Connect with us

comparator timer circuit

Discussion in 'Electronic Basics' started by kell, Jan 19, 2007.

Scroll to continue with content
  1. kell

    kell Guest

    What size cap can an LM393 for example discharge, without a series
    resistor. The cap would be charged to no more than 15 volts. Or is
    that a bad idea, in which case I might ask, how big a cap charged to 15
    volts can you use a 2N7000 to discharge, without any additional
    resistance in series.
  2. The LM393 data sheet does not specify the maximum current
    the output can carry, when not saturated, only a minimum it
    will pass while the voltage is below some maximum. But the
    unsaturated maximum as something around 10 or 20 mA. At 15
    volts drop, 20 mA would produce about 15V *.02A = .3 watts
    of heat in the chip, which is pretty high, to handle
    continuously. But the chip so rated for something like 570
    mW power in an ambient temperature of 25C, so one half might
    survive this. If you add a series resistor that limits the
    current to 20 mA (750 ohms), the chip power would cut way
    down. I wouldn't worry much about dumping a capacitor that
    was half discharged by 20 mA in a second, or about 2700uF.
    It is a poor idea, because of the large variation in
    unsaturated output current from device to device. If you
    intend t0 make only one, it may work fine, if your chip is
    not one with an exceptionally high output current
    capability. Adding the resistor makes it consistent.
    Assume all the energy in the cap (.5*V^2*C) gets dumped into
    the fet die. Look at figure 16 and 17 on page 6 of:

    I think a .001 second pulse is fast enough to assume that no
    heat escapes the die. For the TO-92 case (figure 16) the
    thermal resistance for a single pulse (assumes the die
    starts out at ambient) is about .035 of the 312.5 degrees C
    per watt for the steady state case. In other words, The
    temperature rise during the millisecond pulse will be only
    ..032 of 312.5 degrees per watt 0r about 10 degrees per watt.
    But you need to convert from power to total energy, since
    that is what you know from the capacitance and voltage.

    The peak die temperature cannot exceed 150C. Lets say the
    ambient is 50C, just to be conservative. so the die can
    experience 100C rise during the discharge without failing.
    At 10 degrees rise per watt (averaged over a millisecond),
    and 100 degrees of rise available that means the average
    power during one millisecond could be 10 watts. But 10
    watts dumped over a millisecond is 10 milliwatt seconds
    (joules) of energy. So starting at 15V, .01 joule=
    ..5*15*15*C, so C max is about 88 uF. The reason this is
    lower than for the comparator alone is because all the
    energy will end up in the die, because the discharge will be
    much faster (something like 3 or 4 amperes peak with 10
    volts of gate drive), instead of 20 mA. This leaves little
    time for the heat pulse to spread out from the die. (About
    660 us, so well within the 1 millisecond the estimation was
    based upon.)

    But again, if you added a resistor with more resistance than
    the fet, most of that heat would be transferred to the
    resistor. !0 ohms would be high enough to do that, while
    slowing the discharge to only about 2 milliseconds, or
    allowing a larger capacitor to be safely dumped.
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day