Electronic – 5V and 3.3V power supply

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.

You don't want a switcher. Your input-output difference is so low (for most of the discharge curve the voltage is less than 3.8 V) that even at 90 % efficiency it won't do better much than an LDO.

Have a look at the Seiko S1167:

• Ground current typical 9 µA
• Available in 100 mV steps from 1.5 V to 5.5 V
• 1 % output voltage accuracy
• 150 mA output current
• 150 mV drop-out at 100 mA
• Shutdown input
• SOT23-5 package

edit
Found an even better one in the S1313:

• Ground current 0.9 µA
• 200 mA output current

Seiko doesn't give data for ground current under load, but in my experience you should count on 1 % of load current, so at 1 mA that would be around 10 µA, most likely less than 50 µA.