Ideal components do not exist in the real world. They are only simplified descriptions of components used to create models to help understand the behaviour of components in circuitry. Its not possible to use something that doesn't exist.
It depends how "ideal" your ideal diode is. The first approximation of an ideal diode is that it's a device that allows current to flow one way, and not the other way, while doing this with no losses. The second approximation of a diode implies the inherent 0.7V drop across the device, as well as one way current flow, but with no loss resistance. The third approximation of an ideal diode is a device which allows current to only flow one way through it, with a 0.7V loss across is, as well as a small internal resistance of a few ohms. These all vary from a real diode because these are all linear, in every sense. You can neither calculate the voltage nor the current across/through a diode. Instead, you must use the diode's characteristic curve (given on datasheet), and linearize it. You can get any amount of current to flow through the diode the CORRECT direction, by changing the voltage used. A number of microamps will take perhaps, a third of a Volt.
The simplest way is the Ideal Model representation of a diode. Forward: V=0(short circuit),r'd=0,I=use Ohm's law. Reverse: V=Vbias(Open Circuit),I=0,r'R=0. Ideal model means it is acting as a switch.
No. The purpose of the zener diode is to clip (turn on) at a certain voltage. A capacitor will not exhibit this behavior.
An Ideal pendulum is ones that would be the first choice if you could choose all the materials (ideal materials). A practical pendulum is one that is readily available and reliable (made from things you have right now and are available to use).
The use of the pn diode is for rectification. It can also be used as a switch.
no we not use resistance in place of diode.
Backward diode
Diode is a two terminal semicon. Diode is a Switching device
A: YES but not directly. for instance a discrete differential amplifier one side need to be a transistor to recover gain but the other transistor can be a diode since in real time that is the operative base to emitter current difference. .
if the diode is forward biasedwell practically the current flows in a circuit if and only if an effective resistance is present in the circuit, if we consider the diode to be ideal (barrier potential but no internal resistance) in this case an external resistance is required if we use the approximate model (both barrier potential and internal resistance are considered) we need not use an external resistance the internal resistance itself acts as the effective resistance.if the diode is reverse biased:-the same explanation applies even if the diode is reverse biased but one must take care that the reverse voltage drop on diode should not increase the peak inverse voltage mark the diode would be burnt or damaged if this phenomena occurs.So this can be prevented by adding suitable resistance to the circuit through which the voltage drop on diode can be managed
GtaTauhidul123
In theory A diode clamps 0.7 volts across it but in practice the number is not exactly 0.7 volts because it can not be controlled. This is only the case with Silicon Diodes (clamping 0.7 volts), as Germanium ones, ideally clamp 0 Volts, but in practice it is not exactly Zero. Basically, in practice it is not exactly 0.7 volts but when doing theory one assumes it to be 0.7volts for ease of operation.