Continue to Site

Losses in inductor of a boost converter

Status
Not open for further replies.

ghufran12

Newbie level 5
Dear all

I have designed and develop a simple synchronous DC DC boost converter. The task given to me is to find power losses in the inductor of the converter. I choosed an indirect approach for example I measured the overall losses in the converter, then I found the low side mosfet losses, High side mosfet losses are normally very less because of the body diode of the mosfet so they are neglected, Its also easy to find DC losses in the inductor. In the end I subtract all theses losses from the total losses. remaining are the AC losses in the coil.

Now the problem is that my professor ask me why I am using this indirect approach. He further suggest me to measure the current and voltage across the inductor directly (I have no idea how to do this). Some how my professor agreed with the indirect approach but he said to me that I must have to provide a solid reason why I am not measuring the losses directly in the coil.

I am trying my best to find a reason but still I have no clue. but one thing I know that errors are larger with indirect approach. I am a beginner in power electronics. please share your thoughts on both approaches and if possible something in favour of indirect approach, it would really be a great help. I am also attaching simple diagram of boost converter.
BR
Ghufran

Attachments

• synch.JPG
20.1 KB · Views: 166

Hi,

Now the problem is that my professor ask me why I am using this indirect approach.
...
I am trying my best to find a reason but still I have no clue
Why don't you tell him the same reasons you gave to us?
I have no idea how to do this
...
I know that errors are larger with indirect approach
What could happen?

The direct method is to measure V and I at the inductor.
With a scope, one channel is the voltage the other is the current. Use a suitable current probe.
Now you may multiply V(t) × I(t) to get P(t).
The average of P(t) is the loss in the inductor.
I agree it is difficult to measure. You need the equippment.
To verify the measurement method you should
* use a known load (constant P out)
* use contant V(in)
* and observe I(in) with a current meter (with and without the inductor V and I measurement installed).
Then you easily can see how much the measurement equippment has influence on overall efficiency.

Klaus

ghufran12

ghufran12

Points: 2
Operating cycle has two parts: (a) flux builds in the inductor (Amperes rising), and (b) flux collapses (Amperes falling).

Take an oscilloscope graph of the Ampere waveform for a full cycle. Compare the area under the first half of the cycle with the area under the second half of the cycle.

If efficiency were 100 percent the areas will be equal. In reality we find a smaller area under the second half of the cycle. Measuring tells us the loss of efficiency.

Points: 2

ghufran12

Points: 2
I measured the overall losses in the converter, then I found the low side mosfet losses, High side mosfet losses are normally very less because of the body diode of the mosfet so they are neglected.
How do you determine MOSFET losses? On-state losses can be estimated relative easily, but switching losses can't. It's also wrong to neglect the synchronous rectifier losses.

I presume you are asking about real circuit measurements rather than simulation. Direct current measurement can be made with magnetic or resistive current probes, don't know if you have it available in your lab. If AC losses are small compared to switched power (e.g. < 5%) direct loss measurement by integrating I(t)*V(t), as sketched in post #2, may become too inaccurate.

ghufran12

ghufran12

Points: 2
Hi,

Maybe you should connect a high-accuracy low-value resistor (say 10mOhm) in series with the inductor if you do not have a current probe. With that you can measure the inductor current as voltage across the resistor divided by the resistance. Use a scope so you can see the peaks. With the waveform you should be able to make some close approximation.

I should think that your converter has this resistor already for the current control loop.

ghufran12

ghufran12

Points: 2
Thanks for your quick response. I already gave him the same reason but he was not satisfied. Its my thesis work and if I failed to defend, the result will be deduction in my thesis grade.
I have a question, at this moment I can measure the current at the inductor by using channel one of oscilloscope but how can I measure the voltage at the inductor directly using channel 2. I mean where exactly I need to connect probe, is it like connect the probe +ve to any terminal of the inductor and ground of the probe to the nearest ground of the pcb? I am sorry for this silly question

Hi,

I have a question, at this moment I can measure the current at the inductor by using channel one of oscilloscope but how can I measure the voltage at the inductor directly using channel 2. I mean where exactly I need to connect probe, is it like connect the probe +ve to any terminal of the inductor and ground of the probe to the nearest ground of the pcb? I am sorry for this silly question

Current:
is the current through the inductor. Either left side or right side of inductor.
I don´t know how you measure the current... either by a current probe or via shunt...

Voltage:
Is the voltage across the inductor.
* You may use a differential measurement: [right side inductor voltage w.r.t. GND] minus [left side inductor voltage w.r.t. GND]
* or scope_GND to left side of inductor and scope_signal_ input to the right side fo the inductor. Take care not to cause short circuit via scope_GND

Klaus

ghufran12

ghufran12

Points: 2
Hi thanks for the response. I determine low side mosfet losses in this way
1- for conduction loss, the same inductor current(measured using current probe) flows in the LS mosfet and using rdson conduction losses are calculated
2- switching loss in low side mosfet are calculated by measuring drain to source voltage using oscilloscope(since I am working on low voltage application). for example if rising edge of VDS is actually current fall time and similarly falling edge of VDS is current rise time. and by using this formula

P(low side)is equal to 0.5*Inductor current rms*vout*(rising time+falling time)*fsw

for the high side(synchronous mosfet) conduction losses measurement included but switching losses are neglected. dead time losses for the high side also calculated. not sure about reverse recovery losses for the high side (do you have any idea are they also very less)

Hi,

From post#2 .. with modifications:
The direct method is to measure V and I at the DUT.
With a scope, one channel is the voltage the other is the current. Use a suitable current probe.
Now you may multiply V(t) × I(t) to get P(t).
The average of P(t) is the loss in the DUT.
This is true for every 2-nodes-path.
(R, C, L, D, ....)

For a MOSFET: use Drain_to_Source as the path. (I´d say the loss in the gate is negligible)

You may
********
P(low side)is equal to 0.5*Inductor current rms*vout*(rising time+falling time)*fsw
here I don´t agree.
For switching loss only the current during switching counts.

The RMScurrent may differ in a huge range, depending on duty cycle and CCM vs DCM.

Klaus

ghufran12

ghufran12

Points: 2
Sir Duty cycle is fixed in this work and operation mode is CCM

There's a puzzle because the inductor generates voltage at the very moment you wish to measure power through it (Watts=V x A). This voltage is always great enough to force current to the output stage. Included in that is the supply voltage, because in the boost converter the supply voltage is automatically added to whatever voltage is generated in the inductor. As a result we're uncertain that it's correct to take our voltage reading across the inductor during the second half of the cycle. So we go about making a calculation by an indirect method.

I realize my previous recommendation (post #3) is overly simplified and ought to take additional factors into account.

ghufran12

ghufran12

Points: 2
Thanks I understood the point. I am also searching for more points in favour of indirect approach so in future if you come across something related please inform me. I will discuss the points you mention with my prof.

Although you may succeed at using an indirect approach (taking measurements in other components) I reckon the prof hopes that you grasp inductor characteristics, and what contributes to efficiency or detracts from it.

Examples:

* Flux energy in units of Webers (Henries x Amperes)
* Inductive time constant (L/R)
* Saturation current
* Hysteresis envelope
* 10 or 20 parameters (permittivity, coercitiviy, remanance, etc.)
How flux field builds versus what happens as it collapses

You may find you need to install the inductor in some other circuit so you can measure parameters, then put it back in your boost converter to test it further.

ghufran12

ghufran12

Points: 2
No. prof ask me clearly why not to connect oscilloscope probes directly to inductor and observe the current and voltage to find losses. he further said that indirect approah is as complicated as the direct approach so what is the advantage. And I aid I knew only the indirect way but he is not agreing on that he need reasons

prof ask me clearly why not to connect oscilloscope probes directly to inductor and observe the current and voltage to find losses
Means you have the required tools in your lab but didn't yet use it?

ghufran12

ghufran12

Points: 2
how can I measure the voltage at the inductor directly using channel 2.

Diagram per advice in post #5. Current through low-ohm resistor produces voltage. Be aware of polarity of your waveforms.

Your prof is watching to see if you use a scientific approach in your thesis. The boost converter is not so simple as a buck converter. Our intuitive sense tells us the inductor is doing something unfathomable (converting electrical current to magnetic flux, then back to current), and cannot be quantified in simple terms such as Ohm's law. So we're uncertain whether we should trust direct measurements.

However intuition is not scientific. You can be sure prof asks another student the opposite question he asks you: 'You need to justify your trust in simple measurements at the inductor'.

It helps if your thesis demonstrates you considered trying all paths. We tend to do the obvious, to follow surface thinking by calculating from simple measurements of the inductor. Did you try doing that? Did you get reasonable results? Or did you find it is unreliable to take the obvious path with a boost converter? If so then that is how you justify your use of indirect measurements rather than the direct method.

ghufran12

ghufran12

Points: 2
I am writing my thesis in a company and I have to report my professor in every two months. Company has no issues with the procedure/approach but at the end professor has to give me the grade. I do not know what kind of tools needed for direct measurement, I think possible equipment that I need is digital oscilloscope and I have to ask the company if we have it or not.

I will follow your suggestions. I will do the simple measurement at the inductor and will see the results. Thanks alot

You could have probably avoided the trouble with direct measurement of coil losses if you had determined plausible results with other methods.

Now, as you got the explicit assignment, you need at least try to measure it this way. To check the plausibility and estimate the achievable accuracy, I would refer to loss specifications given by the manufacturer, e.g. the Vishay loss IHLP series calculation tool, or at worst case, data of similar parts form other manufacturers.

ghufran12

Points: 2