Correct voltage translator

Discussion in 'General Electronics Chat' started by Vindhyachal Takniki, May 12, 2015.

  1. Vindhyachal Takniki

    Thread Starter Member

    Nov 3, 2014
    1. I have a motor encode with max speed = 6000 rpm & 3600 pulses at each revolution.
    So max freq = 6000*3600 = 21.6Mhz.
    2. Encode is powered by 10-30V dc. But MCU is powered by 3.3V.
    3. So I need to step down the voltage of encoder pulses.
    4. Currently encoder is powered by 10V. I have a placed a small resistor & 3.6V zener. I am getting step down voltage but there is some rise & fall of pulse at across zener, which I see is comparable when I compare encoder output & zener out.

    5. I can use either optocoupler or any buffer IC for this.
    6. I had looked at FOD8001, 25Mhz, but it require Vcc at input also. and max is 3.3 or 5V only.
    ICPL2631 is 10Mhz.
    7. Avago HCPL-0720, is good, but it require VCC at input with max of 6V. I need to place some regulator for powering input vdd also.

    8. Any other product fit for this?
  2. ScottWang


    Aug 23, 2012
    Fmax = F*3600=(6000rpm/60s)*3600 = 100Hz*3600=360000Hz=360Khz.

    Setup for the voltage of encoder.
    Using two resistors to be a voltage divider, and in parallel a 3.6V zener diode with the low resistor, the current of zener should be about 10mA.

    Attach a clearly photo of waveform.

    Go back to see the first item.
  3. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
    hi VT,
    At 6000 rpm is that 6000/60secs = 100 revs/sec .???
    So 100 * 3600 pps