Talk:DC-to-DC converter

message: a good way to explain the basic idea (which we should do) is to show two caps on top of each other, with two SPDT switches.  Show the 5 or whatever V battery being connected across one capacitor, then disconnected, then connected across the other, and then show that the total voltage across both caps is 10 V. - Omegatron 23:04, Mar 19, 2005 (UTC)
: That sounds like a charge pump and should probably go there.  Do note that a standard voltage-doubling charge pump doesn't actually do this -- rather, it charges the flying capacitor to V+, then connects cap- to V+ and cap+ to Vout, thus using only one flying capacitor. Evand 23:58, 28 May 2007 (UTC)
thread_title:  Charge pump illustration 

message: Give one good reason why we should. --Heron 13:41, 26 May 2007 (UTC)
replies: text: Discussion appears to be taking place here --D0li0 09:41, 27 May 2007 (UTC)
thread_title:  Merge 

message: It's said in the article that DC converters are to be connected to a battery. It's true in portable applications, but all of us use DC converters plugged to a rectifier connected to the AC network in our computers. The DC converters are connected to a power source of positive voltage, and supply DC current with no theorical power losses.--Iruando (talk) 23:43, 20 February 2008 (UTC)
thread_title:  batteries? 

message: The article mentions the following under the header "Non-isolated topologies (using inductors)"
    * Buck (step-down) - The output voltage is higher than the input voltage, and of the same polarity
    * Boost (step-up) - The output voltage is lower than the input voltage, and of the same polarity
Shouldn't this just be the other way around? Step-down results in a lower voltage and step-up in a higher voltage?  —The preceding unsigned comment was added by P.edelman (talk • contribs). 
: Done. You are right; I see someone has already fixed this in the article. --68.0.124.33 (talk) 15:13, 18 August 2009 (UTC)
thread_title:  Step-up and step-down 

message: The DC-DC converters table is unclear and somewhat messy. I think it could be reorganized and redesigned.
I'm not sure how to interpret "They are also used at extremely high voltages, as magnetics would break down at such voltages." It just doesn't sound right.
ICE77 (talk) 05:01, 21 December 2011 (UTC)
replies: text: You are right, the table looks very odd. I'd like to fix it, but in it's current state I can't make heads or tails of it. Added a cleanup-message. Noggo (talk) 10:31, 17 October 2015 (UTC)
replies: text: I went back through the history (in particular https://en.wikipedia.org/w/index.php?title=DC-to-DC_converter&oldid=238809009 and https://en.wikipedia.org/w/index.php?title=DC-to-DC_converter&oldid=152776493) and attempted to restore the table to its original format and merge in additions that have been added. Please take a look and check. SageGreenRider (talk) 19:43, 23 October 2015 (UTC)
thread_title:  Messy table and ambiguous sentence 

message: A transformer doesn't transfer energy by temporarily storing such energy in a magnetic field, because the magnetic field of an ideal transformer contains no energy.  For example, see the book "Electric Machinery" by Fitzgerald 3rd edition, chapter 1: Magnetic circuits and transformers.  Find the  section on "Ideal Transformer" and the following sections that account for the differences between ideal and real transformers.  It's claimed that the properties of an ideal transformer closely approaches those of a real transformer.  Unfortunately, I can't find any statement to the effect that the magnetic field of an ideal transformer contains no energy, but it's clear that it's so.  The energy density of the magnetic field in the core is 1/2 BH = 1/2 B^2/u (since B=uH) where B is the flux density, H is the magnetic field intensity and u is the permeability of the transformer core.  But in an ideal transformer, u (permeability) approaches infinity which implies almost zero energy in the magnetic field since B must exist and be finite in order to generate the flux (= BA where A is the area).  The ideal trasformer must have flux to induce voltages (= Weber-turns per sec) in the windings).  So while the energy flows from the primary to the secondary via the magnetic field, the field never contains any energy regardless of the amount of flux (Webers) in the core or how fast the flux is varying.
Note that is the ideal xformer, H (amp-terns per meter) is zero since the amp turns of the primary exactly cancel the amp-turns of the secondary (the current ratio is exactly equal to the turns ratio).  This permits B=uH to be finite where u approaches infinity and H approaches zero. 
Of course a real transformer has finite permeability and thus contains some energy in the core.  But this energy doesn't flow from the primary taps of the transformer to the secondary taps.  It's represented in the equivalent circuit of a transformer by an inductor which (if assumed to be across on the transformer primary) draws a magnetizing current from the primary feed.  Like any inductor, it draws in energy from the primary and stores it in it's magnetic field and then returns the energy to the primary (120 times a second for 60Hz).  But the energy that crosses from the primary to the secondary isn't stored in this field of this inductor (or in any other field of the transformer).
Another way to look at it is via power.  Let I be rms amps and V be rms volts, 1 be the primary and 2 be the secondary.  Then the power flowing thru the transformer (assuming unity power factor) is E1 I1 = (approximately) E2 I2.  How much power is flowing into the magnetic field of the transformer?  Its just E1 I3 where I3 is the magnetizing current where the mmf of this current (I3 N1 in amp-turns) is the difference of the actual mmf (amp-turns) of the primary winding minus the mmf of the secondary, and this difference is small since the ideal transformer equations show zero difference.  So the power flowing into the magnetic field of the transformer (E1 I3) is far less that the power flow thru the transformer (E1 I1) since I3 << I1.  Thus only a small portion of the power input to a transformer goes into its magnetic field, and that small portion doesn't make it thru the transformer either, as shown by last paragraph.  Note that the above neglects various losses, such as core losses.  Also, power flow into the inductor only happens for part of a cycle and then power flows back out of the inductor so power flow to create and the destroy the magnetic field of the transformer (which is represented in the equivalent circuit by the magnetic field of an inductor) is a bi-directional power flow which in the long run is called reactive power.
An example of a real transformer with finite permeability:  Put a fixed ac voltage on the primary.  Since voltage in Weber-turns per second (= flux-turns/sec.) the varying magnetic field in the core will remain the same regardless of how much power is transferred to the secondary output.  It's the same flux if the secondary is open circuited and no power is flowing thru the transformer and this same flux if the secondary is loaded with maximum load and sending maximum power to the secondary.  Thus power thru the transformer doesn't travel via the flux in the core for if it did, one would expect the magnetic field in the core to increase as power flow increased.  But a small amount of power (as observed with an open circuit secondary) is used to "excite" the core but it's not flowing thru the transformer.  
In conclusion, a transformer transfers power from the primary to the secondary via a varying magnetic field but none of the power being transferred is ever stored (even temporarily) in this magnetic field.  I'm sure that there are a lot of technical people that don't understand this and it should be explained in the transformer article but isn't.  There must be a good reference for it, but where? 66.81.29.245 (talk) 19:36, 2 December 2014 (UTC)
replies: text: Nonsense. Magnetic fields contain energy. See for example sources in Magnetic energy . By the way they also store angular momentum. See Feymann Lectures in Physics for example. SageGreenRider (talk) 01:50, 7 February 2015 (UTC)
replies: text: I agree with SageGreenRider. Storing energy in the transformer is the basic idea behind e.g. a Flyback converter, see . Noggo (talk) 10:11, 17 October 2015 (UTC)
thread_title:  transformer-magnetic 

message: , DC-to-DC convertors are not necessarily boost convertors. You can make a linear buck converter from vacuum tubes. It's wrong to begin the history section talking only about boost converters. Talk to SageGreenRider 22:53, 18 January 2016 (UTC)
replies: text: Thanks. Do you have a source for this having been done in practice? It may have been. I know of an AC/DC TV circuits which used valves to boost the available HT modestly above what can be reached by rectifying the mains voltage; maybe that could be considered a case of voltage conversion? I think these things would have been used rarely, but if they were actually used the text needs some modification (I'd tend to mention the rarity). It would be useful to get other opinions; I don't want to push something that may not be best usage. Best wishes, Pol098 (talk) 21:46, 19 January 2016 (UTC)
replies: text:   There a posting here that contains many historical references. IBM had a vacuum tube switched-mode converter in a computer in 1959. I don't know if it was buck or boost nor its type (DC-to-DC or rectifier or inverter). I tried to read the reference here, see page 60-17 but it wasn't immediately clear to me. This matter was discussed extensively on the talk page of Talk:Switched-mode_power_supply#History Talk to SageGreenRider 22:20, 19 January 2016 (UTC)
replies: text:  Thanks. I've been looking at the IBM PSU manual, and am not going to get very far without careful study, which I'm probably not going to do. I do get the impression that the buck-boost circuitry is involved in stabilising the AC line voltage, not DC-DC conversion, but I may be wrong. I'll continue when I've looked at the historical references. Pol098 (talk) 22:53, 19 January 2016 (UTC)

text: The only other old reference I've found is to a power supply Beckman made for their DU spectrophotometer (originally, 1940s, powered by lots of big batteries, then the 73600 AC supply was designed with some sort of buck technology; the instrument was produced for 25 years, I don't know the date of the PSU but it was thermionic, not semiconductor). I haven't found much about this. Best wishes Pol098 (talk) 23:33, 19 January 2016 (UTC)
thread_title:  Vacuum tubes 

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