The easy answer is here, in TI's literature:
http://focus.ti.com/lit/ml/slvr259/slvr259.pdf
See the bottom of page 1 under "CURRENT LIMITATIONS
AND CAPABILITIES:" It states: "The TPS61020 is capable
of supplyinmg about 350 mA of I/O current with an input
voltage of 1.8 V."
The answer that requires a little more effort is understanding
the information in the datasheets. I'll try to explain
what you can find there regarding output current and how
you can use the information.
First, read the description on the first page for both
the 61030 family and the 61020 family. The 30 family
description states "... is capable of delivering output
currents of up to 1A at 5V at a supply voltage down to
1.8 V". The 20 family description states "Output
currents can go as high as 200 mA while using a single
cell alkaline, and discharge it down to 0.9 V. It can
also be used for generating 5 V at 500 mA from a 3.3-V
rail or a Li-Ion battery." If you interpolate the numbers
shown, with 2 V input you get about 350 mA out using
the TPS61020.
There's more to be found when you look into the circuit
and the datasheet numbers. You need to consider worst case
(battery voltage at ~ .9 v per cell) and use that in your
design, and you need to understand the whole datasheet.
Look at figure 1 (page 7) on the datasheet. It shows the
maximum output current vs the input voltage. Note that with
lower input voltage you get lower max current, and that at
1.8 volts input, it is no where near 1 amp - you can get a
maximum of about 400 mA. You want to allow a little extra,
so that's about 350 mA.
Another interesting specification is efficiency. The 61030
is over 80% efficient at 1.8 volts input with a 1 amp
load, while the 61020 drops below 50 % at that voltage,
even if the load was only about 350 mA. That means that
over 50 percent of the energy taken out of the battery
becomes heat. In your circuit, it is even worse, because
you are boosting the voltage, while the curve is using
Vin = Vout. It's no wonder your 61020 is getting hot!
Ed
