Voltage Converter - Page 1
Forum section: electronics / beginners

author: yashrk
author_rank: Frequent Contributor
timestamp: March 22, 2014, 12:52:50 pm
content: I am learning and experimenting with boost and buck converters and came across few question1. Is there any way to boost current ?2. For this I used a circuit made with LM2576- adj (based on example circuit in data sheet)I provided it 20V and 800mA (from bench power supply) and fixed the output to 5v, I was expecting increase in current as I waslowering the voltage but its showed same 800mA. Is it due to I am using a power supply and not a battery ?3. I know that boost circuit boost the voltage but can it be configured to boost current from the source ?4. How do I control the current flow in a high current application ?Thank You for your time.

author: sleemanj
author_rank: Super Contributor
timestamp: March 22, 2014, 01:04:08 pm
content: Quote from: yashrk on March 22, 2014, 12:52:50 pmDescribe EXACTLY what you measured, where you measured it, and under what circumstances, what was the load.Power ( W=VI ) in will necessarily be higher than Power out due to losses.

author: cyr
author_rank: Frequent Contributor
timestamp: March 22, 2014, 01:05:20 pm
content: What do you mean exactly? Did you draw 800mA at 5V from the output of your buck converter, and the input was drawing 800mA at 20V from your power supply?How did you measure, and what was your load?

author: yashrk
author_rank: Frequent Contributor
timestamp: March 22, 2014, 01:25:09 pm
content: I didn't attached any load to it i just measured it by multimeter

author: sleemanj
author_rank: Super Contributor
timestamp: March 22, 2014, 01:36:42 pm
content: Quote from: yashrk on March 22, 2014, 01:25:09 pmIf you didn't have anything connected, then the output would have no current being drawn to measure, neither would you have anywhere to measure it anyway!The input would have current, but only the quiescent of running the buck converter, and if your buck converter is drawing that much current with no load, something is seriously wrong with it.Once again, describe exactly what you measured, and where you measured it.

author: yashrk
author_rank: Frequent Contributor
timestamp: March 22, 2014, 02:23:09 pm
content: First I selected the voltage of 20v on power supply and then checked it by multimeter, then switched my multimeter on amp mode and checked the amp (I have connected the positive probe to positive of the power supply and negative to negative) no load was attached is there anything wrong in that ?

author: cyr
author_rank: Frequent Contributor
timestamp: March 22, 2014, 02:41:54 pm
content: Yes.What you did was short the output of your power supply and measure the maximum current it will deliver.To measure current you need to connect the multimeter in series with the circuit you want to measure the current through.

author: mariush
author_rank: Super Contributor
timestamp: March 22, 2014, 02:43:04 pm
content: To measure current, you need to put the device you measure with IN SERIES with the device consuming current.+ 20v power supply -----------> multimeter red probe in multimeter current jack , multimeter black probe in ground/COM jack ---------> device consuming current ( + of device , - of device ) ----- power supply ground/-if there's nothing consuming power, then your multimeter will show 0 amps current consumption.If you put the multimeter in series with the LM2576 but in front of it, and then nothing uses power produced by the LM2576, you will probably see just a few mA of power which are used internally by the LM2576 to keep the output voltage on the output wires to a stable 5v.When you connect something to the LM2576 output wires, you will see the current used by LM2576 go up. For example, if your LM2576 will provide 5v at 100mA to something connected to it, this is produced with about 70-80% efficiency.100 % efficiency ---- ? watts80 % efficiency ---- 5v x 0.1 amps = 0.5 watts? watts = 0.5 x 100 / 80 = 0.625 watts.So your LM2576 will probably consume about 0.625 watts to produce at the output 0.5 watts ( 5v @ 0.1a). With 20v input, it means you should expect 0.625A / 20v = 0.031 A or about 31 mA on the multimeter.If you want to measure how much current something uses from the LM2576, you have to put the multimeter in series with that thing.LM2576 output + ------> multimeter red probe in the multimeter current jack , multimeter black probe on multimeter ground/COM jack --------> what you connected to lm2576 ( + of device, - of device ) --------> lm2576 output ground/-

author: yashrk
author_rank: Frequent Contributor
timestamp: March 22, 2014, 05:10:47 pm
content: Quote from: cyr on March 22, 2014, 02:41:54 pmCertainly i want to measure the max output.See what I meant was that max output of my power supply is 800mA @ 20V i.e. 9.6W so if I lower the voltage to 5V (by LM2576) so by calculating i should get 1.92 volts but multimiter attached on the output of LM2576 showed me the same 800mA.And please answer my other questions also. Thanx

author: mariush
author_rank: Super Contributor
timestamp: March 22, 2014, 05:45:10 pm
content: So your power supply can do 20v @ 0.8A or about 16 watts (20 * 0.8 = 16). How did you come up with 9.6 watts?The LM2576 chip can produce a voltage lower than 20v at a particular amount of current with a certain efficiency, usually about 75-80% efficiency. For easier math, let's say 80%. The rest of that 20% is lost as heat and other things in the chip, in the inductor, resistors and capacitors.So if your power supply can do 16 watts, then your LM2576 will probably be able to produce 0.8 x 16 = 12.8 watts (0.8 is 80% efficiency). If you set the chip to 5v output, then your maximum current should be 12.8 watts / 5v = 2.56 Amps

author: sleemanj
author_rank: Super Contributor
timestamp: March 22, 2014, 11:08:27 pm
content: Quote from: yashrk on March 22, 2014, 05:10:47 pmFirstly, shorting an unknown supply through your meter is not a good idea!QuoteYour maths is wrong.Volts = Current * ResistanceWatts = Volts * CurrentLet us assume for the moment that your input supply can give 20v at 800mALet us assume an 80% efficiency.Let us assume for the moment that your DC converter is set to output 5v, and the output is shorted.The max input power we expect to see is 20v*0.8 = 16WThe max output power we could expect to see is 16W*0.8=12.8WThe max output current we could expect to see is 12.8/5=2.56ABUT. You are shorting your outputs, I doubt very much your input is giving you 20v any more, the 2596 could be doing anything or nothing, who knows what the voltage on the output is when you are shorting.Your biggest problem is that your measurement technique is all up the wop.Here is a better McGuyver'd way for you to measure without any fancy equipment.On the INPUT side, use some 5W 1R power resistors in parallel to give a very small resistance in series with the input +. Measure and write down the effective series resistance of this parallel network.On the OUPUT side, use some 5W 1R power resistors in parallel to give a very small resistance in connected to the output +. Measure and write down the effective series resistance of this parallel network. Connect from the other end of these resistors to output -Use your meter in voltage mode to read the voltage across the input resistors, use Ohm's law to calculate the current. Measure the voltage across the input + and -, use W = V*I to calculate the power going in.Use your meter in voltage mode to read the voltage across the output resistors, use Ohm's law to calculate the current. Measure the voltage across the output + and -, use W = V*I to calculate the power going out.Now instead of connecting the output resistors to the output -, put a higher value load resistor in there, and see what happens to all the measurements. Change the load resistor again, and again, make a nice graph of the output. Now you can see how your supply performs over a range of output demands.

author: IanB
author_rank: Super Contributor
timestamp: March 22, 2014, 11:23:16 pm
content: Quote from: yashrk on March 22, 2014, 12:52:50 pmOne important thing to notice here, you have made an error in that sentence.When you set a bench power supply to 20 V and set the current limit to 800 mA, the supply will provide 20 VOR800 mA, but not both at the same time. If it is supplying 20 V the current will not be 800 mA, and if it is supplying 800 mA the voltage will not be 20 V.The power supply has two modes of operation: CV mode (constant voltage) or CC mode (constant current). It will be in either one mode or the other. It will never be in both modes at the same time. For practical purposes, this means the output will never be 20 V and 800 mA. It will either be 20 V or 800 mA but never both at once.

author: AmmoJammo
author_rank: Frequent Contributor
timestamp: March 23, 2014, 02:38:55 am
content: Input voltage was most likely dropping to the same short circuit voltage in both tests... If anything, the "5volt" test should have shown less current due to the slightly increased voltage drop through the regulator.

author: yashrk
author_rank: Frequent Contributor
timestamp: March 24, 2014, 06:08:30 am
content: 1. Is there any way to boost current ?4. How do I control the current flow (current limit) in a high current application ?

author: IanB
author_rank: Super Contributor
timestamp: March 24, 2014, 06:27:48 am
content: Quote from: yashrk on March 24, 2014, 06:08:30 amIf you step down the output voltage you can get more output current for the same input power.QuoteYou measure the current and if the current goes above your limit, you reduce the output voltage.

author: yashrk
author_rank: Frequent Contributor
timestamp: March 24, 2014, 06:34:40 am
content: Thank you guys for your help.Thanks IanB you are always very helpful.