LT3080 regulator question
Yes, you're going to have to generate a negative voltage somehow.
The minimum load could be as simple as a resistor that draws at least 1mA when the output is at zero volts. However, this might be problematic at higher output voltages, because the resistor would dissipate more power and the negative supply would have to deal with the additional current through the resistor. It might make sense to use a transistor to create an active current sink.


If you're using just a resistor, connect it between the output of the LT3080 and the negative supply. Suppose your negative supply is -5V; the value of the resistor would be 5V/1mA = 5000&Omega;. The standard 5% value of 4700&Omega; would give you a little extra margin: 5V/4700&Omega; = 1.06mA.

If you'd like to add an NPN transistor to this, connect the collector to the LTC3080 output, connect the base to ground and connect the resistor between the emitter and the negative supply. The only critical parameter of the transistor is that it needs to be able to handle the maximum voltage that the LT3080 will be producing (Vce(max)). Now, when you calculate the resistor value, you need to allow for the Vbe drop of the transistor. Again, using -5V as the negative supply, the resistor value is now (5V - 0.7V)/1mA = 4300&Omega;. This happens to be a standard 5% value ... or you could give yourself some margin by using 3900&Omega;: 4.3V/3900&Omega; = 1.10mA.