# Voltage regulator - unexpected voltage drop, is this normal?

I am using this voltage regulator.

Following this example:

For input power source, I am using old SATA convertor power adapter, which produces 12V 1.5A max (I tried various power supplies, all similar result).

With no load, everything works as expected, instead of R2 I have potentiometer 0 - 5k OHM. I can easily configure any voltage range from 1 to 10V using the potentiometer.

The problem happens when I connect any load. My understanding of voltage regulators is, that you can configure them to provide certain voltage, and they will hold it as long as you don't exceed the maximum current they can provide (3A in this case). But that isn't happening in my case.

I connected Vout to a simple (strong) LED light, that draws 0.2A at 4.5V just to test it out, and while the LED is working, the Vout immediatelly dropped down to 3.3V (Vin also dropped from 12V to 11.5V, but VR should provide same Vout no matter of Vin according to manual, as long as Vin is at least 1V higher than required output). Measured Il was max 0.2A, never more than that.

Is this a normal behaviour? Why is VR not capable to provide stable voltage on output? I would like to use it to power some USB device, which expects stable 5V output, this doesn't seem very stable to me. My expectation is that it will provide stable Vout as long as I don't exceed 3A Il.

• My suspicion is that your LED light isn't really drawing 0.2 A at 4.5 V. I think the LED wants to draw more current, more current than the LM350 and/or the 12 V supply will deliver. Also the thermal protection of the LM350 will kick in. You should use a load that behaves more like a resistor instead of an LED which behaves like a diode (current varies a lot for only a small voltage change). – Bimpelrekkie Feb 11 at 7:45
• Well the thing is I am meassuring the Il with my multimeter and it really never exceeds 0.2A. Even if it wanted to draw more, VR should be able to work until 3A? The thermal protection could be related on other hand, because the VR package is definitely very hot. – Petr Feb 11 at 7:48
• the VR package is definitely very hot There you go. Then the thermal protection of the LM350 regulates the current down to the 0.2 A you measure. It does that to protect itself. You should use a 10 ohm, 10 W resistor (then you get 1 A at 10 V) as a load or if you don't have that a 12 V 5 W (or less than 5 W) car light bulb might do the trick. Then also the LM350 might get hot and limit the current and voltage ! So mount the LM350 on a heatsink! – Bimpelrekkie Feb 11 at 7:50

For a linear power supply, the power dissipation is $$\V_{drop}*I\$$.