JHiGGs
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« on: April 07, 2008, 06:23:53 PM » |
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mga sirs/ma'am patulong naman po. may balak po akong project na ganito ang requirement power supply 12volts DC tapos pag na detect yung 2 consecutive +12V signal within 3 secs saka lang ma cocomplete yung circuit bale sa hule may motor or solenoid na paandarin. ano ano kaya yung materials na kailangan ko? available kaya lahat sa deecoo recto? medyo marunong po ako mag solder  salamat po. |
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« on: April 07, 2008, 06:23:53 PM » |
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paranz
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« Reply #1 on: April 07, 2008, 06:30:26 PM » |
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JHiGGs
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« Reply #2 on: April 07, 2008, 06:50:00 PM » |
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kala ko serious reply  |
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7_SeVeN_7
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« Reply #3 on: April 07, 2008, 06:51:49 PM » |
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parang di ko yata naintindihan masyado... paelaborate nman po
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E-Gizmo Mechatronix Central: www.e-gizmo.comTel #: (63)(2) 536-3378 Globe +63915-973-7691 Smart +63921-779-0748 Location MapYM: julie.egizmo aka Born2BeWired  |
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paranz
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« Reply #4 on: April 07, 2008, 06:52:05 PM » |
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sorry po... akala ko kasi non-technical thread ito, just by reading the title |
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Ate Shey
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« Reply #5 on: April 07, 2008, 06:52:24 PM » |
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mga sirs/ma'am patulong naman po. may balak po akong project na ganito ang requirement power supply 12volts DC tapos pag na detect yung 2 consecutive +12V signal within 3 secs saka lang ma cocomplete yung circuit bale sa hule may motor or solenoid na paandarin. ano ano kaya yung materials na kailangan ko? available kaya lahat sa deecoo recto? medyo marunong po ako mag solder salamat po. Anong gusto mo? yung may microcontroller na design or yung wala? |
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- Oh Lordy, troubles so hard. Don't nobody know my troubles but God
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JHiGGs
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« Reply #6 on: April 07, 2008, 06:54:55 PM » |
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the simpler the better po
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zer0w1ng
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« Reply #7 on: April 07, 2008, 07:25:13 PM » |
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I'll draw a quick schematic. Adjust R1 and C1 for the 3 second timeout. This is untested, and I have not simulated this. Maybe others can help you find a good RC value using a simulator.  Edit: This may need a freewheel diode across the solenoid to protect the transistors Q7 and Q6 upon turn-off of the solenoid. |
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JHiGGs
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« Reply #8 on: April 07, 2008, 09:59:51 PM » |
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many thanks po mga sirs.
ano ano po ang bibilin ko?
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JHiGGs
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« Reply #9 on: October 11, 2010, 06:04:19 PM » |
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I did this and it does not work please help!
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piona
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« Reply #10 on: October 11, 2010, 06:50:57 PM » |
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love without accomplishing the laws, is complicated - sshrekk
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iyo karpo
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« Reply #11 on: October 12, 2010, 07:25:04 PM » |
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mga sirs/ma'am patulong naman po. may balak po akong project na ganito ang requirement power supply 12volts DC tapos pag na detect yung 2 consecutive +12V signal within 3 secs saka lang ma cocomplete yung circuit bale sa hule may motor or solenoid na paandarin. ano ano kaya yung materials na kailangan ko? available kaya lahat sa deecoo recto? medyo marunong po ako mag solder salamat po. paki post mo ng ung ladder logi o FBD nya. maraming tools dito http://electricuniversity.blogspot.com/2010/07/industrial-automations-ladder-logic-and.html |
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dummy_c
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« Reply #12 on: October 12, 2010, 08:12:53 PM » |
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I think pwede na po dito 555 at flip flop.. Kawawa naman po c TS, dalawang taon na nga. |
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Pro-Engineer
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« Reply #13 on: October 12, 2010, 08:47:30 PM » |
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medyo malabo yung project description...
please explain some more on what you want to do.
drawings will help.
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dummy_c
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« Reply #14 on: October 12, 2010, 09:08:26 PM » |
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Master baka ang gusto nya yung circuit na mag trigger if meron ma detect dalawang 12v pulse within 3 sec. Then yug output po ay solenoid. Tama ho ba sir JHiGGs?
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Ngarus!
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« Reply #15 on: October 12, 2010, 09:20:37 PM » |
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ay uu nga  |
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anthonydp
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« Reply #16 on: October 12, 2010, 10:02:33 PM » |
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ok naman yung circuit ni zerowing. although i would suggest that you put signal conditioning circuit before the clock input if the 12V trigger comes from a mechanical switch. this maybe a simple low pass filter (switch debouncer) to filter any contact bounce that usually associates with signals coming from mechanical switches.
here's an ascii art circuit for the contact debouncer:
normally
open switch
+12V-----o// o-----10K------------> to clock input
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4.7k == 10uf
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GND GND
schmitt trigger naman yung clock input ng CD4040.
you may need to change the 10uf capacitor to lower value if your 12V trigger has pulse width less than 0.1sec
EDIT: added the 4.7k resistor as discharge path. hehehe.
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your question appears to be an XY Problem. why don't you just tell us exactly what you're trying to do? |
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piona
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« Reply #17 on: October 12, 2010, 11:47:01 PM » |
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Master baka ang gusto nya yung circuit na mag trigger if meron ma detect dalawang 12v pulse within 3 sec. Then yug output po ay solenoid. Tama ho ba sir JHiGGs?
Ganyan rin pagkakaintindi ko,  sana makatulong para sa tamang daan ng TS |
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love without accomplishing the laws, is complicated - sshrekk
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labgruppen
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« Reply #18 on: October 13, 2010, 01:26:26 AM » |
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try this circuit. this circuit will turn on the relay if at least two pulses is received within a three second time out. adjust values of R1-C1 to provide three second timeout. pushing reset button will immediately turn off the load. note well that immediately after pushing the reset button, the circuit will only require one pulse to turn on. however after the three second time out period has elapsed after pushing the reset, the circuit will then again require two or more pulses within 3seconds to activate.  i hope this helps. |
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labgruppen
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« Reply #19 on: October 13, 2010, 02:15:39 AM » |
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in hindsight, D1 is not needed and should be left open, that is, no connectivity between the Qoutput to the R1-C1 node. If D1 is omitted, and the reset button is pressed, the circuit will immediately require at least two pulses within 3sec to turn-on (as long as the initial timeout period has passed).
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July 16, 2008, 12:43:50 PM
