Experimentalist
Well-known member
I was wondering if someone could verify my electronic maths for directly driving a TIP120 Darlington NPN from 3.3V output, see datasheet diagram below.
I want to use it to switch a 9W LED powered from a 5 cell Li-Ion battery. If Watts = Volts x Amps then Amps = Watts / Volts = 9/21 = 0.43A . The TIP120 is rated at 5A so I am well within its limits, somewhat overkill, I am only using it because I already have some. As I see it if I directly connect the base to a 3.3V digital output I will get the following current draw from that output:
R1 + R2 = 7000 + 70 = 7070 Ohms
Teensy digital output voltage = 3.3V
I = V / R therefore I = 3.3/7070 = 4.67 E-04 = 0.000467 = 0.467 mA
So as I see it the TIP120 will draw 0.467mA from the Teensy digital output when the output is switched high.
Is my maths correct or am I missing something?
I want to drive this from a Teensy LC which from the datasheet has digital outputs rated at 5mA when driving at 3.3V so I am once again well within limits if my maths is correct.
Anyone?
I want to use it to switch a 9W LED powered from a 5 cell Li-Ion battery. If Watts = Volts x Amps then Amps = Watts / Volts = 9/21 = 0.43A . The TIP120 is rated at 5A so I am well within its limits, somewhat overkill, I am only using it because I already have some. As I see it if I directly connect the base to a 3.3V digital output I will get the following current draw from that output:
R1 + R2 = 7000 + 70 = 7070 Ohms
Teensy digital output voltage = 3.3V
I = V / R therefore I = 3.3/7070 = 4.67 E-04 = 0.000467 = 0.467 mA
So as I see it the TIP120 will draw 0.467mA from the Teensy digital output when the output is switched high.
Is my maths correct or am I missing something?
I want to drive this from a Teensy LC which from the datasheet has digital outputs rated at 5mA when driving at 3.3V so I am once again well within limits if my maths is correct.
Anyone?
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