# Tag Info

1

This is a voltage controlled current sink that has a LED as its load BUt anything as a load would do: - Like I say - ignore the LED and assume it's just a resistor up to some abitrary positive supply voltage that needn't be the same as the op-amp. It works by ensuring the voltage across the emitter resistor is the same as the voltage on the potential ...

0

The main problem with driving P-channel high-side switches from logic levels is that the gate voltage needs to go to the source voltage level for the switch to turn off. That's 60 Volts in your situation, and the output of a logic control is only 5V. The simplest way to fix this is to use a small N-channel signal transistor to pull the P-channel gate to ...

0

If you're measuring resistance, a DMM typically forces a current and measures the resulting voltage. So leakage current in this context makes no sense. If you're measuring a voltage, then maybe he's referring to the input resistance of the DMM which typ is 10Mohm. This will definitely affect your reading if youre probing a high impedance node. That is, if ...

0

You can buy H-bridges with built-in drivers. Just look on Digikey. Their inputs are designed to be driven by a microcontroller and they'll often have a flyback protection diode built in as well -- verify this for your particular part. If you want to roll your own you can do it with 2 N and 2 P MOSFETs but you have to select the parts carefully, paying ...

2

I'll assume dBA is for amperes and not the 'A weighted' dB sound pressure level measurement. Zero dBA will therefore mean 1A RMS and 6dBA will mean 2A RMS. Strictly speaking 2A is closer to 6.021dBA but everyone accepts that if you double the current (or voltage) it increases by 6dB 0dBV is 1V RMS and by the reasons above 2V is 6dBV RMS values are always ...

1

No, a wire fed axially through the centreline of a solenoid would not induce anything into the solenoid coil. The magnetic fields (if both are energized) will be at right angles. There would be no transformer action. At much higher frequencies there will be circulating eddy currents induced in the solenoid winding (from the straight wire) but these won't be ...

0

In answer to your question #1: There is no need for you to measure this, this value in the datasheet is the manufacturer saying, "Do NOT exceed his value" or bad things will start to happen to your device. Typically maximums are not tested for in devices that will then be used in production as these are typically destructive or damaging tests.

1

On Semiconductor published a guide on how to interpret their IGBT datasheets. It says that: The pulsed collector current describes the peak collector current pulse above the rated collector current specification that can flow while remaining below the maximum junction temperature. The maximum allowable pulsed current in turn depends on the pulse ...

2

I'm not sure what you are trying to ask here. The voltage difference between the 5V rail and GPIO pin is 1.7 because 5 - 1.7 = 3.3V When GPIO voltage is measured it is measured from that pin to ground, which is 3.3V,.

2

In order for the zener to "clamp" your voltage at your desired output voltage it will have to burn the excess voltage and dissipate that power. There is no way around that. What you are trying to accomplish seems to be based on an LDO -- or a linear regulator. The biggest problems with LDO's are: 1. efficiency 2. power/heat dissipation. In many cases the ...

-1

Watch this video. Very explanatory. https://www.youtube.com/watch?v=Gwo3pEH7hUE

3

BC337 is a silicon transistor. Silicon PN junction starts conducting between 0.6V and 0.7 volts applied to it with P terminal being more positive than N. In your circuit, when current in 2.7ohm resistor reaches 250ma, it causes 0.675 volts across it turning on BE junction of transistor T2.

0

It's a big read and I hope I understand your basic question - maybe you haven't enabled the opto output on pin 7? Does this explain it: - Look at the first two rows - A high in gives a low out and vice versa BUT only when enable is high. It does say there is an internal pull-up in the data sheet but try it and see. Alternatively, the voltage you may be ...

0

The circuit you have drawn is an AC voltage doubler. It will only work with an AC input. You feed it X volts AC, and you get 2X DC volts out.

1

A couple of points: 1) I guess what you are trying to design here is a boost converter. While it is often called a DC/DC converter, the conversion itself is AC/AC. Basically, you create a square wave out of the DC voltage and then upconvert and rectify it afterwards. Almost no one designs those from scratch. If you need a simple voltage doubler, I would get ...

0

please help me I have some concerns about the current limit circuit (based around the BC337 / R2) and its ability to work with PWM - every time the pulse goes high from U3 (initiating a current flow through the MOSFET), the voltage developed across R2 will activate the BC337 and clamp the drive voltage to the MOSFET. This means the MOSFET will get quite ...

2

Converting a analog signal to a digital number is, oddly enough, done by something called a analog to digital converter, or often A/D for short. There are many ways this conversion can be accomplished, like successive approximation, sigma-delta, tracking, flash, and probably a few more basic strategies. There are lots of variations and tradeoffs within the ...

2

It appears that the reading of the datasheet whereby TLC5940 can sink only 120mA/chip is incorrect: Rather, it seems that the correct reading of the "$I_O$ Output Current (dc)" value on page 2 of the datasheet is as maximum current per channel, which is also the interpretation arrived at by Sparkfun discussion to this driver. The ambiguity of the ...

0

1) It depends on the current draw. USB has a PTC fuse set at 500mA. If your load doesn't exceed that, then it's fine. 2) Current will find the path of least resistance, so that's fine. I would use an optoisolator to separate the two devices. Whenever you switch a large current load, a lot of noise will be induced into your system. 3) Just break the ...

0

Its not so much that they "can't" as much as it is they are just not designed to. Computers use switching regulators that are designed to operate around a certain current and voltage but do not perform well if the current and voltage deviates too much from this point.

9

From the article: As it appears, Haswell's C6/C7 states require a minimum load of 0.05A on the 12V2 rail, and many desktop power supply units (PSUs) just cannot provide that low current, reports The Tech Report web-site. Meanwhile, numerous older PSUs, which comply with ATX12V v2.3 design guidelines only called for a minimum load of 0.5A on the CPU power ...

5

It is easier in some ways to design a high-efficiency switching regulator if you can assume that it has both a minimum load as well as a maximum load, reducing the "dynamic range" it must handle. Many PC power supplies are designed this way, both the main supply for the box, as well as on-board regulators for the CPU and memory. The new chips violate the ...

1

As long as the load is connected, it'll work, but if the load or the shunt becomes disconnected or burns out, expect it to fail as you described (the voltage you see depends on the voltage available to the current source : if that's only 12V, the ADC inputs have enough protection to survive). However, it won't be very accurate. Unless you pay a lot for ...

4

With 156 mA through 19.01 Ohms (load + shunt), the maximum voltage will be 2.966 Volts, which should not be a problem for typical 3.3 Volt microcontrollers. Therefore the voltage dividers may be unnecessary for reading the voltage at either side of that shunt. Further, the Voltage developed across the 10 milliOhm shunt will be a mere 1.56 milliVolts. At ...

2

The circuit you have shown theoretically works but practically it has difficulties in matching the 4 x 30k resistors to get decent accuracy. I'm not saying it shouldn't be considered but an alternative is this: - Place the 10milli-ohm resistor in the ground lead of the 19 ohm load. Then you can directly measure the voltage developed across it into an ADC ...

0

From the equation here we have $I_{ref} = I_c(1 + \frac{2}{\beta})$,so $I_c = \frac{I_ref}{1 + \frac{2}{\beta}}$ where $I_{ref} \approx \frac{12 - 0.6}{100k} = 0.114 mA$. From $I = C\frac{dv}{dt}$ we get $V(t) = \frac{I}{C}t$. Assuming the transistors in the current mirror have a high $\beta$ then $I_c \approx I_{ref}$. The 555 ...

0

In a current mirror the current in the collector that is connected to the capacitor is constant until the voltage on the cap nearly reaches the supply rails then something has to give-out BUT your circuit uses a 555 that discharges the cap before it gets to this point so you can reasonably say the collector current is constant. That collector current is ...

3

The transformer does not violate Kirchhoff's current law in any way. It has four terminals. On the primary side 0.2A flows in one terminal and flows out of another. On the secondary side, a larger current flows in one terminal and out the other. These currents of different magnitudes flow in separate circuits which are electrically isolated from each other, ...

4

Think of a transformer as a gearbox. The rotational speed of the load will be some fraction or multiple of the rotational speed on the driven side, and the torque seen by the drive will be some fraction or multiple of the torque the load side. A "modern" wall power supply will in fact use something more like cross between a continously-variable ...

9

What you are missing is that current flowing in one place can cause a different amount of current to flow elsewhere. What is really being conserved in this case is power, not current. You could just as well ask that since 220 V is going in and 12 V going out, where do the remaining 208 V go? That question is basically identical to the one you asked, just ...

5

Functional Reasons Placing LED's in a parallel-series combination like this is a technique to average out the brightness given constant voltage. It helps present a more even looking illumination cheaply (as opposed to active constant-current circuitry). The Q-point of the two LED's in parallel will be determined by the voltage assigned to the pair by the ...

3

The question is really whether it is worth transforming that power up to 240V to cover the distance. You can establish the feasibility of 24V transmission using Ohm's Law : This page gives the resistance of 2000 feet of 8awg (10mm^2) wire as approaching 1.5 ohms. At 10 amps, that implies a drop of 15v leaving only 9v to run the pump. Now, price out ...

3

Nichrome wire has far, far higher resistance per unit length at a given cross section. It's an alloy chosen for that property. The power dissipated in a resistor is the product of the resistance and current, so a larger resistance at the same current means more power dissipated as heat.

1

When connected in series, the maximum current is as high as that of the weakest link, so yes, 4A. The voltage will be the sum of all cells, so 1V. When connected in parallel, the maximum current is the sum of all cells, 12A in this case. If you do this, you should only connect cells of the same voltage in parallel, otherwise you get unwanted current flows.

2

An "alternative" answer: - 120V x 12A = 1440W BUT because there's a fair to reasonable chance that it is an AC motor, the power factor will reduce the 1440W to something less.

9

There are many unknown factors. Both power and current will vary with load and there is only one way to find out how they relate: measure them. One of the most important properties of AC power you neglected is the power factor cos(φ) with inductive loads: $P = U \cdot I \cdot \cos(\varphi)$ \begin{align} \cos(\varphi) & = \dfrac{P}{U \cdot I} \\ ... 1 int sensorValue = 1; Why are you initializing this variable to 1, when it then gets overwritten in the first line of loop()? delay(sensorValue); Is it intentional to have a varying delay of n milliseconds, where n is the value read from port A0 each time? Also, is that delay required twice, once after setting the output pin high, once after ... 0 Look at the Forrest Mims III book again. It does not claim that resistors must be on the anode and has examples where they are on the cathode. In my 1988 edition of the book, series protection for LEDs is introduced on P. 69: LED DRIVE CIRCUIT - Because LEDs are current dependent, it's usually necessary to protect them from excessive current with a ... 0 As I understand it, you do not generally use a battery with a grid-tied inverter. At least here in the US, the electrical code allows two types of systems: Solar panel → GTI → power line Solar panel → charger → battery → inverter → transfer switch ← power line In the first case, the household loads are attached to the ... 1 There are several factors here. First, the input impedance of the ADC. The ATmega328P uses a Successive approximation ADC. As such, the input is basically the input to a comparator, so the ADC has very high input impedance. The ADC is specified as having a 100 MΩ (that is MegaOhm) imput impedance. However, this seems somewhat suspect to me. Together with ... 5 Robot Power Distribution: An Overview If I understand your question correctly, you are having trouble when you share a power source between your robot's (ex. RC car) propulsion (motors/servos) and control systems (radio/processor). To provide a more quantitative answer, let me introduce Ragobot, something I designed 5 years ago with a team at UCLA. ... 0 simulate this circuit – Schematic created using CircuitLab This is perhaps what you want, although it's not clear what your primary concern is. This will prevent current flow back from the circuits onto the common rail of the + terminal of the battery. But this only protects against the voltages in the sub-circuits that are higher than the ... 1 The voltage source, which you drew, is an ideal voltage source. It's a mathematical abstraction**. You can imagine a caricaturesque character with a scroll and a trumpet proclaiming "The King said ten volts!" Ideal voltage source will provide as much or as little current as necessary to maintain the voltage. In the circuit in the O.P. (10V, 100Ω), the ... 1 There are voltage and current sources. In the case of an ideal voltage source, the source maintains a constant voltage, and the load draws whatever current it needs.I=\dfrac{V}{R}$In the case of an ideal current source, the source maintains a constant current, and the load voltage is induced by that current.$V=IR\\$ Your example falls in the ...

0

The examples you're looking at are assuming that the input is an ideal voltage source, which has zero internal impedance. Real world voltage sources have internal impedance (made of some combination of RLC) elements. The current that flows between a source and load depends on the source and load impedance and the nature of the signal (DC? AC? What ...

0

All those examples you're talking about takes an input voltage and don't show input current for the sake of simplicity, but of course, input current is as important as voltage can be. Ohm's law speaks by itself. I cannot imagine a simpler answer to give you:)

Top 50 recent answers are included