12-volt matrix input->output?
in [Design]
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From: John E. on 24 Feb 2007 15:21 Thus spake Jan Panteltje: > OK, but I'l do only part of the design: Thanks! > So if '2' selected ,as in table above, W goes on. > If 3 is selected, both W and X go on. > If 4 is selected Y goes on. > All diodes 1n914 Ah, diode matrix! In the back of my mind (a crowded place, to be sure) such an idea was lurking. Just couldn't bring it to mind. > Can you do the diodes for 5 and 6? Uh, I think so (c: > Normally 4051 cannot drive a bulb, so we need some amplification. I see no output current spec in the data sheet. Is this simply presumed to be a low value? Not 10 mA? (I'm driving relays with the circuit.) Thanks for making this reasonalby understandable by a good tech, but poor designer. -- John English
From: Jan Panteltje on 24 Feb 2007 15:33 On a sunny day (Sat, 24 Feb 2007 20:21:07 GMT) it happened John E. <incognito(a)yahoo.com> wrote in <0001HW.C205DEBE00CD5B37F01826C8(a)news.sf.sbcglobal.net>: >Thus spake Jan Panteltje: > >> OK, but I'l do only part of the design: > >Thanks! > >> So if '2' selected ,as in table above, W goes on. >> If 3 is selected, both W and X go on. >> If 4 is selected Y goes on. >> All diodes 1n914 > >Ah, diode matrix! In the back of my mind (a crowded place, to be sure) such >an idea was lurking. Just couldn't bring it to mind. > >> Can you do the diodes for 5 and 6? > >Uh, I think so (c: > >> Normally 4051 cannot drive a bulb, so we need some amplification. > >I see no output current spec in the data sheet. Is this simply presumed to be >a low value? Not 10 mA? (I'm driving relays with the circuit.) I remember Ron is < 150 Ohm or so... I think it will not drive a relay. The MOSFETS are only 2 $ or so, and will drive a relay if yu use a flyback diode. +12V | |---------- k | diode [ \ ] relay coil a | |---------- | |---- A' -----| | Power MOSFET | |---- resistor | 100k /// | | <- connect to output 7 of the CD4051 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. >Thanks for making this reasonalby understandable by a good tech, but poor >designer. You are welcome.
From: Joerg on 24 Feb 2007 15:39 John E. wrote: > I want to turn on 4 lamps in 6 combinations. These are controlled by 3 > switches in 6 corresponding combinations of contact closings. > > The truth table looks like this > > 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 > > While it's straight 3-bit binary count input, how to I translate it to the > desired 4 output combinations? What's the best way to implement such an > input-to-output matrix? > > Input from switches is 12 vdc and output requires 12 vdc, milliamp drive. > > I'm ignorant in all things PIC, so prefer to have another solution. > As others have suggested this can be done using a decoder and diode to wire up the desired lamp combination. Instead of analog muxes you can also use the CD4028 and use only the A, B and C inputs. Wire D to ground. http://focus.ti.com/lit/ds/symlink/cd4028b.pdf Digikey has it in stock even in the old DIP version, 52 cents a pop. If you need more than a milliamp or so you could use a driver from the ULN series. -- Regards, Joerg http://www.analogconsultants.com
From: jasen on 25 Feb 2007 03:11 On 2007-02-24, John E <incognito(a)yahoo.com> wrote: > I want to turn on 4 lamps in 6 combinations. These are controlled by 3 > switches in 6 corresponding combinations of contact closings. > > The truth table looks like this > > 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 W = C or B X = B and C Y = ( not B ) or (A and C) Z = A and B > Input from switches is 12 vdc and output requires 12 vdc, milliamp drive. > I'm ignorant in all things PIC, so prefer to have another solution. with those specifications the 4000 series CMOS logic family sound good, they're inexpensive and will run from 12V just fine. > Ideas? it's going to take three chips the easist way would be a 4071 (that has 4 or gates) a 4081 (that has 4 and gates) and a 4069 ( that has 6 not gates - a hex inverter) the three should come in under $2 get ones in DIP package unless you have a goos reason ot to. it'll cost you more in circuitboard than chips google for 4071 datasheet 4069 datasheet 4081 datasheet and it'll find you a PDF that explains how to use the chips. Bye. Jasen
From: John E. on 25 Feb 2007 05:06
Thus spake Joerg: > As others have suggested this can be done using a decoder and diode to > wire up the desired lamp combination. Instead of analog muxes you can > also use the CD4028 and use only the A, B and C inputs. Wire D to ground. > > http://focus.ti.com/lit/ds/symlink/cd4028b.pdf I've looked at the data sheet. Fewer pins to deal with (no inhibit, no input/output control) -- looks more straightforward... -- John English |