No. you need that regulator to accuratly measure the current or you will destroy the laser diode . Blue laser Diodes will Blow on regular 3 volt batteries because a regular battery (aaa) supplies way more current to the diodes than it needs thus burning it out .Most AA batteries can produce 1 amp or more . (Poof)
The voltage regulator known as an lm317 chip has functional uses in electronics. The most common reason for this chip is to help regular direct current (DC) converters and help avoid power shortages and open currents, which can be dangerous.
A lot, the quantity is always changing. Some integrated circuits (ICs) are "general purpose" types, such as the lm317, which are used in lots of different kinds of equipment. Some ICs are very specific, made for only one purpose, such as the chips made for a particular make and type of computer.
The voltage regulator known as an lm317 chip has functional uses in electronics. The most common reason for this chip is to help regular direct current (DC) converters and help avoid power shortages and open currents, which can be dangerous.
A: An lm317 is a negative 1.25v regulator
Both are 1.5A adjustable voltage regulators. The LM317 outputs a positive (with respect to ground) voltage, and the LM337 outputs a negative voltage.
The voltage regulator known as an lm317 chip has functional uses in electronics. The most common reason for this chip is to help regular direct current (DC) converters and help avoid power shortages and open currents, which can be dangerous.
Full form of LM in any IC like LM317 is linear monolithic
The LM317 voltage regulator can be set to any output voltage from 1.2 V to 37 V. You must keep the input voltage between the maximum input voltage and the drop-out voltage for proper operation. The maximum input voltage the LM317 is guaranteed to bear is 40 V. The LM317 is guaranteed to operate when the input voltage is at least 3 volts above the set output voltage.
The LM317 is an adjustable three-terminal positive-voltage regulator capable of supplying more than 1.5 A over an output-voltage range of 1.2 V to 37 V. It is exceptionally easy to use and requires only two external resistors to set the output voltage. internal structure of it show below.
LM317 is a variable-output voltage regulator. I would expect to find it in small home appliances with electronic controls, portable radios, car electronics, cell phone chargers, etc. If you need to find the chip somewhere to do some repairs, you can probably ask a manufacturer to send you a (free) sample. Otherwise, you can buy small quantities online from places like digi-key.
LM317 can be used for regulated dc power supply upto 37V. For more voltage range, a zener can be connected at the reference pin.
A reduction of 36 volts (48 - 12) with 3 amperes of current rating in a series regulator represents 108 watts. That is quite a bit of heat. It would be better to use some kind of a switching regulator instead. Answering the specific question, however... The LM317 is only rated 1.5 amperes, so to get 3 amperes out of it you will need a power transistor that boosts the regulator's rating. You still can't avoid the power dissipation, however, so my original answer of using a switching regulator is apropos. As an alternative to the LM317, you can use the LM150 series, rated to 3 amperes, or the LM138 series, rated to 5 amperes.
wireless head sets have battery's around 3 to 9 volts one of two ways you can do this A replace the battery with recharge type and 12v charger or B remove the battery's fit a socket or hard wire a cable from a simple power supply made from a lm317 or lm338k voltage regulator from any good electronics store time to make it 15 minuets cost $5 lots of data on this chip LM317 on the net easy easy to use it has 3 pins, in, adjust, out, all you need is 2 caps 2 diodes 2 resistors.
A: Follow me on this the lm317 is a 1.25v regulator period. So to make it look like a 11.25 volts is simple assume the the regulator load is 125 ohms so the current through will be 1.25 volts divided by the 125 ohms or 10 ma. now install a resistor from the adj terminal to ground to IR drop 10v which will be added to the 1.25v so 10v divided by 10 ma =1000 ohms from adj to ground will offset the voltage by 10 plus the 1.25 volt give us 11.25 output from output to ground. since we offset the output voltage by 10 volts the adj pin only needs very little current we could have use 1250 ohms and 10k to achieve the same thing. There is a formula to calculate the output voltages but if you understand this you will have no problems. And by the way this scheme can be used for a 5 v,12v any of the three pins regulators. By changing the 1k you will change the output volts since the 1.25 volts is fixed and the 125 ohms if fixed. Furthermore to make it behave like a current source up to an amp or so all you need is a resistor in series with the load and take the output at the adj terminal and good luck.