I really need your help!
For using the NI USB 6008, I have some question want to ask.
I want to using the NI USB 6008 to control a LED array which are consist with 24 LEDs connect in parallel.
To implement this, I need the 6008 to control the current from 0 to 480mA, so that control each LED from 0mA to 20mA.
Can anyone tell me how can I achieve this function using NI USB 6008?
Hope for you response as soon as possible:)
Thank you very much 🙂
You need to read the specification for the USB 6008. The 5V output on the device can only supply 200mA so you will need to use an external power supply source.
I'm assuming here that you are controlling all of the LEDs at once and don't need to control the brightness of individual LEDs (otherwise you'll run out of analogue/digital outputs).
I think you'll need to use a FET to act as a variable resistor to vary the maximum current. You can drive the FET from an analogue output on your USB 6008 but I'm not sure of the exact electronics/circuit needed to do this.
If you just need to dim/turn off the LEDs, you could also consider using PWM on a digital output (connected to a transistor and a current limiting resistor) to vary the brightness.
Thanks for your reply:)
There is a external power supply which can apply 15V.
For the LED array, I need to control its brightness in a very stable condition.So which one should be better? BJT or MOSFET?
Any idea of how to use the 0~5V analog output to control the 0~480mA by drive the FET? I am now holding the 2n3904,2n4403,TIP29A and LT1490CN8.
Anyone will know how to connect the circuit?
Works for me every time I need more muscle from a 6008 - which is usually always.
First, exactly how is the LED array wired? You mention 24 LEDs in parallel but generally LED cannot be wired directly in parallel because even slight differences in forward voltage characteristics of temperatures will cause very uneven current sharing. Do you have resistors in series with each diode? What is the forward voltage specification of the LED at 20 mA?
Second, of the devices you mentioned only the TIP29A can come close to handling the current and power required. The typical current gain at 500 mA is about 65 according ot the graph in the data sheet. Worst case is may be less than half of that. At a gain of 65 the required base current would be ~7.4 mA which is larger than the 5 mA limit for the USB-6008 AO lines. So, you will need to use one of the low power transistors with the TIP29A in a Darlington configuration.
Third, you will need to do some thermal management. With a 15 V power supply you may have 5-7 watts dissipated in the power transistor. It can handle that but will require a significant heat sink to keep the temperature within limits.