12-volt matrix input->output?
in [Design]
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From: Tony Williams on 25 Feb 2007 06:13 In article <0001HW.C205BA3500C4CAEEF01826C8(a)news.sf.sbcglobal.net>, John E. <incognito(a)yahoo.com> wrote: > Switches Lamps > A B C W X Y Z > ----- ------- > 0 1 0 1 0 0 0 > 0 1 1 1 1 0 0 > 1 0 0 0 0 1 0 > 1 0 1 1 0 1 0 > 1 1 0 1 0 0 1 > 1 1 1 1 1 1 1 If you are already using relays then I think this can be done with 3-off 4-pole changeover relays. .-------------------. NC | NC | NC / | / | / / | / | / 12v---A1 | 12v---B1 '---C1 NO---' NO--------------NO---> W output NC NC NC / / / / / / A2 12v---B2 .-------C2 NO NO---' NO---> X output .--------------. NC NC---' NC | / / / | / / / | 12v---A3 .-------B3 .-------C3 | NO---' NO---' NO--+> Y output NC NC NC / / / / / / 12v---A4 .-------B4 C4 NO---' NO---. NO '---------------> Z output /|\ /|\ /|\ | | | A-contacts B-contacts C-contacts A-----. B-----. C-----. | | | ___|___ ___|___ ___|___ | | | | | | | CoilA | | CoilB | | CoilC | |_______| |_______| |_______| | | | 0v----+---------------+---------------+ -- Tony Williams.
From: Jan Panteltje on 25 Feb 2007 06:20 On a sunny day (Sat, 24 Feb 2007 20:33:47 GMT) it happened Jan Panteltje <pNaonStpealmtje(a)yahoo.com> wrote in <erq7fi$7ua$1(a)news.datemas.de>: +12V | |---------- k | diode [ \ ] relay coil a | |---------- | |---- A' -----| | Power MOSFET | |---- resistor | 100k /// | |----< connect to output 7 of the CD4051 | 3300 Ohm pull down resistor. | /// >If you connect the thing as above, you save 4 diodes by connecting all resistors >that hold the gates down to output 7 of the CD4051, so if output 7 goes high, >then all MOSFETS are powered, and all lights go on. >You need these resistors anyways. Small correction, you then also need a 3300 Ohm resistor to ground from output 7 of the CD4051 (see modified diagram above). This because the switch to output 7 is normally open, and _something_ should pull the resistor network to ground if 7 is not selected. With max 2 bulbs on, in the other cases then 7, the divider would be 50k / 3k3. And the voltage it would see about 11.3 V. That gives a max off voltage at the MOSFET gates of about 0.7 V. The current would be (if on) 12 / 3300) = 3.6 mA. The internal switch in the CD4051 is max 150 Ohm and would hardly dissipate anything. Or use a CD4028 BCD to decimal decoder, as Joerg suggested, as it has a true zero output. I personally have a whole lot of 4051 / 4053 switches around, as they can make any logic, modulators, decoders, what not... So that was my natural choice...
From: John E. on 25 Feb 2007 06:41 Thus spake Jan Panteltje: > Or use a CD4028 BCD to decimal decoder, as Joerg suggested, as it has a true > zero output. If I use CD4028, I will not need 3300 pull-down resistor? But the rest of the circuit will remain the same? Thanks, -- John English
From: Jan Panteltje on 25 Feb 2007 06:49 On a sunny day (Sun, 25 Feb 2007 10:05:46 GMT) it happened John E. <incognito(a)yahoo.com> wrote in <0001HW.C206A00600FAA812F058A6C8(a)news.sf.sbcglobal.net>: >Thus spake Jan Panteltje: > >> If you connect the thing as above, you save 4 diodes by connecting all >> resistors >> that hold the gates down to output 7 of the CD4051, so if output 7 goes high, >> then all MOSFETS are powered, and all lights go on. >> You need these resistors anyways. > >Sort of using 4015 output pin 7 as a pull-down resistor for the MOSFET's >gates? > >What MOSFET would you recommend? I might want to power 2 relays in parallel >with each of the 4 MOSFETs... Is this 60v N-channel one, with 1.2 ohm >on-resistance an appropriate choice? : > ><http://www.jameco.com/Jameco/Products/ProdDS/256031FSC.pdf> That one is already 'on' at .8V, I'd rather use one that was on at about 6V :-) >And it includes a diode (which will act as a flyback, yes?) The flyback would be positive, maybe above 60V in this case. That intrinsic diode only limits negative swings. Some MOSFETS have avalanche protection diodes that way (the intrinsic diode then zeners), not this one I think.
From: Jan Panteltje on 25 Feb 2007 06:58
On a sunny day (Sun, 25 Feb 2007 11:41:52 GMT) it happened John E. <incognito(a)yahoo.com> wrote in <0001HW.C206B68D00FFD743F058A6C8(a)news.sf.sbcglobal.net>: >Thus spake Jan Panteltje: > >> Or use a CD4028 BCD to decimal decoder, as Joerg suggested, as it has a true >> zero output. > >If I use CD4028, I will not need 3300 pull-down resistor? But the rest of the >circuit will remain the same? Yes. http://www.ee.washington.edu/stores/DataSheets/cd4000/cd4028.pdf |