Simple Circuit Provides +5V Gate Bias from -48V Input
A small and simple circuit (Figure 1) derives +5V from the -48V rail widely used in telecom applications. Useful for gate bias and other purposes, the 5V supply delivers up to 5mA. A shunt reference (U1) defines -5V as ground reference for a charge pump (U2), and the charge pump doubles this 5V difference (between system ground and charge pump ground) to produce +5V with respect to the system ground.
Figure 1. This small circuit (six components) produces 5V at 5mA from a -48V input.
The shunt reference maintains 5V across its terminals by regulating its own current (IS), which in turn is determined by the value of R. Current through R (IR) is fairly constant, and varies only with the input voltage. IR, the sum of the charge-pump and shunt-reference currents (IR = ICP + IS), has maximum and minimum values set by the shunt reference.
The shunt reference sinks up to 15mA, and requires 60µA minimum to maintain regulation. Maximum IR is determined by the maximum input voltage. To prevent excessive current in the shunt reference with no load on the charge-pump output, use the maximum input voltage (-48V -10% = -52.8V) when calculating the minimum value of R. The maximum reference sink current (15mA) plus the charge pump's no-load operating current (230µA) equals the maximum IR value (15.23mA). Thus,
RMIN = (VIN(MAX) - VREF)/IR(MAX) = (52.8V - 5V)/0.01523A = 3.14kΩ.Choose the next-highest standard 1% value, which is 3.16kΩ.
Guaranteed output current for the charge pump is calculated at the minimum line voltage: -48V + 10% = -43.2V. The charge pump's maximum input current is:
ICP = (VIN(MIN) - VREF)/R - ISH(MIN) = (43.2V - 5V)/3.16kΩ - 90µA = 12mA,where 90µA is the minimum recommended operating current for the shunt reference. Assuming 90% efficiency in the charge pump, the output current is
IOUT = (ICP/2) × 0.9 = (12mA/2) × 0.9 = 5.4mA.Charge-pump input current is halved, because output voltage is twice the input voltage. Power is dissipated via the shunt reference under no-load conditions, so be sure that R can handle the resulting wattage. A 1W resistor suffices in this case.