# Tag Info

14

Ground is 0 V not because we define it like so, but because we measure its voltage with respect to itself. Ground is the point we choose to measure all voltages in the circuit with reference to. So in a strict sense, according to the definition of ground, all points/nodes are grounds! No, because we didn't choose them as the reference for measuring other ...

7

There is no one definition of 'Ground'. 'Ground' is an over-used word that gets used to label several related but subtly different concepts. The best way perhaps to think about it is in an operational or contextual way. That is 'if you want to do thing, then is something like ground useful to you here, and if so, how should it be defined?' The definition of '...

7

I think the term we use for the zero volt point in a circuit should really be called "reference point" or "Common", rather than "Ground", but this misuse of the term "Ground" is so deeply embedded in electronics that we're stuck with it. The point we call "Ground" in a circuit simulator, in your battery-...

6

I feel like Thomas Edison really screwed me when he made everyone believe that current is movement of positive charges. As we're 'clearing things up', let's be very careful in our language. Current is the movement of charges, both negative and positive. Current, aka Conventional Current (because its direction is subject to a convention), is defined to flow ...

4

You'll lose regulation, and the output voltage will be slightly less than the input voltage. The buck converter topology can only reduce voltage, not increase it, because it's essentially just filtered PWM on the power supply. If you want to have an output voltage higher than the input, you'll need a boost converter, and if you want the input to be able to ...

3

Swap the resistor and switch Nothing is perfect. Your switch has resistance. Wanting "0V" is like saying you want something to be exactly one meter long.

3

This is an interesting solution, not a constant current source. An ordinary 2.7k resistor in a given input voltage range. It will be much more interesting to watch at lower supply voltages. The characteristic will have a phase with a negative dynamic resistance between 5 V and 14 V. It's a two-terminal device that exhibits an area of differential negative ...

3

I'll not address the AC behavior of this circuit- I suspect the AC characteristics might not have been 'designed' as such anyway. So ignore the capacitors. You can equate the voltages at the inputs of the op-amp. The voltage at the non-inverting input is simply Vload (we can ignore offset voltage and input bias current- note that the designer has reasonably ...

2

why is there a LC filter at the entrance of the converter? That looks like a noise removal filter. The noise is generated by the 12 volt switching converter (due to high speed switching) on the left and therefore, the LC immediately to its right (its input port) will act as a 2nd order low pass filter and noise/switching artefacts travelling from left to ...

2

No, you can't use just capacitors to do that. They will just sit there and do nothing unless you use the caps as part of a charge pump which can increase voltage. But it would be a bad idea to boost the battery voltage and then use a 7805 as that would increase losses and reduce battery life. If you don't want to bother designing a switching converter, you ...

2

If you want to control a resistive load like a heater in an actual proportional way, you would typical do so with a TRIAC driver. But while a phase cutting TRIAC "sort of" varies the voltage in the sense that it may prevent the load from ever seeing the peak part of the waveform, in reality what it mostly does is create a form of pulse width ...

2

Neither are accurate. yes it must include R4802, but the Zener control has high gain and the PNP has hFE so the negative feedback reduces your node effective input impedance. ( such as if you were to inject noise from some Rs) But since the Zener control current is small, it has negl. Loading effects. (Verify) Ticky Tacky Technical Details

1

For $v(t)=10cos(30 \pi t)$ the maximum the $cos$ function can return is 1. Therefore $V_{p} = 10 \ \text V$ and $V_{p-p} = 20 \ \text V$.

1

If the input is at or near 0V then the transistors will be cut off regardless of the presence of the supply voltage Vbb. If the LED forward voltage is high relative to Vbb - 0.7V then Q1 can saturate, and at some point (maybe when Vce for Q1 falls below 100mV or so) the current will no longer be regulated. It is not possible for Q2 to saturate under normal ...

1

Q2 is approaching saturation as the collector is one diode drop above the base. The saturation region begins where the collector base diode becomes forward biased. Just what constitutes forward bias in a transistor is somewhat subjective; the junction will start conducting at quite a low voltage. If the programmed current was high enough to cause Q1 to heat ...

1

Using the electric potential, the potential energy is given by: U = q * V where q is the charge value of an electron, which is negative: q = -1,6*10^(-19). Taking into example two poles of a circuit, having 5V and 0V as values: For 0V: U=0 For 5V: U=5*(-1,6*10^(-19)) which is a negative value As a law of nature, everything is attracted to the lower energy ...

1

Electrons have, by definition, a negative charge. They are attracted to a positive charge and repelled by a negative one. So, yes, electrons flow from negative to positive charge. We think of them that way when studying physics and electrochemistry. ‘Conventional current’ on the other hand flows from positive to negative. This, er, convention predates ...

1

The Rb is the resistance used for calculating the Voltage division of the 5V between R8403 and "Rb". It is derived by looking down from R8403 looking at the impedances. You first see: R4802 || X. When looking at X you see the SCR (which is infinite when switched off so you dont include it) and you see R4805 in series with a certain Y circuit. So ...

1

The electric field is defined as pointing from positive to negative. Electrons are repelled from the negative and attracted to the positive, hence they move in the OPPOSITE direction that the electric field is pointing. So I guess you could say electrons want to move from the more negative/less positive voltage to the less negative/more positive voltage. In ...

1

In a typical DC/DC converter a proportion of the output voltage is fed back through a voltage divider into the regulator IC, which compares it to an internal reference voltage and raises or lowers the output voltage to make the feedback and reference voltages equal. The block diagram looks something like this:- simulate this circuit – Schematic ...

1

This circuit/sensor uses 81pF as the resonant cap. Thus we can have substantial input capacity of the amplifier (several PF) without serious upset to the energy storage and tuning. To avoid dampening, the interface needs to have impedance >> the reactance of 1Henry at 17,000 Hertz, which is about 100,000 ohms. First, lets bias that sensor to VDD/2 ...

1

In a line of 1/4 watt resistors, they're all the same size with similar construction and packaging, no matter what the value is. The power they dissipate is turned into heat, and no matter what the value is they can dissipate 1/4 watt of heat power before they get too hot and stuff melts. Wires, on the other hand, are sold and used by the foot and you can ...

1

You can also use an IC such as ADE7759 which is a digital integrator with high accuracy no need for analog circuit. You can check also this paper why R-Coils are more common the last 5 years for high current measurments. https://www.analog.com/en/technical-articles/current-sensing-for-energy-metering.html

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