Converter efficiency topology - Page 1
Forum section: electronics / beginners

author: sk614
author_rank: Contributor
timestamp: January 31, 2017, 02:31:15 pm
content: I have a high voltage source (100V) that I need to convert down to multiple outputs (1.8V, 3.3, 5V, 12, 24V). I have two topologies;A) Step down all the outputs directly from the 100V.B) Step down 100V to 24V, and 24V to 12V, 12 to 5 etc...C) Step down 100V to 48or24 and step down the rest from 48or24.How do I calculate the total efficiency for comparisons?

author: Buriedcode
author_rank: Super Contributor
timestamp: January 31, 2017, 03:55:42 pm
content: Efficiency is the the product of many parameters, load current of each voltage, topology, properties of the main components (switch, transformer, output caps and many more!).With that said, your three options, A, B and C, are reasonable, and so provides a starting point for searching for of-the-shelf converters. What you really do need to know is the load current of each voltage rail; how much these will vary. For example if your 24V line will normally draw 20mA, but occasionally draw up to 2A then such a variation can mean in order to meet the 2A capability, at 20mA load the efficieny will probably be terrible. So you'll need a rough range of load currents to work with. Also, size constraints, and any ventilation/heatsinking.A) PC supplies generally do not output 1.8V or 24V, but they can have an input range of 90-260V AC and many off-the-shelf PSU's have a similar input range, with multiple outputs. You could probably find a PSU that will accept 100V AC or DC, and output 3.3, 5, and 12V. With the 1.8V being derived from the 3.3V with a high eff. buck.B) Cascading converters would probably greatly reduce efficiency. This may not necessarily be a bad thing, after all, if you're drawing 10mA @ 12V and the 100V-24V converter has an eff. of 80% and the 24-12V 70%,then its only 0.7*0.8 = 56%... but at such low power levels the absolute power dissipation is tiny. Again.. requires you to know the range of load currents, so you can estimate the rough efficiency of each converter. By cascading them *all* even if each stage is 80% eff. with 4 stages to get to 3.3V, if each converter is 80%, thats ( 0.8 )^4 = 41%. eff. So.. probably unwise.C) This is a reasonable half measure, but once again... you need to know what power rail is drawing the bulk of the current. If your 12V is drawing most of the current then you've got two converter in series which lowers efficiency, and for the higher load, drastically increases losses.tldr;Need to know the sorts of load currents for each power rail before one can do anything. Although *generally* cascading converters will be much less efficient. Right now I would guess you'll find a combination of the above best - one large unit to provide most of the power rails, with a couple of smaller converters for the lower voltages.

author: David Hess
author_rank: Super Contributor
timestamp: February 01, 2017, 01:14:33 am
content: That large a voltage change will require low duty cycles unless a transformer or multiple stages are used.The low voltage outputs will require synchronous rectification for high efficiency.I would probably use a combination of both. Generate the 24, 12, and maybe 5 volt outputs in the first stage directly from 100 volts using a transformer and then use separate buck converters to make the 1.8, 3.3, and maybe 5 volt supplies from the 5 or 12 volt regulated supply. Some of the details depend on the 1.8, 3.3, and 5 volt current levels.