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I have an Arduino providing +5 V and GND. I have a component that requires +5 V, common GND, and -5 V connections. Is it possible to generate -5 V? If not, what input voltage do I need?
I am driving a bias controller for an RF switch PIN diode. I believe typically the largest current drawn is around ±25 mA.
There are various approaches to creating a negative supply rail from only a positive one. Typically this will involve some form of DC-DC converter.
One option would be to use an isolated DC-DC converter module with the positive output connected to GND, resulting in a negative supply rail.
There are also non-isolated DC-DC converters capable of generating a negative supply rail from a positive one - known as "inverting DC/DC converters".
For an inductorless approach, typically for low current applications, charge pumps can also be used to invert the supply voltage, though the negative supply generated will be lower magnitude than the input supply (e.g., you might get -4 V from a +5 V supply, but are unlikely to be able to get a -5 V from a +5 V supply), so you may need a supply of say more than 6 V for this to work.
These are all switching power supplies, so they will inherently generate more noise than a linear supply. For noise sensitive applications, it's usually advisable to generate a voltage a few hundred mV higher than required, and then follow up with a low noise LDO (low-drop-out) linear regulator to reduce noise levels.
Here is a simple inverting charge pump that generates about -800 uA into a 5k load, using a 5V 20 kHz square wave from a typical GPIO pin. You might be able to drive 8 GPIO outputs in parallel to get about -6.5 mA. With lighter load, you can get closer to -5V, but about -4.5V is about the best you can do. Adding a buffer can provide higher current.
Check out charge pumps for low current, some ripple and noise.
Buck–boost converters DC/DC (actually, buck as buck boost can do) for higher current, and still some ripple.
A combination of buck-boost and negative LDO for a clean voltage.
An easy option is always just an isolated power supply. Expensive, but basically one part.
believe typically the largest current drawn is around ±25 mA.(That's absolute maximum.) You get to choose the current for the isolation state, the datasheet suggests 10 mA. Low loss state is from -2 V, larger bias allowing larger signals. I don't see any advantage to regulation here and slim for 5 V instead of 4. Only a leakage current should be necessary in steady low loss state - nano-amperes up to breakdown. The datasheet mentions 150 to 220 Ω parallel 390 - 560 pF to achieve 10 ns attenuation rise&fall time. \$\endgroup\$