OK, but on the small side.The L200 is 40v max though, with capability to handle 60v for 10ms or something like that.The 1n5822 has a peak reverse voltage of 40v so be really careful there, the diodes should have at least 1.5x the peak voltage they'll ever see for safety. So you should use at least 60v rated diodes. With so high input voltage, is there really a point in using shottky diodes? Just use a 0.2$ bridge rectifier.I don't think you really need 10.000 uF, maybe you can just use 4700-5600uF 63v capacitor. And C4 (0.1uF 50v) should really have much larger voltage rating, it's not a good idea to have ceramic capacitors with voltage so close to their peak rating.Anyway, all the above if your transformer is 30v rms, and not 30v ac before the bridge rectifier.I would use a 24v rms transformer, possibly with center tap or two independent 12v windings (and switch between taps at lower voltages to reduce the voltage drop ... 2.85-12v, 12v-24v or something close to that)Also check the datasheet, there's some text there if i remember correctly saying that the maximum vdrop should be less than 30v.. shouldn't get low voltages like 3v with a 40v input but even if you do, at 1A you'd have 35-38w dissipated on the heatsink. You'd need a large heatsink to do that.

author: microbug
author_rank: Frequent Contributor
timestamp: May 16, 2013, 07:46:33 pm
content: Thanks for the info about what rating of transformer I should use; I was unsure of it. I'm not quite sure what you mean by the last bit about maximum voltage drop though - voltage drop across what?

author: mamalala
author_rank: Unknown
timestamp: May 16, 2013, 07:51:55 pm
content: Quote from: mariush on May 16, 2013, 07:36:46 pmI think the voltage rating of the rectifier/diodes should be even higher. Mains is a rather "dirty" source, simply turning on some neons, or the fridge's compressor starting can produce rather big spikes, which in turn means that you will also get spikes on the transformers secondary from time to time. After all, bridge rectifiers are cheap.Greetings,Chris

author: mamalala
author_rank: Unknown
timestamp: May 16, 2013, 07:56:06 pm
content: Quote from: microbug on May 16, 2013, 07:46:33 pmA linear regulator, such as the L200, will simply dissipate the difference between input and output voltage as heat. Plus, due to internal construction such devices can only handle so much voltage difference between in and out. So if the device can handle only a 30V difference, but you supply 40V in and take 3V out, you have a 37V difference, which is too much.Also, using low output voltages at high input voltages means that you rarely can get the maximum possible current out of the device unless you use a really, really massive heatsinking/cooling scheme.Most bench supplies have a transformer with multiple taps, so the regulation circuit gets lower input voltages depending on the output voltage, to avoid that problem.Greetings,Chris

author: microbug
author_rank: Frequent Contributor
timestamp: May 16, 2013, 07:57:19 pm
content: Ah, ok. What sort of setup would I need for that - voltage divider + relays? Also, can you recommend any diode / bridge rectifier that would be better for this, I have some 1N4007s at home - would they be ok?Thanks for your help!

author: mamalala
author_rank: Unknown
timestamp: May 16, 2013, 08:01:26 pm
content: Quote from: microbug on May 16, 2013, 07:57:19 pmIt gets rather complicated pretty fast, a suitable transformer being one of the problem (there are only few available to the hobbyist).In any case, i would recommend that you first start with simple things like the L200 circuitry to gain some experience and knowledge in electronics. Otherwise, as a "newbie", you have a high risk of ending up with a pile of blown parts, on which you spent a lot of money, only to get frustrated at the whole hobby.I was just mentioning regular bench supplies to show you that the problem about in/out voltage differences is common and usually accounted for in "better" designs.Greetings,Chris

author: microbug
author_rank: Frequent Contributor
timestamp: May 16, 2013, 08:03:41 pm
content: Actually, I think it might be a good idea just to go for a lower voltage on the transformer output; maybe 20v; and not worry too much with the heat dissipation (I have a fairly decent CPU cooler I can attach).

author: mariush
author_rank: Super Contributor
timestamp: May 16, 2013, 08:28:43 pm
content: In retail, you won't find transformers with lots of windings on the secondary side. Usually, linear power supplies use custom transformers with up to 4 separate windings and join them together using relays to get voltage only slightly higher than the voltage you want at output, so that there's little voltage drop across the regulator (or pass transistor in case of a linear power supply).I would suggest going for a 18-20v transformer, either with center tap, or with two separate windings of 9-10v on the secondary side.One with two separate windings such as one of the below would work great:http://uk.farnell.com/pro-power/ctfc50-9/transformer-50va-2x-9v/dp/1780856http://uk.farnell.com/pro-power/ctfc75-9/transformer-75va-2-x-9v/dp/1780861As they're separate, you can connect them either in series, or in parallel.In series, you get 18v AC rms at 50 VA or about 24V DC after rectification and about 1.9A of current.In parallel, you get 9v AC rms at 50VA or about 11 V DC after rectification and about 4 A of current.If you connect a side of both windings together, you can just switch between the other two windings sides to get 9v AC rms or 18 v AC rms but then in both cases you only get a maximum of 1.9 A of current because a winding would be unused (wasted) when the switch is on 9v.Either way, you can use 0.1uF 50v capacitors (but you should use 100v rated), you can use more easily available 35-50v electrolytic capacitor.The 50 VA will give you will have 2 x 9v windings, each of about 2.78 A of current (after rectification and conversion to DC), that's about 1.8ASo if you want some voltage less than let's say 8v at 1A, you switch to single winding/both windings in parallel, and you have 11v max at input, the regulator needs about 2-2.5v to work properly therefore 8 + 2.5v < 11v and everything will be fine. Your regulator will only dissipate (11v - 8v) x 1A = 3w and everything is right with the world.You still shouldn't use those 1n5422 as they're too weak, just use regular diodes with 0.6-0.8v forward voltage (rated for 3-4A) or bridge rectifiers like this one:http://uk.farnell.com/multicomp/gsib1504/bridge-rectifier-15a-400v-gsib/dp/1861524Yeah, you get more forward voltage drop but who cares...

author: SeanB
author_rank: Super Contributor
timestamp: May 16, 2013, 08:40:25 pm
content: With bridge rectifiers use a 25A one, lower forward drop of around 0.5v at the 3A mark, plus it runs a lot cooler.http://uk.farnell.com/vishay-general-semiconductor/lvb2560-m3-45/bridge-rectifier-1ph-25a-600v-gsib/dp/2099331this one has lower loss, and a lower than normal voltage drop of under 1V at full load current.

author: mariush
author_rank: Super Contributor
timestamp: May 16, 2013, 09:08:32 pm
content: ^ yeah, but it's 6.3 uk pounds compared to the generic multicomp 15a bridge that's only 0.3 uk pounds.With 6 pounds extra he can just go for the 2x12v instead of 2x9v for the same VA rating (or even up the va rating a bit) and not care about the bridge rectifier.

author: SeanB
author_rank: Super Contributor
timestamp: May 17, 2013, 03:37:39 pm
content: Wasn't looking at price, just to give alternates, and a thought as to selection. I just looked at the lowest voltage drop at the 25A mark with a 400V reverse voltage. There will be cheaper ones in that area with a higher drop.