# Tag Info

61

It breaks up into three simple sections that are each relatively easy to explain: simulate this circuit – Schematic created using CircuitLab The first part is the diode that provides reverse voltage protection. If for some reason the polarity of the input voltage is wired opposite to what it is supposed to be, then $D_1$ will block it and the output ...

56

In most ordinary circumstances, 12 V isn't even enough to feel, let alone cause a shock. However, it's really current that you feel and that shocks you, not voltage. So when we say that 12 V is safe, we're making implicit assumptions about the resistance of your skin. Electrically, you're basically a bag of salt water. Your insides have low resistivity, ...

40

A wrench does not have a resistance of 0.5 ohms, it's way lower. Your basic multimeter cannot measure resistances to better than an ohm or so, the resistance of leads, and the unreliability of contact resistance make it impossible. The way resistances as low as a wrench are measured is to use a 4-terminal Kelvin method. What you do here is to pass a ...

31

It isn't just shocks. A battery can push a lot of current through any piece of metal. This could melt a wrench and cause a fire. It could also cause a ring to get red hot and burn the wearer. It is very easy to get a finger with a ring between the positive terminal of the battery and the body or frame of the car. Always remove metal jewelry when working ...

25

While it is true that you can't actually charge and discharge a battery simultaneously, it is quite common to have a charging source, a battery, and a load, all connected in parallel so it looks like you are charging and discharging simultaneously. If the charging source can deliver more current than the load requires, then the excess current will be used ...

22

Knowing that all sorts of weird stuff can happen in automotive power circuits, and not being especially knowledgeable in those systems, I'd err on the side of caution and use an opto-isolator. simulate this circuit – Schematic created using CircuitLab Pin 1 = Car 12v R1 Pin 2 = Car Ground Pin 3 = NC Pin 4 = Arduino Ground Pin 5 = Arduino 5v R2 Pin 6 =...

18

If by "use the battery" you mean "can the battery be in the circuit?" then yes. If by "use the battery" you mean "draw power from the battery while charging it?" then no. Charging is, by definition, putting power into the battery. This is the opposite of drawing power from it. You can't do both, by definition. Under normal operation, where the battery is ...

18

These are called boost convertors. They take a low dc voltage (such as 5V) and convert it to a higher dc voltage such as 12V. They use an inductor to store energy in one half cycle of a clock and in the other half cycle the energy stored in the inductor charges up a capacitor. The capacitor is naturally charged to a higher voltage than the input voltage ...

18

The text description is right. A wrench will throw sparks and possibly heat up and melt, that's dramatic. The image, however is wrong. It clearly shows a person and a wrench being electrocuted simultaneously. The picture is wrong on 2 levels: a car battery can't electrocute a human in typical conditions no single source can electrocute 2 targets so ...

18

The TIP41A is configured as a voltage follower. The emitter voltage will be equal to the base voltage minus about 0.7 V. The 470 Ω resistor provides the base current to turn on the transistor and pull the base towards the supply voltage. The zener diode will turn on if the base voltage goes above 9.1 V (its breakdown voltage). Therefore the base will be held ...

17

I have a motorcycle battery rated as 8.4Ah (20HR). It also says 8Ah (10HR) and I am not sure what´s the difference. The capacity decreases at higher discharge rates. The figures are telling you that discharge current of $\frac {8.4 \ \text {Ah}}{20 \ \text h} = 0.42 \ \text A$ will last 20 h and $\frac {8.0 \ \text {Ah}}{10 \ \text h} = 0.8 \ \text A \... 16 Perhaps the dimmer circuit won't work without a resistive load. Some thyristor dimmer circuits might not see enough holding current from a rectifier-capacitor load to stay on for the half cycle. You may be able to wire a fat resistor in parallel with the LED lamps and have it work, however it may require some experimentation to determine an appropriate value.... 16 No, that is a bad idea. The two regulators won't have exactly the same output voltage. The one with the higher output voltage will take more current. You can't guarantee that both regulators will put out their maximum current when you try to draw 2x the output current. Another drawback to your approach is that you have a diode drop between the regulated ... 15 No. You don't match rotations per second to PWM frequency. It needs to be much higher. You want it high enough so the motor runs smoothly (motor inertia smooths out motion and motor inductance helps smooth out current), but not so high the switching losses in your electronics is excessive. Often at least 8-10kHz, but you might want >20kHz if audible noise is ... 14 Most certainly a surge (or in-rush) current problem. Electric motors, at start-up (when they are not spinning yet) and especially when driving a load (like turning the pump's impeller against water) draw enormous short-term (surge) current. This surge quickly dies out once the motor "gets going". From wikipedia When an electric motor, AC or DC, is first ... 14 You want the motor inductance to result in reasonably low ripple in current during the PWM cycle. Here is one drive maker's rule of thumb:$ f_{PWM} \ge \frac {0.6V_{SUPPLY} }{L_{MOTOR}\cdot I_{NOMINAL}} \$This formula results in quite a bit of ripple, around 40% peak at the limit and D=50%, so a bit higher frequency might be desirable, especially if the ... 12 How about this for an idea. With the input at 5V no current can flow through the zener (5 + 9 > 12). The output PNP transistor of held OFF by the base emitter resistors (= 4k7 + 2k2) and the output is 0. When the input is pulled down to 0V a small current will flow through the base and the 2k2 resistor. The junction of the two resistors will be a 9V (the ... 12 How would i disconnect and reconnect 12vin every 2 hours to my circuit board? Just go unplug the power every 2 hours, then re-connect it. If you want it done automatically, the simplest is probably one of those timers you can find in home improvement stores. They should be able to be set up to run something for 1½ hours every 2 hours, for example. ... 11 As @KevinWhite said, you can use a voltage controlled switch. Here is an example of a very rough model for a relay: And this is the result of the simulation: The model is very rough because: The excitation coil is modeled only by a resistor, neglecting the inductive component (this may be OK). The switch triggers when a voltage is applied to the terminals ... 11 Read the datasheet of the LM317, on page 9 it states: So when you feed the LM317 14 V it can regulate to 11 V and lower, not 13.5 V. Also there will be 1.25 V across R1 so for 13.5 V you will need to put at least 13.5 + 1.25 + 3 = 17.75 V into the LM317. The ~15 V you're feeding the LM2596 board isn't even enough, there's no need to have that LM2596 ... 11 It will function, but using standard plugs and sockets for the wrong voltage is a spectacularly bad idea. This relies on you being in total control, remembering everything about your setup forever, and making no blunders... Every novice goes "oh, I can handle that"... but that is simply not realistic, and the standards bodies who set electrical ... 10 With some generic parts? The MC34063A is a fairly standard part found in most car usb chargers (And I mean most, even The Dollar Tree$1 car chargers). While it is normally used for buck regulation from 12v to 5v, it can also do inverting and in your specific need, boost regulation by moving the circuit around. The car chargers will have an okay sized ...

10

If your three power supplies are isolated from their input, which is most likely the case if they are line powered, then what you propose should work fine. A isolated power supply produces a voltage difference not some absolute voltage. This voltage difference is usually talked about with the assumption that the negative side will be considered the 0 V ...

9

What your are suggesting can work, but only if the PSU's outputs are isolated from the AC ground and neutral! Once you get the PSU's, do a few continuity checks with a multimeter. With the PSU unplugged, there should be no connection between the DC COM (DC-, 0V, etc) terminal to either the GND or N lines of the incoming AC.

9

A short story Long before solid-state electronics and vacuum tube electronics, lead acid batteries were and are the longest living battery technology. Being a multiple of 2V +/- 0.1 per cell or 2.36V on float charge or 14.2V, the 12V battery was not the 1st common voltage for cars. Initially it was 6V but as gas engines grew , they used 2x 6V batteries ...

8

You can achieve this with two NPN transistors as shown below. As you can see, whenever a 0V signal is given, 12V is seen in the output, and whenever a 5V is given in the input, 0V is seen on the output. Let's see how it works. First of all, let's start with the scenario where the input is 5V, or in other words, HIGH. This will turn Q1 ON and the voltage on ...

8

That's a 27A-size battery, not a battery capable of providing 27 amperes of current. The maximum current from that battery can best be described as "piddly", which is all a remote control needs.

8

You can use a voltage controlled switch to simulate the contacts (Just called SW in LTSpice). Set the operating voltage to match pull-in voltage of your relay. You can use a resistor (and an inductor if desired) to simulate the relay coil. Here are a couple of useful links: Relay Model Voltage controlled switch example

8

I think what all the other answers are leaving out is the internal resistance of the battery. When dealing with large currents and low resistance shorts, this becomes the significant factor limiting current. An ideal battery can be modeled by a voltage source, but real batteries act more like a voltage source in series with a resistance. $$V = IR$$ Let's ...

8

You have just come face-to-face with a fact of life: the starting current for a motor is much larger than the running current. For DC motors the rule of thumb is the starting current is about 3 times the running current. So a 12.5 amp supply should be adequate for a 3 amp motor. But. 1) The current rating for the bilge pump may or may not reflect the ...

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