So lets ignore the hows and whys about the IC etc and assume that as long as the correct current is flowing through the second zener/resistor leg (the 8.2v one) and that the zener is not overloaded by increasing the first zener, I'll be good.
Yes, good.
A) I'm going to replace the relay with a new 24v one.
B) I'll increase the zener which will also mean increasing the resistor in series with the second zener and checking the wattage of the first zener remembering it's only half wave. I am just unsure how much to increase it by or how much to increase the 1u cap by.
The behaviour of the power input circuit is determined by a combination of factors. The mains voltage and frequency can't be changed. Other factors are the input coupling capacitor (the 1 uF one), the zener voltage, the smoothing capacitance, and the load current.
You aren't going to change the load current. It's roughly 40 mA. That's 20 mA for the relay coil, correct? Plus 12 mA for the IC circuitry and 7 mA for the "ON" LED.
Assuming the smoothing capacitor is 220 uF and the period between top-ups from the mains is 20 ms (50 Hz mains, half-wave rectification), the ripple can be calculated from dV / dT = I / C which rearranges to dV = dT I / C which is 0.02 * 0.04 / 0.00022 which is 3.6V peak to peak. That's 15% of the nominal voltage. This would normally be way too much but it's reasonable in this application. There may not physically be room to increase the smoothing capacitor and you rapidly get into diminishing returns. So let's assume you stay with 220 uF.
Since you need a mean rail voltage of about 26V after the drop from the 1N4007, and ripple is about +/- 1.8V, you should be using a zener of 27V or 30V. I would buy one of each and start with the 27V one, measure the DC voltage across the relay coil while the relay is ON, and if it's less than, say, 23V, switch to a 30V zener.
Edit: The following paragraph is wrong. I'm not clear on where I've made the mistake; if anyone can explain it properly, please do. In a later post I show the results of simulation, where it's clear that the input capacitor needs to be at least 2.2 uF.
The last item is the input capacitor. Capacitive reactance (which works like resistance in this application) is calculated as Xc = 1 / (2 pi f C) which is 1 / (6.28 * 50 * 1e-6) which is about 3200 ohms for the original 1 uF value. At 230V RMS, current will be about 72 mA. This is not exact because on positive cycles the capacitor doesn't see the full mains voltage across it, because the power supply is taking roughly 24V for itself, but it's close enough. Half of this current is wasted because of the half-wave rectification, leaving about 36 mA available for the circuit. We want to increase that to 40 mA plus some left over for safety margin and current in the main zener. I would aim for 45~50 mA or an increase of 25~39%. So the capacitor should be between 1.25 and 1.39 uF. So my initial guesses of 1.2 uF or 1.5 uF are probably about right. I will assume you choose 1.5 uF so the current from the capacitor on positive half-cycles is about 54 mA.
Edit: Does what I said in that paragraph, "half of this [72 mA] current is wasted because of the half-wave rectification, leaving about 36 mA available for the circuit", make sense? I don't think it does. (*steve*) perhaps you could comment...? (I will PM Steve and ask him to look at this thread.)
Edit2: I think what I said is valid.
Edit3: There is definitely something wrong with my calculations and my conclusion that the capacitor should be between 1.25 and 1.39 uF. See my later post with the results of simulation, where it's clear that the input capacitor needs to be at least 2.2 uF.
The last parameter is the zener power dissipation. Worst case will be when the relay is not activated, and the 20 mA that it would draw has to be absorbed by the zener. The zener dissipates practically nothing during negative half-cycles of the mains voltage, because at that time, the forward voltage of the zener is only ~0.7V. On positive half-cycles the incoming current will be about 54 mA and the load current will be about 20 mA so the zener will be absorbing about 34 mA half the time. If the zener voltage is 30V (the worst case) this is a power dissipation of about 1W half the time. So you should use a zener rated at 1W at least. A larger part (1.3W or 2W for example) will give a better safety margin and will have better heat dissipation.
Since these changes will beef up the power supply significantly, I would leave the 1k5 series resistor to the IC circuit unchanged.
My sim does not have zeners over 25v sadly so I can't do it myself *sigh* otherwise that would be easy
Can you connect two zeners in series?
I would be interested to see the results of a simulation with those changed component values too.