In this article: http://www.bit-tech.net/article/32/ they use a voltage regulator to not overload the potentiometer. My n00b-ish question now is: why is there a resistor (R1) between the Adjust-pin and the Output-pin of the regulator? I know this might be a silly question for you electronics-gurus, but I'm new to electronics Thanks all
This is to set a voltage divider between the input and output. Since the outputs voltage will be rising when you turn the pot, the resitor would make the input rise also. Giving it a perportional output voltage.
... I don't get it ... shouldn't the voltage on the "adjust"-pin be decreasing constantly because of that connection through the resistor?
/me looking at schematic again.... Ok the pot is connected to ground. Turning it to Minimum will cause the voltage output of the regulator to fall to zero volts. Turning it to Maximum will cause the voltage output to rise to +12volts The resistor between output and adjust will change voltage values as you turn the POT this is because the voltage at both the POT and the OUTPUT will be changing. This will cause a perporional output, meaning the voltage will have a curve to it. If that resisitor was connected to +12volts instead, then the voltage would change in a linear fashion.
erm...okay... and what would happen if you didn't put that resistor there? I looked at the datasheet of the regulator and saw that it's actually necessary
I've tried to explain how adjustable regulators work at the bottom end of this page. Every linear regulator has a fixed reference voltage between the output and adjust pins - for a 317 its 1.25v, for a L200 its 2.77v, for a 7805 its 5v. Put a resistor R1 between the 2 pins and a current I will flow, equal to Vref/R1 amps (Ohms Law) Put a second resistor R2 between the adjust pin and ground, and that same current has to flow in the second resistor (Kirchoff's Law - what goes in=what comes out). This time you calculate the voltage drop across R2, Vdrop = I * R2. The two resistors span ground and output, so the output voltage is Vdrop + Vref. Make either variable, and you can adjust the output to any voltage between Vref and a bit under the input voltage.
nice concise explanation. I hate even thinkinga bout the internals, we analyzed the LM317 (along with a few other gizmos) mt senior year in a course called 'Electronic Circuits II,' and the experience was a painful one. (Two of us passed that course, I was one.) Anyway, there's overcurrent protection if you use this device. That's really a compelling reason to do so. I think there's thermal shutdown as well, I'd have to look at the datasheet. There are some more clever ICs available, but this is a workhorse of a chip.
