-- You shouldn't have to do any of that if you've got your newsreader configured to display incoming posts in a fixed pitch font. There's also a problem in that your newsreader isn't quoting properly, making it impossible to reply to you without some really inconvenient machinations on my end.
-- Thanks. :-) BTW, since it seems you want to build this thing, I emailed you a PDF schematic of what should work in the real world. For some reason it seems my photodump account is broken, so I also posted it to abse in case anybody else is interested.
-- Thank you.
-- Thanks. :-) BTW, since it seems you want to build this thing, I emailed you a PDF schematic of what should work in the real world. For some reason it seems my photodump account is broken, so I also posted it to abse in case anybody else is interested.
-- Thanks. :-) BTW, since it seems you want to build this thing, I emailed you a PDF schematic of what should work in the real world. For some reason it seems my photodump account is broken, so I also posted it to abse in case anybody else is interested.
--- Vcc, actually, but yes.
With a 50mA load, the 7555 drops about 2V across its output, so if you choose a relay with a coil current less or equal to that, a 7555 should work OK.
A perfect fit would be an OMRON G5V-1-DC12:
Farnell has them in stock, and don't forget to put a diode (a
1N914/1N4148 will do) across the coil, with the cathode connected to the pin going to U7-3.Just as an aside, I think you're going to have a tough time sourcing K1 if it's got to have a 240VAC coil and signal level contacts, so one solution would be to use a lower voltage DC coil relay and drive it from a full-wave bridge fed by a capacitor in series with the mains, like this: (Don't forget the fixed-pitch font ;)
MAINS>--[CAP]--+------------+ |K |A [DIODE] [DIODE] | | +---[COIL]---+ | | [DIODE] [DIODE] |K |A MAINS>---------+------------+
If you were to use the 24V coil version of the 12V OMRON relay, that would be the G5V-1-DC24, which has a 3840 ohm coil and a coil current of 6.25mA with 24V across the coil.
Sooo...
Here's the fun part; how do we figure out the value of the cap?
Since the diode drops are insignificant, we can look at the circuit like this:
MAINS>-----+--->240V/50Hz | [CAP] | +--->34VP | [COIL] | MAINS>-----+
Since the coil is seeing full-wave rectified AC instead of pure DC, its RMS value must be equal to 24VDC, which means its peak value is
24VDC * sqrt2 ~ 34VP.We also know that the coil will be drawing 6.25mA so, since in a series circuit the current is everywhere the same, that 6.25mA must also pass through the cap.
For that to happen, the impedance of the entire circuit must be:
E 240V Z = --- = ---------- = 38400 ohms. I 0.00625A
Now, since:
Z² = X² + R²
and we want to solve for the capacitance, we first need to get the capacitive reactance, so we rearrange and solve for X:
Xc = sqrt (Z² - R²) ~ 38208 ohms.
Then, to get the capacitance:
1 1 C = ------------ = -------------------------- ~ 8.3E-8F 2pi f (Xc) 6.28 * 50Hz * 38208 ohmsThat's 83nF.
82nF is the closest standard value, and Panasonic has a nice part, ECQ-E6683JF:Since the proof is in the pudding:
Version 4 SHEET 1 880 680 WIRE -32 -16 -224 -16 WIRE 112 -16 -32 -16 WIRE -224 32 -224 -16 WIRE -32 32 -32 -16 WIRE 112 32 112 -16 WIRE -32 160 -32 96 WIRE 0 160 -32 160 WIRE 112 160 112 96 WIRE 112 160 80 160 WIRE -224 208 -224 96 WIRE -32 224 -32 160 WIRE 112 224 112 160 WIRE -224 352 -224 288 WIRE -32 352 -32 288 WIRE -32 352 -224 352 WIRE 112 352 112 288 WIRE 112 352 -32 352 WIRE -224 400 -224 352 FLAG -224 400 0 SYMBOL res 96 144 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R1 SYMATTR Value 3840 SYMATTR SpiceLine "" SYMBOL diode -48 224 R0 WINDOW 0 -43 33 Left 2 WINDOW 3 -75 65 Left 2 SYMATTR InstName D3 SYMATTR Value 1N4148 SYMBOL diode 96 32 R0 WINDOW 0 46 34 Left 2 SYMATTR InstName D2 SYMATTR Value 1N4148 SYMBOL diode 128 288 R180 WINDOW 0 -48 33 Left 2 WINDOW 3 -77 0 Left 2 SYMATTR InstName D4 SYMATTR Value 1N4148 SYMBOL diode -16 96 R180 WINDOW 0 50 33 Left 2 WINDOW 3 24 0 Left 2 SYMATTR InstName D1 SYMATTR Value 1N4148 SYMBOL voltage -224 192 R0 WINDOW 3 24 96 Invisible 2 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 SYMATTR InstName V1 SYMATTR Value SINE(0 340 50) SYMATTR Value2 "" SYMATTR SpiceLine "" SYMBOL cap -240 32 R0 SYMATTR InstName C1 SYMATTR Value 82n SYMATTR SpiceLine "" TEXT -210 376 Left 2 !.tran .2
Pay no attention to the ground connection, it's not needed (or desired) for your application, but LTspice needs it for a 0 volt reference.
-- JF
-- Good catch! It should be a CD4081B and all the other logic should carry the 'B' suffix as well.
--
Oops...
I forgot about the 24V relay's inductance, which is about 6.6 henrys
with the armature down.
So instead of:
.MAINS>--[CAP]--+-----------+
. |K |A
. [DIODE] [DIODE]
. | |
. +--[3840R]--+
. | |
. [DIODE] [DIODE]
. |K |A
.MAINS>---------+-----------+
we have:
.MAINS>--[CAP]--+--------------------+
. |K |A
. [DIODE] [DIODE]
. | |
. +--[3840R]--[6.6H]--+
. | |
. [DIODE] [DIODE]
. |K |A
.MAINS>---------+-------------------+
and here's the reworked math:
Xl = 2pi f L = 6.26 * 50Hz * 6.6H = 2072 ohms
E 240V
Z = --- = --------- = 38400 ohms
I 0.00625A
Z² = R² + (Xl - Xc)²
Z² - R² = (Xl - Xc)²
38400² - 3840² = (2072 - Xc)²
sqrt(38400² - 3840²) = sqrt(2072 - Xc)²
38208 = sqrt(2072 - Xc)²
38208 + 2072 = 40280
38208 = sqrt(2072 - 40280)²
\
38208 = 38208 Xc
1 1
C = ---------- = -------------------------- 7.8E-8F = 78nF
2pi f Xc 6.28 * 50Hz * 40280 ohms
Closest E24 value is 75nF, and I'd use a Vishay BFC238332753:
http://www.vishay.com/docs/28124/mmkp383.pdf
OMRON reports that if their G5V-1 relays are operated with the coil energized for a long time and no switching is done, the heat from the coil affects the insulation and causes a film to be deposited on the contacts which can lead to problems.
Solution? Ditch the relay.
New schematic at abse.
-- JF
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