![]() All three regions of the transistor have a distinct connection to the external circuit (see Figure 2). The only functional difference between PNP and NPN transistors is the biasing of each PN diode required for the transistor to work. ![]() NPN transistors have a thin p-doped region between two n-doped sections while PNP transistors have a thin n-doped region between two p-doped sections. The most basic type of transistor is the bipolar junction transistor (see Figure 1).īipolar junction transistors (BJT) are composed of three doped semiconductors. They are the basis for digital logic and computation, and they catalyzed a revolution in electronics. Transistors are integral to modern circuitry, with hundreds of millions of them used in modern integrated circuits for computing. Transistors are components of electric circuits that can act as amplifiers and as switches. ![]() The schematic symbol for an NPN transistor with the collector, base, and emitter currents labeled. So even though the power we spent charging the capacitor was not previously lost, it is lost now.Figure 1. In this way it's obvious all the energy stored in the capacitor is now lost, dissipated across the draining transistor. All the current flows in a loop back into the other end of the capacitor: the capacitor when draining here acts like a source. What happens when we reverse this process? Now all the energy stored in the capacitor gets dissipated across the driving transistor draining the capacitance. What is this actually saying? It's saying the total power used in the circuit is the energy lost by the power source ( \$E = \int_0^\infty I(t)V dt\$, simply the supply voltage times the supply current integrated to get supplied energy), minus the energy we're not wasting to dissipation and is being stored in the capacitor ( \$E_c\$). How much depends on the capacitance of the FET, and due to the gate-drain capacitance \$C_ = \int_0^\infty I(t)V dt - E_c The current to charge it comes from the driving circuitry, which has to expend a bit of energy to provide it. The charge required is called "gate charge". In order to switch from "off" to "on" state, its gate needs to be charged to a suitable voltage, which involves both gate-source and gate-drain capacitance. Then you multiply this by the switching frequency (or divide by the period), and you get the dissipated power, which is energy dissipated over one second. Each switching action dissipates a bit of energy. Switching is a transient event, so losses are not usually modeled as "power". If you do this switching at a low frequency the power is lower if you do it at a high frequency, the power is higher. Switching therefore causes power dissipation. So, if this is repeated cyclically, you are taking energy from the power rails cyclically and converting that energy to heat. So, you have "burnt" energy and made it into heat and, when the transistor disengages, the node capacitance re-charges to 5 volts - to do so it has to take energy from the power rails to recharge the capacitance. It also means that the node in question would take no energy to re-establish the 5 volts when the discharge transistor went open-circuit.īut, every node does have capacitance and, initially that capacitance is charged to 5 volts so, in order to discharge that node, you need to remove energy and convert it to heat in the very-low ohmic "on" resistance of that transistor. This would then mean that no energy was needed to change the voltage on that node from 5 volts to 0 volts. Then, imagine that the transistor had to discharge a node from (say) 5 volts to 0 volts - imagine also that the node possessed no self-capacitance. ![]() There's no energy wasted in the driving of this transistor by definition. Imagine you had a near-perfect switching transistor that needed no control energy to cause it to change from an open-circuit to a very-low-value "ohmic" closed circuit (or vice versa). Why does switching cause power dissipation?
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