how to measure efficiency of boost converter ?

[Btrend]

hi all,
how to measure efficiency of boost converter ?
as I know, efficiency= (Vo*Io)/(Vi*Ii),
but should I use rms value, or average value of Vo, Io, Vi, Ii ?
and what is the value readout from the meter ? rms or average ?
thanks

 

[VVV]

The boost is a DC/DC, so you just measure DC values. These are averages, if you will. The meter will display averages for DC.

 

[IanP]

If you consider ideal situation that both, the input and output voltages are pure dc, then you can use dc average.
However, if you want to be very, very accurate, and taking into account that in practice input voltage is an unregulated voltage with 100/120Hz ripple (or any other ripples), the output voltage will have ripples around switching frequency, I think you should use RMS-to-dc conversion and then use formula to calculate efficiency.
Regards,
IanP

 

[Btrend]

i see, thanks.
one more question,
if the boost operate in DCM , should I use average vlaues??
p.s. in DCM the Ii is not contiuous.

Added after 7 minutes:

I think you should use RMS-to-dc conversion and then use formula to calculate efficiency.
IanP

hi IanP,
I don't know what is "RMS-to-dc conversion ", could u elaborate it ?
thanks

 

[IanP]

RMS normally applies to ac signals but you can treat a dc voltage with ripples the same way as ac voltage with dc offset.
Anyhow, rms-to-dc conversion will perform conversion from rms to pure dc which can be then measured by normal means.
Sme more info on rms-to-dc you can find here:
**broken link removed**
https://www.analog.com/UploadedFile...283758302176206322RMStoDC_Cover-Section-I.pdf
Regards,
IanP

 

[VVV]

You measure the average values, regardless of the operating mode. Even if the INDUCTOR current is discontinuous, both the input and output currents are continuous. That is because the capacitors will make these currents continuous, by supplying energy (current) during the time the inductor current is zero.

RMS stands for Root Mean Square. Thus,

Irms=√(1/T*∫i(t)^2*dt)

(square root of the mean of the square).

It is meaningful when you are dealing with power dissipated in resistors due to AC current. Since the power dissipated in a resistor does not depend on the direction of the current, we can write: p(t)=R*i(t)^2. This is the instantaneous power. The average power is the mean of this value:

P=1/T*∫R*i(t)^2*dt=R*1/T*∫i(t)^2*dt

Since we want a formula that links the resistance to the power to be similar to what we know from DC, that is P=R*I^2, we then DEFINE the RMS current:

Irms^2=1/T*∫i(t)^2*dt

Or

Irms=√(1/T*∫i(t)^2*dt)

This is where the formula comes from. With that, the power can be expressed in the familiar way, even for AC currents, with the RMS value for current:

P=R*Irms^2

We can then define the RMS voltage, as: Urms=R*Irms. The formula is of course similar to that for current:

Urms=√(1/T*∫u(t)^2*dt)

As you can see, all this is meaningful in relation to a resistance. In your case, the average values should be used. The input characteristic of your boost resembles most closely a NEGATIVE resistance (the current decreases with increasing voltage). The output current is dictated by the load and it can be almost perfectly DC. The voltages may exhibit some ripple, but the power delivered to the load or drawn form the input is simply the DC current times the average voltage.

 

[Btrend]

You measure the average values, regardless of the operating mode. Even if the INDUCTOR current is discontinuous, both the input and output currents are continuous. That is because the capacitors will make these currents continuous, by supplying energy (current) during the time the inductor current is zero.

thanks for ur detail reply.
but I have measure the Iin from the power supply, it is obviously not continuous!?
see my attachment,
the green line is the Iin=IL in DCM boost

 

[Btrend]

hi all,
Am I wrong in the following derivation ?
derivation of efficiency:
efficiency η=Po_avg/Pi_avg ...............(1)
Po_avg=(1/T)*∫[Io(t)*Vo(t)]dt ........(2)
Pi_avg=(1/T)*∫[Ii(t)*Vi(t)]dt ........(3)
===> η=∫[Io(t)*Vo(t)]dt/[∫[Ii(t)*Vi(t)]dt ] ≠[∫Io(t)dt ]*[∫Vo(t)dt ]/[∫Ii(t)dt ]/[∫Vi(t)dt ] ........(4)

so, if I want to calculate the accurate efficiency, all I need is to measure the average [current*voltage] at the same period (T), and let T as large as possible.
and from eq.(4), I should not use the separated average value.
Am I wrong in any step above ?
thanks in advance

 

[AndrzejM]

In my opinion everything is OK

but there is a small misprint in (4):

Added one hour later:
Sorry , everything in your equation (4) is OK .
your explanation in next post states it clear.

The (4) equation did not mislead me, if it was written like this:

≠[∫Io(t)dt ]*[∫Vo(t)dt ]/{[∫Ii(t)dt ]*[∫Vi(t)dt ]} ........(4)

but of course it is my problem, and my mistake.

 

[Btrend]

in my opinion everything is OK
there is a small misprint in (4):
≠[∫Io(t)dt ]*[∫Vo(t)dt ]/[∫Ii(t)dt ]/[∫Vi(t)dt ] ........(4)

your intention was probably to write "*" instead of "/"
≠[∫Io(t)dt ]*[∫Vo(t)dt ]/[∫Ii(t)dt ]*[∫Vi(t)dt ] ........(4)

oh, I am sorry about that.
in (4) it's like as (A*B)/(C*D)= [(A*B)/C]/D ≠(A*B)/C*D=[(A*B)/C]*D

thanks

 

[VVV]

The input current is not discontinuous. The input capacitor will smooth it, just as the output capacitor smoothes the output current.

Take a look at the picture. As you can see, the output current is merely the average of the diode current. But your load needs a DC current. How does it get it. The output cap supplies the current when the diode current is zero. Notice how the output cap's current is negative when the diode current is zero, that means the cap is delivering to the load. When the cap's current is positive, it is being charged up.

Things are similar at the input, though harder to grasp. The point is that the capacitors cannot conduct the DC component, which is why the currents through them are shifted down by the average of the AC current. In other words, the caps will short to GND the AC component, but not affect the DC. Thus, with ideal caps, the input and output currents are AVERAGES, or DC, which the caps cannot remove (nor should they, or they would be a dead short).

Hope this helps

 

[Btrend]

hi VVV,
ur detail reply is really helpful, thanks again.
based on ur picture and explanation, the input and output values in average are all DC values.
but I still got problem about the derivation and measurement of efficiency.
(1)
the efficiency formula is: η1=∫[Io(t)*Vo(t)]dt/[∫[Ii(t)*Vi(t)]dt ]
and its related formula is : η2=[∫Io(t)dt ]*[∫Vo(t)dt ]/{[∫Ii(t)dt ]*[∫Vi(t)dt ]}
Can I use "η2=Ioavg*Voavg/(Iiavg*Viavg) " to calculate the efficiency ?
but in integral calculus, η1 ≠ η2.
(2)
In real SMPS, Vo(t) is always DC value with some small ripple, and Vi(t) is always DC. So η1 can be approximate as Voavg*∫Io(t)dt/[Viavg*∫Ii(t)dt ] , and this is
η2. Finally, all I need to do in calculating the efficiency is just measured all average values, and use η2.
am I wrong with the above derivation ???

 

[VVV]

Hi Btrend,

I have to brush up on my calculus, since I realize I cannot show mathematically that you just need average values, except for a particular case.
However, that is what you do, you measure average values and calculate the efficiency.

Here is the particular case: assume the output current is constant, which is achievable with an electronic load. Also, most linear IC's will also draw a constant quiescent current when the voltage variations are small, which is the case here, since the only variation is the ripple, which is small compared to the output voltage, or else we would have a sloppy power supply.
Then the output power is Po=1/T*∫Io*u(t)dt=Io*(1/T∫u(t)dt)=Io*Uoave. Both are average values, in fact.
Now consider that the input voltage is "perfectly" constant, which is possible with a good power supply. The input current may have a small ripple, due to the finite value of the input cap and its ESR.
Then the input power is Pin=1/T*∫Uin*i(t)dt=Uin*Iinave. Again, we have two averages.
I will try to see if there is a way to deduce the general formula.

By the way, T refers to the period of the signal (that is, ripple), not some arbitrary length of time, as I think you were assuming.

Regards,
VVV

 

[AndrzejM]

In case of pulsed values the only rigt method to measure it is to integrate the power over the period and divide it by the period.

VVV correctly showed that in case when one value (U or I) is CONSTANT, you can use the other value measured as average.

But !

At first "Btrend" CORRECTLY wrote that:
η=∫[Io(t)*Vo(t)]dt/[∫[Ii(t)*Vi(t)]dt ] ≠[∫Io(t)dt ]*[∫Vo(t)dt ]/[∫Ii(t)dt ]/[∫Vi(t)dt ] ........(4)

However in the next post "Betrend" have changed his mind and declared:

the efficiency formula is: η1=∫[Io(t)*Vo(t)]dt/[∫[Ii(t)*Vi(t)]dt ]
and its related formula is : η2=[∫Io(t)dt ]*[∫Vo(t)dt ]/{[∫Ii(t)dt ]*[∫Vi(t)dt ]}

There are two completly different equations, or may be expression "related formula" is idiomatic and has other meaning from the one that I know, ( English is my second language).

Let me repeat:
η1 ≠ η2
That was correctly stated in equation (4)
Also my first opinion that "everything was OK" was refered to the equation (4) with the "≠" operator.
The equation :
P=[∫Io(t)dt ]*[∫Vo(t)dt ]
is WRONG !
It can be right only if one value I(t) or U(t) = constant .
(tha case presented by VVV)

 

[AdvaRes]

Hi,

P=[∫Io(t)dt ]*[∫Vo(t)dt ] is Wrong I agree.

But as far as I know the average power P=Irms*Vrms.

I faced this problem.
I simulated with Spectre a circuit with VDD=1.1Volt and I plot the Current flowing through VDD pin vs Time. Current can be positive and negative (see figure).

I got the following values using the functions of the Calculators:

RMS(I(t))=Irms= 2.148898839560097E-5 A
Average(I(t))=Iavr= 5.427864339544525E-6 A
average(abs(I(t)))=I3= 6.685111176801027E-6 A

The goal is to calculate the power consumed by the the circuit.

If we use the well know formula of Power we get:

P1=∫I(t)*V(t)dt=∫I(t)dt * Vdd=Iavr*Vdd=5.9706507734989775E-6 Watt
P2=Irms*Vrms=Irms * Vdd=2.3637887235161067E-5 Watt

Normally we should have P1=P2 so the question is:
Why ∫I(t)*V(t)dt≠ Irms*Vrms ?

The average function may result in bad power value because the current can be positive or negative. When the current is negative the Circuit consumes also power since it circulate through resistors. So I though to calculate the power in another way:
P3= I3*Vdd=7.3536222944811297E-6 A.
Could someone Explain why P2≠P3 ?

Thanks in advance.
Advares.

 

[FvM]

But as far as I know the average power P=Irms*Vrms.

Generally not. It's true, if I and V have the same waveform.

 

[AdvaRes]

Generally not. It's true, if I and V have the same waveform.

Thanks for replying FvM.
Could you please explain what do you mean by same waveform ?
Do you mean I and V applyed to a resistive load ?

If so, do we have to consider a circuit (fully built by transistors: a 2 input Nand gate for exemple) as a resistor when we measure its power consumption.
What is the power consumed by the circuit in that case: P1, P2 or P3 ?

Thanks again for you time.
Advares.

 

[FvM]

P1 is the correct expression for the power consumption with arbitrary current and voltage waveforms, could be extended to
multiple circuit terminals if necessary. P3 is correct for a constant voltage and obviously equal to P1 in this case.

P3 could be used e.g. for the supply power of a logic gate. But it's not correct to apply an absolute value if the current sign is
changing, I3 = Avg(I(t)) would be correct.

 

[AdvaRes]

P1 is the correct expression for the power consumption with arbitrary current and voltage waveforms, could be extended to
multiple circuit terminals if necessary. P3 is correct for a constant voltage and obviously equal to P1 in this case.

Ok, mathematically the relation P1=P2 is correct since Vdd is constant. Unfortunately simulation doesn't prove that according to calculation that I've done.

What confuse me is that the difference between P1 and P2 is enormous and I still dont know which one is the real power consumption of my circuit.

 

[FvM]

P3 in your calculation is not correct because of the abs() operation. Avg(I(t))*Vdd is correct.