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.
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| 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.
