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UAF42

WILLIE MCCAIN
Prodigy 235 points
Other Parts Discussed in Thread: UAF42

HELLO I/M USING THE UAF42 FOR A 400 THREE PHASE SYNC CLOCK INTO ZERO CROSSING DETECTORS

NOW FOR RF1 AND RF2 I CAME UP WITH 39.80K AND RFb= 5 MEG, BUT THE VALUES FOR R1&2 AND R3&R4 IS NOT LOOKING CORRECT  CAN SOME ONE SEE WHAT VALUES YOU COME UP WITH IAM TRYING TO GET THE MAX AMPLTD. AND MY SUPPLY VOLTAGEA ARE +15V / -15V AND THE FREQ. OUT IS 400HZ

THANKS,

WILLIE L. MCCAIN

  • 0
    TI__Guru* 93105 points

    Hi Willie,

    • What filter criteria did you establish for the UAF42; cutoff frequency, gain, Q, filter response (Butterworth), etc?
    • Is it a low-pass, high-pass, or band-pass?
    • Non-inverting, or inverting configuration?
    • Can you share your schematic?

    Regards, Thomas

    PA - Linear Applications Engineering

  • 0 in reply to Thomas Kuehl
    Prodigy 235 points
    HELLO MR Kuehl,
    IF YOU GO ON TI'S SITE AND PUT IN THE UAF42 YOU WILL SEE A DESIGN CALLED THREE PHASE SINE WAVE GENERATOR THAT IS WHAT I'M USING THE SANE THING
    BUT AT 400 HZ AND /M TRYING TO GET THE MAX AMPLTD. OUT
  • 0 in reply to WILLIE MCCAIN
    TI__Guru* 93105 points

    Hello Willie,

    Thanks for referencing the sbfa013, UAF42 Applications Bulletin. I ran through the equations provided in the bulletin assuming a 400 Hz oscillation frequency and a 10 Vpk output level. Here are the component values I derived:

    fOSC = 1 (2πRC)

    R = RF1 = RF2

    C = C1 = C2 = 1e-9 F

    Let fOSC = 400 Hz

    Determine the resistance R:

    400 Hz = 1 / (2π∙R∙1e-9)

    R = 1 / [2π (400 Hz) (1e-9 F)] = 397.9 kΩ

    To set the output level use Eq. 2:

    R1 / R2 = R3 / R4 = [(Vo + Vcc) / (Vo + 0.15)] -1

    Let Vcc = 15 V, Vo = 10 Vpk

    R1 / R2 = R3 / R4 = [(10 + 15) / (10 + 0.15)] -1 = 1.54:1

    Let R2 = 1 kΩ, then R1 = 1.54 kΩ

    RFB = 5 MΩ from Table 1

    You mention that you want to maximize the UAF42 output levels. The datasheet indicates that the minimum output voltage is 11 V, with a 2 kΩ load (+/-15 V supply). I used 10 Vpk to be a little more conservative.

    Regards, Thomas

    PA - Linear Applications Engineering

  • 0 in reply to Thomas Kuehl
    Prodigy 235 points
    NOW WHAT WILL BE THE EASEST WAY TO SWITCH FROM 400 HZ TO 60 HZ
    ARE CAN I AD A VERIABLE RESISTOR TO FINE TUNE BOTH FREQ JUST BY CHANGING THE CAP TO SET THE FREQ. AND THAN ADJ THE FINIAL SET FREQ.
    LET ME NO THANKS ALOT FOR YOUR HELP
  • 0 in reply to WILLIE MCCAIN
    TI__Guru* 93105 points

    Hi Willie,

    Posssibly the easiest way to shift the frequency from 400 Hz to 60 Hz would be to add one each, CEXT across capacitor across C1 and C2. Rearranging the frequency equation to solve for the required capacitance at 60 Hz and using the 397.9 k RF1, RF2 resistance from the 400 Hz calculation:

    C = 1 / [ 6.28 (60 Hz) ( 397.9 k) ] = 6.67 nF

    Since C1 and C2 are internally set at 1 nF each, then adding a 5.6 nF across each of them would put the capacitance at 6.6 nF, if they were exact. I would suggest using a low tolerance, quality capacitor for C1 and C2, preferably polystyrene of another film capacitor type. If the frequency is critical you may need to find capacitors that are towards the middle or lower end of the tolerance and then pad the capacitors with lower value capacitances until the frequency is 60 Hz.

    Once you have the frequency set you could use a dual-section switch to switch the external capacitors in and out of the circuit for 400 Hz, or 60 Hz operation.

    Regards, Thomas

    PA - Linear Applications Engineering