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3 switches, 7 lamps (no solid state)
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Benta:
OK, here goes:
First you do a 2:4 demux using two SPDT switches. One switch connected to L, the other to N. Four lamps are connected between the four outputs from the switches in the four combinations possible. Now you can light each bulb independently.

In the second stage you add the third switch that controls two DPDT relays between the first two switches and the lamps. Add a second group of four lamps. The relays switch between the two groups of lamps. Now you have a 3:8 demux. Remove one lamp of your choice, and you can control all seven lamps individually, plus turn them all off.

mikerj:

--- Quote from: bsudbrink on February 28, 2019, 12:09:30 am ---Sorry for the ugly drawing, I don't have access to a real tool right now...
Anyway, is there any reason this wouldn't work:

--- End quote ---

This suffers from a trap that is very easy to fall into when using relay logic.  In any position, one of the eight lamps will be at full brightness, but the lamp it is paired with has a path to ground through the other three pairs of lamps.  This means the paired lamp will illuminate but at lower brightness and the rest of the lamps may also glow dimly if there is sufficient current.
DDunfield:

--- Quote from: Cyberdragon on February 27, 2019, 05:19:35 pm ---i'm trying to come up with a system to switch 7 lamps on (indevidually) with 3 spdt switches and relays (just for funsies).

I'm trying to figure out the most efficient wat to do it. Getting the light to switch on when all switches are on is easy, when all switches are on, none of the secondary switch contacts have power which shuts of a spdt relay that disconnects the common from most lamps and connects it to the remaining lamp. Each switch is connected to a light through the nc contacts of other relays.

What I'm still trying to figure out is the logic that drives the other three lamps.

Aslo, the current circuit requires the same voltage for coils and lamps. So I'd have to have a different circuit using dual pole relays for different voltages (IE LV coils and mains bulbs).

--- End quote ---

If SPDT relays, then 5(**) relays - if you can use DPDT then it becomes three.
**EDIT - apparently I can't count - It's 6 relays if SPDT - still becomes 3 with DPDT **

If S1 cannot be DT then add a relay.
If S2 can be DPDT then subtract two relays (or 1 with DPDT relays)
If S3 an be 4PDT then subtract 4 relays (or 2 with DPDT relays) - ie: no relays at all assuming S2 is DPDT

Dave
ebastler:

--- Quote from: Benta on February 28, 2019, 01:37:13 pm ---First you do a 2:4 demux using two SPDT switches. One switch connected to L, the other to N. Four lamps are connected between the four outputs from the switches in the four combinations possible. Now you can light each bulb independently.

--- End quote ---

Maybe I am misunderstanding, but I think this first stage suffers from the same fallacy as bsudbrink's design: In addition to the desired current path through one lamp, there will be an additional path through two three other lamps connected in series.

I ran out of room on my "DaveCAD" sticky note, hence have not looked at the second stage yet.  ;)

Edit: Just saw your post pop up and disppear again. Your diagram is the same as what I have on my sticky note, and I guess you did spot the unwanted path after posting.  ;)
ebastler:

--- Quote from: DDunfield on February 28, 2019, 01:53:19 pm ---If SPDT relays, then 5 relays - if you can use DPDT then it becomes three.

--- End quote ---

Hmm, your diagram seems to show six SPDT relays?
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