Jay, I don't think you understand how the 317 works as a constant current supply.
The LM317 IS a voltage regulator, that's all it can do, regulate voltage.
How then, you ask, do we make regulate current? The answer is that we understand
Ohm's Law, and
how the LM317 works.
The LM317 always tries to maintain exactly 1.25V between the Vout and Adj pins. So if the voltage between those pins is less than 1.25V, the 317 will raise the voltage at the Vout pin.
Conversely, if the voltage is higher than 1.25V, the 317 will lower the voltage at the Vout pin.
Moving to diodes:
A LASER diode (or 'LD' in LPF slang) like any other semiconductor device, has a voltage drop which is dependant on how much current is flowing through it, as well as on the junction temperature. The catch is that at current levels near (above and slightly below) the laseing threshold, a very small change in voltage will make for a very large change in the current which will flow. So, we need something which can very precisely control the voltage while monitoring the current in order to keep the current constant.
Now to the LM317 as a constant current supply:
In the circuit we've all come to know and love for driving these LDs there is one resistor. Yes, I said one, only one.
But what about the potentiometer some may ask? Any series resistors (defined as: 'all, and only the current passing through one also passes through the other') in a circuit act as one resistor. In the case of a 4 Ohm resistor in series with a 100 Ohm potentiometer we have an approximately 4 Ohm to 104 Ohm variable resistor. This resistor resides between the Vout pin and the load and is called the sense resistor. Tapped just past the sense resistor is the Adj pin of the LM317, if you will recall, it is looking at the voltage drop across the sense resistor and causing the LM317 to raise or lower the voltage at the Vout until the voltage drop across the sense resistor is exactly 1.25V.
The voltage at Vout, provided the regulator is not dropping out (dropping out or regulation for the LM317 is defined as: ‘the voltage across the sense resistor is less than 1.25V’), will be 1.25V (the sense voltage) plus whatever the voltage drop of the load.
Back to the LM317:
The LM317 needs about 2V - 3V just to run it’s internals; we call this the regulator ‘dropout’
Again, the LM317 being a semiconductor device, needs a different voltage depending on how much current is flowing through it. This is why when a voltmeter is connected as the load it will show almost the supply voltage. The voltmeter is a VERY high resistance, so no matter how high the LM317 tries to raise the voltage, so little current flows that the 2V – 3V average dropout is in this case only a few mV.
Finally on to a few examples:
If I supply 7.2V to the circuit when driving a 16X DVD burner diode at 250mA (the sense resistor is set to 5 Ohms), about 2V is lost because of the regulator dropout, there is a 1.25V drop across the sense resistor, and let’s say the voltage drop of the LD at 250mA is 2.8V. So that’s 2+1.25+2.8=6.05V the other 1.15V has to go somewhere, and this is where the LM317 goes to work by wasting it as heat.
Let’s try that example again but with only four alkaline AAA cells to power it. When these cells are brand new, they provide 1.57V each for 6.28V.
If I supply 6.28V to the circuit when driving a 16X DVD burner diode at 250mA (the sense resistor is set to 5 Ohms), about 2V is lost because of the regulator dropout, there is a 1.25V drop across the sense resistor, and let’s say the voltage drop of the LD at 250mA is 2.8V. So that’s 2+1.25+2.8=6.05V this time there’s only .23V for the LM317 to get rid of. But wait, when I applied the load to the battery the voltage began to drop. It seems that under load they can only provide 1.48V per cell. That’s only 5.92V, which means that the LM317, no matter how high it tried to raise the voltage, can only get the sense voltage up to 1.2V and at 1.2V the 5 Ohm sense resistor will pass only 240mA. Our regulator is now dropping out, and this was with fresh cells!
I’m sure you can easily see now why when driving a red LD with a forward voltage drop of nearly 3V you must supply more than 6V to the circuit, and were you driving a Blurry LD which has a forward voltage drop of almost 5V, you must supply the circuit with at least 8V.
Now what would happen if I tried to attach a load to the circuit where the sense resistor had a range of 2 Ohms to 100 Ohms, was set to 2 Ohms, the circuit was supplied with 6V and the load was trying to take 5V? Well, that load sure wouldn’t see 5V, but it might see 3.5V, and with the LM317 doing all it could, it could probably only get the sense voltage up to about 0.31V. 0.31V divided by the 2 Ohm sense resistor (there’s Ohm’s Law again) would only give about 155mA. (Does that sound familiar?)
Anyways, I hope none of this sounded condescending. It’s late and I’m tired, so off to bed with me with one last wish. That all who read this will find great use of it, and ease of understanding in it.
)
