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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 ...


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, ...


12

Instead of saying the consumer (circuit) draws as much as it needs, it is better to say current will flow as much as is allowed by the circuit. How much current a circuit will allow is based on its resistance (or impedance). An LED has a voltage drop, but very little resistance. This small resistance means that more current will flow through it than is ...


11

I am using an infrared LED in a circuit and it seems like it is drawing around 15 mA at 5 V. That is unlikely unless it has a built-in resistor. Figure 1. The I versus V curves for a range of typical LEDs shows that the forward voltage of an IR LED is about 1.25 V at 20 mA. Source: LED IV curves.* Note that if you were to apply even 2 V to the IR LED that ...


8

It is powered from +/-24V and there are regulators for +/- 12V on top of the schematic. U4 is wired as non-inverting amp with a gain of 2. We got 8 CPC1918 OptoMOS relays, presumably activated by sending enough voltage into the "PADx" inputs to light the internal LED. These select feedback resistors between 10R and 10Meg (R6,8,10...). So we can just look ...


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 ...


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 ...


6

How to analyze an unknown circuits operation If the circuit diagram makes no sense to you then you need to incrementally approach the circuit and derive the functionality of each block in it. If the schematic represents something that has been successfully built, then you have to assume it's both workable and designed for the task. If you assume it has not ...


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 mesure of the maximum output current. I hope I succeeded in providing a nicer answer. The maximum output current stated customarily in datasheets ...


5

Not every gate needs current for it to operate. MOSFETs are an example of voltage controlled devices whereas BJTs are current controlled devices. A high input impedance means that no significant current will be drawn and as such components can be chosen accordingly. It also means that the voltage between the divider won't change according to current draw ...


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 "...


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 ...


3

Basically, there are two currents flowing in the common line, one for each bulb, but since they are equal and they flow in opposite directions, the effectively cancel each other out. If you draw the current for each bulb lit individually, and then superimpose them, you can see this. simulate this circuit – Schematic created using CircuitLab


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 ...


3

Power supplies are available in a wide range of voltage and current ratings. If I have a device that has specific voltage and current ratings ... is it OK to go lower voltage, or should it always be higher? What about current? I don't want a 10 A supply to damage my 1 A device. Voltage: The specified voltage of device and the voltage of supply should ...


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


2

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


2

The formula for the voltage divider's output \$V_{out} = V_{in}\cdot \frac{R2}{R1+R2}\$ only applies if you're not drawing any current of the output node. This because the formula above presumes that the same current flows through both the resistors R1 and R2. In other words, to have a stable output voltage Vout you either have to take the required output ...


2

Yes. It is your body surface area and dielectric fluids of capacitance that acts the interface to dielectric of freespace air and not the ground conduction or your feet.


2

Its easier to see the symmetry if you redraw the circuit. If the circuit is symmetrical, then the voltages of the batteries will be at the same potential because the voltage will be equal due to symmetry of the currents. An equal voltage means no current flow. However, in the real world this would not be possible, there would be some mismatch (it's really ...


2

I got confused because my understanding of current is that the consumer draws as much as it needs and the only requirement for the power source regarding current is that it must supply at least that. This is an extremely common misunderstanding. There is nothing fundamental about electronics that makes this case. It just happens to be a property of some ...


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 ...


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 ...


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 ...


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

If an op-amp amplify the difference between the V+ and the V-, how can it work if for the principle of virtual ground V+ = V-, and so if V+ = V- there aren't difference to amplify. You are right, In real life, the op-amp alone (without any feedback loop) is nothing more than a differential amplifier. The output voltage is the difference between the \$+\$...


1

this looks like homework, so I'm only going to give hints Write the expression for the total complex impedance of the circuit for all three cases (R, L or C) Assume a fixed voltage as input, calculate the magnitude of the current for all three cases Identify the case, where the current actually goes up EDIT based on the work done. There nothing special ...


1

Try doing the problem again with a much lower restance for the IC input, you'll find you get a very different answer. That is why it matters. Most (modern) IC inputs (not power supply pins) can be modeled as a large resistor. Again, this problem is talking about an input pin, not a power supply pin. The input pin could be one of the sense terminals of an op-...


1

Per the datasheet, the absolute maximum collector current (Ic) is 30mA. If you need 50mA you will need a beefier optoisolator. Possibly something like the LTV-825 From the LTV-825 datasheet, figure 3, you can see that you can drive a collector current of 50mA with a Vce of ~2V with an input current( If) of 4mA. Without knowing your application (drive side ...


1

Assuming that you are using this as a switch and not in linear mode ... The collector current is only rated for 30 mA absolute max, so you can't get 50 mA on the output. You either need to pick another part or add a transistor. If you can handle the higher saturation of a darlington, I would just do that. Now you have plenty of gain and the design is easy, ...


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