Semiconductor Devices: Introduction To Diodes

Рет қаралды 2,956

Күн бұрын

Пікірлер: 25

@kevinklocke7796 Жыл бұрын

Thank you for your content, Professor Fiore! Commenting on the final circuit. D2 is reverse bias, so relatively no current would go through that loop. 0.7v would be at D2 cathode. D1 is forward bias, so it should drop 0.7v. Knowing that, the 1k resistor should drop 9.3v. Current through that loop should be around 9.3mA.

@ElectronicswithProfessorFiore Жыл бұрын

Yep!

@Muhammad-Salman-Saeed Күн бұрын

Dear professor, where i can find the soloutions of all the given problems in this book?

@ElectronicswithProfessorFiore Күн бұрын

Answers to the odd-numbered exercises are the appendix. Evens are not given because many professors use them for graded work.

@myawoo Жыл бұрын

What is the criteria for selecting knee voltage in the diode I-V curve?

@ElectronicswithProfessorFiore Жыл бұрын

There really isn't a specific mathematical criterion. It's more of a practical, empirical thing given typical usage. That's why you'll sometimes see texts refer to the knee as 0.6 V and others as 0.7 V. (Plus, there is some variation due to the diode design itself). In most designs, the precise value is not important.

@myawoo Жыл бұрын

@@ElectronicswithProfessorFiore OK. Does university lab experiment ask students to find knee voltage?

@ElectronicswithProfessorFiore Жыл бұрын

@@myawoo It might, but typically they would ask you to plot the curve and then give an estimate. See Exercise 3 in my Semiconductor Devices lab manual (it's a free PDF download; follow the links given in the video description, above).

@myawoo Жыл бұрын

@@ElectronicswithProfessorFiore 👍👍👍

@fishclips6888 5 ай бұрын

I've got a quick question now if we got 2 circuits both got the same resistance and battery first one got a silicon diode and the second has a germanuim diode and then we placed a voltmeter on both diodes and showed 0.5 on both of them does that mean the silicon one has a current of zero??? (Since the voltmeter show rms value)

@ElectronicswithProfessorFiore 5 ай бұрын

First, RMS is only valid for AC signals and you have a battery (DC). In any case, it would be odd for a Ge diode to have a 0.5V drop across it as the barrier potential is closer to 0.3V. If you measured that, I'd assume something was wrong with the circuit or the diode.

@fishclips6888 5 ай бұрын

@@ElectronicswithProfessorFiore it was a question in our a-level exam 3 days ago and the question was so odd since the choices were "Ac1 is less than ac2" other choose was "ac1 equals zero" idk if there's a way to contact u and show u the question

@ElectronicswithProfessorFiore 5 ай бұрын

@@fishclips6888 From your description, it sounds like there are AC sources in there along with the battery. If there are AC sources, that changes everything. If it's literally nothing more than a battery, a resistor, and a diode in series (forward biased), then my initial answer stands. I will add that if the sources are AC instead of batteries, then the waveform across the diode will be a clipped sine wave. For them to have the same RMS voltage, that would imply that the source potential in Ge circuit is larger (because the forward part is smaller, that means the reverse part would have to be larger in order to create the same RMS value).

@fishclips6888 5 ай бұрын

@@ElectronicswithProfessorFiore nope both of them were connected to a battery and the question didn't even state the values of anything beside the voltmeter read

@ElectronicswithProfessorFiore 5 ай бұрын

@@fishclips6888 Are the diodes reverse biased? That's the only way I can see this making sense. It remains odd that they are referring to RMS values for a DC circuit. Ignoring that, and assuming that the diodes are reverse biased, then each of them showing 0.5 V (DC) tells you that the batteries are each 0.5 volts. The currents would be so small that the voltage drops across the resistors would be inconsequential (unless they're huge, as in many mega ohms). Thus, all of the voltage would drop across the reverse biased diode. Also, it is odd that they would refer to a battery source as "AC1" or "AC2". Let me back up because maybe I am making some unwarranted assumptions based on your initial comment. This does have me intrigued. First, are the two circuits identical except for the fact that one uses Si and the other a Ge diode? Second, does the circuit consist of just a series loop containing a battery, a resistor, and a diode, and there is a voltmeter shown across the diode? Third, what is the direction of the diode (to keep it simple, is the "diode symbol arrow" pointing toward the short vs. long bar of the battery symbol (or if it's using a generic round symbol for the battery, is it pointing toward the minus terminal of the battery)? {Or more technically, is the cathode connected to the minus battery terminal?)