Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Register Log in

Boost Converter Efficiency More than 100 %

Status
Not open for further replies.
M

mengghee

Full Member level 3
Joined
Nov 29, 2005
Messages
163
Helped
8
Reputation
16
Reaction score
5
Trophy points
1,298
Location
United Kingdom
Activity points
3,105
hiye,

i have build a boost converter and measure the input and output current and while i was trying to draw a graph with excel. i notice there are some point where efficiency goes higher than 100%. isn't this impossible ? thank you


regards,
mengghee
 

mengghee,

Yes, this is impossible. Otherwise, you'd be a millionaire! You can not use the DC input voltage and current measurements to measure input power. If the input voltage is truly a constant DC, then you can multiply this DC voltage by the true RMS input current to get the input power. If the input voltage is varying with time, then you must calculate the input power by integrating the product of instantaneous input voltage and current. Most digital 'scopes have this capability.
Regards,
Kral
 

    M

    mengghee

    Points: 2
    Helpful Answer Positive Rating
kral,

i don't quite get you. the input current of the power supply is from a bench power supply. so i would assume it's a pure dc. but the output current of the boost converter on the other hand might have ripple. but you were suggesting me to find the input voltage rms ? to get the input power ? should i do it on the output power ? thank you.


regards,
mengghee
 

mengghee, did you get perpetum mobile ?)

Speaking seriosly , i agree with Kral. Most probvably input data you rely on are erroneous.
 

    M

    mengghee

    Points: 2
    Helpful Answer Positive Rating
mengghee,
If your input voltage source is a bench supply, it probably is a pure DC voltage. However, the input current will not be constant. You must measure the true RMS value of this current in order to calculate the input power. The current meter on your bench supply gives you average current, not true RMS current.
Regards,
Kral
 

    M

    mengghee

    Points: 2
    Helpful Answer Positive Rating
Hi mengghee
Efficiency never reach 100%.
because [input power = output power + lossed power, as heat ].

I think, for better measurement, you can use large capacitor in input and output.
because with use of capacitor just you need measure DC values of input and outputs.
I recommend you to test your circuit one more time, carefully.

Regards,
Davood.
 

    M

    mengghee

    Points: 2
    Helpful Answer Positive Rating
i agree with you all, i know that achieving 100% efficiency is totally impossible. how can i measure the current current ? also ... what kind of measurement equipment ? should i measure the input current and voltage from digital oscilloscope ? and also ... output as well ? thank you


regards,
mengghee
 

Mmm, may be that is amateur way :
if you can connect 2 shunts to input and output and register data continously by analog data loggers at a sampling speed much more than your load frequency behaviuor, then dividing sum of all measurements of input power to sum of output power should give you efficiency .
But why are trying to calculate efficiency under dynamic load ? Is it requirement spec ?

Added after 3 minutes:

Yet , momentary measurements (let say sampled) can give you efficiency of more than 100 %. That is about your internal energy storage discretes give power to output being not charged from input . But that is momentary and not relevant to integrated measurement over considerable interval of time which is of course for sure less than 100 %
 

    M

    mengghee

    Points: 2
    Helpful Answer Positive Rating
well, it's my uni project and ... i would like to measure it's efficiency to show it in my report. i do not know how to find shunt and data logger. is there any way i can do it with the method kral mentioned ?


regards,
mengghee
 

How fast is your load behavious . periodic or random ?
If periodic - what is the frequency ?

Added after 11 minutes:

As example i am looking to LM3485 data sheet for efficiency measurement . The graph is built on current versys efficiency for given constant otput voltage . National semi does not say that current is dynamic , i suppose that it is constant . Would not it better to you to find a current source (just simple reostat variable rezistor) and plot graph for constant current versys efficiency . Simple measurement with multimeters will suffice .
 

    M

    mengghee

    Points: 2
    Helpful Answer Positive Rating
I have to plot a graph of vout vs duty cycle, and also graph of vout vs vin with vout varying at 4 different load. 20 ohm, 40 ohm, 100 ohm and 200 ohm with output voltage of 10v.

regards,
mengghee

Added after 1 minutes:

if it is a constant current with a constant load, the voltage and input and output (especially ) will still have ripple and i still need to measure their rms values
 

Have a look there :

https://www.calex.com/pdf/2modular_guide.pdf
https://www.gaia-converter.ca/content/application_notes/terminologyandbasic.pdf

I guess that measurement tool yo use for output power does not filter incoming
voltage so ripple and overshoots (its amplitude is much more ) affecting your measurement.

The first doc says measurement is sufficient with integrating multimeters .
Mine (cheap multimeter) for simple UPS without load gave about 380V ! When i got a
scope and found that overshots are to blame and of course my multimeter which does not have a filter on input. if you have old good analog multimeter - that is cheap way to go if you do not expect high accuracy in measurement.

Regarding ripple - if it is about 50 mV, for 5 V output
it is just a percent of output voltage you can sacrifice if no RMS is available under hands .
 

    M

    mengghee

    Points: 2
    Helpful Answer Positive Rating
oscilloscope is not accurate measurement device.
for measuring voltage and current you can use multimeter.
if ripple is low (in compare with nominal value). you dont need to calculate RMS value.
simply multiply input current by input voltage to get input power,
and multiply output voltage by output current to get output power.
Regards,
Davood.
 

    M

    mengghee

    Points: 2
    Helpful Answer Positive Rating
well artem, davood, that is the way i go about it. i used the dvm to measure input output current and voltage. that's exactly how i got my efficiency more than 100%

i used 2200uF as my filter capacitor and my output is 10v. i think it is a rather huge capacitor. and my frequency is 50khz. shouldn't provide too much ripple.



regards,
mengghee
 

Hi mengghee,

could you please give us your schematic and measured values?

regards,
Davood.
 

    M

    mengghee

    Points: 2
    Helpful Answer Positive Rating
Hi mengghee,

you said output voltage is 10V. but from table value it is not correct.
for example in line 1 ouput current is 0.241 and Vout=10 then output power will be 2.41 W
and in your schematic i dont see input filter capacitor.
i attached your schematic and write on it my sugesstion.
capacitors 0.1 uf are for high frequency, and 2200 uf and 1000uf for low frequency filteration.

regards,
Davood.
 

    M

    mengghee

    Points: 2
    Helpful Answer Positive Rating
mengghee,
It's OK to use a dvm to measure input current, but it must be capable of measuring true RMS current. Most dvms don't have this capability. The problem is that, even though the input Voltage is constant, the input Current is not.
~
Let me give you an example:
Suppose the input current waveform is a 50% duty cycle square wave with a peak value of 1 Ampere. A dvm will give you a reading of 0.5 Amperes. This is the average input current. However, the true RMS current is 0.707 Amperes.
Regards,
Kral
 

    M

    mengghee

    Points: 2
    Helpful Answer Positive Rating
Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top