# What is the function of LM317T?

## What is the function of LM317T?

The LM317T is an adjustable 3-terminal positive voltage regulator capable of supplying different DC voltage outputs other than the fixed voltage power supply of +5 or +12 volts, or as a variable output voltage from a few volts up to some maximum value all with currents of about 1.5 amperes.

## What is LM317T voltage regulator?

The LM317 is an adjustable 3−terminal positive voltage regulator capable of supplying in excess of 1.5 A over an output voltage range of 1.2 V to 37 V. This voltage regulator is exceptionally easy to use and requires only two external resistors to set the output voltage.

Is LM317 a transistor?

The LM317 is an adjustable three-terminal positive-voltage regulator capable of supplying more than 1.5 A over an output-voltage range of 1.25 V to 32 V. By using a heat-sinked pass transistor such as a 2N3055 (Q1) we can produce several amps of current far above the 1.5 amps of the LM317.

What are the features of LM317?

Features

• Adjustable 3-terminal positive voltage regulator.
• Output voltage can be set to range from 1.25V to 37V.
• Output current is 1.5A.
• Maximum Input to output voltage difference is 40V, recommended 15V.
• Maximum output current when voltage difference is 15V is 2.2A.
• Operating junction temperature is 125°C.

### What does LM stands for in LM317?

Maximum output current. 1.5. A. IL(MIN) Minimum load current.

### What is the drop out voltage of a LM317?

When used as a voltage regulator, the LM317 will drop a minimum of 1.7v across itself (Vin-Vout >=1.7v); this is known as the “dropout voltage”.

What is LM317 module?

The LM317 DC-DC Adjustable Voltage Regulator Power Supply Module is an adjustable 3-terminal positive voltage regulator. It is capable of delivering more than 1.5 Amp current over an output voltage range of 1.25 to 37 Volts. The LM317 requires only 2 resistors to set the output voltage.

How do you calculate the output voltage of LM317?

Using the formula for the output voltage, VOUT= 1.25V (1 + R2/R1). Being that R1=240Ω, our equation is now 5V= 1.25V (1 + R2/240Ω), so R2=720Ω. So with R2 being a value of 720Ω, the LM317 will output 5V if fed an input voltage greater than 5 volts.

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