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Voltage Regulators

Discussion in 'Electronic Basics' started by Joe, May 8, 2007.

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  1. Joe

    Joe Guest

    Hello,

    Has anyone used the AP1501 voltage regulators.
    (http://www.diodes.com/datasheets/dsAP1501.pdf)

    I have a 3.3v and a 5.0v output version of these v-regs. I was
    wondering if I really need the inductor and capacitors for this v-
    reg? And also would a 39uH inductor work instead of a 33 uH on the
    3.0v-reg? Would the inductor value be different for the 5.0v
    regulator? What would the capacitor values be for the the Cin and Cout
    shown in the datasheet?

    Im sorry for all the questions, but any help in this matter would be
    appreciated.

    Thanks,
    Joe McKibben
     
  2. Chuck

    Chuck Guest

    Hi Joe,

    My initial reaction is that 39 uH may be
    near the edge of the chip's frequency
    range (~+/-15% of 150kHz).

    Cin can probably be 0.1uF. It is there
    to filter transients.

    Cout is determined by establishing the
    ripple you're willing to see in the
    output and the maximum load current. The
    output from the chip seems to be a
    "square wave" between ~0Volts
    and Vreg. You choose Cout so that at the
    desired load resistance, it discharges
    no more than the maximum p-p ripple you
    want between the 150kHz pulses.

    With luck, someone with direct
    experience with the chip will offer some
    more detail.

    Chuck
     
  3. I haven't.
    Since these are switching regulators that output pulses, not
    steady voltages, the LC output filter is an essential pert
    of getting a DC output from them. Since they also draw
    current pulses from the upstream supply, if that supply
    feeds anything else, you may need to worry quite a bit about
    how you filter than side, also, so the operation of this
    regulator does not trash the upstream supply for other users.
    Almost certainly. The manufacturers are always pushing the
    required inductance down to the absolute minimum value,
    because storing energy in inductors at low loss is spendy.
    All those values on the data sheet tend toward the absolute
    minimum that will work. When selecting these capacitors the
    equivalent series resistance (ESR) and ripple current
    ratings are as important as the capacitance. The input
    capacitor sees worse ripple current than the output
    capacitor, because it does not have a series inductance
    smoothing the pulses into a triangle current wave.

    The actual value of capacitance, ESR and ripple current
    rating involves the peak current you expect the regulator to
    supply and the maximum ripple voltage you can tolerate.

    Can you help with any of these specifications?
    No apology necessary. This is the right place for such
    questions.
     
  4. Joe

    Joe Guest

    Theses regulators are a little over kill for my application. On the
    3.3v reg. max current might be 2mA (dual axis gyro and tri-axis
    accelerometer). And on the 5.0v reg. approx. 150mA (SX-
    Microcontroller). Not sure how much ripple voltage I can tolerate, but
    I'll try to found out.

    Thanks for the help,
    Joe McKibben
     
  5. I assume that is a typo and you mean that the load on the
    3.3 volt regulator may be as high as 2 amperes.

    Does anything else (other than two of these regulators) run
    on the up stream source? Are there any RF sensitive devices
    anywhere near by?
     
  6. Joe

    Joe Guest

    For the 3.3v it might be close to 10mA. Accelerometer (http://
    www.robotshop.ca/PDF/ADXL330_0.pdf), Gyro (http://www.robotshop.ca/PDF/
    IDG-300_Datasheet.pdf)
    And I dont think there are any RF sensitive devices to really worry
    about.

    Im not sure what you mean by on the up stream, but I have a 12volt
    battery. It supplies the current for two motors and the motor
    controllers. The 12 volts also supplies the voltage to the
    regulators. I think I might use a seperate battery source for the
    regulators to eliminate any voltage spikes from the motors.

    Joe McKibben
     
  7. For any load less than about 100 mA, I would go with a
    linear regulator, like an LM317. The small efficiency gain
    you get with a switcher is not worth the trouble for lower
    current loads. The 3.3 volt regulator could get its input
    power from the 5 volt output. This will give you
    essentially the same efficiency as a switcher producing 3.3
    volts from the 12 volt source, with less output noise,
    assuming the 3.3 and 5 volt loads share a common negative rail.
    Yes, from a power flow standpoint, that is up stream of
    these regulators.
    Those will generate some noise and will need their own
    storage capacitors to help the battery supply the large
    current spikes they need. Something like 1000 uF at 16
    volts for each motor.
    That shouldn't be necessary, especially for switching
    regulators, since they contain their own LC filters.

    For the 5 volt 2 amp regulator, I would use a 470 uF 16 volt
    Panasonic type FM electrolytic cap on the input and a 330 uF
    6.3 volt across the output. Digikey sells these. They have
    a low ESR and high ripple current rating. I would probably
    also parallel the output cap with a Panasonic V series
    (stacked film) 1 uF 50 volt to take care of the high
    frequency edges that couple across the inter winding
    capacitance of the inductor. In addition, I would add a
    ferrite bead on a lead in series with thew 12 volt input,
    upstream of the input storage cap, to keep the motor noise
    out of the regulator supply.
     
  8. Joe

    Joe Guest

    Cool, thanks for all the help. Ill try to get a linear regulator for
    the 3.3volt. and the other components I need.

    Joe McKibben
     
  9. The LM317 adjustable linear regulator requires a couple
    resistors to program its output voltage. An input and
    output capacitor is also recommended. See the data sheet.

    http://www.fairchildsemi.com/ds/LM/LM317.pdf
     
  10. Joe

    Joe Guest


    Why wouldn't I just get a fixed reg. instead of an adjustable?

    Joe McKibben
     
  11. 5 volt linear regulators are common as dirt, but 3.3 types
    are a lot harder to find. I would avoid low drop out types,
    since they are harder to stabilize against oscillations.
    The 317 is very common and is available in a low power
    version that comes in a TO-92 transistor package.

    But a type that does not require setting resistors (and is
    not LDO) would be:
    LM78L33ACZ
    http://www.st.com/stonline/products/literature/ds/2145/l78l05ab.pdf
     
  12. Joe

    Joe Guest

    Do you know of a good place to find the LM78L33ACZ v-reg? I cant find
    it anywhere.

    Joe McKibben
     
  13. No. Hence the LM317 suggestion.
     
  14. Joe

    Joe Guest

    Oh, ok thanks. I found some surface mount versions of the LM78L33, but
    that doesn't help me me to much.

    How do you choose the resistors for the adj regulator? I went through
    the formula under the typical application drawing using 3.3volts, but
    I am not sure what do with the value. I need to do some brushing up on
    my DC circuits understanding, well probably more than brushing up.

    Thanks again,
    Joe McKibben
     
  15. There are generally two constraints. The series pair of
    resistors must consume at least the regulator minimum output
    current, if the regulator must work all the way to zero
    external load. And the divider voltage must be the desired
    output voltage minus the nominal difference voltage between
    the output and the reference pin.

    A secondary constraint is that the range of current passed
    through the reference pin does not distort the divider
    voltage too much.

    For the LM317L
    http://www.onsemi.com/pub/Collateral/LM317L-D.PDF

    the maximum value of the minimum output current allowed if
    you expect the regulator to regulate is 10 mA, so if you
    will not always have at least that much external load, then
    the divider must consume it. Since the nominal difference
    between the reference pin voltage and the output voltage is
    1.25 volts, there will be that much voltage across the top
    resistor of the reference divider, so it must be no more
    than 1.25/.01=125 ohms. If you use the typical value of
    minimum load current or 3 mA, instead of the worst case
    (usually okay for a single system that will not be
    manufactured in high volume) that resistor rises to
    1.25/.003=417 ohms. Usually, any value between those will
    work. I notice that the example on the sheet uses 240 ohms.

    Then the bottom divider resistor must drop the rest of the
    desired output voltage, or 3.3-1.25=2.05 volts, in this
    case, while it also passes the reference pin (adjustment
    pin) current of about 50 uA. Since the first resistor is
    delivering 3 to 10 mA, the additional current from the
    reference pin may be within the tolerance effect of the
    resistors.

    Ignoring the reference pin current and assuming you use 240
    ohms as the upper resistor
    (delivering about 1.25/240=5.2 mA), the lower one would be
    2.05/.0052=394 ohms.

    All this process is described in detail starting on page 8.
     
  16. Joe

    Joe Guest

    Ok, thanks again. That data sheet is different than the first one you
    posted for the LM317. It is more helpful than the first.
    Now I just have to get the LM317, capacitors, inductors, and resistors
    and I can finally get this thing hooked up.

    Thanks for being patient with me,
    Joe McKibben
     
  17. Before you send for parts or at least before you solder
    anything together, perhaps you should send me a schematic of
    the system, so I can correct any misunderstandings. We have
    covered a lot of ground, since your first post.
     
  18. Joe

    Joe Guest

    OK, I'll get one drawn up.

    Joe McKibben
     
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