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FrozenGate by Avery

Lpc826 not bright help

A shunt resistor is just a normal resistor being used for the purpose of measuring the current in a given circuit.
 





It looks to me that, in your photo, the test diodes are backwards +/-. Also, to test current, measure voltage from one side of test resistor to the other... not to ground. Ohm's law: 1mV=1mA accross 1 ohm.
 
The diodes I'm sure I solder the positive side to the resistor and the resistor goes to the adjust pin of the lm317 and thanks for the information
 
OP's not followed their drawing correctly as far as I can tell from the picture - the 3.9 ohm resistor is not connected between the Adj and Vout pins but from Vout to ground.
 
I will try to simplify my observation. In the photo of test diodes/resistor, connect the all black end of diodes to resistor. The end with a silver stripe on it will then connect to battery negative.
 
I like you Shiraiei :)

It reminds me of 10 years ago when drivers really weren't available and lots of the members were doing the same thing and killing diodes left and right. Its a little bit of a learning curve, but don't let that get to you. Its a great feeling when you get your first laser built.
 
Immo1282 said:
A shunt resistor is just a normal resistor being used for the purpose of measuring the current in a given circuit.
Click to expand...
A shunt is just a parallel circuit. This means that your shunt resistor is parallel to something else.....in this case, the meter.


I would fix the LM317 in place and use only the resistor and cap as this is all you need to make this circuit work. If it is only supplying 200 mA it will likely not heat up too fast. At higher currents the LM317 would need heat sinking to keep it from overheating.
 
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You've got the meter in current range while trying to measure voltage. This shorts out the sense resistor and puts full input voltage into your load, which will kill it soon if it hasn't already. You've not metered anything I requested, and aren't taking many suggestions here seriously.

You are in desperate need of a remedial course in electronics. I say that not to be patronizing, but to be realistic. If you don't have this basic underlying knowledge, building circuits will only frustrate you, and lead to expensive mistakes.

Immo1282 said:
having to factor in the loading of a moving-coil voltmeter into your measurements may not be helping you so much...
Click to expand...

That isn't necessary, since all the points in this circuit are low impedance. 20kohm per volt for this meter means 25kohm is in parallel with the ~4 ohm. That will have no real effect on anything.
 
Hi so I just made a new driver and it's made up of a lm317t ,3.9ohms resistor,a 16mf capacitor,and a diode.
And I connected it to a dummy load and when I try to measure the voltage between the resistor and diode in the dummy load nothing seems to be showing up on my multimeter.
Anybody knows what's happening here?
 

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So I tried it out with my 12v power supply and same results.no voltage between the resistor and diode.But when I measure the voltage that the driver puts out it seems like its more than 10 v which I guess is 12 v.
I would be very grateful if someone could take a look at these pictures and see what's wrong.
 

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Alright update so the few posts I made up there regarding about me not being able to measure the voltage on the dummy load is because of me soldering it at a wrong polarity which I corrected it now .

The voltage across 4 diodes = 3v
The voltage across the resistor is too little for me to even see I think it's about 0.20 volts??
 
Now you are getting somewhere. Assuming your test resistor is 1 ohm, .2v means you will have 200mA to laser. A better DMM may be in order. They are not expensive. If I were you, I would wait for some of the real experts here at LPF to chime in before moving forward.
 
Alright thanks for the reply I'll wait for more people to see and see if it's okay to use the laser diode
 
Glad you're starting to get more sense from your circuit. Using the 3.9R resistor you've got there, you should expect to see around 320mA of output current. It's kind of hard to see in the picture what the tolerance on your load resistor is - as if it's not exactly 1 ohm you may get a slightly different reading. I'd reccomend picking up a cheap digital multimeter, which will let you measure resistances as well as voltages/current. If you use I=V/R to calculate the actual current with the measured values of your load resistor and voltage across it, you'll be able to see if it's really producing the 320mA that I'd expect from using a 3.9R resistor with an LM317.

On an unrelated note, I will say please use the Edit button at the bottom of each reply you make to add new content to a thread if you made the last comment... For those that have notifications set up, it does tend to be quite annoying if you multiple-post in that way :)
 
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