The simplest analog circuit I can come up with is this:

V1 represents the temperature sensor output value.
The values of R1 and R3 may need to be adjusted specially if you use other transistors (you can use variable resistors to find out the correct values then replace them with fixed value resistors).
You may also need a voltage divider on the base terminal of Q1.
This is the output signal analysis.

This assumes you are using a common anode RGB LED.
The output of voltage of the transformer will be proportional to the input voltage. If it is 15 VAC at 220 VAC in, then it will be 13.6 VAC at 200 VAC in, for example.
You seem to have a lot of headroom, so should be able to tolerate significant power voltage sag. In fact, you seem to have so much that your supply is quite inefficient. At 15 V out, the peaks of the waves will be 21.2 V. Assuming a full wave diode bridge, you loose about 1.4 V leaving 19.8 at the peaks. This is the level that the capacitor after the full wave bridge will be charged to twice per line cycle, which is every 10 ms for 50 Hz power.
The 7805 regulator needs about 7.5 V in to make reliable 5 V out. There is a lot of room between the 7.5 V minimum the regulator needs and the 19.8 V peaks the cap gets charged to. The cap voltage will drop between the peaks depending on the cap size and the current. Usually things are sized for a few volts drop. For example, let's say you have a 1 mF cap and the current draw is 100 mA. That would only sag 1 V over 10 mS, which brings the lowest voltage into the 7805 down to 18.8 V.
We can work this backwards and see what the minimum necessary line voltage is in this case. You need 7.5 V into the regulator minimum. The sag between peaks is still the same because it is only a function of the current draw and the capacitance. That means the cap needs to be charged to 8.5 V at the peaks, which requires 9.9 V AC peaks before the full wave bridge, which is 7 V RMS. That is 47% of the output at 220 V, so your minimum input limit is 103 V.
Of course you should always leave some margin because stuff happens, but any vaguely reasonable "220 V power" line isn't going to sag as low as 103 V unless something is very wrong.
As for another solution, get a "universal input" power supply. These are designed to work with any of the worldwide house power. They are usually specified for something like 90-260 V AC, 50 or 60 Hz. They are switchers, so are a lot more efficient than your linear regulator. This is really a better answer anyway than a big iron line power transformer type power supply.
Best Answer
At least some (I don't know about all of them) AVRs have a temperature sensor on-chip, which is connected to the ADC.
A number of PIC devices also have an internal temperature indicator, like the PIC16F15xx and 16F18xx.
Further reading
AVR122: Calibration of the AVR's internal temperature reference, Atmel application note,
Using the AVR internal temperature sensor,
PIC16F15xx datasheet, p.133
AN1333: Use and Calibration of the Internal Temperature Indicator, Microchip application note,