![]() A typical 5V AVR can source/sink up to 20-30mA/pin being TTL, and the SAM based arduino's like the DUE have two kinds of pin capabilities low and high current pins, high current pins which can only source 15mA/sink 9mA(low power CMOS) so keep this in mind if you're not using an op-amp as a buffer.ģ) While BJT's are great at amplifying small signals with low distortion, and precisely controlling high currents, BJT's make for poor switches however because even if saturated, they still have Vce voltage drops over 2V, this means significant power dissipation at high currents, which means significant heat production. Microcontroller pin currents do vary with sourcing/sinking ability and different MCU families will have different capabilities. If the main supply voltage varies, a current sense resistor needs to be used for feedback to compensate. BJT's and particularly the Darlington configurations allow you to control precisely an output current in the 0-10A+ range with typically less than 2mA from an MCU with a simple current set resistor to the base connected to a microcontroller pin.Ģ) For precision using a PNP Darlington, the base current is referenced to ground, a microcontroller pin can still be used, the output is just turned low to ground the base resistor. In addition BJT's can make excellent and cheap constant current sources, making a simple but precise constant current source for sensitive current controlled devices like LED's. ![]() BJT's inherently have higher bandwidths than FET's and are generally cheaper for identical current carrying. A BJT functions best as a linear device which is precisely CURRENT controlled. EDIT: I made mistake,I mean why is the current higher,not voltage.1) Power FET's and Darlingtons are two different animals. I would think that the first higher left side transistor have zero collector > emiter resistance and voltage drop,I imagined that the voltage at the base of the second lower transistor should be just as high as voltage at its collector,most transistors I read datasheets would blow up becose they cant handle anywhere near as much voltage at their base as they can on collector. Almost all the energy goes collector > emiter but very little through base > emiter.I am assuming that both collector > emiter and base > emiter resistance is zero,I know this is not the case in real world,but this is extremly primitive idealized simulator,I dont see anywhere shown that there are any resistances in this bipolar transistor,it acts like there is voltage divider thing going on between base collector and emiter. ![]() I am looking how the current is flowing through it,how the voltage is at various points.The specific question I have is,why is the voltage and current at the base of the second transistor so low? I mean the transistor that is little bit lower and on right side,considering there is 5V voltage source and single 300 ohm resistor before it,I would expect the base current and voltage be higher. I am noob,still learning the basics,right now I am trying to understand how Darlington transistor pair works.Its probably simplest thing in electrical engineering after two resistor voltage divider yet I cant understand it.Specificaly I am looking at live graphical animation in the preset of falstad free online circuit builder and simulator. ![]()
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