25

Yes, actual electrons do move through conductors and yes, they cross boundaries like connections between two wires. While electrons do have mass, there is no net addition of electrons through a conductor. If N electrons are added at one end you will get N out the other end for a net mass change of 0. Keep in mind that electrons move in a reverse direction (...


11

Just a short answer to get you thinking ... When we connect two conductors together in a junction, do electrons actually move from one conductor to another? Electrical current is the flow of charge. This is stated neatly in the equation $$ I = \frac {dQ}{dt} $$ where I is current, Q is charge and t is time. Now consider a DC circuit with a switch. When you ...


5

Electrons have an appoximate mass of \$10^{-30}\text{ kg}\$. Even so, when current flows in a loop, the electrons bump in one end and bump another out the other end of the conductor, so no change in mass occurs. Electrons can jump across contacts when the voltage is different when contact is made. When the current is high enough you can hear it and maybe see ...


5

\$ v(t)=L\large\frac{di(t)}{dt}\small+Ri(t)\$ \$ v(t)=L\large \frac{d}{dt}\normalsize 2\:cos(100\pi t)+50(2\:cos(100\pi t))\$ \$ v(t)=0.1\times (-200\pi \:sin(100\pi t))+100\:cos(100\pi t)\$ \$ v(t)=-20\pi \:sin(100\pi t)+100\:cos(100\pi t)\$ we have: \$100\pi t=100\pi\times 0.0025=\large \frac{\pi}{4}\small\equiv \normalsize 45^o\$ hence, \$ v(0.0025)=-\...


5

You're missing a lot of basics missing here. The first and most egregious is thinking that the motor doesn't affect current draw. By doing what you did, you basically assumed that the motor has no control over how much current will flow through it. This is false since it draws more current when it needs to apply more torque. The motor doesn't need a resistor ...


5

Current mirrors are quite common in audio power amplifiers as well: Source TR10-TR11 pair in the schematic above (bottom-left) form a current mirror to make the collector currents of TR2-TR3 differential pair equal. Fun fact: A class-AB power amp is a bigger (more powerful) version of a transistor-operational-amplifier.


4

The reasoning goes a bit like in circles. The more base current flows, the more collector current flows. The more collector current flows, it brings down collector voltage, and it reduces base current. So there is an equilibrium where voltages and currents are stable, so just the right amount of base current is in balance with just the right amount of ...


4

simulate this circuit – Schematic created using CircuitLab Figure 1. The universal op-amp and variations "with bits left out". What is Ag, Rg, or Ai? This has been addressed in some of the other answers. Also, what is the purpose of these types of Op-Amps? There's nothing special about the op-amps. It's the way they're used that's special ...


4

Inductors are internally represented as voltage sources, so unless you have some series resistance in there, somewhere, the circuit sees two voltage sources in parallel, which is not possible. The same for two current sources in series. You can either add some minor resistance (something like 1 mΩ, or so) in series with any of the two, so that the local loop ...


3

The below picture should shed more light on the way these currents circulate in the circuit: The differential-mode currents can be quite easily predicted with a simulator considering the below scheme while common-mode types are more difficult to predict/model as they involve PCB layout, stray capacitance with heatsinks (if any) or inter-winding capacitance ...


3

Certains pins of the ESP are boot mode selection straps, and are internally weakly pulled high or low to boot normally, but these can be overridden externally to select other boot modes (to load firmware etc). The LCD data, R/W and RS pins all have internal weak pull-ups to LCD supply voltage. These weak pull-ups of the LCD module override the weak pull-...


3

You have no model for the opamp, every component needs a model: add the line .lib opamp.sub for further reading on adding models see these links https://www.analog.com/en/technical-articles/ltspice-using-an-intrinsic-symbol-for-a-third-party-model.html How do I import a spice model into LT spice?


3

The leftmost part is a closed circuit, and that shorts the vertical 1k resistor on the left. You are left with two 1k’s in parallel, so 500 ohm, that goes in series with the 200 ohm resistor, making it 700 ohm. (answering because you tried in the comments)


3

The ideal op-amp is a device that Draws no input current Tries to do whatever is needed at the output to force its two inputs to the same voltage. If there is negative feedback, and the output doesn't hit either supply rail, it will succeed, this is called operating linearly.1 That universal diagram you have is intended to cover most op-amp applications. ...


3

in a recent lecture in my AC Circuits course, the professor put this slide up and hastily explained the Universal Op-Amp. Since I already struggle with circuits, let alone operational amplifiers, I did not understand his explanation very well. I think it's his pet circuit configuration. If I've ever seen that presented, I haven't found it useful to ...


3

"A transistor with a large gain will cause a large voltage drop over the collector resistor. Because of this, the collector voltage and the base current will decline." Imagine the circuit running just fine with some collector and base currents and the collector voltage sitting at, let's say, \$3\:\text{V}\$. This means that the resistor to the \$+...


3

Consider the case where transistor current gain \$ \beta = 220 \$... Thus, collector current is 220 x base current, and current through the 1k collector resistor is dominated by collector current. Since 220k base resistor is 220 x collector resistor, voltage drop across each of these resistors would be very nearly the same....because their ratio is the same ...


3

It means that the output voltage of the op-amp will not depend on what is connected to it. In other words, on a perfect op-amp, it means the op-amp will supply whatever current is required to keep the output voltage to what it should be. If you connect a 1kohm "load" or a 10kohm "load" the output voltage will be the same regardless, the ...


3

Current mirrors are frequently used in linear integrated circuits such as op amps. They are used to provide a stable current for biasing transistors and transistor pairs such as differential pairs.


2

During (+) cycle the Zener diode is reverse biased (Vz = 5). The other 5 V will be divided on the two resistors so the voltage across R3 is around 2.5 V. During the (-) cycle the Zener diode is forward biased (Vf = 0.3 ~ 0.7). Assuming Vf = 0.6 V so -9.4 V will be divided on the resistors, thus the voltage across R3 will be around -4.7 V.


2

Where does the number 11 come from? Figure 1. The current path when the left transistor is on. If we ignore the voltage drop across the left transistor when hard on then we have a potential divider given by the two resistors. $$ V_{U2} = \frac {470}{4k7 + 470} {V_{CC} } = \frac 1 {11} V_{CC} $$ Why does one need to add the emitter voltage to the BE-voltage?...


2

It's not universal op-amp, it's universal op-amp formula, which is the professors way of saying that this is a generalization of all of the previous circuits that you've been shown. The key takeaway is that by applying this you can get any Vout that is the weighted sum of some inputs Vi (V1 through Vn), using any weights Ai you wish. Basically, you can think ...


2

The two transistors in a current mirror should have equal characteristics; therefore, current mirrors are almost never built from discrete transistors, but are used inside chips where the transistors are next to each other on the same die. As a newbie, you would probably never design a circuit with a current mirror. In practice, current mirrors are most ...


2

I saw a real life use of a discrete current mirror a couple of months ago in this question. Here's the circuit diagram from the question: The circuit originally came from this thesis (see page 14.) There's a current mirror in the upper left corner. It is used to convert the voltage at Vin to a current to control the charge current (and thus the frequency) ...


2

Here is a trivial, but real life example. I had an existing design of a product with a motherboard having a built in current pwm sink for controlling the backlight of a single screen on the front panel. For a new product I needed to add a second screen without changing the motherboard PCB in any way. The question was how to control the brightness of the ...


1

Essence. This professor's creature can be called "universal summing-subtracting op-amp circuit" (not "universal op-amp" since the op-amp is just one of the circuit components). It reminds me of the distant past when analog computers tried to compete with digital ones... but soon lost that battle (analog computers consisted of building ...


1

Using non-passivated non-transformer oil < 1GOhm will produce too much leakage. The capacitance will also resonate in the circuit with high reactive currents. The surfaces of each part and the oil must be void of contaminants like fingerprints and moisture absorption into the oil. Also the solder joints are too sharp thus increasing the e-field gradient ...


1

The op-amp produces an output voltage (\$v_O\$) that forces the inverting-input voltage to virtually equal the non-inverting-input voltage. It does this using negative feedback via R2 and R1: - This happens because the open-loop gain of the op-amp is very, very high; sometimes in excess of 1 million. So, if you think about that sort of gain magnitude, for ...


1

In the model shown, output voltage depends on the voltage generated across the internal resistance R, due to the gm of the two input ports. There is no output resistance show - the output is modelled as an ideal voltage source. Thus, this circuit can (in theory) drive a load of 0.1R (in reality it probably cannot). A more realistic model would of course have ...


1

You need to do this: - Hopefully, that should work. The gain value for F3 should be 3 and not "F". You should not short I1.


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