It is not clear to me what you are asking, but I assume you are wondering why Vy = 5 V when DA and DB are off. In this case there's no voltage drop across the resistor R when the current through it equals zero, so by Ohm's Law the voltage equals 5 V on both sides of the resistor.

The DC current through Rg must equal zero because there is no path for it to flow. It cannot flow through the capacitor, and it cannot flow into the gate of the transistor. So by Kirchhoff's Current Law at the gate node, the current must equal zero.

In later lectures I introduce the CVD (constant voltage drop) model as well as the exponential p-n diode equation. The ideal diode model is only used at the start of the course.

@@wtimothyholman Thank you for your response and time, I will be continuing the course in that case as I needed help with understanding CVD models especially, thank you once again for your lecturing as well.

You've just described what is known as the "two capacitor paradox", which has been debated and discussed for many decades. There's an excellent article on Wikipedia that I recommend you read. But in short, such a circuit is impossible to build, as it would require wiring and switches without resistance or inductance. You could remove the resistance by using superconductors, but you cannot remove inductance. That inherent inductance between the capacitors would prevent the voltage from changing instantaneously across either of them.

R_sig represents the Thevenin equivalent resistance of the circuit that is connected to the input of the amplifier. Note also the V_sig voltage source that represents the Thevenin equivalent voltage. It is a more accurate representation than using an ideal voltage source alone at the amplifier input.

Yes, that's correct. Ideally, in saturation the MOSFET drain current Id = K'/2 * (W/L) * (Vgs - Vt)^2 (i.e. the square law equation), and Id has no dependence on Vds. But because of r_o (the drain-source resistance), the drain current will increase slightly as the magnitude of Vds increases. If not for r_o, transistors would behave like ideal dependent current sources, and we could build amplifiers with infinite gain. With actual transistors that's not possible.

You are correct. That has been pointed out to me before, and I've added a correction in the comments for the video. However, kudos to you for actually doing the calculations for yourself and realizing the error. There's a valuable lesson in that: always check the numbers yourself.

BJTs are still used in modern integrated circuit design, e.g. in BiCMOS processes. But even bulk CMOS processes have vertical PNP transistors that designers can take advantage of. BJTs also have advantages over MOSFETs in terms of lower noise and higher transconductance. The dominance of MOSFETs in digital ICs is what makes them so common, but RF and precision analog designs will still use BJTs. That makes them worth learning about. Many older textbooks have a "BJT first, MOSFET second" approach because historically that is in order in which they were invented. Textbooks like Sedra & Smith have (in my opinion) a more sensible approach of analyzing the MOSFET first, since it is used so much more often.

These videos are intended to be viewed full screen on a laptop or monitor. However, you can pinch to zoom on a phone or tablet while watching them.

Go to Lecture 6 for a more in-depth discussion of the reverse breakdown effect and the Zener diode which utilizes it.

Saying Vcb < 0 is equivalent to saying Vbc > 0. In this case I used the convention in the textbook. The authors prefer to express the collector-base and base-emitter voltages as positive voltages for normal operation. So in the forward-active region of operation, the BJT will have Vcb > 0 and Vbe > 0. Regardless, you can always flip the order of the subscripts and switch the direction of the inequality sign to obtain the same results.

real sht this guy is awesome!!!!!!!!!

If an ideal diode has VD > 0, then it cannot be turned off with ID = 0. On the other hand, if an ideal diode has VD < 0, then it cannot be turned on with ID > 0. That is the contradiction. If an ideal diode is off, then VD < 0 and ID = 0. If it is on, then VD = 0 and ID > 0. I recommend that you watch the next lecture with additional examples of ideal diode circuits.

You are correct. Of course, for that matter the demodulator also needs a resonant tuned circuit between the antenna and the diode in order to work. In practice, the input resistance of the next stage of the receiver (not shown) would supply the leakage resistance across the capacitor. This is a basic "arm waving" diagram to illustrate the concept, but it lacks sufficient detail to create a working design. That is deliberate, as resonance and filtering are not topics covered in this course, but instead in a parallel course taught at Vanderbilt University.

Unfortunately I can't share the equation sheet because it contains copyrighted material. I copy and paste figures and tables from the Sedra & Smith textbook onto a handout that I distribute to my students for their exams, but I cannot make them available outside of the university. If you have a copy of the 8th edition of Sedra & Smith's "Microelectronic Circuits", look at the following: Fig. 5.11, Fig. 5.19, Table 5.1, Fig. 5.12, Table 5.2, Fig. 5.20, Table 6.1, Table 6.2, Fig. 6.13, Table 6.3, Table 7.2, Table 7.3, and Fig. 14.1. That is what you need.

@@NeighborhoodSecurity-Nashville Thank you for taking the time to give a detailed response! I have the 7th international addition (because it was less expensive). Unfortunately the tables use the t model for some of the equations which are different than the hybrid version. I may just take the equations you derived and manually edit the tables. Mark

Unfortunately, I cannot provide the equation sheet or the exams because they contain copyrighted material from the Sedra & Smith textbook. If you have a copy of that textbook, look at the different tables in each chapter. That is what I included on the equation sheet.

That is a very good analogy.

The 2 A current flowing through the 5 ohm resistor is left over from the previous initial condition calculation. It is not the current through the resistor for t > 0. I didn't finish erasing everything until I began using source driving to calculate the equivalent resistance across the inductor beginning at 18:47. At that point everything was erased and the current was redefined through the resistor.

@@wtimothyholman Perhaps I didn't ask the right question. I understand the 2A through the 5 ohms resistor is the leftover the from initial condition. Actually, after playing around with a similar circuit where I replaced the inductor with an independent current source of 1A (going up), and replaced the dependent source with just a wire, and put a R resistor on the switch. After letting R -> 0 ohms (to mimic a perfect wire), the result showed that the switch is NOT a short wire. This goes against my intuition because I thought the switch acts like a short, thus, 2A also goes through the switch. It was the reason why I thought there should be no current going through 5ohm resistor according to KCL. Now that my confusion has been resolved. It leads to another question why the switch isn't a short wire. There must be a something I'm not understanding about what a short wire actually is.

@@nhitc6832, your question isn't clear to me. The switch -is- a perfect short circuit (i.e. wire) when it is closed. The switch is open for t < 0, and so the only path for the current from the 2 A source is through the 5 ohm resistor during the initial condition calculations. The switch is closed for t > 0, at which point the node voltage on the left side of the 5 ohm resistor is 0 V because of the short circuit to ground caused by the closed switch.

That negative sign can only happen if the voltage and current of the inductor are defined opposite of the passive sign convention. In other words, the current flows into the negative end and out the positive end of the defined voltage polarity across the inductor. Occasionally you'll see a problem or example where someone violates the PSC and does this, but I never do it in my lectures. There's no point in confusing students this way. I always stick to following the PSC for resistors, inductors, and capacitors.

In the lecture example, the direction of I1 was given as part of the problem statement, where it is the controlling current for the 20*I1 current-controlled current source. The direction is therefore fixed. Unlike other types of problems, it is not a current whose direction you can arbitrarily select. In circuit problems using dependent sources, the polarity of the controlling voltage or the direction of the controlling current will be specified at the beginning. You can't alter them without altering the problem and changing the answer. But note that you COULD flip the direction of I1, so long as you also flipped the direction of the 20*I1 current-controlled current source. If you do that, you'll get the same answer as the original problem.

Before watching my Electronics lectures, I recommend that you first watch my Circuits lectures. You'll also need some background in physics, linear algebra, and calculus to understand some of the concepts introduced in both courses. As an EE, you'll be required to take the following core courses during your first through third years at a university: (1) Physics (in particular, the fundamentals of electricity and electromagnetism) (2) Mathematics (linear algebra, first-year calculus, differential equations) (3) Circuits I (DC and AC linear circuits, nodal analysis, mesh analysis, op amps, phasors, etc.) (4) Circuits II (frequency-dependent circuits, filters, Laplace transforms, etc.) (5) Electronics I (transistors, diodes, amplifiers, small-signal analysis) (6) Digital systems (Boolean math, combinational digital logic, state machines, computer organization) (7) Programming (using structured languages such as Python, C, or Java) My KZbin lectures cover (3) and (5), provided you have sufficient background in (1) and (2). You can check out Khan Academy for those. Watch (3) first, then (5). Good luck!

@@wtimothyholman yes Prof , I did study (1) (2) and finished few weeks ago I can say I am pretty good at Mathematics and good at the basics of Electricity , Circuits , diodes , logical gates , resistors , both DC and AC circuits and RLC circuit , and I do fix some simple PCBs from time to time , thank you so much for your advice , and Good luck for you 😇

You are thinking of the diode as if it is a resistor, but that is not the case. When the ideal diode is on, it behaves like a perfect wire with zero resistance (although the current can only flow in one direction). And like a wire, the ideal diode can therefore conduct any amount of current with zero voltage across it.

@@wtimothyholman I'm still not getting it because from what you said and with that theory if there is no voltage source even hooked up to the ideal diode it would start conducting current because the Vd is essentially 0? Is this thinking wrong and if so please correct me Its really bugging me that I cant understand this. Thanks

If there's no independent voltage source or current source connected to a circuit containing an ideal diode, then no current can flow through the diode. A ideal diode either behaves like an open circuit when "off" (and Vd < 0), or an ideal (one-way) wire when "on" (and Id > 0). When an ideal diode is turned on, the voltage across the diode is zero. You can replace the diode with a wire when calculating other voltages and currents in the circuit. Whatever current flows through the diode will be a function of those other components in the circuit. But by itself, an ideal diode cannot create current on its own.

@@wtimothyholman oh ok. I get it now. When the diode is on it behaves like an ideal wire and the voltage is 0. Thank you so much. Now I can go to bed peacefully.

If you want problems to work along with solutions, I would recommend a copy of Schaum's Outline, available on Amazon. Schaum's Outline of Electric Circuits contains lots of example problems using the basic circuit techniques needed for the analysis and design of transistor and diode circuits.

Thank you! Too many textbooks tend to make supernodes and supermeshes sound far more difficult than they actually are.

You could, but only if the phase angle difference is 180 degrees or less compared to the waveform you're comparing it to.

"Microelectronic Circuits" by Sedra and Smith is the book I have used in my courses. Any recent edition will do.

If you mean filter design with op amps, frequency response, poles and zeros, and Laplace Transforms, then I do have the notes for such a course and may begin work on it later this summer, depending on my schedule.

@@wtimothyholman great!! hope that you find the time to record. Thank you again for helping us 🙏

You are correct. Using vL = L*(d iL/dt) is faster, but you could write the following equations and solve for vL and i2 simultaneously: 2e^(-12.5t) + vL/5 + i2 = 0 and i2 = (vL - 3 i2) / 2 No matter which method you choose, you will get the same answer. In fact, solving a circuit two different ways is the best way to check your answer to confirm that it is correct.

@@wtimothyholman thank you sir, had my final for the course today and went decently well, I was in a bad situation after midterms, but performed better ever since i started following your lectures, will surely recommend them , thanks again!

You should use the standard Thevenin source driving technique. Look at Lecture 19 beginning around 1:08:00 to see the analysis for the input resistance.

Rsig is part of the input source, not part of the amplifier circuit. Rsig is the Thevenin equivalent resistance of the circuit that is connected to the amplifier input.

In the first part of the lecture, an ideal voltage source is connected at the base in order to simplify the analysis. Later on, a more accurate Thevenin equivalent circuit of Vsig + Rsig is used. In actual circuits, any input source will have some amount of intrinsic Thevenin resistance, but it can sometimes be ignored if it is small. For example, a good laboratory signal generator will have a Thevenin impedance of 50 ohms. If the input resistance of the amplifier is large, then that 50 ohm impedance will have very little effect on the overall gain of the circuit.

In such cases, the Thevenin resistance is determined between a given single node and ground. The single node is node (a), and the ground node is the second node, or node (b). It is just not explicitly shown as it was in Circuits I. Note that when the 1A driving source is connected to calculate Rth, it is connected between the single node and ground.

@@wtimothyholman understood!

It is difficult for me to make this type of recommendation. You seem to be taking what is called a “survey course”, which covers a lot of different topics in electronics. The problem is that my lectures build on each other, so that you usually need to watch the previous lectures in order to understand the next lecture in the series. I’d recommend you watch Lectures 8 through 20 if you need to learn small-signal analysis. If you don’t, then you can stop at lecture 16. For ideal op amps, check out Modules 37 through 43 in the Circuits I lectures. For digital circuits, the only lecture available is Lecture 22. If you need a more comprehensive review of digital circuits, you’ll need to get it from another source.

I've taught several EE subjects in my career, but right now I only have Circuits I and Electronics I recorded and posted online. I do plan to add more lectures when I teach my Advanced Analog Circuits courses this fall. I may also add some specialized lectures in different subjects this summer.

@@wtimothyholman will wait for it sir ❤❤