# Tag Info

19

It does not appear in your equation because this equation assumes you're using the battery at its output voltage during the whole usage without conversion. This is not the case here, because you're using a step-down converter. So, to build the correct equation you: get Vavgbat: the average voltage of the battery during the whole discharge cycle: the ...

15

You need to know the power factor, which is highly depedant on your application. Then you simply use this formula: $$P = S * PF$$ where P = actual energy S = apparent energy PF= Power Factor This answer tells us, that power factor in general public households can be expected to be greater than 0.9.

15

If you add the resistor across the battery, so it's in parallel with your existing circuit it will draw an additional 300 mA (I = V / R), assuming the battery is able to supply it. This is not what you want. You could reduce the current draw by adding the additional resistor in series with your current circuit.

15

If you add another resistor in parallel with the LED and its limiting resistor it will not change the current going through the LED, just draw more current from the battery. That MIGHT drop the battery voltage a little bit which would reduce the LED current, but that really is a secondary effect. You effectively have two independent circuits attached to the ...

15

While, in theory, high impedances would reduce power dissipation for the same voltage swing, there are several important issues in practice. 1) It's the power, not the voltage, of a signal that determines signal to noise ratio. If you must swing the full rail, then you'd win by increasing the impedance. However if you launch a specific power, then low ...

14

While there's no reason why you couldn't design electronics with constant current sources, there's a few good reasons why we don't. Batteries and mains power are supplied more as constant voltage sources than constant current sources so it's simply more convenient to use what we have. The other reason is that a constant current source is always burning power....

12

Since the forward voltage of the diode varies slightly according to the current through it, more accurate value of the current can be found out by - Plot forward V-I chara of the LED (from its datasheet). Plot the load line of the series Resistor for the given value of R. Find the meeting point of the two graphs. It should be the operating point of the LED ...

11

If a car is travelling at 40 mph it doesn't mean it travels at 40 mph per minute. If a TV is consuming 140 watts it means it consumes 140 joules per second. What you get charged for by your utility company is joules and that is oddly (but acceptably) converted (without mathematical error) to watt seconds or more conveniently watt hours or kilowatt hours. ...

11

Use Ohm's law: $${220\:\mathrm{V} \over 220\:\mathrm{k\Omega} } = 0.001\:\mathrm{A} = 1\:\mathrm{mA}$$ And power is the product of voltage and current, so: $$0.001\:\mathrm{A} \cdot 220\:\mathrm{V} = 0.22\:\mathrm{W}$$ A 1/2 W resistor could connect across 220V just fine without burning up. The LED is also a diode, and lets current flow only in one ...

11

Because these microcontrollers can operate at 8Mhz and 16Mhz, is the power consumption a linear function of frequency? i.e. does operating at 8Mhz consume 0.2mA * 8 = 1.6 mA of current? First of all, the 328 can run at many more speeds; its maximum clock speed is 20MHz, and at least down to 32kHz is supported, possibly lower. Anything in between is also ...

10

The resistor power dissipation is safe as Phil Frost stated. However through my 54 year old eyes it looks like a SFR16 resistor that has a Voltage rating of 200V. I would always put 2 resistors in series when using SFR16. The actual resistor in the photo could be some generic or copy or ripoff so the voltage rating could be bad. If the LED is just a standard ...

10

You didn't bring up something else -- frequency. You don't mention the country, but I suspect that if they are supplying a mains voltage of $220\:\textrm{V}_\textrm{AC}$ then it's probably also $50\:\textrm{Hz}$. Just using a transformer alone to drop the voltage to US levels doesn't change the frequency and not all devices work properly (or well) on the ...

10

There's different way these things can fail, and in most, they might still be using power. I've seen all three: failure modes where the bulb "shut down" and used totally insignificant amount of powers, failed bulbs that used way less power, since they had no emitter to drive, and failed bulbs where the built-in supply was constantly trying to drive a ...

9

Yes it will lower your consumption, the real question will be how many people will it take to do it..? j/k

9

My question is, why does voltage not feature anywhere in my battery life calculation? Because your calculation is missing one aspect. You can use two kinds of voltage regulators: linear or step-down switch mode. Now, with linear, the Energy per Charge (= Physical definition of Voltage) that is "too" much is just converted to heat (and subsequently, lost)....

8

One solution is to use an instrumentation amplifier to measure the voltage drop across a shunt resistor. These are designed to offer an extremely high input impedance to both inputs of the amplifier (in excess of 1 giga-ohm), while allowing you to amplify this signal by relatively large factors (1000x is not uncommon). Note that the fact that there is a ...

8

Most power sources are constant voltage, and not constant current. If you take the two main sources of electrical energy, which are batteries and rotating generators(regardless of size), the one thing in common is that their voltage is fixed theoretically to a certain value and can be controlled. For example, a standard AA dry-cell battery has a voltage of 1....

8

The I2C data and clock lines draw power when they get pulled low. Because then power is sunk through the pull-up resistors. While a line is pulled low it will draw 5V/4.7k$~\Omega \approx$ 1mA. Assuming 5V VCC and 4.7k pullup resistors. The clock line will have a 50% duty cycle. The data line is low at least 1 out of every 9 clock cycles (every ack for a ...

7

Yes, it's called an ammeter. It measures current. Any common "multimeter" will have ammeter capability. For the test, run the unit from a fixed voltage supply and put the ammeter in series. Since voltage is fixed, current is proportional to power. You can then experiment with different software options to see what draws what current. However, while I'...

7

This was going to be a comment, but I felt it ran too long, so I expanded it a bit more... some of you guys know me by now ;-) This seems like a brilliant opportunity to include some energy harvesting methods, next to a safety backup cell to keep the proc active if all else fails. If the measurement isn't complex you could consider buffering a couple, ...

7

555 timers consume on the order of single milliamps of current (the lowest is around .5 mA as far as I know). That's far from the lower limit as far as timing-capable ICs go, though, as even a small microcontroller can easily work with consumption around 200 Î¼A and some (if not most) RTCs consume mere nanoamps (not that those will help in this particular ...

7

Figure 1. LED current vs forward voltage curves. Note that red has much lower $V_F$ than green and blue. Red LEDs have a lower forward voltage than green and blue. If connected directly to the same supply as the green and blue they would draw a very high current and would be destroyed. To prevent this we normally add series resistance. On a matrix ...

7

For an ideal transformer, power in equals power out: it's 100% efficient. For a realistic AC line step-down transformer, power in is a little bit more than power out: it's not quite 100% efficient, but it should be close. Thus, if you plug a 300W load into a step-down transformer (assuming the transformer is rated for more than 300W), expect it to draw a ...

7

Get a \$20 Kill-a-watt and measure the actual wattage during load and idle. Then determine the average amount of time you'll be in each mode and from that you can get a good estimate as to how much power it'll use. At the upper end, you know that it can't use more than 19V*2.1A ~= 40Watts (ignoring powerbrick inefficiencies). 40 Watts * 24 hours * 30 days =...

7

Your thinking is correct, as long as you can achieve a higher speed with the same pull up resistors.

7

This termination resistance closes the current loop between two differential signal lines and creates a decent amount of power loss so that low voltage amplitudes are required to reduce power consumption. I think there's a misconception here about how transmission lines work. The purpose of receiver-end termination is to dissipate all the power, in order to ...

6

A watt is a joule per second. It is a measure of power. A joule is a measure of energy. To calculate power consumed by an electrical device for a specified amount of time you multiply the power by the time. You rarely if ever see things represented in watts per unit time. See more: at this wikipedia link

6

If you want to calculate the amount of energy consumed by your device in a certain amount of time, multiply the power rating of your device by the period of usage. For example: If you have a 100W lamp that is operational 24 hours for a full month, how much energy does it consume? How is that useful? First, for energy consumption Q =Power * time = 100 * ...

6

In general, the amount of power an FPGA "consumes" is related to the function it is performing and the frequency it is operating at. That's why you can't find an authoritative number; because "it depends". I suggest that you play around with the power estimation tools, figure out what the input fields mean, and then you'll have a much better understanding of ...

6

The phrase you are looking for is "Vampire Power." Most phone chargers will have it, but will be minimal, like a few cents a year per cell charger in energy usage. Computer power supplies will be worse. And anything that is on a standby mode, like sleeping computers, TVs, game systems (like the PS2/3, Xbox 360/One, Wii), microwaves, etc. will consume power ...

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