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Sore subject. I've been perma-banned from three electronics forums because I don't have photographic capability. I can scan schematics. That's it.
The middle and the back should change too but that all depends on how you have it plugged into the breadboard. I think you want the longest dimension of the POT parallel to the longest dimension of the breadboard and the pot on the divided side if you have the standard breadboard.
Here is a breadboard pic. http://wiring.org.co/learning/tutorials/breadboard/ Note the break in the parallel busses. A lot of people think that there are 4 bussed parallel to the longest dimension. Note that there are eight busses.
Let's ask what are your resistor values? Something about 220-270 and what else?
The regulator sometimes needs a minimum load to work, otherwise it tends to stick close to Vin.
a LM317 need about 2 volts more than the desired output to work and can not go below 1.24 volts, so it's probably not what you want to be using.
what is you power supply, what what is your goal?
If the input is 18V and the output is 2V at 0.5A then the poor little LM317 will be dissipating (18V - 2V) x 0.5A= 8W which is a lot of heat so it needs a pretty big heatsink.
When the trimpot is set to 1k then the 240 ohm resistor has 1.25V across it so its current is 1.25V/240 ohms= 5.2mA and the 1k trimpot will have a voltage across it of 5.2mA x 1k ohms= 5.2V. Then the maximum output voltage from the LM317 is 1.25V + 5.2V= 6.45V.
Why is the input voltage so much higher than 3V? All it does is make lots of heat. For a maximum output of 3V the input should be about 5V then the heating when the LM317 output is at its 1.25V minimum is (5V - 1.25V) x 0.5A= 1.875W and only a small heatsink will be needed.
Adjustment to 0V is hard to obtain with a single supply. One way to get it is to create a 3V reference and use a pot as a voltage divider and buffer the output with an OP amp.
A true 0-3 V "power Supply" is going to be harder.
In your post #26 you pretty much describe a textbook LM317 regulator circuit with the diode protection. Between common and the LM317 input what do you read for voltage? LM317 input to common?
Ron
Your 100uF main filter capacitor is MUCH TOO SMALL. For a load of 0.5A it should be 2200uF or more.
The LM317 is spec'd with a minimum allowed load current of 10mA. The more expensive LM117 has a minimum allowed load current of 5mA.
The LMx17 develops 1.25V between its output and its ADJ pin which is also across the resistor. Ohm's Law calculates the resistor to be 1.25V/5mA= 250 ohms so the LM117 can use 240 ohms. This resistor in series with the pot is the load when there is no additional load. The LM317 needs 120 ohms so that the output voltage does not rise without an additional load.
Assuming that you have the more expensive LM117 since you are using its 240 ohm resistor, the resistor current is 1.25V/240 ohms= 5.2mA. When the pot is at maximum it is 1k ohms so with the 5.2mA in it then its voltage is 5.2V. Then the output of the LM117 will be 1.25V + 5.2V= 6.45V.
If you replace the 240 ohms resistor with 120 ohms for an LM317 then the resistor current is 1.25V/120 ohms= 10.4mA. When the pot is at maximum it is 1k ohms so with the 10.4mA in it then its voltage is 10.4V. Then the output of the LM317 will be 1.25V + 10.4V= 11.65V.
You have a 1N4007 rectifier diode parallel with the 240 ohm resistor. Maybe its polarity is backwards? It is not needed in your circuit, it is used to discharge an additional filter capacitor sometimes used from the ADJ pin to ground. You do not have the capacitor.
TI's datasheet shows an LM117 in a 0V-30V regulated supply. It has a +35V supply and a -10V supply. The resistor is 150 ohms, not 240 ohms and the pot is 5k ohms, not 1k. It uses a 1.2V regulated reference IC connected to the lower terminal of the pot so that the output of the circuit can go down to near 0V.
Nope, the diode is not needed. A protection diode is needed from input to output when there is an output capacitor more than 25uF and the input or output is shorted.The diode in parallel with the 240 Ohm resistor, connected to the ADJ terminal is to provide protection from the 1.0 uF capacitor that is in parallel with the load resistor.
Nope, the diode is not needed. A protection diode is needed from input to output when there is an output capacitor more than 25uF and the input or output is shorted.
Page 7 of your datasheet says the diode across the 240 ohm resistor of an LM117 protects it when there is a capacitor of 10uF or more from the ADJ pin to ground (that you do not have) and the output voltage is higher than 25V.