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MC33063A - Efficiency at light loads

Stefano Infante
Expert 2320 points
Other Parts Discussed in Thread: MC33063A, TL317, LM317L, MC34063A

Hi all!

I was looking for a cheap switching regulator, capable of powering a light load (30 mA max) at 15 V, with 18-36 V input.

I found this MC33063A, but all information I can get on the data sheet are about efficiencies at heavy loads (500 mA or so). I am strictly limited in the available power, so I cannot drain more than about 35-40 mA at the input power source.

Can anyone give some help?

Thanks in advance,

Stefano

  • 0
    TI__Guru**** 164726 points

    Hello Stefano,

    If the input current can be a little greater than output current and vout is less than Vin(min), I suggest a linear regulator.

    The TL317 or LM317L can handle 36V input. The output voltage can be set by two resistors. The adjust pin current is 50uA typical, so I suggest setting the feedback current about 5mA. This will make input current be 5mA higher than output current load regardless of input voltage. This will be cheaper than a MC34063 circuit.

    Regards,
    Ron Michallick

     

  • 0 in reply to Ron Michallick
    Expert 2320 points

    Hello Ron,

    I am not sure about how to use TL317. My application needs a regulated output of 15 V (or, better, 17 V), with an unknown input of 18 to 36 V (nominally 24 V). The current drain at the output will vary from about 2.7 to 25 mA (I made the calculations again).

    If I use TL317 with R1 = 470 ohm, R2 = 5.9 kohm (Figure 1 in the data sheet) my output will be about 16.9 V; the current flowing on the two resistors will be 16.9 / 6370 = 2.65 mA, which is acceptable according to the data sheet (the minimum output current should be 2.5 mA at maximum). Is it all ok?

    Then, I don't understand what you mean with "feedback current"... may it be the 2.6 mA current I previously  mentioned? If it is, then I should use resistors with about half the values (say, 236 ohm and 2.94 kohm) in order to obtain a minimum current of 5 mA.

    And what about the power loss? If the input voltage is 36 V, being the output at 17 V, is the total current drawn by TL317 made only by those 5 mA + the current needed by the circuits powered by TL317 itself?

    Thanks for your help, and sorry for my confusion...

    Stefano

  • 0 in reply to Stefano Infante
    TI__Guru**** 164726 points

    Stefano,

    Using the 470 ohm and 5.9k ohm resistor are good. It satisfies the minimum current meaning the voltage will be accurate even with no external load.
    I only suggested 5mA to further lower the effect of the adjustment pin current on Vout.

    Yes, feedback current is the current in the two resistors.

    With a 36V input and 17V output the power dissipation (loss) on the TL317 is (Vin-Vout)*Iout. (note: Iin & Iout are the same);  Assuming 30mA application load. Power loss =(36V-17V)*(30mA+2.6mA)=619mW. At this VIN we are about 50% efficiency. I suggest the "D" or "PS" package as they have better thermal impedance.

    With a 18V input, there is not enough headroom (the TL317 drop out voltage is about 2V) so the output would drop to about 16V, but the power loss will be very low.
    If the output was programmed to 15V by R1 & R2 then drop out would not be an issue.

    The MC34063A solution would be much better than 50% efficient at 36V input, however it could not produce a full 17V output with 18V input.

    Regards,
    Ron

  • 0 in reply to Ron Michallick
    Expert 2320 points

    Hello Ron,

    first of all thanks for your help.

    I can certainly go for a safer Vout (16 or even 15 V are fine, 17 V was a choice dictated by the simple rule "higher is better", but it isn't a strict requirement); and I think TL317 could still be a good choice, but I would like to explore the MC34063A solution, if it would have better efficiency. How could I estimate its efficiency in the 2.7-25 mA Iout range?

    Regards,

    Stefano

  • 0 in reply to Stefano Infante
    TI__Guru**** 164726 points

    Stefano,

    For MC34063A
    When VIN=18V, Vout =15V, Iout = 2.7mA; efficiency should be at least 36% (6.2mA input)
    When VIN=18V, Vout =15V, Iout = 25mA; efficiency should be at least  63% (33mA input)
    When VIN=36V, Vout =15V, Iout = 2.7mA; efficiency should be at least 24% (4.6mA input)
    When VIN=36V, Vout =15V, Iout = 25mA; efficiency should be at least  58% (18mA input)

    For TL317
    When VIN=18V, Vout =15V, Iout = 2.7mA; efficiency is 42% (5.3mA input)
    When VIN=18V, Vout =15V, Iout = 25mA; efficiency is 75% (27.6mA input)
    When VIN=36V, Vout =15V, Iout = 2.7mA; efficiency is 21% (5.3mA input)
    When VIN=36V, Vout =15V, Iout = 25mA; efficiency is 38% (27.6mA input)

    Ron.

     

  • 0 in reply to Ron Michallick
    Expert 2320 points

    Ron,

    thank you very much for your help. This is more than enough to take my decision.

    Regards,

    Stefano