I have been working with this circuit, and i am wondering what R4 is for?
I see R1 and R2 is for make a steady voltage at 6 V
And R3 is to control that there is max 0.6 A running.
Im pretty new at the electronic side, so a well explanation of whats going on would be great.
I hope you can help.
Just a quick notice, i use it to power a circuit, where i use a IR transmitter/reciever, to control a motor, for when it have to be on/off. So i dont use it on a battery like on the picture, and i have dropped the Capacitor aswell, because the current are delivered from a standard 12V power supply (computer) so i think the transients will be sorted out in that.
Yeah i thought so aswell, but i have tried double the resistor or make it the half, but neither does any difference for the total current able to run, but when i put in a 1 ohm resistor parallel to R3 i found that the current limit where doubled, so i could take 1.2 A.
An LM337 (any voltage regulator) needs some capacitance on the input and output for stability.
Usually ranging from 1 to 10uF. The datasheet could help you select value and type.
That 1000uF was just there as a smoothing cap in case the power comes from a transformer and bridge rectifier.
Not needed if you feed the circuit with DC.
Panoxio:
when i put in a 1 ohm resistor parallel to R3 i found that the current limit where doubled, so i could take 1.2 A.
Yes, R3 is a current sense resistor. The 2N2222 detects the resulting voltage across that resistor.
When you (accidently) short or overload the supply, then the voltage across R3 could briefly spike higher than normal.
R4 makes sure that does not fry the transistor.
With 12volt in and 6volt out there is 6volt across the LM337.
That will generate 6*1.2 = 7.2watt of heat with a load of 1.2Amp.
That would need a heatsink (expensive) the size of a fist.
One of the reasons linear supplies are being replaced with switching regulators.
A 12volt to 6volt switching buck supply that can deliver 2Amps is the size of a stamp, and costs >US$1 on ebay.
Leo..
This link will show you how R1 and R2 give you your needed voltage output.
R3 acts as a current sense resistor.
If the current through R3 causes the volt drop across it to approach 0.7V'
The base emitter junction of the transistor will begin to conduct.
The base current will cause transistor to pull the voltage across R2 down.
This will result in the output of the LM317 dropping and so the output current will drop.
I'm not sure why your first change in sense resistor value didn't work because you need about 0.6-0.7 volts on the base of the transistor to turn it - on which pulls the adjustment pin of the LM317 low, lowering the output voltage which in turn lowers the charging current.
Knowing the above, you can calculate the need R for any current by substituting the 0.6 volts into R = E/I so R = 0.6/desired current. For example, to have 200ma max current, R = 0.6/0.2 = 3 ohms. This is not exact due to the transistor turn on point but is close enough for crude current limiting.
