I can give you some pointers on the battery side of things, but can't help you with respect with the design.
The MRSM201A has a max current consumption of 12ua, with a voltage of between 1.6 and 3.5 volts.
The bluetooth module requires 4.8mA, with a voltage of 2.35 to 3.3 volts.
So your battery has to supply 4.8 mA + 0.012 mA = 4.812 mA.
The battery you describe sounds like a 2032 coin cell - 3 volt lithium primary cell with approximately 225 mAh capacity. The 225 mAh capacity is measured at a discharge rate of 0.20 mA, and the cell starts at 3 volts, but when the battery is depleted, it's voltage is only 1.9 - 2.0 volts. The bluetooth requires a minimum of 2.35 volts, so you won't be able to use the full capacity of the battery. However, from this datasheet: http://www.farnell.com/datasheets/1496885.pdf you can see that you will only lose a tiny amount of capacity, not enough to really affect the calculations to a great degree.
So: 225 mAh รท 4.812 mA = 46.75 hours
But, if you look at that datasheet, you can see that at a discharge rate of 0.2 mA, they got approx 1050 hours of discharge or 210 mAh (This battery has a slightly lower capacity than the one you quoted, energizer makes one with 240 mAh, but this manufacturer had the better data sheet). Next they discharged at a rate of 0.769 mA for 250 hours for a capacity of 192 mAh. As you can see, because of the high internal resistance of the coin battery, as you increase the discharge rate, the effective capacity of the battery drops as some of the energy is lost to heat. Your project requires 6 X the discharge rate of the 0.769 mA, so the coin cell will not even last the 46.75 hours. No where near the 6 months. Note that this does not include the current required to light the LED!
AA batteries have approximately 2500 mAh capacity and a much lower internal resistance. They start out at close to 1.6 volts each so you would need to use two in series and they are depleted at approximately 1.0 volts, which is a little too low for the blutooth. You can look up a datasheet for the AA battery and do the same calculations to see if they will give you enough capacity before their voltage drops too low.
Hopefully someone else can help with the resistor and LED component of your circuit. I don't understand how the MRSM201A can drive the LED when an LED typically requires at least 20 mA current and the MRSM201A only draws at max 12 microamps.
Good luck!
I just did this and got a reasonable result. One problem though, Pspice ploted the current and the voltage on the same axis. Since the amplitude of the voltage is 10 and the amplitude of the current is 4mA, the current looks like a flat line. You can either select the current and multiply it by 1000 or plot it on a separate Y axis.
Best Answer
That is the symbol for a relay coil. The relay should be a DC type with a coil voltage equal to the DC power supply.
24 V DC is common in industrial control systems and, since your switch is rated for 300 mA max., the minimum coil resistance can be calculated from \$ R = \frac {V}{I} = \frac {24}{0.300} = 80 \ \Omega \$.
The switch is NPN type which means that there is an NPN transistor internally that will connect the black and blue wires internally when the switch is 'on'. This may also be called a 'switched negative'. The relay '+' terminal is always connected to the positive supply.