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10-14-2010

# Power-supply circuit operates from USB port

Stefano Palazzolo, Senago, Italy;
Edited by Martin Rowe and Fran Granville – EDN

Get 1.25V to 3.75V from an adjustable regulator

Every PC has a USB (Universal Serial Bus) port that can supply 5V±5% at 500 mA for peripherals. Powered USB hubs also provide this power. You can use a USB port to power an external circuit, which is useful when you have no other dc source available.

A USB port has VBUS, the power pin; a return pin, GND (ground); and two signal pins. If you need just a simple 5V supply, you can tap the power pins from a USB connector, but you should place a 10-μF filter capacitor between the ground and power-supply pins.

 Figure 1. Resistors R2 and R3 set the adjustable voltage regulator’s output at 1.25 to 3.75V.

You can, however, use an adjustable voltage regulator to get voltages of 1.25 to 3.75V, a range that many circuits use. The circuit in Figure 1 covers that range. You use R3 to change that range, as the following equation shows:

VOUT = 1.25V × (1 + R3/R2)

The 1.25V in the equation occurs because the LM1117-ADJ linear regulator generates 1.25V between the VOUT and the ADJ (adjust) pins. Resistor R2, therefore, has a constant current that passes through resistor R3; the IADJ (adjusted current) is generally small enough to ignore. Selecting 100Ω for R2 sets its current to 12.5 mA. If you use a 200Ω potentiometer for R3, you get a voltage range of 1.25V when R3 is 0Ω, causing a short, to 3.75V when R3 is 200Ω.

To prevent circuit damage if the output becomes shorted or when you don’t know the load, you can add a current-limiting circuit that keeps the maximum current at 500 mA. A polyswitch fuse or pair of transistors can easily implement this current-limiter site at the power-supply input line.

The filter capacitor shouldn’t exceed 10 μF. That level keeps the inrush current under control in the absence of a current-limiting circuit. Generally, capacitors of 1 to 10 μF work best.