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250mA constant current Laser Diode Driver

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kokos

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Hello everybody. I would like to design a 250mA constant current Laser Diode Driver with very low noise, 0-5v, 0-250mA, ESD protection, slow start and to be able to monitor the current. I would like to monitor the voltage in terms of current. I found the attached circuit online and I would like to ask you here how I can change this circuit into my requirements?

Thanks
kokos
 
Voltage regulator, IC2, can provide up to 500 mA for the diode, so it is good as is. Rs needs to be changed to 10 ohms (2.5/.250) to set 250 mA diode current. Q1 can't handle 250 mA, so look at 2SJ377 or similar instead. Q2 can't handle 250 mA, so look at IXTP08N50D2 or similar.
 
Thank you very much all for your help. I will have a look and come back to you again for more questions or to upload my final circuit.
 
Thank you ccurtis for your reply. Where i can connect a 12 turn trimmer so i will control the current? Actually i want to monitor the voltage in terms of current, how i can do that?

Thanks
 
250mA constant current 0-5V Laser Diode Driver

Could you please tell me if everything is okay with my design? I want to have 0-250mA output, 0-5V output, Low noise, slow start, ESP protection and current monitor.

Thanks very much
 
Thank you ccurtis for your reply. Where i can connect a 12 turn trimmer so i will control the current? Actually i want to monitor the voltage in terms of current, how i can do that?

Thanks

Rs sets the current, so that is the resistor you need to trim to control the current (in the original circuit). What voltage do you want to monitor and why do you want to monitor it in terms of current? It's easier to monitor voltage in terms of voltage. The current through the diode is equal to the voltage across Rs divided by the value of Rs.
 
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Rs sets the current, so that is the resistor you need to trim to control the current (in the original circuit). What voltage do you want to monitor and why do you want to monitor it in terms of current? It's easier to monitor voltage in terms of voltage. The current through the diode is equal to the voltage across Rs divided by the value of Rs.

Thank you very much ccurtis for your answer. So in the original design, I have to change the Rs (fixed resistor) with a trimmer in order to control the current across the diode right? and if i do that i will have from 0-250mA across the diode?

Also as you told me the other day i have to replace the Q1 and Q2.
Also can you please tell me the LD-EN of the TPS7201 where is connected? Also is the ODL output of the IC5 connected to the DL enable input of the TLC070?

Thank you very much for your quidance.
 
Thank you very much ccurtis for your answer. So in the original design, I have to change the Rs (fixed resistor) with a trimmer in order to control the current across the diode right? and if i do that i will have from 0-250mA across the diode?

Yes. The current through the diode is 2.5/(value of Rs).

Edit: Be aware that the power dissipated by Rs will be .625 watt at 250 mA. If you want to use a .5 watt trimmer, replace TL1431 with LM4041. Diode current is then 1.23/(value of Rs); .3 watts at 250 mA.

Also can you please tell me the LD-EN of the TPS7201 where is connected? Also is the ODL output of the IC5 connected to the DL enable input of the TLC070?

LD_EN is grounded to turn the laser diode on. "ODL" is really DL. The "O" is the symbol for the output terminal. DL is the label for the terminal. DL of IC5 is connected to DL of the TLC070.
 
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Yes. The current through the diode is 2.5/(value of Rs).

Edit: Be aware that the power dissipated by Rs will be .625 watt at 250 mA. If you want to use a .5 watt trimmer, replace TL1431 with LM4041. Diode current is then 1.23/(value of Rs); .3 watts at 250 mA.


LD_EN is grounded to turn the laser diode on. "ODL" is really DL. The "O" is the symbol for the output terminal. DL is the label for the terminal. DL of IC5 is connected to DL of the TLC070.

Thank you very much ccurtis for your help. I will order today all the components and first do it on a strip board and see what I will get. I will come back to you with the final design once i make sure that it is working.

Thanks again for your help.
 
Yes. The current through the diode is 2.5/(value of Rs).

Edit: Be aware that the power dissipated by Rs will be .625 watt at 250 mA. If you want to use a .5 watt trimmer, replace TL1431 with LM4041. Diode current is then 1.23/(value of Rs); .3 watts at 250 mA.

LD_EN is grounded to turn the laser diode on. "ODL" is really DL. The "O" is the symbol for the output terminal. DL is the label for the terminal. DL of IC5 is connected to DL of the TLC070.

Hello ccurtis. On the original design the TL1431CD doesnt require an external series resistor connected between the supply voltage and its input? Cause i tried it without resistor and it is not working (not giving 2.5V) it is giving 6.74V which is the output voltage of the LT1763. Is it possible if i will adjust the TL1431 output to 5V and then connect its output to Terminal 1 of a 1K trimmer to regulate it from 0-5v and from Terminal 2 of the trimer to the Rs (20R) to have 0-250mA current? The terminal 3 of the trimmer is connected to ground. is that corrent?

For Q1 you told me to use 2SJ377 but i couldnt find it so I have chosen IRF9520NPBF, what do you think about that? Also if i dont have the exact values of the resistors of the comparator circuit LM393 (example 37.4k , 53.6K,) can i use resistors close to these values?

Thanks
 
Hello ccurtis. On the original design the TL1431CD doesnt require an external series resistor connected between the supply voltage and its input? Cause i tried it without resistor and it is not working (not giving 2.5V) it is giving 6.74V which is the output voltage of the LT1763.

An external resistor is not required. The voltage ACROSS the TL1431CD should be 2.5V. The voltage at the top of the 2.8K resistor, referenced to ground, should be the output voltage of the LT1763 minus 2.5V. If it is not, there is something wrong with the connections or parts, or some other error.

Is it possible if i will adjust the TL1431 output to 5V and then connect its output to Terminal 1 of a 1K trimmer to regulate it from 0-5v and from Terminal 2 of the trimer to the Rs (20R) to have 0-250mA current? The terminal 3 of the trimmer is connected to ground. is that corrent?

I'm not exactly sure what you are proposing to do. If you adjust the output of the TL1431 through the use of its control terminal, leaving everything else connected as shown in the schematic, it will have the effect of adjusting the current in the diode, to a point. The laser diode needs about 2.5 volts to operate, so if you adjust the voltage across Rs to 5V, the laser diode will have at most 6.7V minus 5V, or 1.7 volts to operate, which is not enough.

For Q1 you told me to use 2SJ377 but i couldnt find it so I have chosen IRF9520NPBF, what do you think about that? Also if i dont have the exact values of the resistors of the comparator circuit LM393 (example 37.4k , 53.6K,) can i use resistors close to these values?

That FET should work okay. You can use resistors close to those values. The idea is to put the + terminal of U5A at a slightly positive voltage when Vb reaches its operating voltage.
 
An external resistor is not required. The voltage ACROSS the TL1431CD should be 2.5V. The voltage at the top of the 2.8K resistor, referenced to ground, should be the output voltage of the LT1763 minus 2.5V. If it is not, there is something wrong with the connections or parts, or some other error.

I'm not exactly sure what you are proposing to do. If you adjust the output of the TL1431 through the use of its control terminal, leaving everything else connected as shown in the schematic, it will have the effect of adjusting the current in the diode, to a point. The laser diode needs about 2.5 volts to operate, so if you adjust the voltage across Rs to 5V, the laser diode will have at most 6.7V minus 5V, or 1.7 volts to operate, which is not enough.

Thanks ccurtis I have managed to get 2.5v across the diode and 4.2v at the top of the 2.8k resistor.

What i want to do and i am trying to regulate the voltage is that i want to monitor the voltage across the Rs and then divide it by the value of Rs so i will be sure that i am providing the correct current to the LD before i connect it. So I want to connect a variable resistor somewhere on the circuit (if you could advise me where please) so i will achieve 0-5V and the same time 0-250mA on the LD.

I have developed the circuit today but it is not actually working and i am not sure why. I want when the voltage across the LD reaches 2V to have about 100mA, on this what i have designed today when i have 2V the current is about 40mA. I achieved that by replacing the fixed resistor Rs by a 10R trimmer. Could you tell me what might be the problem?

And also, Q1 and Q2 can you please confirm how to connect them? For example for Q1, G is connected to the output of the TLC070, S is connected with the negative input of the op amp and the Rs and D is connected to the LD is that correct?
And for Q2, G is connected to the output of the IC3B, D to Q1 D, and S to the ground is that correct or I made a mistake here?

Thanks, i appreciate your help
 
What i want to do and i am trying to regulate the voltage is that i want to monitor the voltage across the Rs and then divide it by the value of Rs so i will be sure that i am providing the correct current to the LD before i connect it. So I want to connect a variable resistor somewhere on the circuit (if you could advise me where please) so i will achieve 0-5V and the same time 0-250mA on the LD.

You can put a variable resistor in place of the LD as a test of the current before inserting the diode. Say you want 2v across the test resistor at 250 mA then set the test resistor to 2/.25=8 ohms and insert it in place of the LD. You should measure 2V across the variable resistor, as long as Rs is set to provide 250 mA.

I have developed the circuit today but it is not actually working and i am not sure why. I want when the voltage across the LD reaches 2V to have about 100mA, on this what i have designed today when i have 2V the current is about 40mA. I achieved that by replacing the fixed resistor Rs by a 10R trimmer. Could you tell me what might be the problem?

The current is set by adjusting Rs, independent of the voltage across the LD. If you attempt to achieve 2V across the LD at 100 mA it is not possible because the voltage across the LD is purely a function of the current through the LD. Do not worry about the voltage across the LD. It is the current through the LD that you want to control.

And also, Q1 and Q2 can you please confirm how to connect them? For example for Q1, G is connected to the output of the TLC070, S is connected with the negative input of the op amp and the Rs and D is connected to the LD is that correct?

That is correct.

And for Q2, G is connected to the output of the IC3B, D to Q1 D, and S to the ground is that correct or I made a mistake here?

G is connected to IC5B. Otherwise, that is correct.
 
You can put a variable resistor in place of the LD as a test of the current before inserting the diode. Say you want 2v across the test resistor at 250 mA then set the test resistor to 2/.25=8 ohms and insert it in place of the LD. You should measure 2V across the variable resistor, as long as Rs is set to provide 250 mA.



The current is set by adjusting Rs, independent of the voltage across the LD. If you attempt to achieve 2V across the LD at 100 mA it is not possible because the voltage across the LD is purely a function of the current through the LD. Do not worry about the voltage across the LD. It is the current through the LD that you want to control.



That is correct.



G is connected to IC5B. Otherwise, that is correct.

ccurtis thank you very much for your help.

So if I understood correctly, since I have a reference voltage of 2.5V due to the TL1431, I can use a variable resistor 10R as the Rs in order to controll the current through the LD, and I can achieve 0-250mA, right? Now for monitoring , I can use a variable resistor at 8R in place of the LD and regulate the voltage to be 2V and then regulate the RS to reach the desirable current of 100mA across the Variable test resistor right?

This is where i am finding difficulties, to monitor the voltage and current in order to regulate it to the desirable values before to connect the LD.

Thanks for your help once again.
 
So if I understood correctly, since I have a reference voltage of 2.5V due to the TL1431, I can use a variable resistor 10R as the Rs in order to controll the current through the LD, and I can achieve 0-250mA, right?

No. Current through the LD is 2.5/Rs so with a 10 ohms variable resistor the current range is 250 mA minimum. Current reduction requires a higher resistance value.

Now for monitoring , I can use a variable resistor at 8R in place of the LD and regulate the voltage to be 2V and then regulate the RS to reach the desirable current of 100mA across the Variable test resistor right?

Rs controls the current through the test resistor. The voltage across the test resistor is equal to the the current through the test resistor times the resistor value. With an 8 ohm test resistor and a 250 mA current, the voltage across the test resistor should be 2V.

This is where i am finding difficulties, to monitor the voltage and current in order to regulate it to the desirable values before to connect the LD.

Again, you only need to verify that the proper current flows through the 8 ohm test resistor by measuring the voltage across the test resistor to be 2V. When that is done you can insert the LD and the same current will flow through the LD. The voltage across the LD may not be 2V, but the current will still be 250 mA. Edit: This circuit does not regulate voltage. It regulates current, which is what you want for a LD. The way to test the circuit it is to insert different values of test resistor (within limits) in place of the LD and verify that the current remains the same through each test resistor value by measuring the voltage across the test resistor and dividing that voltage by the test resistor value to calculate the current.

Thanks for your help once again.

You're welcome! I looked at your IRF9520NP again and am concerned that its GS threshold voltage range is not ideal, but it may still work for you. The threshold voltage should be slightly less, more like -1 to -3 volts instead of -2 to -4 volts. The FET given in the schematic CAN handle the 250 mA current. It would be a better choice.
 
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No. Current through the LD is 2.5/Rs so with a 10 ohms variable resistor the current range is 250 mA minimum. Current reduction requires a higher resistance value.



Rs controls the current through the test resistor. The voltage across the test resistor is equal to the the current through the test resistor times the resistor value. With an 8 ohm test resistor and a 250 mA current, the voltage across the test resistor should be 2V.



Again, you only need to verify that the proper current flows through the 8 ohm test resistor by measuring the voltage across the test resistor to be 2V. When that is done you can insert the LD and the same current will flow through the LD. The voltage across the LD may not be 2V, but the current will still be 250 mA. Edit: This circuit does not regulate voltage. It regulates current, which is what you want for a LD. The way to test the circuit it is to insert different values of test resistor (within limits) in place of the LD and verify that the current remains the same through each test resistor value by measuring the voltage across the test resistor and dividing that voltage by the test resistor value to calculate the current.



You're welcome! I looked at your IRF9520NP again and am concerned that its GS threshold voltage range is not ideal, but it may still work for you. The threshold voltage should be slightly less, more like -1 to -3 volts instead of -2 to -4 volts. The FET given in the schematic CAN handle the 250 mA current. It would be a better choice.

Thank you ccurtis i will try it and come back to you. See below what i am trying to make

https://www.electro-tech-online.com/custompdfs/2012/06/0009-D01.pdf
 
No. Current through the LD is 2.5/Rs so with a 10 ohms variable resistor the current range is 250 mA minimum. Current reduction requires a higher resistance value.

View attachment 64731

Rs controls the current through the test resistor. The voltage across the test resistor is equal to the the current through the test resistor times the resistor value. With an 8 ohm test resistor and a 250 mA current, the voltage across the test resistor should be 2V.



Again, you only need to verify that the proper current flows through the 8 ohm test resistor by measuring the voltage across the test resistor to be 2V. When that is done you can insert the LD and the same current will flow through the LD. The voltage across the LD may not be 2V, but the current will still be 250 mA. Edit: This circuit does not regulate voltage. It regulates current, which is what you want for a LD. The way to test the circuit it is to insert different values of test resistor (within limits) in place of the LD and verify that the current remains the same through each test resistor value by measuring the voltage across the test resistor and dividing that voltage by the test resistor value to calculate the current.



You're welcome! I looked at your IRF9520NP again and am concerned that its GS threshold voltage range is not ideal, but it may still work for you. The threshold voltage should be slightly less, more like -1 to -3 volts instead of -2 to -4 volts. The FET given in the schematic CAN handle the 250 mA current. It would be a better choice.

ccurtis what do you think about the attached circuit? It is only the part which drives the LD. I have used a Variable resistor right after the RS as you can see, and a Test resistor in place of the LD. By trimming the RS i get from 0-5 volt, and from 0-250mA at the outpu. Atleast in the software i get these values. Do you think in practice it will be correct? I am using a 100R variable resistor.

Thanks
 
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