New answers tagged

1

You are correct in your analysis of what's happening. Fortunately, the fix is easy. Change your output circuit to simulate this circuit – Schematic created using CircuitLab Now the lower transistor's base current will pass harmlessly to ground and not through the LED. A couple of things to remember: 1) As Transistor says, you've got the battery ...


1

Your analysis of the problem is correct. You have chosen a poor AND gate circuit - but one that is promoted in many tutorial sites. A couple of comments on your schematic: The battery is connected backwards. The convention is to have the positive rail at the top, ground at the bottom. Circuit should generally read from left to right with current flowing ...


0

Add a resistor across the red LED. Maybe 1K depending on Vb. But it would be better to move the red LED and series resistor to the collector of T1, ground the emitter of T1.


1

As to DNKguyens reply it was answered in a couple of ways first off when the transistor sinks the voltage there is none left to flow through the led theoretically playing the Devils advocate we know that there is likely .7 volts left which is far too low (especially) with the resistor is series with the led to make it gate and fire or turn on. So in short ...


0

The answer here begs another question. First off is this just a mental exercise or are you trying to do. Work? If what you are doing is to turn the led off an on this is a bad way to do it, because it uses current all the time and generates waste heat. Now if you are supplied with unlimited supply power this might not be an issue, although if your supply is ...


0

The transistor and LED are both nonlinear semicoductors, they do not obey Ohm's law. When the transistor is fully on, the output voltage across it drops to perhaps 0.3 V, below even the base voltage. The diode needs a minimum of around 0.65 V (depending on the material used) before it begins to conduct, so it cannot do so. When the transistor is fully off, ...


1

I'll have a go at the layman's answer. Your circuit can be considered as have only two conditions, When the transistor is ON and the other when it is OFF. In the ON case, the transistor resistance drops very low, and can be considered a straight piece of copper wire between the ground and diode/resistor junction (or if you want to get mathematical, then a ...


0

The gate or the switch is the deciding factor for current to flow in either directions. When the switch in open the base gets no signal and so the transistor is off and all the current has to flow thru the LED bypassing the transistor. But when the switch is closed the base gets a biasing voltage and starts to conduct so much that all the current flows thru ...


0

a transformer is a feedback system, where the core implements a comparison of flux from the primary current with the flux from the secondary current. If this comparison changes, then the primary voltage will slightly vary, in the direction that acts against the source impedance of the external energy to restore the primary/secondary voltage ratio. Again, ...


0

A Transformer Cannot Regulate Voltage or Current Well, a transformer can regulate the voltage, increasing and decreasing it. To regulate something, is to keep something a constant (does not change), even if the operating conditions have changed. For example, a voltage regulator keeps its output voltage unchanged, no matter how the input voltage or load ...


0

What you're groping around for is a current transformer. Yes; that is a thing. And it's at the heart of how discharge lighting works. Arc Discharge lighting (neon, fluorescent, metal halide, mercury vapor and the sodium lights) is a sort of "arc light" going through a special mix of gases. This technology is much older than incandescent lights, and Tesla ...


1

There's yet another way of looking at it, that involves looking at the characteristic curves of the parts involved. Perhaps that helps you figuring out what is going on. First, figure out how much base current there will be through the transistor when it is on. The base-emitter junction has the characteristic curve of a diode, so you should be able to ...


25

tl; dr version: The transistor's Vce(on) is lower than the LED's Vf, so when the transistor is on the LED is well below its Vf threshold and so doesn't conduct. LEDs, Vf and Hue The LED, like all diodes, has a forward anode-to-cathode voltage, Vf. The LED will not conduct until the Vf threshold is reached, after which current climbs rapidly. This Vf ...


1

a transformer can regulate the voltage , increasing and decreasing it . That's not how we usually describe it. When we say some circuit regulates a voltage, we mean that it produces the same output voltage, even if the input voltage or the output current changes. Insensitivity to input voltage is called line regulation, and insensitivity to output current ...


1

Varying the transformer voltage only adjusts the current for a fixed load but it's not regulating the current. Typically we like that fact that a transformer outputs a fairly constant voltage for varying loads. This is how the supply to your house works. You and your neighbours are merrily switching stuff on and off through the day and night and you expect ...


4

Roughly speaking, when the transistor is on, you can replace it with a piece of wire... and the LED will be totally shunted. So, all the current will flow (be diverted, steered) through the "wire". You can remove the LED and all the elements except the 1 k collector resistor and the power supply. Your circuit will consist only of two elements - the resistor ...


18

The mathematical truth is that the current will actually split between the LED and the transistor, but the current that flows through the LED will be effectively zero. The LED only starts to draw significant current when the voltage across it gets relatively close to the normal operating Vf (typically around 2-3V for visible light LEDs). The transistor ...


0

Current is the amount of electric charge passing through a cross-sectional area per unit time. A voltage can be thought of (but isn't!) a force. Electric charge is carried by electrons. Voltage provides an energy differential to cause the electrons to move in one direction overall. With that picture, we can now see that what is being converted into work isn'...


3

Yes, a resistor will slow down the rate of current flow (for a given potential). But that will cause a "traffic backup" which will also slow the rate of current flow into or out of any junction feeding that resistor to that same slow rate. Basically, in steady state, electrons can't "bunch up", because the increased charge concentration will repel them away ...


0

if the sum of electrons entering a node is not longterm zero, then eventually there will be arcing. Does your phone arc, even if you leave it on?


2

how can I estimate the average current from the RMS current? Generally you can't. For instance, think about a regular AC power waveform; the current will be a sinewave and a sinewave has an average value that is always zero (despite having a significant RMS value). However, if you want to understand how your half saw-tooth RMS value converts to "average", ...


1

If you don’t need USB communications, the DP/DM lines can be left open. If you do, you need only the standard pull-down on DP or DM so the host can determine if you’re a low-speed or full-speed/high-speed device. Regardless, the downstream-facing port will deliver its rated current no matter what the endpoint devices does. If your device exceeds that rating ...


0

To start with, current (A) and power (W) are two different things. How did you figure out that you need 900mA of current for your device? Did you take into account the efficiency of your power supplies to apply back to the USB power feed (eg. 5V)? As you mentioned correctly, a standard USB 3.0 port, like the one you can find on a computer, can only deliver ...


1

As you can see, if using 1st equation, we can get output power 50W, while if using 2nd equation, we can get output power 96W. No, you can't use the 2nd equation meaningfully to obtain output power. The secondary RMS value of 3.97 amps is the current flowing through the secondary winding and, it may have an average (DC) value that is somewhat related to ...


2

Some relationships: - cos(wt) = sin(wt + 90°) sin(wt) = cos(wt - 90°) Take your pick on which you use because \$\theta\$ can be made to be 90° degrees at will.


-2

Well, we have the following circuit: simulate this circuit – Schematic created using CircuitLab I used Mathematica to solve this circuit. The code I used is the following: FullSimplify[ Solve[{I1 == I6 + I5, I6 + I2 + I9 == 0, I8 == I2 + I1, I9 == I3 + I4, I3 + I8 + I7 == 0, I4 == I7 + I5, I1 == (V5 - V6)/R1, I2 == (V4 - V6)/R2, I3 == (V4 - ...


2

If a fuse burns out when too much current is flowing through it, then how does turning on all knobs to their full setting burn out the wall-plug? Turning on more rings will draw more current (in the same way that turning on more water taps will draw more water current). If the wall-plug burns out then it is inadequate for the job. Shouldn't turning on ...


0

It may not be enough, depending on what you are doing on the tablet. It may try to pull more current than just 1.5A. And the tablet may need the battery sensor connection, as its not just power and ground. Some devices are finicky about that (for example, a Mifi 7700 hotspot). You can certainly try it to see if it will run or not.


3

Your correct course is to use a larger relay with a higher current rating. Simple as that. The issue is thermal, and is caused by current. The voltage doesn't really enter into it. So using a lower than intended voltage doesn't help. Voltage matters to the relay's ability to interrupt an arc. At your low voltage, that's not the question; it's the ...


3

Some switches and relays are rated for higher current at lower voltage, however this relay is specifically limited to 10A in the datasheet, even at 24V. If you exceed the maximum rating, especially by a factor as large as 50%, you can expect unpleasant things to happen.


4

The current rating determines the \$I^2R\$ losses of the closed contact hence, operating at 15 amps when the rating is 10 amps is going to result in early failure of the relay. The voltage rating determines characteristics of the contact when it opens. There is usually no trade off.


0

@Andyaka Can you show me how to do that? Consider this simple modification: - I've created V2 (36 volts) just to make the picture easier. Then if you temporarily remove R1 at the purple squares, you have two voltage sources and two potential dividers: - V1 feeding R2 and R4 = 18 volts in series with 2 ohm (R2||R4) V2 feeding R3 and R5 = 12 volts in ...


1

Addressing your points in order, 1) I wonder if the solution they have is using a transformation of the source. You can convert the 2nd circuit you have to the following one via source transformation. The voltage source in series converts to a current source in parallel with that same resistor. The current source's rating is calculated using V = IR, where R ...


0

You have two good options: Get yourself a current limited power supply that can deliver 40A. These are pretty hard to find actually, but I can personally recommend this one. Good option too because it's a useful piece of equipment for other projects, and not terribly pricey. Get a DC electronic load that can sink more than 40A and use your battery pack to ...


0

So what you need is a "DC electronic load". You can do it yourself, your principle schematics seems good, (example) but it will probably be a hard project. Or you can buy a laboratory DC Electronic load. Some electronic loads have related batteries features (capacity measure, ...). It exists Power Supply + Electronics loads in the same device in order to ...


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 ...


1

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.


-1

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}...


0

For correct operation of an NPN bipolar transistor the Collector should be more positive than the Emitter. If the voltage is reversed then the Collector and Emitter swap roles, the Collector effectively becoming an Emitter and the Emitter a Collector. In this 'reverse' mode the current gain is very low, but the Collector-Base junction is usually larger than ...


-1

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 ...


1

In general this will not work. simulate this circuit – Schematic created using CircuitLab Figure 1. Really simplified schematics of (a) what you've got and (b) what you want. Most bench power supplies are of the form of Figure 1a. They are designed to provide current, not sink it. The output stage will have beefy transistors from the positive power ...


1

For safety reasons, AC current measurement typically performed using either a current transformer (CT), or a Rogowski Coil, or a Hall Effect sensor. For various reasons—not the least of which are equipment safety and personnel safety, current measurement in an AC power distribution line is typically not performed by inserting a series resistance into ...


1

A homework, I guess. About 200 years ago it was generally accepted that static current distibution generates a magnetic field which has vector curl equal with the current density. The most common form of that law is this: So, present H as vector field and calculate its curl. Nabla is for rectangular coordinates, but cylindrical coordinates fit better in ...


1

The boiler will always put a DC current on the bus, so if nothing is connected, the voltage on the bus will be between 24-48v. Transmitting Data: The thermostat needs to “push down” the bus voltage to either around 6v (for transmitting a low level or idle) or around 17v (for transmitting a high level). U1 will select either D7 or D6 to put 4v3 or 15v to the ...


1

Each buffer has active current limiting with a soft transition towards either rail. This is common practice for Op Amps to maintain low output impedance and low error load regulation , yet be short circuit protected.


0

Note that this is a typical value. A transistor isn't a linear device. The 0.5 ohm DC impedance is probably the incremental resistance for Rl=2K to ground as seen at the top of the chart. For a short circuit, you'll be on the flat part of the curve, and the current will probably not change much from 0.5V (overloaded) to 0V (short circuit). It will ...


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

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 ...


0

The increase is the capacitor charging up. The decrease or negative current is the capacitor discharging which is why the current is decreasing. The horizontal line is the time that the capacitor holds the charge for before it discharges. The job of a capacitor is to store charge/energy in a electrostatic field between two plates.


3

what is the physical meaning behind this negative current or what has happens to the charges to make the current negative in this interval ? A negative current is a current flowing in the opposite direction from whichever direction you decided positive current flows in. If this graph is about charge on a capacitor plate, then positive current is current ...


Top 50 recent answers are included