5 replies to this topic

[Eric Burdo]

I didn't have any shift registers available, so I wrote this to control the 7 segment display without using a shift register. http://brick-labs.co...shift-register/ Video, picture and source code available.

[CW2]

I didn't have any shift registers available, so I wrote this to control the 7 segment display without using a shift register.

FYI, given the typical forward voltage ~2 V for a red LED and 41 Ω resistor the display segment is drawing ~30 mA (no wonder it appears so bright on the video). Although it is probably no big deal for the display to run on current close to absolute maximum rating (typical value is 10-20 mA depending on type), it is dangerously high for Netduino - it can provide only 16 mA on three pins (2, 3, 7) and 8 mA on the rest (with the exception of analog pins that can provide only 2 mA). I would recommend you using a current booster, or increase resistance of the current limiting resistors to get below safe 8/16 mA per Netduino pin.

[Eric Burdo]

Hi CW2 Thanks for that... is there a good place to learn how to calculate the resistors, and learn what would be the optimal resistor for this instance? Based on what you said... I'm more concerned with making sure I don't pull to man mA's from the Netduino, then I am feeding too many to the LED display. Right? I'm a software guy, learning hardware. So some of this stuff is new to me...

[CW2]

Thanks for that... is there a good place to learn how to calculate the resistors

Hm, I don't know any special place I could recommend, just pick one returned by Google search for 'LED resistor calculator'.

, and learn what would be the optimal resistor for this instance?

The formula is simple: ResistorValue [Ω] = (SupplyVoltage [V] - LedVoltage [V])/MaxCurrent [A], in your particular case (typical LED segment current 10 mA, 2 V red) R = (3.3 - 2)/0.01 = 130 Ω. If you don't know voltage of your LED display (usually referred to as Forward Voltage or Voltage Drop), you can use typical values, based on the color.

Based on what you said... I'm more concerned with making sure I don't pull to man mA's from the Netduino, then I am feeding too many to the LED display. Right?

As you have the display connected directly to Netduino, it is basically the same thing. But in general yes, Netduino is certainly more expensive

[Cosivox]

FYI, given the typical forward voltage ~2 V for a red LED and 41 Ω resistor the display segment is drawing ~30 mA (no wonder it appears so bright on the video). Although it is probably no big deal for the display to run on current close to absolute maximum rating (typical value is 10-20 mA depending on type), it is dangerously high for Netduino - it can provide only 16 mA on three pins (2, 3, 7) and 8 mA on the rest (with the exception of analog pins that can provide only 2 mA). I would recommend you using a current booster, or increase resistance of the current limiting resistors to get below safe 8/16 mA per Netduino pin.

Why is drawing ~30 mA? Assuming the led works with ~2v and ~15mA then we can calculate the mA's drawn applying 3.3v:

Calculated resistance of led = 2/.015 = ~133.33 ohms, plus the resistance applied of 41 ohms = 174.33
so the total of mA's drawn is = 3.3 volts / 174.33 ohms = ~18mA

Please correct me if I'm wrong

[CW2]

Why is drawing ~30 mA? Assuming the led works with ~2v and ~15mA then we can calculate the mA's drawn applying 3.3v:

The LED consumes all the current that comes through the resistor: I = (V_CC - V_F)/R, where V_CC is the supply voltage, V_F is LED forward voltage, so for the sample values above I = (3.3 - 2)/41 = 31.7 mA.

To have 15 mA for the LED, you'd need current limiting resistor R = (V_CC - V_F)/I = (3.3 - 2)/0.015 = ~87 Ω.