A small and simple circuit derives 5V from the –48V rail that telecomm applications typically use (Figure 1). Useful for gate bias and other purposes, the 5V supply delivers as much as 5-mA output current. A shunt reference, IC₁, defines –5V as ground reference for a charge pump, IC₂. 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. I_S is a function of the value of R. The current through R, I_R, is reasonably 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 as much as 15 mA and requires 60 µA minimum to maintain regulation. Maximum I_R is a function of 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) to calculate the minimum value of R. The maximum reference sink current, 15 mA, plus the charge pump's no-load operating current, 230 µA, equals the maximum I_R value, 15.23 mA. Thus, R_MIN=(V_IN(MAX)–V_REF)/I_R(MAX)=3.14 kΩ.

Choose the next-highest standard 1% value, which is 3.16 kΩ. You calculate the guaranteed output current for the charge pump 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)=(43.2–5)/3.16–90 µA=12 mA, where 90 µA 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=(12/2)×0.9=5.4 mA. You halve the charge-pump current, because the output voltage is twice the input voltage. Be sure that R can handle the wattage under no-load conditions. A 1W resistor suffices in this example.

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