ripple rejection improvement - LM317

[akamilrazak]

lm317 power supply

Hi guys, I want to ask u all about what i am doing right now. as far as i can see all LM317 circuit will never produce a 0 volt minimum voltage if u take the volt from the voltage output pin.. I have try to adjust the output. i take output from the Adjust pin of LM317. so i want to ask if there was any risk doing that. i can get the voltage minimum of 0 volt.. nice to try.. ^^

 

[jblueink]

lm317 power supply circuit

pls can u be more specific about ur question.

 

[pranam77]

power supply lm317

Reading your posting i guess you are trying to bring down the default lower level 1.2 volts of LM317. It is possible with this given project sourced from www.edn.com. Good luck

This is an LM317 based adjustable voltage regulator with a maximum output of 3V and 1.5A. The output voltage depends on the VIN, R1 and R2 values, so the circuit can be modified to use with a maximum output of greater than 3V. The maximum current output is also independent of the circuit design, it is related to the package options.In this circuit, LM317T, which is capable of transferring up to 1.5A, is used.

Since the internal reference voltage of the LM317 regulator is 1.25V, the output voltage can be also minimally 1.25V. One way to overcome this problem is using a reference voltage source built on two diodes. But this approach is mostly suitable for 1.5V to 15V regulators since the sensitivity becomes poor for the low voltage outputs less than 1.5 . On the other hand, diodes have temperature dependent forward voltages.

Another way is using LM185 or AD589 adjustable voltage references. But they are expensive and in this application they require both zero adjustment and matching.

In this circuit, the key idea is solving the problem with an inexpensive and easy way by using a temperature stabilized constant current source. The current source value is calculated by the equation;

I = (VF - VEBO) / (R5 + R6)

Where VF is the LED's forward voltage that is about 2V and VEBO is the Emitter-Base voltage of Q1 transistor which is about 0.67V. As a result the constant current source creates about -1.25V on resistor R3.

R6 sets the current of the current source so it operates as a zero adjustmend resistor. When the resistor R2 is set to zero, adjust R6 to see zero voltage at the output. R5 protects the transistor BCW33. Red LED is used as a light indicator.

R2 adjusts the output of the power supply. LM317 regulator's output voltage can be calculated easily by using the equation;

VOUT = VREF (1 + R2 / R1 )

But in this circuit it changes to;
VOUT = VREF (1 + R2 / R1 ) - VR3

Where VR3 is the voltage on R3 that is equal to the reference voltage of LM317, namely 1.25V. So the equation becomes;
VOUT = VREF (R2 / R1 )

When we adjust the R2 to 1.2K then the output becomes nearly 1.56V which is a typical battery voltage level that can be used in development projects.

Source: www.edn.com

 

[Audioguru]

lm317 circuit

Hi Pranam,
Your circuit has an error. Its output voltage will rise when the load current is low because R1 is not 120 ohms.
R1 must be a max of 120 ohms because the idle current of the LM317 is a max of 10mA which is the reference voltage of 1.2V over the 120 ohms.

 

[pranam77]

using lm317

Hai Guru
In fact i was thinking of the same before posting. But as it was in my collection and i had it from EDN magazine, i thought it may be correct. Thanks for the effort.