yes
No freaking way but it would make an excellent low level switch if forced at beta of 10
When p-n junction of a diode reversed biased then majority carriers are not able to cross the junction and are attracted in respective regions.So current becomes approximately zero.But because of minority carriers a reverse current keeps flowing.It is called Reverse Saturation Current.And due to attraction towards sides,charges go away from junction.So width of depletion reason increases.
Asking about biasing of the emitter alone does not make sense. When you talk about bias, you talk about a junction, such as emitter-base or emitter-collector or base-collector. In a bipolar junction transistor (BJT) both the emitter-base and emitter-collector need to be forward biased, otherwise you are operating the BJT in cutoff mode. Certainly, if you intend to operate the BJT as a switch, then reverse bias for emitter-base (actually, zero bias) could well be one of the valid states, corresponding to a cutoff condition for emitter-collector. However, operation in linear mode, the other normal way to use a BJT, requires that both the emitter-base and the emitter-collector be forward biased. Of course, depending on the ratio of emitter-base to emitter-collector versus hFe, you could also be saturated, which is a non-linear mode, i.e. an on switch.
Current flows in a reverse biased diode because diodes are not ideal. They do have leakage current and a breakdown voltage in reverse, just as they have a breakdown current in forward and a non-linear and non-parallel forward voltage to current curve. It is also possible that you are looking at a zener diode. A zener diode is specifically design to conduct at a certain voltage in reverse.
if a diode is in forward biased the diode acts as switch is on and when we apply the diode in reverse biased then it work as the switch as off.
A transistor acts like a valve or gate that opens and closes, and allows a current to flow. Since the amount of current that flows is controlled by another input, they can be used to make amplifiers. Carbon microphones and vacuum tubes have the same property, and so have also been used to make amplifiers historically.
Saturation mode is the condition wherein the base-collector junction becomes forward biased, as opposed to reverse-biased in the case of active mode. It is necessary for the base-emitter junction to be forward biased, and thus a base current will be flowing. Typically the base current is much higher than it would be in active mode, and the effective Hfe of the transistor drops rapidly. These conditions apply to both NPN and PNP transistors equally. In practice, the collector-emitter voltage of a transistor in saturation is very low, less than 0.1 V, but this depends on the specific transistor. Some high-power transistors will only saturate to 0.4 V. Saturated transistors sometimes begin to overheat or smoke, although saturation is not always a fault condition. When a transistor is used as a switch, this means it alternates exclusively between cutoff and saturation.
when a semiconductor is doped with p-type and n-type impurities, a pn junction is formed which acts as a diode and prevents the charge carriers to flow to either side of junctionpn juction diode is a semiconductor device that allows current to flow only to one direction.
The easy answer - it's not always forward biased. Both it, and the collector-base junction, must be forward biased to pass current through to the collector. Whether NPN or PNP the relative bias (voltage) on the base determines the conduction from emitter to collector. NPN: if the base is positive, relative to the collector and emmiter, the transistor conducts. PNP: if the base is negative, relative to the collector and emmitter, it conducts. For either transistor arrangement, draw two diodes connected either by their anodes or by their cathodes. The base is the region between them. In an NPN, a positive voltage on the anode, compared the to the cathode(s), will forward bias both, allowing current to flow. The same applies to a PNP with a relative negative voltage being the 'switch', turning both on. bob 02/07/2009 The first paragraph is incorrect. The collector-base junction will be reverse biased for normal operation. The only time an NPN base will be biased more positively than the collector is when it's operating in saturation mode. The second paragraph is also misleading. It implies that current flows (for NPN) from the collector to the base and then from the base to the emitter. Emitter current is base current plus collector current. The collector-base junction is normally reverse biased, so little current would flow. Here's a link with relevant info: http://www.nationmaster.com/encyclopedia/Bipolar-junction-transistors Dennis
An ordinary semiconductor diode uses a P-N junction, but when reverse biased it takes a period of time to remove the current carriers from that junction to create the depletion region that blocks reverse conduction. A Shockley diode instead uses a P semiconductor-metal junction, which removes the current carriers much faster from the semiconductor allowing the device to switch much faster. It also has a much lower forward bias voltage than an ordinary diode. In many ways it is similar to the previous point contact diodes (a piece of semiconductor like galena or germanium with a metal "cat's whisker" point contact) in operation, but is more reliable and easier to mass produce.
nissan micra 2006. how to change a reverse switch?
At a junction.