[SOLVED] efficiency of a dc-dc

[imfresh]

hello,all.
i'm simulating a buck dc-dc by candence, and i am not sure how to measure the efficiency. someone told me measure the rms of input current, then multiply vdd, and this is input power. but i don't understand, why rms, not average?
and another question is the other control part's of the system also consum current, so this part's current how to measure, also rms?or avg?

i'm so confused that wish someone can help me.

 

[manojdharap]

To calculate the Power, you need to use instantaneous current & instantaneous voltage and then multiply it to get the I/P power, but since its not convenient to measure this, we will normally use average power I/P, which will be calculated by Irms * Vrms. If other control part is consuming the current, use dc-dc IC i/p to calculate I/P power. If you can give schematic, it will be useful..

 

[imfresh]

If other control part is consuming the current, use dc-dc IC i/p to calculate I/P power.

thank you for your reply, manojdharap. I am a beginner of dc-dc convertor, so I am just simulating a basic current mode one and it is integrated except L, C, compensation net. there are current sensing circuit, osc, comparator, driver and so on. these control circuits' power is constant. I can't understand this sentence:"If other control part is consuming the current, use dc-dc IC i/p to calculate I/P power." Can you give me more detail. Thanks a lot.

 

[FvM]

we will normally use average power I/P, which will be calculated by Irms * Vrms.

The formula would be incorrect for almost any power measurement, it's the expression for apparent power and only identical to real power for either pure DC current and voltage, or AC with unity power factor and no harmonics.

The first expression, product of instantaneous current and voltage is correct for all cases. But as long as the input voltage is almost constant, you can simplify it to P = Vdc*Iavg. Using Irms instead of Iavg would overestimate the input power. You can also use Iavg for the output power measurement, as long as the output voltage is fairly a DC voltage (e.g. ripple < 5%).

 

[imfresh]

thank you, FvM. What you said is very helpful for me. I also found that when I used Irms instead of Iavg, the efficiency was lower. There is still a question: when the load current is very small, current flowing through the inductor will be reversed. Then the Iavg is incorrent to calculate the efficiency. How to calculate the efficiency under this condition?Thanks again.

 

[FvM]

when the load current is very small, current flowing through the inductor will be reversed. Then the Iavg is incorrent to calculate the efficiency.

No, it's still correct. Assume an unloaded synchronous converter with a bipolar triangular input current, Iavg will be near to zero. This correctly represents the fact, that the converter doesn't consume power in this situation, except for some losses. Typically, you want to have an input capacitor with your DC/DC converter, that absorbs the AC currents, and the external current won't actually reverse.

 

[imfresh]

No, it's still correct. Assume an unloaded synchronous converter with a bipolar triangular input current, Iavg will be near to zero. This correctly represents the fact, that the converter doesn't consume power in this situation, except for some losses. Typically, you want to have an input capacitor with your DC/DC converter, that absorbs the AC currents, and the external current won't actually reverse.

thanks again,FvM. But when I use the calculator in the candence to measure the Iavg, I found the efficiency >100% in small load current. Then I think this method is questionable.

 

[rvionics]

You will need to measure input average current (before input filter capacitor inudctor i.e. close to input source) & input average voltage , multiply them & that will be your input power.
The switching currents will be averaged by your input filter that is why you should measure average inpur current at the input source & do not have to worry about waveform of the current.

 

[dick_freebird]

The simplest is to keep port currents on the input supply
and a load sense element, and use Calculator math to get
Pin as v(vin)*i(Vvin), Pout as v(load)*i(Vload), Eff=Pout/Pin.
This will be instantaneous power based. You can wrap an
intreg() around both terms and get a steady state efficiency
number that's reasonable.

 

[imfresh]

thank you ,all. I found the reason why the efficiency I simulated was above 100%. It is my fault. I calculated the efficiency too early when the system is not steady.Thanks again.