New answers tagged

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Yes, with a circuit called a voltage doubler or multiplier - a circuit that uses capacitors and diodes in a ladder structure to ‘stack’ the voltage. The amount of current is limited by the capacitor sizes used, so it might be appropriate for your relatively modest current requirement. Don’t run it directly from the line - use an isolation transformer. This ...


2

Yes technically this is possible ... although I would not recommend bringing this to practice without a firm understanding of the basics! You say low current, but depending on who you ask this might be different by orders of magnitude. (ECG specialist vs welding engineer. You get the idea.) Can you elaborate on the current requirement or application? There ...


1

You can use an isolated or a high-side current sensor. LEM makes some very good ones (however they are not cheap) that are non-contact- just run a wire through the sensor. There are ICs that work from the high-side shunt and pass a current proportional to the sensed current. For example, the Si8540. You do have to ensure that there are no transients on the ...


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You could measure currents in that range by using special IC's designed to measure current, for example the ACS724LLCTR. Here you can find all the current sensors offered by digi-key.


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Buttons on remote controls are often arranged in a matrix (grid of rows and columns) to conserve I/O pins on the microcontroller. The rows are polled while the microcontroller observes the voltage on the columns to determine which, if any, buttons are being pressed. It is likely the button you are trying to measure the voltage of isn't seeing a voltage for ...


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If you charge them as individual packs like the picture bellow, you should not run into particular issues. But ! The safe discharging of the 4 batteries that you separately charge will rely on the quality/precision of your battery charger. Since the only way your complete battery is balanced, is by reaching the maximum voltage of each cell to that of the ...


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Well, the other people who answered your question have shown why it is the case. Now use the math so show what the equations tell us. We have the following circuit: simulate this circuit – Schematic created using CircuitLab Using KCL, we can write: $$\text{I}_\text{i}=\text{I}_1+\text{I}_2\tag1$$ Using KVL, we can write: $$ \begin{cases} \text{I}_\...


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The definition of a current source is that the output current will not vary with a varying load. For a resistive load, this means that \$V_{load}\$ can vary while \$I_{source}\$ does not vary. We can solve this using Ohm's law. Provided the source is in its compliance range (the range of output voltage it can support), then the dynamic output resistance \$...


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1.Why can we seem the current source as a open circuit,and voltage source as a close circuit? Short answer: By definition. A real voltage source is modeled as an ideal source with small output impedance connected in series. And, a real current source is modeled as an ideal source with large output impedance connected in parallel: Look at the left-most ...


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Why can we see the current source as a open circuit and voltage source as a close circuit? If you had a 1 amp current source, that could be approximated by 10 volts in series with 10 ohm and improved by 100 volts in series with 100 ohm and improved again by 1000 volts in series with 1000 ohms. Do you see where this is going? Why can we know when RL=Rth=...


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According to the picture, the 6.8 V is across the series combination of both Zener voltage and the resistor. With out further context, I can only go ahead with the image. As per the image, the equations derived are right. It is the representation of Zener diode as a whole. Hence the statement, the Zener diode has 6.8 V across it. When you split Zener ...


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Black dots do NOT indicate nodes. They are visual markers. A node is the complete interconnection between circuit elements. There are four nodes in your circuit. An essential node a node where three or more elements meet. There are three essential nodes in your circuit. The bottom-most node in your circuit is one of three essential nodes.


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Well, using a different method that makes it a bit more clear maybe. We are trying to analyze the following circuit: simulate this circuit – Schematic created using CircuitLab Using KCL, we can write: $$ \begin{cases} \text{I}_5=\text{I}_1+\text{I}_4\\ \\ \text{I}_4=\text{I}_2+\text{I}_3\\ \\ \text{I}_6=\text{I}_1+\text{I}_3\\ \\ \text{I}_5=\text{I}...


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I am assuming that you are using a NPN BJT in the diagram. Please connect a resistor of about 10k(higher) between base and the ground(battery negative) of the circuit. Connecting the pulldown resistor provides a known state instead of leaving it floating. Currently, When the yellow and Gray parts are not connected, the base of the transistor is floating.


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It's very simple: A pull-up resistor pulls the voltage up to the "high" logical level (5V, 3.3V or whatever is used as a "high" level) when the is no signal driving the input. A pull-down resistor pulls the voltage down to the "low" logical level (0V or close to it) when the is no signal driving the input. This is achieved because the input impedance of ...


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A pull-up or a pull-down resistor, when connected to a point in a circuit, provides that point with a weak default voltage which dominates unless some other stronger component forces it to something else. A pull-up means it pulls the point towards some positive voltage (typically the supply voltage), for example +5V, while a pull-down pulls the point ...


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where does \$ \frac{−10}{2}\$ come from? I don't understand! When the author tries to find \$V_X\$ voltage he simply converts the voltage source to a current source. Also, you should always track the units, in this case, you will see that \$\frac{10V}{2\Omega} = 5A\$ has a unit of amperes. $$ V_X = (2A - 5A) \times 4\Omega||2\Omega =-4V $$ where does ...


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The left-hand side of R1 is connected to ground (or more strictly, to the circuit's 0 V reference). Your professor has chosen to show current flowing from the output of the op-amp to ground, which is how we usually think of (conventional) current flowing.They also know that Vout = +2 V so current will be following in that direction. Although it's usual to ...


2

It's arbitrary which way you consider the current to flow as long as you're consistent. So if you pick current flowing from ground through R1 to R2 to the output, that will work too. The resulting values will be negative in this particular case (if Vin was < 0 the signs would be positive).


3

Your question and pictures don't make a whole lot of sense. These terms do not apply to circuits in the general case. An "open circuit" is a circuit or portion of a circuit in which there is no path for current to flow. An open switch is one in the "off" position. The impedance may not truly be infinite. But it is much higher than desired. High enough that ...


2

The reason why you measure an AC voltage on the box when you have the box/housing grounded might be stray currents. Your multimeter has a very high input impedance and thus it displays such voltages from stray currents. If you put an oscilloscope probe on your desk and do not connect it, you might also see some 50Hz AC voltage on the oscilloscope display. ...


1

I thought input capacitors are used only for providing a constant DC supply during power interruptions to the downstream circuit. In the first image, it is writter as "Add cap for filtering high peak currents". I dont understand how a capacitor can filter high peak currents. Please explain with an analogy. Capacitors can't provide DC current when the power ...


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SMPS switch at high frequencies to reduce the magnetics. Because of this switching, the current in some traces are discontinuous; i.e. the current switches from some very high value to zero at high frequencies. These currents need to be delivered by your source which can be at some distance. The distance (pcb trace, cable etc) translates itself into ...


1

I dont understand how a capacitor can filter high peak currents The input capacitor provides an unambiguous source of energy during the switching cycle. The level of energy required by the switching circuit can be high over a short period of time and, this can collapse the external input power supply if a capacitor is not used. When that "instantaneous" ...


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The simplest answer is that some components are polarized, that is to say they are only designed to work in one direction. If the flow of current is moving in the opposite direction, they will not work as expected. For instance if you were to add a diode ( a component that only allows current flow in one direction) to a simple circuit that connects the two ...


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The problem with using the batteries is that they will always be charging. Draw it out and notice how current will enter the positive terminal of the battery. That's opposite to the way batteries are usually used and means the battery is charging, instead of supplying. Always charging means the batteries will eventually pop. Unfortunately, to convert ...


-1

The thermal resistance of your transistors, if you only use a small copper area for the tab, is 104°C/W. If you ask them to dissipate 0.9W you should expect that the die temperature will increase by about 90°C above the ambient temperature. With a large copper area (50mm by 50mm of 2oz copper) you can get the thermal resistance down to 42°C/W, ...


0

Well, we are trying to solve the following circuit: simulate this circuit – Schematic created using CircuitLab Using KCL, we can write: $$ \begin{cases} \text{I}_2=\text{I}_\text{a}+\text{I}_1\\ \\ \text{I}_\text{a}=\text{I}_2+\text{I}_3\\ \\ \text{I}_1=\text{I}_\text{b}+\text{I}_\text{c}\\ \\ 0=\text{I}_\text{b}+\text{I}_\text{c}+\text{I}_3 \end{...


6

That's a crystal. Crystals don't have a supply. They are passive elements. You might be confusing this with an integrated oscillator, maybe, which contains a crystal, an inverting amplifier and an output buffer in one package. Those need a supply voltage and are useful as standalone oscillation generators. The crystal you link to needs to be driven from ...


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Why does the sign matter? It matters because many electronic components only work with current in one direction. simulate this circuit – Schematic created using CircuitLab Figure 1. Polarity sensitive components. D1, a diode, has a symbol which shows the direction of current flow (in the direction of the arrow). If the polarity is reversed current ...


0

Yeah, don't think so much about the actual numbers of the voltage (magnitude), but more so how the voltage is referenced to ground and how current flows in loops. Your power supply is looking for a high voltage at 6.6V and a lower voltage at the GND pin, and current flows from high to low potential. If you switch the two potentials then current would flow ...


3

Using your example of a hard drive, the ground reference is typically connected to the chassis of the computer, which is connected to the mains earth, which finds its way to earth potential in a building. So the chassis is at 0V. The supply voltages are something like +3.3, +5 and/or +12V. If you put a reference voltage somewhere else, perhaps the +3.3V ...


2

That is because there are devices in a computer that need both +12V and -12V, referenced to a GND in the middle. RS232 transceivers are the classic example.


1

Why couldn't -6.6v or -12v work? I thought when it came to voltage, only the magnitude of the difference of voltages matter? The square of the magnitude of the difference is proportional to power, essentially you are asking why isn't it enough to just make sure each device gets the right amount of power. For something like a heating element that just takes ...


2

1) Around but less than 4 days (500 Wh / 5 W = 100 hours = 4 days). Because the conversion efficiency is less than 100%, you will not get the full time. 2) it won't hurt the battery itself but who knows what the 5V circuit will do? In any event it is not adviseable to exceed the limits of the stuff you buy unless you are willing to take risk. Most likely ...


2

There's no initial conditions nor any turning switches, one must assume this is a steady state problem where starting transients can be assumed to have died off. You have three different sources which all have different frequency. This can be solved quite easily with superposition. You have 10V DC voltage source. To calculate its effect to Vo(t) remove ...


2

I would use a finite element analysis. Break the bar into small cubes of material and then model each cube as six resistors, all connected together at the center and one connected to each face. Assume that your two points are at the exact centers of two cubes. Use SPICE, or construct the resulting resistor network and calculate the current through each ...


2

You can use mesh equations for this (the most general and fundamental way), but the simpler way is to successively work your way from the bottom left up to the top right combining resistors in parallel and series. The rule for series it that the same current must flow through both components. Similarly, the rule for parallel is that the voltage across both ...


0

The "passive" in passive sign convention means that you are assigning positive voltage drops to passive components (i.e. loads). Negative voltage drops then occur across sources. Why you ask? Why not positive voltage drops across sources and negative voltage drops across loads (i.e. active sign convention which you will probably never see because no one ...


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Homework, only guidance will be given. The right half has no connection to the left half except R3. Thus there's no current in R3 => Vc=Va+5V. The situation would be different if there were for ex. a GND connection in the right half of the circuit, but there isn't, so the voltages at points b,c,d and e ride on what sources V1 and V6 generate to point a. ...


1

If you take the usual reference direction (right to left), you have \$V_Y + V_{R_2}-10 + V_{R_3} = V_X\$, or \$V_X-V_Y = V_{R_2}-10 +V_{R_3}\$. You have \$V_{R_2} = (0.2)(10) = 2\$ and \$V_{R_3} = (0.2)(5) = 1\$, hence \$V_X-V_Y = -7\$.


3

I think you know that \$I=I_1=I_2=I_3=I_4=200\:\mu\text{A}\$ as this is a series circuit and the currents must all be the same. (I'm agreeing with your computation of the series current magnitude and its sign as it relates to the direction shown in your schematic.) But let's modify your schematic slightly: simulate this circuit – Schematic created ...


0

The voltage source is 10V but when you probe it across the positive terminal and ground you will not measure 10 V. So, the second way of solving is wrong. The currents are equal but the equations are wrong due to the 10 V confusion.


5

The positive terminal of the source isn't at 10 V relative to ground, it's at 10 V relative to the negative terminal of the source.


0

First, notice that there is zero current flowing in the circuit, therefore there is zero voltage drop across each of the resistors. \$\small V_{AB}\$ is the voltage of A above B, so start at B and call this 0V. To travel from A to B, you must first go down through 20 V, and then up through 10 V, hence: \$\small V_{AB}=-20+10=-10\:V\$ \$\small V_{CD}\$ is ...


2

The relation between charge Q, voltage V, and capacitance C is Q = V*C. If something has high voltage, and low charge, then it has low capacitance. The capacity of the human body with respect to ground is often quoted as 100pF. This is an approximate, ballpark, figure. When you calculate the capacitance of two conductors in a given geometry from first ...


1

Voltage is always measured between two points. So Va = 10V does not make much sense except when a GND is defined somewhere and all measurements are implicitly referenced against that. But there is no GND in your circuit. Vab or Vac, and Vdb are okay because it tells you which two points you are measuring between. In this case, since you are only interested ...


1

You have to pay attention to the signs of the voltages. Look from A to B and count the voltage from 0V. Across R1 is 0, so we have 0V at the next node Across V1 is -10V so we have -10V at the next node Across V2 is +20V so we have +10V at B Similarly Vcd is -30 + 0 -10 + 20 + 0 = -20, which disagrees with your solution, so either it's Vdc or the ...


0

You can use superposition. Replace each pair of voltage sources in turn, leaving one, with a short and calculate the voltage at the T-junction for each, then add all three voltages. So you'd be adding the voltage due to the 10V battery, the voltage due to the 2V battery and the voltage due to the 15V battery to get the total. The difference between ...


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If you are studying circuits you look in you text book or online how to solve circuits like this with loop mesh equations. Very easy with just two loops like this circuit.


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