Boost Converter Efficiency More than 100 %

[Miguel Gaspar]

You must remember that eficiency is the ratio betwen input power to output power. But there are losses due to devices action so:

eficiency = output power/( output power + losses power)

and always is less than 100%

 

[mengghee]

hiye everybody,

i have read on the paper from the site recommended by max0412. in that page, the inductor current waveform looks like a square wave. i have tried obtaining the waveform with an oscilloscope current probe. but the waveform looks more like a sawtooth wave. so i thought it was wrong, but when i try to obtain the current waveform in pspice, the waveform still looks like the sawtooth. why is this ? and from the current waveform. how can i get the rms value ? by using a digital oscilloscope. is it possible i can calculate the rms value ? of input and output ? voltage and current ? thank you

regards,
mengghee

 

[artem]

Most probably you are talking about push pull based coild handling , but have you noticed the figure 30 on that document ?

Also RMS measurement is mentioned for AC supplied , for DC doc recommend either finding integral trhough probe or put a low pass filter between inp0ut and converter then measure curernt before that filter .

 

[Kral]

mengghee,
Many digital 'scopes have the capability of calculating true RMS value. Other alternatives are:
~
Capture the waveform and do the integration numerically.
~
Try to fit a know mathematical function, such as a sawtooth or exponential, and do the integration analytically.
Regards,
Kral

 

[max0412]

Is your converter designed for "DCM"? That would be why you see a pulsed ramp wave.Use a DSO with differential probes over a current sampleing resistor at the input of your SMPS.From this you can determine the "Average current"(I=V/R).The Integrals to solve for average waveform values "sawtooth " should be in your class text.This is the way I did mine in college.You can also determine inrush current which sounds like what is happening in an other thread you posted.

 

[Kral]

I disagree with the main point of the response by max0412. You do not want the average current. You need the true rms current. As I showed in the example in my earlier post, average current does not, in general, = true rms current. There are exceptions of course, such as when the current is DC.
~
Having said this, I agree that you can use a DSO, but the DSO must be capable of computing true RMS from the current waveform.
Regards,
Kral

 

[Davood Amerion]

Dear Kral;
I saw it 's document. it is correct.
also your equation is correct , but when voltage along with current are depend to each other.
for example in resistor.
in resistor when we double it's voltage, current is doubeled too.
but in this specific situation, when current doubld voltage not change (it is fixed), thus we need average value.

Regards,
Davood.

Added after 17 minutes:

Dear Mengghee
Do you check my suggestion (circuit)?
any way,
you said
"in that page, the inductor current waveform looks like a square wave"
your switching regulator is buck boost. but that wave for you mentioned is for step down switching regulator.
in step down convertor we have 2 mode of operation.
1- continues current mode
2- discontinues current mode
that wave form you mention is for continus current mode.
if inductor value is high enaugh convertor going to work in this mode, and current wave form
going to like square wave.
but if inductor value is small, inductor current can reach to zero, and wave form would be sawtooth.
in boost convertor current must be discountinus, and it can not operate in continus mode.

Regards,
Davood.

 

[mengghee]

hiye everybody,

ok, first of all, i think i need to make sure if the way i am doing it is correct. to measure the input current, where should i measure it ? the input before the inductor or after the inductor ? i do have access to a digital oscilloscope and current probe. and i do think the digital oscilloscope can calculate avrg, rms, etc.
if i plot my current measurement after the inductor, i will get a sort of square wave. and if i plot the current before the inductor, i ill get a ramp. i suppose the ramp is because my mode of operation is ccm.

there is a problem, in the document, the input current ( AVERAGE) is measured. what do you guys reckon that i need ? average or rms ? thank you.

regards,
mengghee

 

[Kral]

mengghee,
It doesn't matter. The current is the same (except for some stray currents due to distributed capacitance) on both sides of the inductor. It's the voltage that will be different on the output side of the inductor.
~
You must use the voltage at the input side of the inductor. Unless the internal impedance of the battery is quire high, this voltage will be the battery DC voltage.
Regards,
Kral

 

[mengghee]

hiye,

well, i have tried it on the simulation, i don't suppose in the simulation pspice we have stray capacitance in there and it looks a bit too much difference for some stray capacitance occuring. and i am using a variable lab power supply instead of a battery. the sawtooth type wave occur before and after the inductor in simulation. the pulse square wave type occurs only at mosfet current. doesn't occur in inductor current in my simulation. sigh*

thank you

regards,
mengghee

 

[max0412]

Please check this page **broken link removed**

Particular attention should be paid to equation 14&15
"current drawn from the source flows through the inductor".

In short the method I describe earlier.Place the current sampling R before or just after the inductor.If you have a quality current probe and take measures to minimize the errors that they can introduce this is also valid.Use the average value.

 

[Davood Amerion]

Hi mengghee
I think you must measure average value as in max0412 's document mentioned (www.national.com/appinfo/power/files/f5.pdf ).
but as i said before , it is better to use filter capacitor before connect current probe.
see my (2) last post above.

Regards,
Davood.

 

[mengghee]

hello,

I don't know much about filters. in the document, it says that if i wish to use a cheap ammeter i should use the filter but if i have a current probe it should be fine. plus i have an access to digital oscilloscope. so i should use the digital oscilloscope to calculate the average current ? not true rms or something like that ? thank you ?

max0412,

I have tried the method u have mentioned, i have also access to a current sensor available in the project lab which i assume is more like a standard resistor which helps with sampling current. results i have obtained is similar with using a current probe.

thank you,
mengghee

 

[rauol]

According to law of conseravtion of energy getting 100% efficiency is impossible.
sometimes multimeters can be erranious. for dc voltage you dont need a True RMS multimeter. twist all the wires if the multimeter and REMEMBER to bypass the input of the Boost regulator with a Good electrolytic capacitor for accurate measurement.