URGENT!!! ***** using boost converter with VOUT lower than VIN**** - Page 1
Forum section: electronics / beginners

author: SnakeBite
author_rank: Regular Contributor
timestamp: March 19, 2014, 02:12:27 pm
content: Hi,i started to design a DC/DC booat converter that take 12V battery voltage and convert it to higher voltage. this specific converter chip i used is DUAL PHASE with means that it has 2 output that can be driving separately. one of the output voltage is 30V@6.5A and that other i want it to be 5V@4A can i do it?! i don't really want to replace the IC because i spend A LOT of time designing the 30V@6.5A output and unfortunately only NOW i realized that i can't get 5v out because i use 12v supply and it's a boost converter and not a buck converter.the IC i'm using is thehttp://cds.linear.com/docs/en/datasheet/3788fc.pdfthanksIdo

author: BravoV
author_rank: Super Contributor
timestamp: March 19, 2014, 03:01:32 pm
content: When it comes to asking help especially in switching power supply problem ...... post your schematic AND the real photo of your working circuit.

author: Skimask
author_rank: Super Contributor
timestamp: March 19, 2014, 03:02:43 pm
content: But it'sURGENT!

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: March 19, 2014, 03:08:51 pm
content: Quote from: SnakeBite on March 19, 2014, 02:12:27 pmSince it is urgent, the answer is NO.With a Boost converter there is always a path through the diode or the top MOSFET to the output.Jay_Diddy_B

author: madires
author_rank: Super Contributor
timestamp: March 19, 2014, 03:24:42 pm
content: The IC you've selected is specifically designed for boost converter operation. So you have to add a 5A buck converter IC.

author: Falcon69
author_rank: Super Contributor
timestamp: March 19, 2014, 03:55:48 pm
content: Wouldn't two resistors work as a voltage divider?

author: madires
author_rank: Super Contributor
timestamp: March 19, 2014, 06:23:17 pm
content: Quote from: AcHmed99 on March 19, 2014, 04:20:40 pmWhy do you want to put the 5V buck converter behind the 30V boost converter if you can power the buck converter directly by the 12V battery?

author: sigxcpu
author_rank: Regular Contributor
timestamp: March 19, 2014, 07:25:51 pm
content: Because that is too simple

author: lapm
author_rank: Frequent Contributor
timestamp: March 19, 2014, 07:44:30 pm
content: Why do something simple when you can over engineer it to include complexity that could be avoided..

author: hlavac
author_rank: Frequent Contributor
timestamp: March 19, 2014, 08:42:14 pm
content: Buck behind boost = double conversion losses.Feed that buck converter directly from the battery to avoid losing power.

author: Nerull
author_rank: Frequent Contributor
timestamp: March 19, 2014, 11:48:19 pm
content: So in English, when someone is walking behind someone else, they are walking before them?They're antonyms. Behind means after. Most people would put the voltage source 'first' in a circuit, everything follows behind that. That matches conventional current flow.

author: dr.diesel
author_rank: Super Contributor
timestamp: March 20, 2014, 12:18:03 am
content: Anything truly urgent would include a proper explanation of why the rush, whichdoesn'tinclude at whits end.

author: madires
author_rank: Super Contributor
timestamp: March 20, 2014, 02:07:41 pm
content: Quote from: AcHmed99 on March 19, 2014, 11:39:38 pmI think you missed two important issues. The first one is that boost converters got a lower efficiency than buck converters in general. And the second one is that the boost converter has to deliver the power for the 30V load and the buck converter, i.e. it needs to be redesigned for +20W at least.

author: madires
author_rank: Super Contributor
timestamp: March 20, 2014, 02:14:40 pm
content: Quote from: AcHmed99 on March 19, 2014, 11:55:47 pmSo 'behind' is specific to your point of view? If I'm looking from the source (battery) towards the load, the load is behind the battery? But if I'm looking from the load towards the source, the source is behind the load?

author: cosmos
author_rank: Regular Contributor
timestamp: March 20, 2014, 03:16:10 pm
content: It might be possible to fit a SEPIC structure to your second DCDC channel while not changing the IC.SEPIC can step both up and down.You would at least need a new (dual winding) inductor, capacitor and a diode.

author: johansen
author_rank: Super Contributor
timestamp: March 21, 2014, 06:29:47 am
content: To buck 30 down to 5 volts requires a bigger inductor than 12 into 5 for the same switching frequency.

author: Kjelt
author_rank: Super Contributor
timestamp: March 21, 2014, 08:55:54 am
content: Quote from: AcHmed99 on March 19, 2014, 11:39:38 pmLets ask all those medals back from those brave soldiers who fought behind enemy lines, in your definition they were fighting their own sideBehind = after = post.

author: madires
author_rank: Super Contributor
timestamp: March 21, 2014, 03:46:58 pm
content: Quote from: AcHmed99 on March 21, 2014, 11:40:37 amIt's similar in German. 'behind' is 'hinter' and the butt is also called 'Hintern' commonly ;-)

author: free_electron
author_rank: Super Contributor
timestamp: March 21, 2014, 06:22:31 pm
content: root causes are- not reading the datasheet- not realising 5v is less than 12 volts thus requiring a buck ...