Rectifier diodes. What data points generally determine efficiency? - Page 1
Forum section: electronics / projects

author: pauljmccain
author_rank: Contributor
timestamp: May 12, 2015, 10:56:07 pm
content: Apologies if this should be in the "beginner" section! Seems like a beginner question, but I searched quite a bit, and there doesn't seem to be a clear answer.I'm dealing with a design problem for a DC power control module that requires a bridge rectifier. I need to rectify 5 amps of DC power at 12V. The whole thing needs to be as compact as possible. Additional heatsinking is not an option. The goal is to reduce heat from being generated to begin with. Currently, the diodes used on the product are overheating.I'm trying to find rectifiers that are more efficient. I'm looking through datasheets, and of course, as we know, diodes are rated in Vf, barrier V, current capacity, etc. Some, but not all, have thermal resistance figures.So, some questions!What data points should I focus on to select as efficient a rectifier as possible, instead of buying a dozen of each to make, test, then dispose of a dozen prototypes? Does a lower Vf, lower Vrrm, and higher current rating imply that the diode will be more efficient?Does a higher current rating for comparable package size mean that the diode is more efficient, or just capable of handing higher loadifthere's enough heatsinking?Thanks very much to anyone who can offer any insight.

author: Alex Eisenhut
author_rank: Super Contributor
timestamp: May 12, 2015, 11:03:55 pm
content: Vf and TrrUse ideal diode controller and MOSFET.

author: free_electron
author_rank: Super Contributor
timestamp: May 12, 2015, 11:04:41 pm
content: Vf vs If. Vf is not a constant. it is dependent on temperature an current through the diode.

author: pauljmccain
author_rank: Contributor
timestamp: May 12, 2015, 11:13:15 pm
content: Quote from: Alex Eisenhut on May 12, 2015, 11:03:55 pmYes, considering a MOSFET rectifier for the next layout but it's still going to be a tight squeeze. Currently tons of boards on the shelf, need to just switch out components.Does Trr matter if it's rectifying a constant power source? DC to DC. Ridiculous, but this is just for user convenience and to prevent any chance of electrolytics blowing.

author: technix
author_rank: Super Contributor
timestamp: May 12, 2015, 11:18:43 pm
content: Quote from: pauljmccain on May 12, 2015, 10:56:07 pmDiodes turn all power they consumes into heat so if you want minimum heat, shoot for minimum power. From the rules of physics we know that the power of electronics, and hence heat output for resistors and in this case rectifier diodes, equals to its current multiplied by its voltage drop.You have a fixed 5 amperes of current, so to minimize heat you need the least forward voltage drop you can get out of your diodes or diode equivalents. Currently the lowest forward drop possible is attainable using active rectification using 4 MOSFETs and some kind of controller likeLT4320. This will give you almost zero loss. If you have budgetary limitations and such FET bridge is not an option, you can use Schottky diodes.Usually you need some safety margin just in case things go awkward in extreme cases, so we need, say, 10A and 40V rated parts. If you are using MOSFETs with a controller,IRF540NandIRF4905are good options with at least 30A current rating and 55V voltage rating. If you need Schottky diodes,MBR1045can be a good option.Or if you are really gilding the lily, you can put FETs and Schottky rectifiers in parallel to eliminate the recovery time of the FETs' body diodes, as the Schottky will effectively prevent that body diode from ever conducting with enough voltage in the first time.Also, since youa re talking high currents and lots of power, put a fuse in line would be a very smart idea - it is very nasty to handle a blown power FET, which may indicate lots of blown components downstream - maybe even your main 40-bucks-a-pop MCU.

author: technix
author_rank: Super Contributor
timestamp: May 12, 2015, 11:21:35 pm
content: Quote from: pauljmccain on May 12, 2015, 11:13:15 pmRecovery time means your diode is still conducting even it is reverse biased - essentially a short circuit and burn a lot of power! Schottky rectifiers don't have this though.

author: Paul Moir
author_rank: Frequent Contributor
timestamp: May 12, 2015, 11:48:20 pm
content: Is "doesn't do anything" an acceptable outcome with reverse polarity? If so, there's a trick with a PMOS FET you can see on page 7:https://www.infineon.com/dgdl/Reverse-Batery-Protection-Rev2.pdf?folderId=db3a304412b407950112b418872b2614&fileId=db3a304412b407950112b41887722615&ack=tOtherwise I think it's just Schottky diodes as a solution due to the lower Vf, but don't forget to check Vf at the current you're running like free electron said. Reverse recovery time doesn't enter into it since it's DC. An ideal bridge is going to be pretty expensive without much reward.

author: langwadt
author_rank: Super Contributor
timestamp: May 13, 2015, 01:01:07 am
content: Quote from: pauljmccain on May 12, 2015, 11:13:15 pmThe only Vf vs. If matterit doesn't take much to do it with a fethttp://provideyourown.com/wp-content/uploads/tech/p-ch-fet-circuit.png

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 13, 2015, 01:49:15 am
content: Doing that as small as possible, you probably want to investigate an "ideal diode". There are MOSFET FWB controllers available for this.Tim

author: calexanian
author_rank: Super Contributor
timestamp: May 13, 2015, 05:52:23 am
content: I think you are looking beyond his question. The pertinent factors are average forward voltage drop, and capacitive junction losses at higher frequencies. Your voltage drop times the junction current at the given conditions will yield your loss expressed in watts. A standard rectifier will have a given voltage drop. A shottkey rectifier will have a significantly lower voltage drop, therefore less wasted wattage. The others are talking about what are known as active rectifiers that use fets and driver circuitry to reduce waste. Good for low voltage high current rectification but for the sake of your question not relevant.

author: Liv
author_rank: Regular Contributor
timestamp: May 13, 2015, 08:00:41 am
content: Quote from: Paul Moir on May 12, 2015, 11:48:20 pmIt's true, LT chips are very expensive. So I do synchronous rectifiers on the components of the trash.

author: mzzj
author_rank: Super Contributor
timestamp: May 13, 2015, 09:45:45 am
content: Quote from: Alex Eisenhut on May 12, 2015, 11:03:55 pmNot to forget leakage current if you are using Schotky diodes. Just as a random example SS10P2CL schotky diode in "12v 2A application" gives you ~0.3v forward voltage drop @125C junction temperature and 0.3*2= 0,6W forward conduction loss. Leakage current is something like 15mA so 0,015A*12V = 0,18W leakage current loss.

author: pauljmccain
author_rank: Contributor
timestamp: May 13, 2015, 12:24:53 pm
content: Great info all around, thanks to everyone who replied! Lots of good ideas for how to move forward here.

author: macboy
author_rank: Super Contributor
timestamp: May 13, 2015, 02:20:50 pm
content: Quote from: pauljmccain on May 12, 2015, 11:13:15 pmPower in the diode will be IFtimes VF. Lowest VFwill be from a Schottky. These are "nominally" about 0.3 V vs. about 0.6 V for a silicon. But it is important to consider what the VFis at 5 A, since that is your operating point. VFincreases with IF, and will be lower for larger diodes (use bigger ones than needed to reduce VF!).To prevent dumb users from destroying the circuit, you can also use a reverse-biased diode across the power terminals, preceded by an appropriate fuse or other protection device (circuit breaker, polyfuse, etc.). This will short circuit the input if connected backwards, tripping that protection device, and preventing reverse power from reaching the circuit (note that ~ -Vf of the diode will reach the circuit, usually without consequence). The advantage is absolutely no impact when used correctly, no voltage drop or power dissipation. The disadvantage is replacing blown fuses due to oops moments.

author: macboy
author_rank: Super Contributor
timestamp: May 13, 2015, 02:35:41 pm
content: ...Or try a single in-line rectifier instead of a bridge. The thing won't work connected backwards but you only have one diode dissipating power instead of two.