EV Charging Cables and EV Charging Plugs Explained [2024 Update] > 자유게시판

In effect, this circuit gradually fades in and fades out the light when the switch is toggled. When the voltage is close to the positive rail, the situation will reverse; in effect, this arrangement is an inverting switch. In other words, practical resistor-based voltage dividers can be considered medium to high impedance sources, and should not be driving low impedance loads. These are less dangerous, and with some, the voltage carried is so low that there is virtually no chance of shock. This has an interesting consequence: when the signal is a sine wave of a very low frequency, much lower than the capacitor charge time - the output voltage will, with a slight lag (phase shift), simply follow the input. As with any project, there are a few hints, tips and guidelines that will help to prevent any issues arising and ensure that the process is successful every time.

There is no clear distinction between an electric wire and an electric cable. There is one key difference inductors oppose changes in current for a longer time when the current is higher; whereas a capacitor has a more pronounced effect when the current that charges or discharges it is lower; were we to eliminate the upstream resistance, the capacitor would charge instantaneously. X some time later, the resulting electromagnetic field propagates a 2X voltage swing across the capacitor; but this time, the reverse-biased diode will not conduct, and will prevent the already deposited charge from exiting the capacitor (except through any externally connected load). A current will be allowed to flow through the lightbulb, discharging the capacitor in a familiar manner, with discharge time governed by capacitance and the equivalent resistance of the driven device. Because of the need to limit current, the NPN arrangement shown in column B is universally a bad idea: it will likely destroy the transistor; its PNP equivalent is equally disastrous. Also note that the upper transistor, connected to the positive rail, is PNP or p-type MOSFET; and the bottom one, grounded, is NPN or n-type MOSFET.

In NPN and PNP circuits, note the use of a resistor to limit the base-emitter current: the current flowing through this path must be controlled, because the corresponding junction is essentially a normal, forward-biased diode - and will conduct as much current as you supply, possibly destroying the transistor in the process (and certainly making it misbehave). It can be alternatively fixed by adding a large pull-down resistor from the gate to the ground, to dissipate the deposited charge when the switch is opened, but at the expense of lowering input impedance. The problem with this is that all real-world loads will develop some voltage across them in normal operation; this raises the emitter or source voltage accordingly - perhaps close to, or even above, the driving base / gate voltage. In all cases, the driving voltage applied to the base (or gate) must be high enough to trigger the transistor; that is, at least 0.6V in BJT, and at least 1-2V for most MOSFETs. Therefore, the resistors need to be picked with the expected loads - and the acceptable voltage swings - in mind. When turning on a lightbulb or a LED, this is not a problem - but when driving inductive or capacitive loads (including the gates of MOSFET transistors), or encoding information as voltage levels, it may be more desirable to offer two-pole operation, where the output can be switched between two low-impedance rails.

This causes the switching to be far more erratic, especially with inductive loads. To receive far away FM stations, you should probably get an FM Yagi. As the frequency increases, however, the signal will get more and more attenuated - and the phase shift will become greater - as the source can't charge or discharge the capacitor quickly enough. The capacitor will never charge past the voltage across the terminals of the lightbulb (which can be calculated from the lightbulb's resistance, as per Ohm's law) - and will discharge through it when SW1 is opened. The circuit on the left is, essentially, a band-pass filter: the capacitor needs the signal to change slowly enough to charge it up to an appreciable level - and above this frequency, serves as a shunt; but when the current is not changing fast enough, what are electric cables the inductor will begin conducting and will discharge the capacitor. A more interesting case is what happens when both switches are switched on at the same time: assuming the non-linearity of the lightbulb is negligible (which is fair if the capacitor is large enough), initially, the capacitor will offer a low-impedance path, with very little current flowing through the bulb; but as the charge builds up, the voltage across its terminals will begin to rise - and more current will flow through the bulb.