# comparator timer circuit

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

1. ### kellGuest

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. ### John PopelishGuest

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:
http://www.fairchildsemi.com/ds/2N/2N7000.pdf

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.