home *** CD-ROM | disk | FTP | other *** search
- Newsgroups: sci.electronics
- Path: sparky!uunet!mcsun!sun4nl!dutrun!donau!dutecag.et.tudelft.nl!knop
- From: knop@dutecag.et.tudelft.nl (Peter Knoppers)
- Subject: Re: Simple transistor project (Q)
- Message-ID: <1992Aug13.100326.3698@donau.et.tudelft.nl>
- Summary: Use a 555 timer.
- Sender: news@donau.et.tudelft.nl (UseNet News System)
- Nntp-Posting-Host: dutecag.et.tudelft.nl
- Organization: Delft University of Technology, Dept. of Electrical Engineering
- References: <1992Aug13.062418.999@cs.uri.edu>
- Date: Thu, 13 Aug 1992 10:03:26 GMT
- Lines: 65
-
- reynhout@cs.uri.edu (Andrew Reynhout) writes:
-
- >Hello. I've been having some problems with a circuit, and I'm wondering if
- >someone out there can help me...
-
- > Basically, I need to take a 5v low-current logic signal, and on the
- >transition from low to high, cause a relay to close for about half a second.
- >This is my attempt (forgive the ASCII graphics...)
-
-
- > +----------------------------------+
- > | 1/6 7404 |
- > | |\ INVERTER |
- > logic in >--+----+-----| O-----+ |
- > | |/ | |
- > R1 | B Q1 | B Q2
- > 3.3k ----- 2N2222 ----- 2N2222
- > | / \ / \
- > GND C / \ E + C / \ E +
- > +5v >-----+ ->----|>|---+---+ ->--+----|<|-------+
- > D1 | | D2 |
- > 1N4148 | | 1N4148 |
- > --- C1 | |
- > --- ~10uF | relay |
- > | +---coil-------+
- > | (5v / 80mA) |
- > GND |
- > GND
-
- Let us get the steady-state of this circuit first.
- If the logic input is high then
- -> Q1 is turned off
- -> Q2 is in a funny state. It will act as a diode conducting current from
- the logic input (which is supposed to be low current) to the relay coil.
- -> if the logic input can supply sufficient power the relay is turned on,
- otherwise you are overloading the source of the logic input signal.
- This is problem 1.
- If the logic input is low then
- -> Q1 is turned on
- -> Q2 is turned off
- -> the relay is off
- This is intentional.
-
- To cure problem 1 you could add a resistor between base and collector of Q2
- and insert a diode in de connection between the base of Q2 and the logic
- input. The diode should "point to" the logic input. I don't see any need
- for diode D1, nor for resistor R1.
-
- On an up transition of the logic input the capacitor is supposed to discharge
- over the relay. The resistance of the relay coil is (evidently) about 60 Ohms.
- The capacitor is 10 uF. This yields a discharge time in the order of 600 us
- (micro seconds). No physical relay will operate on such a short burst.
- This is problem 2.
-
- Increasing the capacity of C1 to, say 15000 uF, might work, but it would
- probably kill T1 or your power supply.
-
- I suggest that you obtain an application note for the 555 timer and use
- that chip. The 555 can deliver enough current to drive a relay (you will
- still need the diode over the relay coil) and the 555 can be driven directly
- by a low power logic signal.
-
- Success!,
-
- Peter Knoppers - knop@duteca.et.tudelft.nl
-