No, you don't need supression caps or varistor/thermistors .....

Depending on the size of those capacitors after the bridge rectifier, you will get a relatively large initial pulse of current, could be up to 5-10 A. So your 200mA better be slow blow/delay fuse or you'll go through a few fuses.

so 24v rms ... after it's rectified it's 33.9v ... but don't forget you have voltage drops on the diodes, about 0.4-1v on each, so take out 0.8-2v ...and you're left with about 31v. On low loads like I said the transformer will output more than 24v rms but at high load, it's gonna be more accurate, close to like this 24v ac rms.

The capacitor after the bridge rectifier should be rated for about 50v because 31-33v is awfully close to 35v without even considering that 10-15% extra at low loads.

Capacitance wise I would guess about 820-1500 uF to have a low voltage ripple of a few volts .... approximate formula is :

capacitance = current (1.2a) x duty cycle (about 0.7) / [2 x V ripple (how much you find acceptable to let the voltage sag) x ac frequency (50hz for eu, 60 hz for us) ]

assuming max 33.8v and you don't want it to drop below 30v then => c = (1.2 x 0.7) / (2 *3.8* 60) = 0.84/456 = 0.0018 F = ~ 1800 uF

^ that's at the peak of 1.2a, if you'll only pull 100-500mA naturally the voltage will not sag down to 30v

so realistically with about 1200-1500uF 50v (cheap capacitors) at 1.2a maximum output, you'll have about 28v before the linear regulators. Double or triple that (put capacitors in parallel or use 3300uF-4700uF 50 capacitor to get as close as possible to 33.8v you want.

Next, the LM317 and LM337 will have a voltage drop on them of about 2-2.5v at 1.2A so you'll get at most the input voltage - that 2-2.5v drop.

You

**WILL NOT** get 1.2v at 1.2A, not even 5v at 1.2A with this regulator.

The difference in voltage is dissipated by the regulators as heat - ( 30v input voltage - 1.2 output voltage ) x 1.2v = 28.8v x 1.2a = 34.5 watts. You'd need a monster of a heatsink to dissipate that much (or a cpu cooler with a fan blowing on it to keep it cold).

If you check the lm317 and lm337 datasheets (should be the same) - you'll find that the maximum power dissipation is limited to about 15w for the classic 3 pin to-220 regulator and maybe about 25w for that to-3 (large round metal case) IC.

So if you want 5v with 30v input ... that's 15w / (30-5) = 15/25 = maximum 0.6A on 5v with the lm317 in to-220 package... and the heatsink will be super hot.

Even so, you'll need a quite expensive heatsink (with about 1.2-1.5c/w rating or less) if you want this to work without fans, and maybe 2-3c/w heatsink if you'll constantly blow air on it. \

See

http://www.ef-uk.net/data/heatsinking.htm (the explanations section at the bottom explains how to do the math manually based on data from datasheet of the linear regulator)

Linear regulators are just not designed to work with such big voltage differences, traditionally the voltage differences are at most 5-10v, not up to 30v. Linear power supplies use multiple taps, like 0-8-16-24-32v and have some circuitry to connect the next winding when needed so that at any point, only about 5-8v are dropped on regulator.