Measuring efficiency of a buck regulator?

Discussion in 'General Electronics Chat' started by spinnaker, Nov 29, 2012.

  1. spinnaker

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    Oct 29, 2009
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    I am going to build a 12 - 5V step down buck regulator based on the MC34063. Once I am done. Assuming I get it working. :) I would like ti measure it's efficiency. How would O go about doing that?
     
  2. KJ6EAD

    Senior Member

    Apr 30, 2011
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    Just measure voltage and current in and out then apply the formula:

    %\ efficiency\ =\ \frac{voltage_{out}\ \times\ current_{out}}{voltage_{in}\ \times\ current_{in}}\ \times\ 100
     
    Last edited: Nov 29, 2012
  3. bountyhunter

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    you have to measer the DC current in which means you need to install a L-C filter.... a small inductor and a HUGE capacitor at the input pin of the buck converter. You have to get the input current to DC so the meter will read it correctly. Then calculate input power (volts times current). Measure output voltage and current, calculate power. Eff is ratio of output power to input power.
     
  4. takao21203

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    Apr 28, 2012
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    You can use a DMM and measure the input current only. That will be incorrect, however, you can choose a coil which produce the lowest input current.
     
  5. crutschow

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    A non true-RMS meter (the usual inexpensive multimeter) will measure the average of any ripple current plus the DC component of the input current which is what you want. So all you should need is a capacitor on the input to provide acceptable accuracy. The input capacitor that is normally part of a proper design should be adequate.
     
  6. bountyhunter

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    Not usually true. Switchers these days are cranking above 100 kHz, some above 1 MHz. TRMS meters don't measure accurately that high.

    Most of the time the meters lean more toward catching peaks which means it will give a false higher reading. The easy way to clean it up is with a small L-C filter into the input port and measure the current on the outboard side of the L. I have had to do this a few dozen times.

    Again, not usually true. Most app circuits recommended by the manufacturer spec the absolute min size cap for functionality, they don't care how much current ripple is going back up the line. The reason is marketing always insists that the "recommended design" component cost be as cheap as possible.
     
  7. spinnaker

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    OK so what do I need to do here? There seems to be difference of opinion.
     
  8. JMac3108

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    Aug 16, 2010
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    If your circuit is well designed, there won't be much ripple current coming from the input wire - it should be coming from your input capacitor. If this is the case just measruse the input current without adding an external filter.

    You need to take care when making efficiency measurements. The usual mistake people make is to measure the output voltage at the load. It should be measured directy at the output cap. You don't want the voltage drop of the wiring to your load included in the measurement.

    Also, measure the efficiency at the actual load the circuit will see. Efficiency will vary significantly over load. Better yet, measure the efficiency at a range of loads and generate a plot of efficiency vs load.
     
  9. spinnaker

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    My input will be coming from a battery.
     
  10. bountyhunter

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    Make an L-C filter. You can wrap maybe 20 turns of wire around a pencil to make an inductor. Put a big cap near the input of the device. Insert the ammeter on the power source side of the inductor.
     
  11. crutschow

    Expert

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    I was referring to a non-TRMS meter while measuring in the DC current mode. It measures the average value of the current.

    Have you actually tried to do measurements without the LC filter and gotten significant errors in the measured DC current?
     
  12. bountyhunter

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    YES. It's why we always made sure use a filter to get the input current to smooth DC.

    Not really. It might measure the average of an AC sine wave, but the input ripple to a switcher is not sinusoidal and not purely AC. It looks more like a sawtooth with asymmetric teeth width riding on DC level. What meters read in that case is not usually accurate. In most cases, the meters get the DC level and then pick up the peak of the AC on top of it, so it reads a little high.

    The problem is that if you don't filter it, the data can't be trusted to be correct. It might be "close", depends on how much ripple is there. That depends on how good (and large) the input caps used are. If you want to know it's right, it has to be DC.
     
  13. bountyhunter

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  14. bountyhunter

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    Here's a cut from a TI app note on measuring sw converter efficiency. It shows adding a big cap to smooth out the sawtooth current ripple to DC.

    next paragraph after the fig posted:

     
    Last edited: Dec 1, 2012
  15. spinnaker

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    I have some 220uh coils on hand. Will they do? What is a "big cap"? How many mfds?
     
  16. bountyhunter

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    That coil will work. Big cap means bigger the better, at least a few thousand uF or however much is available.
     
  17. THE_RB

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    It will run about 80-85% efficiency, with a typical inductor. A nice hand wound iron toroid might get you 85-90% with that IC.

    Those figures are assuming you use a decent external switch. If you use the 34063 internal switch transistor in buck you will get more like 75-80%.
     
  18. spinnaker

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    How do you hand wind and inductor if you do not have an LC meter? Is there a formula? Where do you get the cores?


    What do yo mean by "decent" external switch"? Is there another transistor that can be added to the 34063 circuit?
     
  19. Potato Pudding

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    Jun 11, 2010
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    Post 2 answered the original question.

    Accuracy of measurement is a concern, but that is all that you really need to take away from the rest of the posts.

    Just be careful with your measurements.

    So watch:
    If you have one value that is measured to only one or two significant digits.
    You don't add any measurement or rounding bias.
    You don't create circuit modification warps. For example the shunt resistance for a current meter might inject a major change into the input resistance for the circuit.

    If you are trying to do fine tweaking and performance comparisons then really sweat all of this. If you just want a rough number like approximately 86% efficient then take your measurements and make the calculation.
     
  20. THE_RB

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    If you don't have an inductance meter you should probably just use off the shelf inductors.

    I meant another transistor that is used as the main switch, this will have better performance than thetransistor inside the 34063. Circuits are all shown in the IC datasheet.

    The 34063 operation and its efficiency is well known, as is your PWM duty (12v->5v) and efficiency will depend on the losses from the pass transistor type and inductor type mainly, as it is expected that you would use a schottky diode so that is also known.

    The figures I gave you will be pretty close and save you doing a lot of testing. If you want tips to maximise the efficiency please post your circuit and photo and specs for your inductor, we can probably suggest some tips to gain you a few percent. However the 34063 is not the right chip to get 95% type efficiencies from, if you want something like that there are much better chips.
     
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