# Simple Circuit Provides +5V Gate Bias from -48V Input

### Abstract

In this design idea a small circuit with six components derives 5V gate bias from the -48V rail widely used in telecom applications. The MAX6138 shunt voltage reference and the MAX1683 charge pump are featured in the design.

A similar version of this article appeared in the September 19, 2002 issue of *EDN *magazine.

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.

The shunt reference maintains 5V across its terminals by regulating its own current (I_{S}), which in turn is determined by the value of R. Current through R (I_{R}) is fairly constant, and varies only with the input voltage. I_{R}, the sum of the charge-pump and shunt-reference currents (I_{R} = I_{CP} + I_{S}), 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 I_{R} 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 I_{R} value (15.23mA). Thus,

R_{MIN} = (V_{IN(MAX)} - V_{REF})/I_{R(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:

I_{CP} = (V_{IN(MIN)} - V_{REF})/R - I_{SH(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

I_{OUT} = (I_{CP}/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.