ron said:
I am measuring it after the 3 pots that have been connected in parallel.
I am sorry but you have to tell me what you mean by DMM setting.
Always 350mA? Are you saying that even if I put the resistors in parallel, the current will measure 350mA unless I plug the laser diode in?
Hmmmm....so it dissipates 1 Watt for 100mA shunting? But Gazoo uses a 33 ohm 1/2 Watt resistor and he says it dissipates roughly 100mA.
Hey, really appreciate your help.
Your setup should be something like below. You will have to use the actual LD if you want to "dial in" pot, otherwise the circuit will not perform as expected when you put the LD back in.
Let's start from the beginning. The 7135s have a range of current outputs, so determine what yours puts out. Use Pic1 below to calculate yours. Set up the circuit and measure the voltage drop across the resistor. That voltage, divided by the resitor value will give you the actual output of your 7135. You can use a different value resistor but choose one close to the value shown to keep the power dissipation at a reasonable level (don't use a pot for this step).
Pic 1
OK, let's assume you have step 1 done and you know the know the actual output of the 7135. (I'll use 350mA for the rest of the explaination). Now you need to decide what current you want your LD to run at - let's use 200mA for this example. You have to apply Kirchoff's current law. See Pic 2. The current being sunk by the 7135 is 350mA, so the current being fed into it by the two legs (your inserted resistance and the LD) must total 350mA. Since you want the LD current to be 200mA, the current through the resistor must be 150mA (insert your actual/desired values when you do this).
Pic 2
Now comes the more difficult part. LEDs/LDs are non-linear, dynamic devices - that means that their voltage and current are not always proportional (think of the diode curves you have seen on this site).
Ideally, you would have a 200mA constant-current source that you could test the LD you are going to use (the properties vary between devices) at 200mA. What you are looking for is the voltage across it when it's running at 200mA. You would then calculate the resistance required at that voltage to shunt 150mA and use that as your resistor value. If you can't do that, we will move on to another method.
Let's make a guess at the voltage across the LD to be 2.85V when it's warmed up and running at 200mA. Since the voltage across the resistor is the same as the LD, you can get a rough idea of the size of resistor needed. You want the resistor current to be 150mA so the resistance would have to be *around* 19ohms (2.85V/0.15A).
Now you have to do some testing. Set up the circuit in Pic3 (use your actual LD). Use secure connections because is a wire in your resistance loop disconnects the LD will see the full 350mA. Ensure the wiper contact is connected to one of the legs to reduce the risk of "wiper bounce" (intermittant contact). Measure the voltage across 1 ohm resistor to calculate the current in that leg (DO NOT insert your DMM in the circuit and measure current that way). Start with the pot around 10 ohms and gradually increase the value until you have a 0.15V (150mV) drop across the 1 ohm resistor.
When you get to that point:
- ensure the pot does not get moved,
- power down the circuit,
- disconnect one end of the resistance loop,
- measure the resistance across the pot AND the 1 ohm resistor
Find a resistor as close to that as you can get and replace the pot/resistor combo (remember that choosing a smaller value will result in less current to the actual LD and just the opposite with a higher value). Test the circuit again with the new resistor - since you know its value, you can calculate the current through it by measuring the voltage drop across it (I=V/R). The LD current will be the 7135 current (from step 1) minus this current.
Pic 3
BTW
I was asking about the DMM because, depending on how you use it to measure the current, the results may be skewed.
(Especially with the fact that the LD power output will reduce once it warms up).