Turning LM25085 into an electronic load - Page 1
Forum section: electronics / general

author: king.oslo
author_rank: Frequent Contributor
timestamp: July 30, 2012, 08:09:59 pm
content: Hello there,I am still working on my first power supply. I decided to make a switcher with the capability to source and sink current.Please see the attached schematic. Current will be sunk and sourced through the header called "Binding post". The relays will direct the current the way I want it.Whilst sinking current I reckon I have to input an alternative signal for FB-pin and the differential amplifier: VFBneeds to fall when input voltage rise. Second, the output of the differential amplifier will become negative whilst sinking current, so I need to rectify this as well.So my questions:First, what about the RT and VIN pin? I reckon VRTneeds to fall when VINrises? Otherwise the One-shot-module will not function properly? But what about the VINpin?Second, any additional constructive criticism is welcome.Thank you for your time.Kind regards,Marius

author: tom66
author_rank: Super Contributor
timestamp: July 30, 2012, 09:39:39 pm
content: You cannot readily sink current with a switching converter; unless you have somewhere to store the energy in.

author: king.oslo
author_rank: Frequent Contributor
timestamp: July 30, 2012, 09:57:59 pm
content: Current is stored in the coil before it is passed to GND. Did I miss something?Thanks.M

author: tom66
author_rank: Super Contributor
timestamp: July 30, 2012, 10:21:06 pm
content: Quote from: king.oslo on July 30, 2012, 09:57:59 pmHow are you dissipating the load power? It has to go somewhere, as heat usually.

author: king.oslo
author_rank: Frequent Contributor
timestamp: July 30, 2012, 10:34:45 pm
content: It is sunk:Binding post -> Rsense -> caps -> inductor -> PMOS -> GND.How does that sound?M

author: tom66
author_rank: Super Contributor
timestamp: July 30, 2012, 10:44:39 pm
content: Quote from: king.oslo on July 30, 2012, 10:34:45 pmSo, where does all the heat go? Into PMOS? Otherwise, it is just a dead short.

author: king.oslo
author_rank: Frequent Contributor
timestamp: July 30, 2012, 10:59:15 pm
content: The pmos will switch at 300khz. When it is on, current stored in the inductor will go to gnd. When it is off, the inductor will charge up.M

author: tom66
author_rank: Super Contributor
timestamp: July 30, 2012, 11:09:54 pm
content: Quote from: king.oslo on July 30, 2012, 10:59:15 pmYou aren't dumping the heat anywhere. Current will flow through the inductor, but then it turns off. Inductors don't like sudden current interruptions (=HV/breakdown), so you need a return path. This return loop must be where heat is dissipated; either that, or on the input through some resistors or similar.My suggestion: Forget about using the switching regulator to make a current sink. It won't work reliably as the regulator is compensated and designed for a specific feedback configuration. Once you start overdriving the feedback with an op-amp you introduce additional phase shifts which shouldn't be there and they'll cause a whole lot of trouble for you.I would just add a powerful MOSFET array, and a typical current sink control (you could share the high-side measurement.)

author: king.oslo
author_rank: Frequent Contributor
timestamp: July 30, 2012, 11:24:59 pm
content: Thanks for your suggestionI will keep it in mind. So for the sake of learning, when the MOSFET switches off, the magnetic field of the coil collapses, and current flows through the diode, and all is sweet? Or not?What exactly is the problem with this? Are you saying high voltage will appear at the inductor/PMOS node which will destroy the MOSFET?Could high speed opamps cure problem with phase shifts?Thanks.M

author: tom66
author_rank: Super Contributor
timestamp: July 31, 2012, 06:32:58 am
content: Quote from: king.oslo on July 30, 2012, 11:24:59 pmIn a typical (non-synchronous buck converter):- high side switch turned on- current flows through high side switch (MOSFET)- current flows through inductor- current flows through output capacitor (charging it)- high side switch turned off- current flows through inductor- current flows through output capacitor (discharging it)- current flows through buck diodeQuote from: king.oslo on July 30, 2012, 11:24:59 pmNo, it probably won't destroy the MOSFET, but it won't work as an electronic load. It will be a dead short to your device under test.Quote from: king.oslo on July 30, 2012, 11:24:59 pmPossibly, but I would think high speed op amps could introduce even more problems (noise sensitivity and higher frequency oscillations being some.) The chip is compensated (stabilised) for a few particular configurations - operating it outside of this range can lead to anything happening.

author: jahonen
author_rank: Super Contributor
timestamp: July 31, 2012, 06:42:13 am
content: If you push current into the output of a synchronous step-down regulator, it causes regulator input voltage to rise. It turns into a step-up regulator, working from output to input. Unless there is an overvoltage lockout, you'll just destroy the regulator due to overvoltage.Regards,Janne