For determining the maximum current, this graph of a typical BSP129 gives guidance:
Bear in mind this is a typical plot and individual BSP129s may deviate significantly from it. The VGS=0 line indicates the maximum current available, which is about 130mA. To get for example 100mA the VGS is about -0.2V so divide those numbers (volts/current) to obtain the resistor value at the source terminal, which works out at 2R.
A single BSS129 won't handle over 130mA but the current source circuit (FET plus resistor) can be paralleled to go beyond the limits of a single transistor.
Bear in mind this is a typical plot and individual BSP129s may deviate significantly from it. The VGS=0 line indicates the maximum current available, which is about 130mA. To get for example 100mA the VGS is about -0.2V so divide those numbers (volts/current) to obtain the resistor value at the source terminal, which works out at 2R.
A single BSS129 won't handle over 130mA but the current source circuit (FET plus resistor) can be paralleled to go beyond the limits of a single transistor.
Thank you abraxalito. So connecting gate straight to source will max out the regulator's current at around 130mA.
Am I correct in assuming that we limit the current just to avoid unnecessary waste (heat and power) - or is there a performance downside to running the reg hot?
Am I correct in assuming that we limit the current just to avoid unnecessary waste (heat and power) - or is there a performance downside to running the reg hot?
If you don't limit the current in some way, a shunt reg can't work. So the current source keeps a shunt reg a practical solution and normally we attempt to minimize waste heat by setting the current source a safe margin above the current consumption of the load. There's no performance downside for running hot other than the excess heat and its consequences for reliability, there may even be a performance upside (lower output impedance) from running hotter.