Power-supply circuit operates from USB port
Get 1.25V to 3.75V from an adjustable regulator.
Stefano Palazzolo, Senago, Italy; Edited by Martin Rowe and Fran Granville -- EDN, September 23, 2010
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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.
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.| Read More Design Ideas |
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.
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