Resistor values for transistor and optocoupler

Discussion in 'The Projects Forum' started by onlyvinod56, May 16, 2013.

  1. onlyvinod56

    Thread Starter Active Member

    Oct 14, 2008
    362
    1
    Hello,

    My project title: multicarrier multilevel inverter.

    A sine wave is to be compared with four triangular waves as shown in the link.

    http://share.pdfonline.com/9a20a61666394ab698d97e05b7ec048c/31.Multi-level%20Inverter%20Capable%20of%20Power%20Factor%20Control.htm

    The comarator i used here is LM339. This quad comparator IC is wired to give the four pulse trains from four triangulars.

    the outputs of 339 are pulled up with four 10K resistors.
    The supply used for 339 is +/- 12V. see the attachment.

    @ A:the pulses are swinging from +12 to -12

    @ B:the negative part of the pulses are avoided. now the pulse amplitude is from 0 to +12V.

    NOW AFTER CONNECTING THE 12V PULSE TO THE TRANSSITOR AND OPTOCOUPLER SECTION, THE PULSE VOLTAGE IS DROPPED TO 3v.

    UNDER THIS SITUATION:

    @ B: the pulse voltage is 3V

    @ C: the pulse voltage is less than 1v

    @ D: the pulse is inverted w.r.t 'C' but the voltage is less than 1V.


    These 1V magnitude is not sufficient to ON the optocoupler.

    My question is why the voltage is droping to less than 1V?

    How to choose the base resistor value for - the transistor and for the optocoupler diode section.?
     
  2. wayneh

    Expert

    Sep 9, 2010
    12,086
    3,025
    The biggest problem is probably that your un-shown pull-up resistor is too big. The comparator does not source a voltage, all of the "output" signal passes thru the pull-up.

    Current is flowing across R1 and then R2, R3 and R4. By ohm's law, this drops the voltage.

    I see no function for R1. Try eliminating it. R2 could be a much higher value, since it only needs to pull the transistor base low.
     
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