Help with simple circuit please

--- 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

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

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:

That's 83nF.

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

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.