AMC7135 blues. (Gazoo, need your assistance)

Gazoo!!!

I took your advise and purchased the AMC. It really small and great for my flashlight. But there is a problem:

I wired it up but changed a couple of things:

1.) Instead of putting a 33 ohm resistor like you did, I put a 100 ohm pot. It is wired in parallel. The reason I did this was to try and tweak the pot to see what resistance I need to get an output of 160mA.

2.) I substituted my laser diode with a normal LED.

The problem I have is that no matter what level of resistance I set the pot to, the current does not seem to reduce. The LED brightness remains the same all the way. If I set the pot to a lower resistance, the pot does get hot quickly, but even then, the LED brightness (and the current measured using a multimeter) does not change.

Any idea of what's happening?

Thanks for your time man.

Ron

Your pot likely cannot handle the power you are putting through it and is burning out.

Hmmm...interesting.

I thought that was the case too, so I wired 3 pots in parallel to help distribute the load. Still does not help. The current does not want to reduce (when measured with a multimeter), but can feel the pots getting warm a bit.

Thanks for your help man.

Do you know that you have to put the resistance in parallel with the load to reduce the current? The 7135s are constant current, so putting in a series resistance has no effect. A resistance in paralle will divert some of the 350 (or so) mA away from the LD.

ron said:

Hmmm...interesting.

I thought that was the case too, so I wired 3 pots in parallel to help distribute the load. Still does not help. The current does not want to reduce (when measured with a multimeter), but can feel the pots getting warm a bit.

    Thanks for your help man.

Click to expand...

Yes, as I mentioned in the last 2 posts, I did put the resistors in parallel. That's why I'm so perplexed.

Initially, I wired one 100 ohm pot in parallel. Did not work, and the pot was starting to get hot. So I wired another 3 pots in parallel to help distribute the load. Still no result. The current does not seem to reduce.

Any ideas?

Where are you measuring the current and how are you measuring it (including DMM setting)? The current out of the 7135 will always be 350 mA. Even if the pot is shunting, say 100mA, then it is dissipating 1 Watt! Very few pots would be rated for that amount of power.

ron said:

Yes, as I mentioned in the last 2 posts, I did put the resistors in parallel. That's why I'm so perplexed.

Initially, I wired one 100 ohm pot in parallel. Did not work, and the pot was starting to get hot. So I wired another 3 pots in parallel to help distribute the load. Still no result. The current does not seem to reduce.

    Any ideas?    

Click to expand...

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.

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.  

Click to expand...

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.

BTW, the power rating for the resistor can be calculated by:

P = V * I

or

P = I^2 * R

Example:
If I=150mA and R = 19ohms
P = 0.15 * 0.15 * 19 = 0.4275 Watts (dissipated in the resistor). Therefore you would need a 1/2 Watt resistor.

Paul

Paul,

I really appreciate your effort man. Wow...I got quite a bit to test out before I can get this laser working.

I wont be able to handle it now though, because I have just been called up for an exam (damn Med School). But I will give this a shot, and get back to you on it. Many thanks man.

Just 2 small questions:

1.) What does DMM stand for? At least I can go google it and find out more if I know whats the fullname.
2.) Do you think that voltage is a concern for this LD? I mean...if the current is regulated at e.g. 200mA, the voltage would still be the same as the batteries (3 batteries = 4.5V). Isn't that too much for the LD?

Dude, thanks again. I will try to get working on it as soon as I can. Take care.

ron said:

Paul,

    I really appreciate your effort man. Wow...I got quite a bit to test out before I can get this laser working.

    I wont be able to handle it now though, because I have just been called up for an exam (damn Med School). But I will give this a shot, and get back to you on it. Many thanks man.

    Just 2 small questions:

1.) What does DMM stand for? At least I can go google it and find out more if I know whats the fullname.
2.) Do you think that voltage is a concern for this LD? I mean...if the current is regulated at e.g. 200mA, the voltage would still be the same as the batteries (3 batteries = 4.5V). Isn't that too much for the LD?

 Dude, thanks again. I will try to get working on it as soon as I can. Take care.

Click to expand...

DMM = digital multi meter

ron said:

Paul,
2.) Do you think that voltage is a concern for this LD? I mean...if the current is regulated at e.g. 200mA, the voltage would still be the same as the batteries (3 batteries = 4.5V). Isn't that too much for the LD?

 Dude, thanks again. I will try to get working on it as soon as I can. Take care.

Click to expand...

You are welcome.  Good luck with the exam.

If you take care of the current to the LD, you don't have to worry about the voltage.  The 7135 sinks the power above what the LD/resistor will draw.

e.g. #1 assume:
4.5V battery source
2.85V across LD/resistor combo (therefore 1.65V across the 7135)

Power from battery = 4.5V * 0.35A = 1.575 Watts
Power sunk by LD/resistor combo = 2.85V * 0.35A = 0.9975 Watts (assuming a 150/200mA load split, Resistor gets 0.4275W, LD gets 0.57W)
Power sunk by 7135 = 1.65V * 0.35A = 0.5775 Watts

e.g. #2 assume:
6V battery source
2.85V across LD/resistor combo (therefore 3.15V across the 7135)

Power from battery = 6V * 0.35A = 2.1 Watts
Power sunk by LD/resistor combo = 2.85V * 0.35A = 0.9975 Watts (assuming a 150/200mA load split, Resistor gets 0.4275W, LD gets 0.57W)
Power sunk by 7135 = 3.15V * 0.35A = 1.1025 Watts (not good, the SOT-89 version of the 7135 is rated for 700mW)

Note that for the 2 examples, even though the source voltage changed, the current seen by the LD did not. That's the beauty of using current regulation.

BTW, the reason for step 1 in the previous post is that the range for the 7135s is 300 to 380mA.  That's a pretty big variance.

Hahahaha..... DMM = Digital Multi Meter.

  Quite stupid of me not to notice that...heh.

Perhaps its because I only have an analog multimeter. Hopefully that is adequate.

  Hey Paul, let me get a couple of things straight:
   
   1.) DO NOT place the multimeter leads and measure the current flow or resistance across a resistor with the LD connected. But its ok to measure the voltage across it. Correct?

   2.) Once I get the correct resistance value, can I not keep the pot there instead of replacing it with a resistor of the same value? I mean...My initial plan was to use a pot and another resistor in parallel in such a way that I can adjust the pot to absorb as much as 200mA or as little as 150mA. This will mean that I can have a laser with a customizable power range that I can adjust when I want to. Wouldn't that be cool?   (Especially with the fact that the LD power output will reduce once it warms up).

ron said:

   
   1.) DO NOT place the multimeter leads and measure the current flow or resistance across a resistor with the LD connected. But its ok to measure the voltage across it. Correct?

   2.) Once I get the correct resistance value, can I not keep the pot there instead of replacing it with a resistor of the same value? I mean...My initial plan was to use a pot and another resistor in parallel in such a way that I can adjust the pot to absorb as much as 200mA or as little as 150mA. This will mean that I can have a laser with a customizable power range that I can adjust when I want to. Wouldn't that be cool?   (Especially with the fact that the LD power output will reduce once it warms up).

Click to expand...

1. In general, I find this the best way to measure the current. Using a DMM in current mode inserts an additional resistance in the circuit which is later removed so the circuit parameters are changed and may no longer be valid when you remove the DMM. The voltage reading mode has a very high impedance so it usually has almost no impact on the circuit under test.

2. It's OK as long as you keep within the pot's power rating.

Using a DMM in current mode inserts an additional resistance in the circuit which is later removed so the circuit parameters are changed and may no longer be valid when you remove the DMM.

Click to expand...

This depends on the meter really. If you want to use it in-circuit, always use the maximum range (usually 10 or 20 amps) that connects using the positive lead in an extra socket. The shunt in a modern DMM is just a piece of wire really, including the leads it should have less resistance than the 1 ohm sense used above.

In this example i would still advice to use the sense resistor soldered in place though, since any disconnection of the current meter would otherwise result in the LD getting the full 350 mA.

Btw, make sure the laser housing does not touch ground - the AMC is a sinking current driver, and accidentily touching the laser casing to ground will destory it immediately.

Benm said:

Using a DMM in current mode inserts an additional resistance in the circuit which is later removed so the circuit parameters are changed and may no longer be valid when you remove the DMM.

Click to expand...

This depends on the meter really. If you want to use it in-circuit, always use the maximum range (usually 10 or 20 amps) that connects using the positive lead in an extra socket. The shunt in a modern DMM is just a piece of wire really, including the leads it should have less resistance than the 1 ohm sense used above.

In this example i would still advice to use the sense resistor soldered in place though, since any disconnection of the current meter would otherwise result in the LD getting the full 350 mA.

Btw, make sure the laser housing does not touch ground - the AMC is a sinking current driver, and accidentily touching the laser casing to ground will destory it immediately.

Click to expand...

Very good reminder about the ground - that could be an issue for a lot of people.

I use both methods for measuring current. The reason I didn't mention it in this thread is that many folks on the board do not know when it is OK to use this method as well as the effects of using the different current ranges on their DMMs.  In the example I used, 1 ohm represents about a 5% change in the intended resistance value.  Cheers,

Paul