# Tag Info

## New answers tagged current

1

Motors when starting draw from 3x ( soft-start 3 phase type) to 5x typ to 8x with Start-Cap surge current to 12x for high-efficiency BLDC motors. (10x typ) Your inverter is unable to handle the startup load. It is intended for Blenders, Vacuums and Power Tools and not an air-conditioner compressor which has a much longer startup time. For industrial use, ...

0

If you're not too concerned about the effects of self heating, then a circuit like the one shown in fig. 1 might be useful. simulate this circuit – Schematic created using CircuitLab Figure 1. Resistor R1 is the current sense resistor. If you know R1's resistance value (by measuring its resistance with an ohmmeter), and if you measure with a ...

3

John D has answered the inverter overload issue. I'll answer another part. With 40 amp hours shouldn't it be able to provide 40 amps of current at any moment (more than needed for surge)? To understand the maximum current from the battery you need to consult the datasheet. 40 Ah is an energy rating (sort of - a better energy measure would be Wh, watt-...

2

Your compressor is likely run by an AC induction motor, and they take very large surge currents to start. Your inverter is rated at 2kW, and no matter how much battery capacity you have it will cut off when it reaches its overcurrent limit. If your mains voltage is 120VAC 40A would be 4080W, well above the rating of your inverter. (Even worse if your ...

1

If your motor is not too big the DCR is significant . In your example 16VDC and 2 ohms the prospective stall current can only get to 8 amps .Low voltage mosfets that will handle this are cheap and easy to find .Adding External DC resistance like 1 ohm will waste power and reduce maximum motor speed when loaded .You were discussing motor inductance which is ...

1

Your understanding is good. Adding a series R increases the R/L slew rate of current but to lower final short circuit current so this does not protect the driver. Slew rate is the inverse of Tau=L/R ! The only way to provide OCP (over-current protection) is to sense current with say a 60mV shunt resistor ( e.g. 60mV/6A = 0.01 Ohm ) then amplify ...

0

@Kurt E. Clothier answered nicely what is the positive and what is the negative cycle. The second question asked to compare the two rectification types. Very short answer is: FULL WAVE RECTIFICATION keeps the positive half-wave as-is; and turns the negative half-wave into positive one = no power is lost (we kept both half-waves) HALF WAVE RECTIFICATION ...

0

Sorry for the answer to an old question, but it could use some additions. While Andy is technically correct, it is in fact recommended that current is measured as OP shows in 2nd picture, ie from one sensor, NOT between them. The reason is twofold: the current only flows through the resistors, so if you measure between them you do not minimise the ...

0

Some motor driver chips may include thermal overload protection to guard against moderate over-current conditions such as those caused by a stalled motor or a shorted winding, but require that peak fault currents be limited by something outside the chip. If a motor driver used to drive a jammed 24 volt motor whose resistance is 0.5 ohms, it might overheat ...

2

this link shows a precision current reference circuit with approximately a 10ppm stability http://www.ti.com/lit/an/sbva001/sbva001.pdf

6

Edit: Following the suggestion of Chris Stratton and lucasgbc, I tried to remove an inaccuracy I did due to a too quick reading of the datasheet and to emphasize the reasons of the results the asker found in his measure of the maximum output current. I hope I succeeded in providing a nicer answer. The maximum output current stated customarily in datasheets ...

15

Recently, I bought the H-Bridge VNH7070BAS from ST because it would drive up to 15A It will not drive anything like this level of current for anything more than a few micro seconds before the automatic current limit circuit operated. The internal transistors together contribute circa 0.1 ohm impedance in the current path and, at (say) 10 amps continuous, ...

-1

I returned this power supply and purchased another 10A12V power supply. It is able to supply 10A as expected.

1

Yes. Increasing the load on your battery, either by adding more fans and pumps or replacing the existing fans/pumps with fans/pumps that require more current, will lower the voltage available to all of these devices, including the cooler module. The reason is that your battery is not an ideal voltage source, so as the load current increases the terminal ...

1

Some suggestions. Use only a pure sine wave signal when testing and calibrating an AC measurement system. As a general rule do not use arbitrary wave forms like triangle, square, powered/spinning motors, aperiodic signals, etc. for testing / calibration purposes. Ensure your ADC sampling rate (samples per second) is high enough, ensure your circuit grounds ...

1

I think your statement about stepping down from 5v to 3.3v is confusing but it looks like your issue is solely with the output of the ACS712 AC current sensor which nominally outputs 0V-5V to a microcontroller ADC corresponding to a range of -20A to 20A through the sense inputs of the ACS712. In a normal setup with ACS712 providing your current reading to ...

1

... what factors determine this current value? Cable size. Connector rating. Fusing.

1

This entirely depends on: Country Protection devices fitted Building code In the UK a standard socket will provide 230V rated at 13A

1

How to block reverse current from battery to charger output? I assume here that you are already using the recommended layout of the LTC4416 AND that V1 is the AC-DC supply AND that V2 is the battery supply for your Power Path Switch. Your Power Path Switch circuit will look something like this: Since V1 is also used to drive the Charger DC-DC convertor ...

3

An ideal diode would be the best solution with a suitable low RdsOn FET. D4185 RDS(ON)< 20mΩ(VGS= -4.5V) \$3.19 simulate this circuit – Schematic created using CircuitLab

0

You have presented a number of solutions, but skipped over the most obvious one. You're using a DC adapter. This implies you have 120 VAC (or maybe 240 VAC) mains for the charging current. Just find a normally open relay with a coil the is rated for the same as the mains current and has contacts that can handle the DC current. Connect the coil to the same ...

1

It can often help to flip it upside down and think of conductance instead of resistance. Conductance has a unit of Siemens or S. Conductance in Siemens is 1 / resistance in ohms. C E = I Conductance * voltage = current E = I / C So for instance a 10 ohm resistor has a conductance of 0.1 Siemens. A 5 ohm resistor has a conductance of 0.2 ...

0

The ATtiny13 is spec'ed to draw ~0.2uA without WDT and ~4uA with WDT @3V @25C so if you are seeing 100uA then something else is going on here. There is a lot happening in the above code. I'd scale it back to just the minimum code needed to put the chip into power down mode and then measure how much current that uses without anything connected to the chip ...

0

Well first off I would say that an LED is an active device, which means voltage and current are not linearly related like a passive device such as a resistor. The resistance of an LED is not static, it changes according to the device's IV (current-voltage) relationship. That is more of an aside, since Ohm's Law is not as straight forward as it may appear ...

1

A string of LEDs is a bunch of LEDs in series and this means that they share the same current. If the forward volt drop across each LED is (say) 2 volts, then 5 LEDs requires a supply of at least 10 volts by the way.

5

You have a number of misunderstandings. Ohm's law applies to resistors. LEDs are non-linear elements and therefore Ohm's law doesn't apply. Elements in series have the same current flowing through them. If you have 3 LEDs in series and one has 50mA flowing in it, so do the other two. Total current? 50mA. Your source has to have enough voltage to "...

1

Multimeters measure resistance by applying a known voltage across and measuring the current, or vice versa, a known current being applied and measuring the voltage. You can't measure devices that are on this way because the voltage of the device is almost always higher than the voltage being applied, overriding it. On top of that semiconductors can have ...

0

You may also want to consider building a discrete high-side current source. Instead of multiplexing the output of the current source, you can add 18 parallel PMOS pass elements and multiplex the gate drive signal. In this case the multiplexer can be a jellybean component like the CD4051.

2

You can't measure resistance between VCC and GND when power supply drives voltage between VCC and GND. The multimeter is confused because there is an external voltage source. Also, measuring resistance of anything that is not a resistor will give weird readings, as they can have nonlinear behaviour like LEDs or semiconductors. At most you can measure if ...

1

What could be causing this? When the board is off this is a little unusual because most often the leakage current through transistors or loads when the power is off is in the 1-100kΩ range for most of the boards I have with voltage regulators on them. If you had something with a high side switch or a relay this could explain why the numbers are so high. ...

0

You could use three inverting one-of-eight decoders eg. 74HC138 to drive 18 3.3V (or less) drive P-channel MOSFET gates. For example, AO3401 which have Rds(on) of typically 60m$\Omega$ (85 max) with 2.5V drive so they will drop less than 15mV at 150mA. The three address bits on each decoder go to the three LS bits of your address, and the remaining two ...

0

Everything becomes conducting at a high enough voltage. If you are standing on a dry wooden floor and touch mains live you will probably not get a shock. On that same floor and you touch a high current 20kV PSU you are in a lot of fiery trouble.

0

As mentioned, you've used a resistor divider (R1 and R2) to level down the voltage, but did you check the datasheet for the burden resistor? So the resistor divider between Vout and the ground should have an equivalent resistance greater than or equal to the burden resistor ( R1 + R2 ≥ Rb ) I don't know why they have not mentioned the resistor in the ...

2

I'm having trouble understanding the flow of current from a live wire to the floor via a human. Some current is through conduction but also some current is due to capacitive coupling between body and earth. The human body has typically 100 pF capacitance to ground. Reliance on a conductive path does not mean current can't flow due to capacitance. In ...

0

Sometimes things goes wrong. Floor may become dirty, wet. A human can touch some conductive element connected to large capacity (Earth or other big machine). The most frequent occurrence is wet floor. Wet mud is conductive medium.

0

AN 8-channel analogue mux for 150mA would be rare and thus expensive. I would go for 18 LDO's with an enable input. You can place the current sense at the input of the LDO's. It would also prevent voltage drop. For calibration you can switch them all off and measure the current which you have to subtract. The problem for this solution is that I am ...

2

If you can find a multiplexer with a high current limit then yes, that is possible. Many chips have a 50mA limit, and multiplexers are typically not used for switching current. It would be better to find a high side switch configuration with a logic level input like this (There should also be integrated high side switches if you need compactness with the ...

0

The Resistors are dissipating nearly 5 Watts, and the LED's another 2 Watts. It's really unlikely that power dissipation in the copper is your issue. Your plan won't help.

0

I was wondering about answering this question, as you seem to be demanding others to do your work rather than putting in some effort to show some working. But I felt that as you are clearly very early stages of learning about this, I shall give you the benefit of the doubt. You may be new to this subject, but you should be aware of Ohm's Law: V=IR. You ...

0

The Ohm’s law states that the current through a conductor between two points is directly proportional to the voltage across THE (same) two points. It’s applicable to all circuits and transformer is not an exception. A mistake that led to contradiction is that (decreasing) current is measured not between the same points, where (increasing) voltage is. Current ...

0

You are confusing the "Lossless Transformer's" function with the resistor's function. The resistor's function is to convert the applied voltage and current flow to thermal energy for dissipation. The transformer's function is to convert an applied input voltage and current to another voltage and current with NO DISSIPATIVE LOSSES. For 10 Watts input at the ...

0

Don't muddle up the voltages and currents on different sides of the 24V power supply. If the power supply is putting out 24V 13A, then the current going in will be slightly more than 1.3A. Maybe 1.5A assuming 85% efficiency in the power supply. So provided that the switch is on the 240V side of the power supply, it will be well inside its rating. The ...

0

Several of your questions seem to depend on electrical building codes, which are very location-dependent. I won't try to answer those but I wanted to give some general feedback. You keep saying "technically 13A" and "technically 312 watts" as if you don't believe those numbers and you expect we will tell you to ignore them. Respect those numbers. Even if ...

7

The "left" side of the transformer (the side the voltage is applied to) obeys Ohm's law (technically a generalised form that describes impedance instead of just resistance). The currents and voltages that don't seem to obey Ohm's law happen on the other side of the transformer, in an electrically isolated circuit. Ohm's law doesn't describe how two circuits ...

3

The transformer uses the shared-flux of the core as a negative feedback mechanism. The primary and secondary fluxes ALMOST perfectly cancel, with the residual called the "Magnetizing flux". If the magnetizing-flux becomes too small, then more energy is taken from the primary (the energy source) and the core flux is again adequate to produce what the ...

23

Ohmls Law states V = IR. That means when we increase voltage we must also increase the current (I). That is true when feeding a resistor. But transformer increases the current while decreasing the voltage or decreasing the current while increasing the voltage. A transformer is not a resistor so you can't use Ohm's law on it. How does it happen? A ...

8

"when we increase voltage we must also increase the current(I)" while R is constant. You should look at the transformer from a power perspective: P=I*V and Power In = Power Out, Now, if you have 10V in and 1 A then that is 10W, so then power out is 10W If you have 10 times the number of turns on the output compared to the input side then you will get ...

2

There may be currents entering/leaving the nodes via the spurs. This doesn't matter, you don't need to find these, just solve the isolated mesh with the information given. $\small R_2$ has $\small 10\: V$ across it, with + on the left and - on the right. The $\small 5\:V$ source has + on the right and - on the left, since its voltage is -5V. The ...

4

After 30 years working with motors, I've never heard that term. For fixed field (permanent magnet or shunt-wound) brushed DC motors, and BLDC motors without speed control, the torque is fairly linearly related to the current, but is offset by the mechanical losses from ideal i.e. the line doesn't intersect the torque and current axes at zero. Normally the ...

-1

Neither of your answers is correct. If $I_x$ is -10mA then the total voltage drop across the two resistors is 20V and KVL is violated. If $I_x$ is 5mA then KCL is violated at the junction between the two resistors. Your problem is nonsensical as shown.

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