15

According to the reference manual: This apparently can reduce the typical current draw from uA to nA. It's not clear if they make assumptions about whether the pins are tied high, low or floating for these comparisons.


6

You need a wheel. A bicycle wheel should work, and the flashing light makes sense on a bike. Some other wheel may work, depending on what do you have at hand. Maybe a fan would work. Buy a magnet and reed contact like these. Connect battery, LED, resistor and the reed contact in place of the switch. Now just approach the magnet to the reed contact, it ...


6

Floating inputs into a Schmidtt can draw lots of microAmps, possibly even milliAmps. Hence the Schmidtt is disabled. For minimum propagation delay near the 2 thresholds, a Schmidtt needs a very fast amplifier built into the circuistry. Pleasantly, a simple two_FET inverter (Nchannel + Pchannel) provide the speed, but will consume LOTS OF CURRENT near the ...


5

but won't be more efficent to just cut de power with a transistor controlled by the MCU? Start looking at the datasheet! If standby mode consumes no more power than the leakage of your FET, then the FET is pointless. There's probably no leverage to be had here because of your 1-day interval, but a GPS receiver takes time to lock on to enough satellites to ...


4

Here is an approach to thermal isolation while maintaining some mechanical strength -(the chip is an ovenized buried zener voltage reference- it would have a urethane foam insulator over it in actual operation): As well as the wide routed slots, notice the thinner ones to the left and right and the slots to the top and bottom. The copper, even 34\$\mu\$m ...


4

The valve works by attracting an iron armature into a solenoid. This means that when un-energised, there's a large airgap in the magnetic circuit. When it's energised, the airgap is much smaller. The large initial airgap means that a large current is required to develop enough force to move the armature. When the airgap is smaller, less current is needed to ...


3

Because you've done some adequate research into a solution and have put together a half decent requirement I'm making an answer... The ADG819 2:1 multiplexer operates from 1.8 volts to 5.5 volts and has a maximum D-S leakage current of +/- 10 nA across the full -40 C to +125 C temperature range. On-resistance is 0.8 ohms maximum and this would result in a ...


3

You need to work out a power management strategy: Will each circuit be powered all the time? Can the MCU be put in power-save modes and if so, how often? Should you use extra components to power off power-hungry parts completely? For example, you might be able to only power the RTC and wire it up to power up the voltage regulator at the right times, for a ...


3

Indeed a normally closed switch means that a current must flow otherwise you cannot detect when the switch opens (and there is no more current flow). Fortunately there are several ways to minimize the current consumption. The simplest way is to use a very high value pull-up (or pull down if you prefer) resistor, a value of a couple of Mega-ohms will reduce ...


3

Regarding the latter, how can the CPU enter sleep mode and keep the OS running? I am open to any suggestions. You use co-processors. Phones are doing this for years now. They have one or more big ARM Cortex-A cores for the *unix based operating system. And one or more smaller ARM Cortex-M like processors for the background stuff. Like counting your steps ...


3

The datasheet is providing a promise that 10 uA is the worst-case (non-switching) current across the -40 to +85 degC temperature range. As you mention, there are hints both in semiconductor physics and elsewhere in the datasheet that the high temperature condition is likely to be critical for power usage. Notice that as the temperature range extends upwards (...


3

In active mode the current consumption of the ICs are around 20 mA and according to the datasheet the efficiency of MAX640 is around 90 %. Hence the total current consumption will be around (20 mA/ 90 %) = 22-23 mA, right? No, that doesn't make any sense. Look at the graphs in the data sheet typically this one: - With a load current of 20 mA (not ...


3

The capacitor's voltage drops as current is drawn from it. Assuming constant power draw, the voltage curve (green line) should look like this (simulated with LTspice):- In this example a 2F capacitor is discharged from 13.5 V to 0 V at a constant power of 5 watts, reaching 3.5 V in 34 seconds. At this point the capacitor has delivered 93% of the theoretical ...


3

A current limiter simply won't work - you can't get energy from nowhere, and causing the supply to "brown out" during transmit is likely to at best yield poor on-air results; more likely cause a crash and hopefully restart of the entire system. To the best of my knowledge, all LoRa radio chips have programmable output power, so if you want to limit ...


3

As others have said, a current limiter is not what you need here. In fact, this is the very cause of your problem: a current limiter will decrease/block the voltage as it approaches the current limit, which is exactly what the battery is doing already. The Tadiran TL-5903 indicates it can supply up to 200mA pulses, but it will not maintain the rated voltage ...


2

The voltage divider you mention is the easiest way but it doesn't mean it's the appropriate. however, if you want to go for it, to reduce the current the resistors draw, use large values (Ohm's law, I=V/R), just make sure the relation of both resistors complies to your ADC resolution vs the battery voltage. There is also Battery Chargers IC that provide ...


2

The key is the Vcc_H and Vcc_L suppl that the op amp is using. It's referenced to the 150V supply, not ground. And then this OP_A whose power supply voltages are "hanging under" the 150V rail, gets fed the voltages across the current sense resistor. Since these current sense voltages are tiny, they are also close to 150V so nothing fries. Then the op-amp ...


2

If keyboard components aren't rated for LED light through them, don't buy them. They are likely to be opaque, which means you could put a searchlight down there and not get any light through. Anyway you would be unable to power it off USB, and you'll need to get under 2.5W for that, and will want to get under 500mw if possible. How to size Let's take ...


2

LEDs do have a maximum current rating, but it does not mean they must be run at full power all the time. For this application at least two things can be used. Buy better, more efficient LEDs so that they light up brightly enough at lower current, say 1 mA or less. You can also divide the LEDs into multiple groups and multiplex them so that only one group is ...


2

Cell towers only transmit around 10 watts usually. Sometimes up to 50 or so in urban areas. Your phone can transmit up to 2 watts. Transmit power does obviously have a big effect on the range of a signal, but it is not nearly as much as you would think due to the 1/r2 relationship of radio waves propagating out from the source. If your transmitter puts out ...


2

Nothing is wrong with your approach except for being somewhat vague. For example i understand the intent behind "using the button that will latch the DC-DC converter enable pin", but not how a button (I assume you mean momentary switch) can "latch" anything. In any case, I'd recommend taking a look at TPL5111 chip, which was specifically designed for ...


2

Let's do a quick analysis: Note that the current into a linear regulator (like the LM7805) will be the same as the load current. You need 1 W at 5 V, that means a current of 1 W / 5 V = 0.2 A That current needs to flow through the 100 kΩ resistor, that means a voltage drop of 100 kΩ × 0.2 A = 20 kV across the resistor! That's not going ...


2

Flashing requires some kind of active device. Either a semiconductor like a transistor, or a electromechanical one like a relay, or something exotic like high voltage discharge devices. Here is how you can make a LED flash using a relay and a capacitor: https://www.youtube.com/watch?v=T_az4omJx0k This is how the turn signals on cars often works. But if ...


2

Look toward Beaglebones. They run linux on the beefy ARMs, but also have Cortex-M cores that you have access to for lower level stuff like interrupts


2

I once designed a non contact telephone line tracer. The specification was to be able to detect a 1Vpp 1kHz sinusoid signal on a 1 metre length of wire at a distance of 1 metre. The input was a probe tip connected to a static protected FET opamp with a high Q filter. I imagine a 2 metre range could be achieved by pushing the voltage up to about 9V. The ...


2

from what I know, I should drive LEDs by constant current, not constant voltage Isn't that why you chose the TLC6C5912 LED driver: - So, choose a voltage boost converter to give you circa 30 volts and you should be good to go. The one you indicated in your question (TPS61096A) might do but, bear in mind that it runs from a 5 volt rail normally and, has ...


2

Assuming an IR remote receiver as used in TV/PVR appliances... An IR receiver should be continuously-powered if asynchronously-received light pulses are expected...a receiver hobbled by reduced-power cannot detect signals reliably. Choose one that has been designed for low DC-current from its supply. When no IR light pulses are detected, these devices pull ...


2

No, but USB is 5.0V so if you add a suitable regulator you could power a 3V 50mA device. Inexpensive modules are available from China. You could even use a ~$1.50 USB->RS232 module based on chips such as CP2102, which have a 3.3V regulator on-board and ignore the serial port. If you directly apply 5V to a 3.3V device you run a high risk of permanently ...


2

This really depends on the implementation of the microcontroller and cannot be answered in a general way. Some parts will always be affected by the supply voltage, for example the dynamic power consumption of GPIO when they are switching. Some microcontrollers have internal voltage regulators for the core or other parts of it so they will internally run on ...


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