If you're stuck using linear regulators, you can put big power resistors in series with the inputs of the regulators to drop the voltage and share some of the heat dissipation. You're dissipating the same amount of power either way, but it might make it easier to fit on your board or whatever.
If you're using an LM7833, for instance, and supplying 150 mA, the datasheet says the dropout is 2 V, so the input voltage has to always be above 5.3 V. From a 14 V supply at 150 mA, this is a 2 W, 56 Ω resistor. The resistor just needs air circulation around it, not a heatsink, and then your regulator only needs a 500 mW heatsink instead. The highest power dissipated in the regulator will be at the current when resistor and regulator are both dissipating the same power, which in this case is about 95 mA.
Use full wave rectification, not half wave. HW uses transformer poorly, may not be good or TEC, has no obvious advantages except the cost of 3 diodes.
If you want to operate it at full power with no control of cooling level then 12V is fine. LM350 regulator needs about 3V headroom. So 12V out from regulator = 15VDC in min.
Full wave rectified transformer will give ABOUT Vmax DC ~= 1.4 x AC voltage.
Or VAC_min ~= (Vdc + dropout) / 1.4
So 12V + 3V = 15V
VACmin ~~~~= (12 + 3) / 1.4 =~ 11 VAC if ~= NO ripple voltage
ie 12 VAC transformer will give 12 VDC after regulator if well smoothed.
More is better, so maybe 14 VAC - 15VAC will allow regulator headroom plus some ripple allowance.
If all you want is to run it at full power then a transformer, bridge rectifier, smoothing capacitor and series resistor are all that is needed. Resistor drops excess voltage. A 10 VAC transformer will be about enough (1-VAC x 1.4 = 14 V with smoothing and ripple) and 12 VAC will definitely be enough. Resistor dissipates 2.5 Watts per volt of drop. A length of Nichrome wire adjusted to provide correct voltage to Peltier is one option or 0.4 Ohms per volt of drop select-on-test.
Regulator WILL need heat sink - how much depends on transformer. At say 3V regulator drop - about the minimum you should figure on, the dissipation V x I = 3V x 2.5A = 7.5W. Say allow 10 Watts. More if transformer is of excessive Voltage.
Heatsink can be selected using degree C (or K) per Watt for commercial heatsinks. For a 10C rise at 10 Watts you need 1 C/W heatsink which is "very large indeed".
If you want heatsink at almost cool enough to touch (almost) say 60 C the if ambient = 30 C worst case heasting delta T = 60-30 = 30C so heatsink = 30C/10W = 3 C/W
Even that is largish. Going other way, 10C/W is common so 10 W = 10W x 10 C/w rise = 100C./ Add ambient + Tsink = 100C rise + 30C = 130 C.
You rally don't want 130 C heatsinks.
so somewhere between 3 C/w and 10 C/W leaning towards 3 C/W end.
Fans and Peltier together are OK. Fan load is small compared to Peltier load.
Fans could run from smoothed DC before regulator - maybe with a dropping resistor of their own suited to VFan an Vdc.
Strip board construction OK but keep wires short and heavy. If running higher currents along a piece of stripboard you can solder wire to the strip for longer high current leads or use a wire link from points to be joined.
Main mains caution is DON'T PLAY WITH MAINS whn not needed. eg here all the circuitry is LV apart from mains feed to transformer primary. Do the primary side wiring well. Insulate as required. Then leave it alone.
Best Answer
OK. You've got 3 V when new.
Yellow warning lamp! As far as I remember the original 555 timers draw a large current at the instant of output change. Consider the use of the CMOS version.
40xx is CMOS not TTL. CMOS should work OK but check the datasheets for minimum working voltage. TTL works in a voltage range around 5 V.
Two things to watch for:
Sounds good.
Hook it up directly. You won't over-volt and any kind of regulator will cost power.