Good day folks. Messing around the shop and sometimes not having stuff to actually do, I thought might as well learn something, so: constant current source using an LM317....I know the general principle of operation and even built one a while back, but can't figure out a couple of things and there don't seem to be any clear-cut tutorials. I don't just want it to work without knowing HOW it works. I can't sleep well at night if something's puzzling me, so let's work this out ![Biggrin](https://cdn.badcaps-static.com/images/smilies/biggrin.png)
1) what exactly are those 1.25v we divide by the resistance and why that number ?
2) what happens if the load is disconnected ?
3) what's going on with the power dissipation in the resistor ?
Have a look HERE at the very first diagram.
1) I understand that 1.25v divided by R equals the current we try to achieve, which leads us to my first question: what exactly is 1.25v ? Why that number ? Of course, I had a look at the datasheet and saw it's the reference voltage on the non-inverting input of the op-amp, but the anode of that internal diode doesn't go to ground like in a traditional zener regulator, but to the ADJ pin....how does this work and what's the mechanism that keeps the current constant anyway ? Indeed, drawing 125mA through that 10ohm resistor in the example DOES equal 1.25v and I don't have to take their word for it: the math doesn't lie.
2) this one is easier and makes more sense: I reckon if the load is disconnected, the voltage goes all the way up to VCC, or really close, since the non-inverting input is driven high straight off the input pin and opens the transistor. Correct ?
3) nobody seems to mention that resistor: take the first example again: 125mA...ok, but what about the voltage drop on that resistor ? How do you calculate its power dissipation ? I KNOW the formula of course, V x I = P, I don't mean that, I mean where do I get my numbers. Going back to 1) again: 125mA through a 10ohm resistor gives us that elusive 1.25 again. Solving for the formula above, would this be 1.25V x 0.125A = 0.156W ? Let's say 0.15w to round it out...is this correct ?
The thing that's really confusing is the absence of GND in the LM317....it just floats there...
Cheers and thanks.
![Biggrin](https://cdn.badcaps-static.com/images/smilies/biggrin.png)
1) what exactly are those 1.25v we divide by the resistance and why that number ?
2) what happens if the load is disconnected ?
3) what's going on with the power dissipation in the resistor ?
Have a look HERE at the very first diagram.
1) I understand that 1.25v divided by R equals the current we try to achieve, which leads us to my first question: what exactly is 1.25v ? Why that number ? Of course, I had a look at the datasheet and saw it's the reference voltage on the non-inverting input of the op-amp, but the anode of that internal diode doesn't go to ground like in a traditional zener regulator, but to the ADJ pin....how does this work and what's the mechanism that keeps the current constant anyway ? Indeed, drawing 125mA through that 10ohm resistor in the example DOES equal 1.25v and I don't have to take their word for it: the math doesn't lie.
2) this one is easier and makes more sense: I reckon if the load is disconnected, the voltage goes all the way up to VCC, or really close, since the non-inverting input is driven high straight off the input pin and opens the transistor. Correct ?
3) nobody seems to mention that resistor: take the first example again: 125mA...ok, but what about the voltage drop on that resistor ? How do you calculate its power dissipation ? I KNOW the formula of course, V x I = P, I don't mean that, I mean where do I get my numbers. Going back to 1) again: 125mA through a 10ohm resistor gives us that elusive 1.25 again. Solving for the formula above, would this be 1.25V x 0.125A = 0.156W ? Let's say 0.15w to round it out...is this correct ?
The thing that's really confusing is the absence of GND in the LM317....it just floats there...
Cheers and thanks.
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