Microcontrollers. Assembler Maths. Op-Amps. 555 Timer Tutorials. LED Lights & Accessories - SUPER BRIGHT LEDS. Bipolar junction transistors as switches : Worksheet. Question 1: Solid-state switching circuits usually keep their constituent transistors in one of two modes: cutoff or saturation. Explain what each of these terms means. "Cutoff" refers to that condition where a transistor is not conducting any collector current (it is fully off). "Saturation" means that condition where a transistor is conducting maximum collector current (fully on). Follow-up question: is there such a thing as a state where a transistor operates somewhere between cutoff (fully off) and saturation (fully on)? Notes: In all fairness, not all transistor switching circuits operate between these two extreme states. Question 2: Explain the function of this light-switching circuit, tracing the directions of all currents when the switch closes: Notes: Ask your students to explain what possible purpose such a circuit could perform.
Question 3: If switch SW2 were opened (and switch SW1 remained closed), what would happen to the currents through R1 and R2? Question 4: Question 5: Bipolar Junction Transistors. Capacitance. Capacitor Voltage Change. This curve fitted equation can then be used in LTSpice to model the Y5U capacitor, with its capacitance versus voltage relationship. To model a non-linear capacitance in LTSpice, it's necessary to write an equation relating charge (in Coulombs) versus bias voltage. This mathmatical relationship is written into the value of the capacitor (instead of so many uF or pF) as Q=f(x) where X is the pre-defined variable in LTspice representing the instantaneous voltage across the capacitor.
More generally, the charge Q stored in a capacitor is: C(V) is the relationship between capacitance and applied voltage, in this case, as determined by our 7th order polynomial fitted to the measured C versus V data: In a theoretically perfect capacitor of constant value, of course, the relationship between charge and voltage is simple; Q=CV where Q is the charge in Coulombs, C is the capacitance in Farads and V is the voltage in Volts. This is the case since the integral from 0 to V of CdV is simply CV.