Hot answers tagged

4

This won't work as it required the current to be constant. In reality it is somewhere between 5nA and 550µA, depending on the state within the IC. That's what the voltage regulator is for. It permanently adjusts the voltage drop over the regulator that way the output voltage is both independent from the input voltage as well as from the current drawn.


3

At this current level, if you have 5-20mA to spare, you could regulate it with a 3.3V zener and a resistor of about 160 ohms. The zener goes from 3.3V to ground, and the resistor goes from 5V to 3.3V. If you're running other things off the 3.3V supply, take those currents into account as well. When selecting a diode, make sure that over all current and ...


3

The diodes D1 and D2 will each have a very similar voltage drop across them. Therefore they will cancel each other out, and as D1 is connected to 0V, there will also be 0V below D2.


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 ...


2

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 ...


2

The body diode of the NMOS is conducting, clamping the voltage to 620 mV when the PMOS is turned on (also conducting). You should rotate the NMOS 180 degrees as shown in left picture. But to draw a schematic correctly, you should actually put Vcc on top, and GND at the bottom (right picture). Note the pictures for the mosfets in LTspice have the both an ...


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 ...


1

The hose pipe analogy. The amount of water flowing per second is the current. The pressure is voltage, and resistance is the inverse of the diameter of the pipe. To increase total amount of water you either need to increase the pressure, or increase the diameter of the pipe. As for danger... a tiny diameter pipe at high pressure is not going to do you a ...


1

voltage requires electric fields; once the fields are in place, no further electron movement is required. amperage is moving electrons.


1

I'll provide a different approach than Jan did (he shouldn't have computed "RMS" for power), which may be a little more directly understandable. I get Jan's average power value, though. (I see he just deleted his posted answer.) Assume \$t_1=5\:\mu\text{s}\$, \$t_2=7\:\mu\text{s}\$, and \$t_3=13\:\mu\text{s}\$ (\$t_0=0\:\text{s}\$). We also know these ...


1

Ok, No, this will not work. Right now your strip's + is connected to ground, the most that would happen is that the strips power connections would both be grounded. But even worse is the simple fact that your switches have NO ground connection, so turning them on would have no effect. I edit this shortly with a diagram showing how to fix it, DONE. ...


1

You you can use a ultra low power voltage monitoring IC. This IC or similar IC will keep the MCU at reset state even down till 0.7 V of \$V_{IN}\$ . As soon as the Voltage drops below the internal reference, the MCU will be held under reset. This will make sure that MCU will work normally when the input voltage is back in its valid range.


1

In my opinion there are no problems here as your maximum output is 2V and the opamp is able to handle up to v+(2.5V The OPA344 and OPA345 are optimized to operate on a single supply from 2.5V and up to 5.5V with an input common-mode voltage range that extends 300mV beyond the supplies. When the datasheet states that VCM must be between -0.3 and -1.8 ...


1

In a perfect world, the current would be zero yes because once the capacitor "fills up" its charge, it'll behave like an open circuit and thus the current would be zero and your voltage V(t) would be equal to V(t0). Think of that as a taller water tank emptying itself out into another water tank. There will be equilibrium between the two water tanks. (...


1

120V-277VAC are typically rms voltages, so I will assume you want to measure the rms value. All three of your methods should be able to do that with a sufficiently fast ADC. Method 1 is able to capture the entire waveform without needing a bipolar ADC input, so it is suitable for use with ADCs that can only measure positive voltage. It relies on the ...


1

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=...


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

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 ...


Only top voted, non community-wiki answers of a minimum length are eligible