How to calculate input power of buck converter if efficiency is unkown?

Thread Starter

mike _Jacobs

Joined Jun 9, 2021
223
I have seen the question of how to calculate input power of a converter explained many times as
Power_out/efficiency = Power in

That seems reasonable to me if you know the efficiency.

However, if you have no idea what the efficiency of a converter is that you just breadboarded i dont see how you can calculate the power in

The power output can be calculated simply with Power_out = Vout X Iout

Since the input current of the converter is discontinuous how do you measure the power input of a converter?

I would of said that Power_In = I_Average X Voltage_in

But how are you supposse to measure I_average?? When i say I_Average I mean AVERAGE not RMS.

What is the generally accepted practice here?
 

Ian0

Joined Aug 7, 2020
10,054
What is the generally accepted practice here?
Guess!
Seriously, assume that the efficiency might be about 80% to 90% and work from there until you get some figures that are more accurate.
The average input current is easy to measure, just measure the current between the power supply and the input capacitor: the input capacitor will average the current for you.
As @panic mode says, if you change the output current, you will get a different value for efficiency.
The only value for efficiency that you can be certain of is the value at zero output current, and it is exactly 0%.
 

Thread Starter

mike _Jacobs

Joined Jun 9, 2021
223
The average input current is easy to measure, just measure the current between the power supply and the input capacitor: the input capacitor will average the current for you.

How do a capacitor average the current for you.
It will average the voltage not the current.
The current is discontinuous because the FET opens and stops the current flow.
I dont see how a capacitor does anything for you there.

Also, i cant go measure it since im working with a simulation. I dont have a measuring tool in the sim that will give me the average current
 

Ian0

Joined Aug 7, 2020
10,054
How do a capacitor average the current for you.
Because it stores charge.
The current which becomes the output current of the buck regulator will tend to be supplied from the lowest impedance source, which will be the input capacitor, so the output current from the capacitor is pulsating.
if the power source is of a higher impedance than the capacitor, and the voltage does not drop too far when output current flows, the current will be rather more continuous, and rather less pulsating, so it will have been averaged.
 

Thread Starter

mike _Jacobs

Joined Jun 9, 2021
223
Because it stores charge.
The current which becomes the output current of the buck regulator will tend to be supplied from the lowest impedance source, which will be the input capacitor, so the output current from the capacitor is pulsating.
if the power source is of a higher impedance than the capacitor, and the voltage does not drop too far when output current flows, the current will be rather more continuous, and rather less pulsating, so it will have been averaged.
I put one in for kicks and it made literally no change. I dont understand where you are going here.

The capacitor cant sink any more current if its charged. The FET opens and there is no current sink so there is not going to be any current smoothing.
 

Thread Starter

mike _Jacobs

Joined Jun 9, 2021
223
hi mike,
This simple sim may help to show you the averaging of the current.
The Vin is a square wave, note the effect of the smoothing capacitor.
thanks for the effort

Can someone please explain how the current is continuous if the FET opens and there is literally no path for the current to flow.
 

crutschow

Joined Mar 14, 2008
34,697
Also, i cant go measure it since im working with a simulation
What simulator?
LTspice can measure the average current, for example.
The capacitor cant sink any more current if its charged. The FET opens and there is no current sink so there is not going to be any current smoothing.
Of course it can.
In a real circuit with a finite source impedance, the capacitor discharges when the FET conducts, supplying charge (current) as it's voltage drops.
When the FET is off, the capacitor recharges.

In a sim, if you are using an ideal voltage source, then the capacitor doesn't do anything since all the current is always supplied by the zero impedance source.
 

Thread Starter

mike _Jacobs

Joined Jun 9, 2021
223
What simulator?
LTspice can measure the average current, for example.
Of course it can.
In a real circuit with a finite source impedance, the capacitor discharges when the FET conducts, supplying charge (current) as it's voltage drops.
When the FET is off, the capacitor recharges.

In a sim, if you are using an ideal voltage source, then the capacitor doesn't do anything since all the current is always supplied by the zero impedance source.
So i found this on the web
1713895917867.png

I did the math and put a 500uF cap in. It had no effect whatsoever on providing a non discontinuous current into the converter
 

panic mode

Joined Oct 10, 2011
2,818
thanks for the effort

Can someone please explain how the current is continuous if the FET opens and there is literally no path for the current to flow.
which FET? show the circuit....

three basic passive components are RLC. two of them can store energy (L and C). there are designs using L and C to create different voltage source for example. designs using capacitors tend to be low power, often used in communication transceivers. designs using inductor are suitable for large powers.
 

Ian0

Joined Aug 7, 2020
10,054
What do you mean which side? Its a parallel cap between the positive and negative input rails
Assuming you would be measuring the current in the negative supply, are you measuring it in the wire between the capacitor and the power supply, or in the wire between the capacitor and the buck regulator?
 

Thread Starter

mike _Jacobs

Joined Jun 9, 2021
223
ok need to switch gears. Apparently im bumping into discontinuous mode.

I am trying to the following equation for the minimum inductance.

1713904205041.png

However, i can see now that my inductor current is going negative and running dry.

This is a very standard ideal converter using ideal componennts in the sim. So there is little to no nuances like RDS or diode drops or so on.

I know this states this is the inductor minimum. I put a factor of 1.2 on it. Shouldnt that be enough to keep me out of discontinuous mode?

I compared this equation to a TI calculator i found online and this apears to be off by an order of magnitude.
 

WBahn

Joined Mar 31, 2012
30,250
I have seen the question of how to calculate input power of a converter explained many times as
Power_out/efficiency = Power in

That seems reasonable to me if you know the efficiency.

However, if you have no idea what the efficiency of a converter is that you just breadboarded i dont see how you can calculate the power in

The power output can be calculated simply with Power_out = Vout X Iout

Since the input current of the converter is discontinuous how do you measure the power input of a converter?

I would of said that Power_In = I_Average X Voltage_in

But how are you supposse to measure I_average?? When i say I_Average I mean AVERAGE not RMS.

What is the generally accepted practice here?
If your approach is to measure the input power, then you need to measure the average power, which may or may not be proportional to the average current.

Your starting point is the definition of average power over some time interval.

\(
P_{avg} \; = \; \frac{1}{T} \int_{0}^{T} p(t) dt
\)

For electrical systems like you are talking about, that becomes

\(
P_{avg} \; = \; \frac{1}{T} \int_{0}^{T} v(t) \cdot i(t) \; dt
\)

Now you get to apply conditions that are relevant for your specific situation.

IF v(t) is constant (i.e., a fixed and steady DC value), then you can move it out of the integral and you end up with average power is the product of the voltage and the average current.

But if the voltage is not constant, then you need to be more careful. If both the voltage and current are sinusoids, there is some simplification you can do involving the amplitudes of both and the relative phase between them. If the voltage varies (such as being supplied by an AC source) and the current draw is complex, then your best bet is to sample both waveforms quickly enough to be able to get samples of the instantaneous power and simply averaging them.
 
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