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| deth502 |
Posted: January 03, 2013 12:38 am
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![]() Sr. Member ![]() ![]() ![]() ![]() ![]() Group: Trusted Members Posts: 365 Member No.: 14,050 Joined: March 23, 2008 |
ok, working on my project. i have a vibration sensor as such:
http://www.engineersgarage.com/contributio...-diagram?page=2 the one i was trying to sub in a cheap avail 555 for, but couldnt get it to reliably trigger. ill take my chances finding more 4538's. anyway, the alarm is going to trigger a flashing led. im using this circuit: http://talkingelectronics.com/projects/Fla.../LedFlasher.gif ok, so the lights not flashing. the flasher works. i can hook that to anywhere from 5v to 12v and other than the different rate for different v-in's, it works fine. now, the cd4538 is putting out about 4.8v (from a 12v supply) so i think the v's too low to flash the circuit. so i put a transistor in as a switch to get more poser to the flasher circuit. i put the transistor in and i measure about 6.5v to the flasher now (while i was expecting closer to the 12v through the transistor as a switch?) but anyway, still no flash. i thought maybe it was the power supplying the flasher circuit, so i scoped it and, while its not totally clean and there is some ripple, its around or less than .5v, so i see no reason that should affect the flasher? (was thinking it might be a low duty cycle square or something) any ideas why it wont flash on the signal from the sensor? (again, works fine if just hooked straight to the +v) |
| Sch3mat1c |
Posted: January 03, 2013 01:46 am
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![]() Forum Addict ++ Group: Moderators Posts: 18,144 Member No.: 73 Joined: July 24, 2002 |
Think of the output pin of any CD4xxx chip as a switch to +V/GND plus about 200 ohms series resistance, and all will become clear. If you want to run a load heavier than just a few mA, you need a driver, a switch or something.
Or an enable. Example: you could disconnect the 330k resistor from +V, moving that to the CD4xxx output instead. When the output is low, the transistor doesn't get any current and nothing happens; when the output is high, the 330k slowly charges the cap until the transistor turns on and the circuit kicks over. 330k won't draw even 1mA, so the CD4xxx is as happy as can be. Hmm, I'd recommend a 100 to 1k ohm resistor in series with the 10uF cap; omitting this (as shown) will deliver a rather substantial current pulse into the NPN base, which could destroy it over time. Always good practice to keep currents controlled. This also has the effect of stretching the pulse duration, keeping the LED lit longer. The 22R controls LED brightness. Tim -------------------- Answering questions is a tricky subject to practice. Not due to the difficulty of formulating or locating answers, but due to the human inability of asking the right questions; a skill that, were one to possess, would put them in the "answering" category.
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| deth502 |
Posted: January 03, 2013 02:50 am
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![]() Sr. Member ![]() ![]() ![]() ![]() ![]() Group: Trusted Members Posts: 365 Member No.: 14,050 Joined: March 23, 2008 |
ok, so, ill add in the resistor to the base of the flasher npn, i see your point there. and knowing that that resistor is the led resistor helps, im trying to use a high intensity led and was thinking that it was just run full on without the resistor, overdriving the led, since it has a very short duration as set up (about .1 sec per 1.0 sec)
the output of the cd4xxx, upon activation, actully pulses low, so the pin is connected to the ground of the flasher. given that, if im thinking correctly, eliminating the 330k ----------- wait a minute. set up with the transistor switch, the low output pulse then switches the high through a pnp, so i can still eliminate the 330k and go right to the transistor coming out of the cd4xxx. ill give that a shot and see what happens. just typing as im thinking this through........ like i said, the pat that gets me is that the flasher works hooked up hot, but being powered by the sensor, the light still comes on, it just wont flash, thats the part i cant wrap my brain around. lemme try that mod......... |
| deth502 |
Posted: January 03, 2013 02:56 am
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![]() Sr. Member ![]() ![]() ![]() ![]() ![]() Group: Trusted Members Posts: 365 Member No.: 14,050 Joined: March 23, 2008 |
ok, for some reason, when i switch out the 330k with a jumper wire from the transistor switch connected to the cd4xxx pin, the light wont light at all.
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| deth502 |
Posted: January 03, 2013 03:01 am
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![]() Sr. Member ![]() ![]() ![]() ![]() ![]() Group: Trusted Members Posts: 365 Member No.: 14,050 Joined: March 23, 2008 |
alright, i put the led in the flasher straight to the +v, and then the base of the npn to the output and then the light came on, but still no flash. i do, however, think i killed the led. over about 15-20 sec it kept getting dimmer and dimmer till the light was barely visible.
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| MacFromOK |
Posted: January 03, 2013 03:11 am
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Forum Addict ++ ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Spamminator Taskforce Posts: 12,952 Member No.: 5,314 Joined: June 04, 2006 |
Sounds like it. Over-current or connecting one in reverse (usually around 5V or more) can do that. -------------------- Mac *
"Basic research is what I'm doing when I don't know what I'm doing." [Wernher Von Braun] * is not responsible for errors, consequential damage, or... anything. |
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| deth502 |
Posted: January 03, 2013 03:26 am
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![]() Sr. Member ![]() ![]() ![]() ![]() ![]() Group: Trusted Members Posts: 365 Member No.: 14,050 Joined: March 23, 2008 |
all i can think is that the 330k must have something to do with the led as well, since when that was in there, i tried it several (well over 50) times (at each signal to the piezo, the led comes on for about 5-6 sec). when i put the led right to +v, and put the base of the npn to the +v from the transistor switch from the cd4xxx pin, and omitted the 330k in there, i hit the piezo once, hmmm, looked like that got dimer at the end. hit it 2x, yeah, thats dimmer. 3x, now its really dim. 4x, barely on. its dead now. ill repeat this part again, this is whats really killing me, this is why im so perplexed. 1) the flasher works fine when connected right to +v. 2) the sensor is putting out clean and ample +v, 3) when the flasher is powered by the sensor, the light stays on steady. WHY?!?!?!?! what is different in the power right from the rail and the power from the sensor?!!?!?!? why would the flasher care where the power is coming from??!?! |
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| MacFromOK |
Posted: January 03, 2013 05:08 am
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Forum Addict ++ ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Spamminator Taskforce Posts: 12,952 Member No.: 5,314 Joined: June 04, 2006 |
Hmmm... I'm thinking the flasher circuit probably starts with the LED "on." If so, your sensor vibrating probably keeps turning it on/off much faster than it can flash "off," and this makes it appear to be staying "on" all the time. My best guess anyway. -------------------- Mac *
"Basic research is what I'm doing when I don't know what I'm doing." [Wernher Von Braun] * is not responsible for errors, consequential damage, or... anything. |
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| deth502 |
Posted: January 03, 2013 05:48 am
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![]() Sr. Member ![]() ![]() ![]() ![]() ![]() Group: Trusted Members Posts: 365 Member No.: 14,050 Joined: March 23, 2008 |
by "sensor", i mean the censor circuit. the piezo gets amplified through the transistors, then signals to the cd4xxx, which then latches on the (low pulse) the output for the r/c determined time.
like i said, i scoped the power out of the cd4xxx that powers the flasher, and while there may be some ripple, its still a fairly pure dc voltage. no ac or big dips in it or anything. this makes absolutely no sense to me. |
| MacFromOK |
Posted: January 03, 2013 06:12 am
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Forum Addict ++ ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Spamminator Taskforce Posts: 12,952 Member No.: 5,314 Joined: June 04, 2006 |
The flasher circuit uses BJTs, which are current driven. You might check the current it uses when flashing properly, then see how much it's getting when triggered by the sensor.
The old "divide and conquer" routine... -------------------- Mac *
"Basic research is what I'm doing when I don't know what I'm doing." [Wernher Von Braun] * is not responsible for errors, consequential damage, or... anything. |
| Sch3mat1c |
Posted: January 03, 2013 11:36 am
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![]() Forum Addict ++ Group: Moderators Posts: 18,144 Member No.: 73 Joined: July 24, 2002 |
Noooooo! Don't do it without a resistor! That will do two things: it forces the NPN on almost constantly while the output is high, burning the LED, and it forces the capacitor to discharge through the CMOS output diodes, which will probably cause latchup at this current. The base voltage will go negative while the LED is off, this is what causes it to generate off-time.
Since your output is active-low, you can flip the flasher circuit upside down, P for N, N for P, like so: ![]() Tim -------------------- Answering questions is a tricky subject to practice. Not due to the difficulty of formulating or locating answers, but due to the human inability of asking the right questions; a skill that, were one to possess, would put them in the "answering" category.
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| deth502 |
Posted: January 03, 2013 10:57 pm
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![]() Sr. Member ![]() ![]() ![]() ![]() ![]() Group: Trusted Members Posts: 365 Member No.: 14,050 Joined: March 23, 2008 |
i tried that, but being the brief flash, around .1 sec every 1 sec, depending on v+, it only draws current when the led is lit, so i cant get a read on it with the meter. the refresh rate on the meter isint fast enough to stabelize, but i was getting spikes around 25-30mA max, so i figured it at about 50mA to be safe. |
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| deth502 |
Posted: January 03, 2013 11:22 pm
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![]() Sr. Member ![]() ![]() ![]() ![]() ![]() Group: Trusted Members Posts: 365 Member No.: 14,050 Joined: March 23, 2008 |
i dropped the transistor switch between the 2 circuits that was being used to switch the state, and used this flasher ^^^ hooked right to the cd4xxx output. it worked fine. thanks! im still bothered as to why it didnt work before though. it got the power from the rail, it worked, it got the power from the ic, it didnt. it boggles my mind. but, it is working now. i just have to test and finalize (might have to change a few component values with the timing) and ill get it all together. thanks again |
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