Hi all,
The OPT8241 CDK provides the reference for Laser diode driving,By the way,what the part of "HEX_LED ckt" is used for?
Can you provide me with some reference about LED drive circuit?
The input voltage is 5V,LED 4715S,FOV 100deg,Max dis 8m or farther.
can 2 LEDs do this (4715S VF 3V,IF 1A)? Does "Path Delay Correction" needed here?
Hi,Subhash:
I list the result of the tool as below.
If a depth resolution of 1.5% instead of the requirement of 1% is used,the peak power will below 10W.
according to the datasheet of 4715S,the IF = 1.8~1.9A when the duty cycle is 50%,and VF is about 3.1~3.2V ,so the power should be about 5.7W.
Maybe,there shold be three LEDs for better extensibility.
Regards,
Zuan.
Zuan,
The LED datasheet has 2 parameters on which the maximum current is dependent. It is the pulse time and the duty cycle. At 24MHz, the pulse times will be so small, they are out of the lower limit of the data given. You would have chosen another limit in the System Estimator tool called "Integration Duty Cycle".
Integration duty cycle is (Integration Time / Quad Time) for the usual timing waveform depicted in the diagram above. As the high frequency modulation has a very short pulse time, you should consider the Integration Time as your pulse time and decide upon the current according to that. You can actually drive the LED at double that current to get the peak optical power. You just need to make sure it is less than the absolute maximum current of 5A for SFH4715S. The LED can get heated at such high currents and duty cycles. It is advisable to use good heatsinking or an aluminum substrate PCB for dissipating such high heat levels
About the lens, do you think you can use an 87 degree Diagonal FoV, F1.1 lens which is used in the Tintin CDK? This would greatly improve your performance spec/reduce the number of LEDs required.
About the ambient, is your application in outdoors sunlight? You would have to enter the expected amount of light spectral power around the 850nm wavelength. This light will affect the way your system behaves.
In absence of sunlight, it is preferable to have fewer sub-frames if Thermal Noise is more than the shot noise. So I suggest the following configuration:
This requires Illum Peak Optical Output Power of 6.6W.
Integration Time = 5.5ms
Ipulse according to the datasheet = ~1.25A
LED peak current = 2 * Ipulse = 2.5A
LED peak power for 2.5A from datasheet = 2.5 * 1070mW = 2.675 W
Number of LEDs = 3
In your calculations, I noticed that you are taking the electrical power of the LEDs. The tool can calculate the Optical power of the LEDs. So, you would have to refer to the datasheet to see how much current has to be driven for the required optical power.
Please let me know if the choice of lens is acceptable to you. I feel that the system will be more optimised for this lens. Also, the system response is expected to change a lot if it is in presence of ambient sunlight. Please consider that as well.
Regards,
Subhash
Subhash
Thank you,I have got it.
The pulse times are so small that It seems to be continuous ,while the equivalent current is half of the peak current.Then I can consider the Integration Time as the pulse time in a Quad Time cycle.
application 1 : FOV 100 deg & 5m
Integration Time = 8.33ms
Ipulse according to the datasheet = ~1.25A
LED peak power for 2.5A from datasheet =1.25*2* 1070mW = 2.675 W
Number of LEDs = 2 (5.34/2.675) but the Number 3 is better suited than 2; If lower the depth resolution or the max distance abit, 2 LEDs would be ok.
application 2 : FOV 70 deg & 8m
Integration Time = 5.56ms
Ipulse according to the datasheet = ~1.3A
LED peak power for 2.5A from datasheet =1.3*2* 1070mW = 2.782 W
Number of LEDs = 2 (5.03/2.782) .
Now can you tell me about the circuit for driving 2 or 3 LEDs?
Zuan
Zuan,
You could use a similar Inductor + MOSFET + Illuminator configuration as the Hex_LED circuit in the OPT8241 CDK. Except, you would be having 3 LEDs instead of the 6 LEDs. So the schematics, in short, would look like this:
Each LED will have a Vf a little greater than 3V. Combined Vf of the 3 LEDs plus the overhead to drive the parasitic inductance will approximately between 10V to 12V.
The power source that supplies current to ILLUM_HEX_LED has to be a constant current source. This current source would have to sustain an output voltage of approximately 0.5*Vf. This effect is because the inductor L1 and the MOSFET U4 together make something like a boost converter with 50% duty cycle. So you will need approximately 6V as the current source output voltage. For 2 LEDs, you will need a lower voltage as the current sosurce output voltage.
We would recommend that you use TPS2559 as the current regulator. You can refer to the 'Linear' page in the schematic 'OPT8241-CDK-EVM IB 10-50' to see the implementation of the same. You might have to use a buck converter or a boost converter to supply required input voltage to TPS2559. To improve efficiency, you could set your desired current levels and then carefully adjust and minimise the voltage headroom across TPS2559 in a prototype PCB.
Please note that the system estimator tool assumes that all of the LED power is going into the FoV of the camera. It does not account for the light which lost outside the FoV. You would have to build in that factor of safety in your actual system design. Also, the LED output radiance is not constant across the FoV. The center will be more brightly illuminated as compared to the corner of the FoV. You will see a proportional variation in the performance.
Regards,
Subhash
