![]() The transistor is now said to be switched “ON” (conducting). If we now forward biased the base terminal with respect to the emitter by using a voltage source greater than 0.7 volts, transistor action occurs causing in a much larger current to flow through the transistor between its collector and emitter terminals. As the base terminal is grounded, no current flows from the collector to the emitter terminals therefore the non-conducting NPN transistor is switched “OFF” (cut-off). The technique is the same for a 3.3 V digital output, just that the voltage across the resistor will be lower, so a lower resistance is needed to get the same base current.When the base terminal of the NPN transistor is grounded (0 volts), zero current flows into the base therfore Ib = 0. ![]() Consider the B-E drop to find the voltage drop across the resistor, then calculate the resistor so that you get the desired base current. To drive a NPN transistor, whether darlington or not, from a 5 V digital output, put a resistor between the output and the base. That again is a bad idea since you don't know what current the digital output can supply, which might damage the digital output and/or the transistor. If you are actually driving the base with a digital output running from a 5 V supply, then the darlington is loading the digital output to bring its voltage down. In any case, applying 5 V between the base and emitter of a darlington transistor will blow out at least one of the B-E junctions. The load will also see the power supply voltage reduced by the saturation voltage. This makes a darlington inappropriate for switching high current, since this current times the high saturation voltage will be dissipated as heat. However, one drawback of this approach is that the saturation voltage will be much higher than for a single transistor. The advantages of two transistors arranged in a darlington configuration is that the overall gain is the product of the two individual gains. You didn't provide a link to the TIP120 datasheet, so I'll take it on face value this is a darlington transistor, presumably NPN. Though if your motor current is << 1A then Zetex/Diodes Inc have some transistors that can be directly driven by a micro (characterized at 1A Ic with 10mA Ib as well as the more usual 100mA which is beyond most micros).Īgain, use parametric search and check the datasheets with a fine-tooth comb. Simulate this circuit – Schematic created using CircuitLab If you want to use BJTs then the best method is to use one transistor to drive another as below: Consider undervoltage lockout if you use such a small package since the consequences of it turning partly on can be disastrous. For example, it may be rated for 2.5V drive, but Rds(on) quoted with 4.5V drive.įor example, a quick search comes up with the Si3442CDV. The parametric search is sloppy enough that you may have to go through several datasheets to be sure. Also pick one with sufficient voltage capability (3.3V + one diode drop for the flyback diode), sufficient maximum current for the stalled condition. Better 0.1 ohm to account for temperature. For example, for a 1A motor you might want < 150mV drop so Rds(on) < 0.15\$\Omega\$ with 3V drive. For 3.3V drive you should pick one that is rated with 2.5V or 1.8V or 0.9V Vgs and look for a guaranteed Rds(on) at < 3V Vgs that is sufficiently low. You can search distributors using parametric search. The voltage drop is too high (At 3A the voltage drop could be as much as 2V, but at 2A it's typically a bit over 1V- even so it takes about 1/3 of the motor voltage typically). See for example these tutorials on how to use nFETs as a switch:Ī Darlington is not a very good way to drive a motor from 3.3V. This is especially problematic if you are operating at high currents and low voltages. Between collector and emitter of a Darlington transistor you will always lose around 1.2 V, no matter how much current you are switching. MOSFET can often be just as easy (or even simpler) to use and losses at the transistor will be smaller. ![]() ![]() If you switch a low voltage load, e.g.: a single LED, a motor running at 3.3 V volts, then bipolar transistors and especially darlingtons will be very ineffective. Using a Darlington transistor can be effective for loads with a high voltage that are switched often, but are only switch on for a short time. ![]() You might need to reduce the resistance your base resistor a bit at 3.3V. The resistor between your micro controller and the TIP120 will convert the voltage to a current. You do not apply a voltage to a bipolar transistors base, but you apply a current to the base. ![]()
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