However, a power MOSFET can produce far lower 'on' voltage at 3 A, say, down in the tens of millivolts. Vce(on) of 2.0 V 3 A, that's a third of the drive power to your 6 V motor lost in the drive circuit and a sixth lost for a 12 V motor. Vce(on) of 2.0 V when Ice is 3 A, 4.0 V when Ice is 5 A. For small transistors, the Darlington Vce(on) might be 1 V instead of 0.3 V for a single driven BJT. The downside of a Darlington pair is that the second BJT's lowest Vce 'on' voltage is much higher than a single BJT. Here, your Darlington pair is on single silicon and package with some pull-down resistors for driving from low strength sources like logic gates. The final gain is theoretically the two BJT gains multiplied, though in practice it's not as much as that. If you look in the OnSemi TIP120 datasheet, you'll see that it's a Darlington transistor with the below internal schematic.ĭarlington BJT pairs use the current gain of one Bipolar Junction Transistor (BJT) to drive a second BJT harder. Also, with a MOSFET, will it still be able to work as intended and which would be the best option? My other question is: After reading a bit in other threads about this transistor, I saw some people recommending using a MOSFET instead of the TIP120 (but I don't fully understand why) and I'm not sure if I should replace the transistor or just keep using it. While this works for a 6 V motor (as I have tried it), I wanted to know if this could also work with the motor I originally mentioned (with a 12 V, 3 A charger instead of 6 V) and if I need to change either the diode or the resistor to more fitting ones. The resistor is 2.2 kΩ and a IN4004 rectifier diode is used.
#Tip120 transistor arduino code
More info on the code or schematic is here.
The basis of the schematic (as I understand it) is to control how much voltage the motor will receive based on the value of the potentiometer. I'm doing a project where I'm controlling how much voltage a 12 V, 3 A motor receives with a TIP120 Darlington transistor like in this picture:
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