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now comes the question. [...] under no load condition divide points voltage is correct and varies as i change the output voltage. but when I load the circuit and begin to draw current voltage at divide point increases. actually when I load the regulator, regulator's output drops slightly. I can see this on multimeter. but voltage of dividing point increases rather than decreasing. its funny, just like ohms law got inversed. that is as voltage across trimpot drops voltage on divide point increases. (it should be decreasing right?). got any idea what the heck is going on?
Given that it works OK unloaded, it's a safe bet that the load (how large?) alters the functionality of the circuit - the most likely possibilities...
Bad layout, with eg. the 10k trimmer close to the regulator and some wire resistance to where the load is, but you didn't mention how much it rise, when the voltage at the load drops how much, which makes it a bit hard to see if that's a realistic scenario.
What does the writing on the regulator say (verbatim)?If you want a better answer, provide better info - you are the only one knowing what exactly is in front of you (and personally, I'm much more inclined to solving electronics problems, than guessing on what you didn't care to mention or drag it out of you bit by bit).
You can remove D1 and D2, they are not needed in your circuit.
C1/C3 should be mounted as close to the input/output terminal as possible.
Connecting the free end of the 5k potentiometer to the wiper/ground will ensure a measure of control when the potentiometer ages and the wiper might lift from the carbon(?) track when adjusted. With the free end unconnected, it will spike up to a few volts below the input voltage (and that's an annoying thing if you feed eg. a microcontroller board from it).
I use thick wires but problem begin to arises when i draw about 100 mA. I don't think wire resistance is the cause. I uses AC current carrying wire used in house wiring. voltage at load drops about 10-20 mV. voltage rise at divider is few milivolts and increasing as load increases.
regulator says:LM317TWK00P0017chinaand has ST logo.
QuoteYou can remove D1 and D2, they are not needed in your circuit.D1, D2 is recommended in datasheet as protection against capacitor discharge through regulator
Figure 18 shows the LM317 with the recommended protection diodes for output voltages in excess of 25 V or high capacitance values (CO > 25 µF, CAdj > 10 µF). Diode D1 prevents CO from discharging thru the IC during an input short circuit. Diode D2 protects against capacitor CAdj discharging through the IC during an output short circuit. The combination of diodes D1 and D2 prevents CAdj from discharging through the IC during an input short circuit.
5K potentiometer is temporary just for testing purposes. I intend to use variable voltage (using a DAC) to replace potentiometer
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