Assuming it's the circuit from post #15, then the TL431 is directly across the supply with just a Vbe drop of the transistor in series with it.
Could probably do with a resistor from the TL to the base?.
It definitely isn't wired wrong.
It gets hot when the tip30 and the 47r 5w resistor is drawing half an amp.
Steve
Yes, at a panel current of 0.5A, the TL431 will get quite hot to the touch. I considered its power dissipation with the orginal panel current of 300mA when I ran the simulation. At higher panel currents, the problem is that the Hfe of the TIP30 at Ic=500mA is only 30. Making Q1 a Darlington, or using a PNP with a higher Hfe would reduce the power dissipation in the TL431 at higher currents.
If the Emitter current of Q1 is 500mA (the entire panel current because the battery is fully-charged), the current out of the base of Q1 is Ic/Hfe or about 500/30 = 17mA. At the same time, the current through R3 is Vbe/R3 = 0.7/680 = 1mA, so the total current into the cathode of the TL431 is ~18mA.
The voltage from the TL431 cathode to the TL431 anode is the battery voltage - Vbe = (14.2-0.7) , so the power dissipation in the TL431 is I*E = 18mA*(14.2-0.7) = 243mW. The thermal resistance of the TO-92 package from junction to ambient is 140degC/W, so the temperature rise of the TO-92 package would be 140*0.243 = 34degC, which will not burn it up. It is still well within its maximum operating temperature. The circuit as-is is fine up to the original 300mA Panel current, begins getting a bit marginal at 1/2A.
I am not in love with Nigel's suggestion, because putting a resistor between Q1 base and TL431 cathode will create a voltage divider with R3, thereby reducing the shunt regulator's loop-gain by the divider ratio. This would flatten out the nice sharp knee that this circuit has.
To reduce the power dissipation in the TL431 at shunt currents of >0.3A, I would do one of the following:
1. Replace the TIP30/32 with a TIP142 PNP Darlington. It has a Hfe of 1000. That reduces the TL431 cathode current to 2.5mA, which reduces the dissipation to (14.2-1.4)*2.5m = 32mW, which makes the temperature rise 140*o.032 = 4.8degC.
2. Make a homebrew Darlington by adding small PNP ahead of the TIP30/32 base. Results similar to above.
3. Use a PFET in place of the TIP30/32. Now the TL431 cathode current is determined only by the threshold voltage of the PFET divided by R3, so about 3/680 = 4.4mA, so the dissipation is (14.2-3)*4.4m = 50mW, so the temp rise is 140*0.05 = 7degC.