I'd like to power cycle a RaspberryPi0 each 1h. I plan to use a TPl5110 for this, which can be powered by around 2.5-5.5V. I want to deliver 5V to the Vdd-Pin using a powerbank. However, once the RPI0 sends the done signal to the TPL5110-IC, too little current (&lt;50mA) is drawn from the powerbank, which shuts the powerbank off. Thus, I am thinking of using a Backup-Battery with 3x1.5V AA (alkaline) batteries to keep the TPL5110 powered on all the time. Will it work as simple as adding a standard silicon diode after the 3x1.5V backup batteries, as sketched here:  I want to avoid current flowing from the backup battery to the RPI0 if it is powered on (at least as much as possible to avoid draining the backup battery low). EDIT: my requirements are different from this question:  since I do not just want to use a diode with forward voltage drop all the time since it consumes too much power for my application. Thank you!
121000
I have a project running with MCU supplied with a 24V power supply regulated via an SMPS to 9V also, I have Li-ion battery pack 9V  how to connector these two sources to run the MCU with battery pack when the power failure occurs?  or   what is the best way to switch from battery to power supply and power supply to battery?
5942
How do you maximize the cycle count of your battery?  There seems to be a debate between:   Leave plugged in 24/7 but do a full cycle once a month Charge to ~100%, drain to ~10%, repeat.   NOTE: Taking your   Which method is better and why? How much of a difference does it really make?  Here are some of the better sources I've found about battery behavior so far. Even still I feel they don't conclusively answer the question above. See my elaborations below:           Elaborating on Option 1  Now IF Apple engineers were smart and optimized for AC draw when plugged in to not use up your charge cycles, then it seem logical to assume that you are using effectively none of your 1000 charge cycles while plugged in. If that's the case, it would seem that the LiPo drop from holding at 100% would be vastly outweighed by the fact you're not using any charge cycles.  However, this is a purely speculative assumption and I could find no evidence to either confirm or deny.  If this is NOT true, and the AC adapter only goes to charging the battery, then there would be no difference. Your battery would drain to 99%, then back up to 100%, then back down to 99%, etc. Those micro-charge cycles would add up at the same rate as 100% -> 0% -> 100%, and you would get no gain. In this case the negative effect of holding LiPo batteries at 100% would outweigh everything else.  Elaborating on Option 2  There are lots of good reasons why Option 2 is the way it is:   Because of known LiPo chemistry issues, holding a 100% charge for a long time causes the battery to degrade Apple specifically recommends not to leave it plugged in all the time  Draining a battery all the way to zero all the time is bad for the cell (Which is why ~10% is used) Heat is a killer, and when plugged in AND charging, you get extra heat that causes damage.   There are a couple reasons why I'm challenging Option 2:   It seems silly that the most intuitive use case for thousands of people that use their macs as primary computers is wrong. If the hypothesis in option 1 is true, you're just burning unnecessarily through your finite charge cycles. It's a pain to have to remember to keep plugging in and plugging out and be worried about whether or not your device is plugged in.
36893
I have an application in which an RFID reader demands 200 mA of current when it attempts to read an RFID tag. For some reason the current switcher power supply either can't supply enough current or can't supply a clean enough voltage and I end up with a poor performance on the reader.   I want to use a standard LM7805CT regulator in place of the switcher, but how large of a capacitor should I use? The datasheet suggests 0.1 uF, but I'm not sure that is large enough since the reader gets turned on and off a lot.  I am regulating down from a 24 V battery (it is pushing the limit of that particular regulator, but I will add active heatskinking to it).
24924
I've got 15 devices each with a current of 20mA and I am interested in running them all for 10 seconds. How would I go about working out the amount of mAh a battery would need to provide for this?   I understand that to find the mAh to run something for an hour you take the current and multiply it by 1000 and so my understanding of his problem was that you could get minutes by multiplying by 100 and then seconds by 10. So What I thought I needed to do was something like this:  15 x 20mA = 300mA/0.3A   0.3 x 10 = 3  3 x 10(seconds of runtime) = 30mAh   However I am not sure if this is right, so I would be grateful if someone could look it over and point out my mistakes.
52457
Imagine I have a 10V battery with a capacity of 1Ahrs connected to a 100 Ohm resistor.  I'll get 0.1A (V/R) at a power of 1W (current x voltage) for a total of 10hrs. As a watt is defined as J/s my battery produces 3,600J (1x60x60) in total.  Now, imagine I get a second, identical battery and connect it in series to the first one. I will effectively have a battery of 20V with a capacity of 2Ahrs, all connected to the same 100 Ohm resistor.  I'll get 0.2A at a power of 4W (current x voltage) for a total of 10hrs. This will generate 14,400J (4x60x60).  So, from doubling my energy source I have quadrupled the energy output!  I've obviously made an error somewhere either in my calculations or conceptual understanding, and I would be most grateful to anyone who could set me right.  Thanks!
39949
I want to supply constant voltage to my LED's but the supply varies from 3.7V to 4.2V (Li-ion battery). I cannot use 3.3V voltage regulator as it will need at least 4.3V voltage at the input but my battery can supply only 4.2V max. so is there any other way I can supply regulated 3.3V to my LED.   RGB LED = 3.3V, 60mA Battery = 3.7V - 4.2V, 3000mA Number of LEDs = 24 The LED's are connected in parallel
116306
I’m currently working on a solar powered boat with a modest sized team of people. For our competition we are only allowed to have 3 12v batteries on board but we are allowed to have up to 52v at any point in our setup. We are hoping to kick up the speed of our outboard motor for the sprint portion of the completion and are looking into using a boost converter for this.   After doing a little research I’m a bit confused about how they fundamentally work. From what I can tell they take in a voltage and current and put out a higher voltage with a lower current with some level of efficiency. But it seems to be independent from the load( in this case the motor) which doesn’t make much sense to me because I thought the current would be determined by whatever my load is. How do loads actually interact with boost converters?( The simpler the better.)  For reference we are using the Saietta 119r motor system and I’m considering using the MPS MPQ3425 boost converter.
113772
I have this sim800L module   It has a 3.7v - 4.2v voltage requirement. So I created a simple buck regulator using MIC4576 regulator according to it's schematics and adjusted it's output to 4.2 volts. It can provide 3A of current. And i powered it through my laptop charger, so there's no shortage of current in the circuit.  But the module(sim800l) resets as soon as it finds a network(it's my guess). If I run it without sim, it runs fine and doesn't reset.  I've also tried 2 18650s in parallel to be able to provide atleast 3A to the circuit. But it still resets after a few seconds.  Is there anything obvious I'm missing?
16477
I have got two power supply options on my board:   USB 5.0V DC Power Jack of 5.0V   My intention is to design a or-ing power supply option, as either of them will be used at a time. Possible solutions are shown below:    Using oring diodes   or using two p-mos    Possible issues observed with the above options:   Using diode is cheap and good option but an ordinary diode gives a drop of 0.7V and a schottky diode will give a drop of 0.16 to 0.20 volts, which is also not acceptable because my processor  recommended operating power supply is 5.0V and minimum supply voltage is 4.75. With a 0.2V with schottky diode the 5.0V comes down to 4.80V and the space between recommended and supply fed is very less. Then I thought to use p-mosfets which serves the purpose of oring along with reverse polarity protection, but assume a case when both the supplies are fed together by mistake and there is minor difference between USB supply and DC jack say 4.9V for USB and 5.10V for DC Jack. Then, the DC jack supply may pump the current in USB Jack, which may damage USB.   Is there any other better and cheap option I should use for or-ing supply, or is there something wrong with my analysis?
69545
I have an old HP dv7-7210sp laptop and I want to use it as a server. Is it safe to have it turned on 24/7? The server will be used with Nextcloud to backup my data.  Should I remove the battery from the laptop or keep it as a backup power source?  Also, is it safe to run HDD 24/7 or should I enable APM so it spins down after some time of inactivity?  Can the laptop burn if I leave it running?
14661
I am doing a project where my electronics runs on 5V which I wanted to power from a 7.2v lipo regulated down to that voltage. I'd really like to include the opportunity to charge the lipo without removing it from the project. What kind of circuit/components do I need to do it properly?  Say, would putting a battery charging ic at the battery's terminals in parallel to my system be enough to achieve what I need?
62759
As many of you probably know, a normal light emitting diode, while very handy as a cheap and ubiquitous indicator, can double as a power generator. The mechanism is the same that is exploited in photovoltaic cells, aka solar panels.  When an LED is exposed to light it produces a current flowing from anode to cathode. This current can be used to directly power a device, or to charge a capacitor so that a periodic load can use the stored charge all at once, when needed.  To give you a better idea of the values involved, the short circuit current ranges from tens of nanoamps up to a few microamps, while the open circuit voltage is somewhat less than the standard forward voltage of the LED.  I am currently in the process of choosing a range of LEDs to test them and see if they are suitable for my application, but I am kinda stuck, mainly because using a led to power something is convenient only money wise, while a photovoltaic cell or even a photodiode is a better choice under many other aspects. No manufacturer that I have seen list the photogeneration current in the datasheet, and this seems fair enough to me, but since I want to avoid testing thousand of LEDs I would like to buy a selected range of them, so I need to better understand the underlying mechanism that leads to photogeneration.  Let's just write down the diode equation: $$ I_D = I_s(e^{\frac{V_D}{V_t} - 1})-I_G $$         &ndash; Schematic created using   Where \$I_s=J_s\cdot A\$ is technology dependent, and die area dependent. \$I_G\$ is the generation current, and is the parameter I am interested in maximizing.  Here are some things I have thought of:  LED color since the LED can absorb a lightbeam only if the energy it carries is greater than the gap, choosing LED with low energy (i.e. low frequency) gap is a must, to be able to absorb a wider portion of the spectrum. This lead me to choose red or infrared LEDs.  max current max forward current is proportional to die area, and power dissipation capabilities. Since photogenerated current is proportional to area, choosing an high current LED seem wise to me.  forward voltage this is somewhat related to the open circuit voltage and is not currently an issue for me, and I believe it has no relationship with photogenerated current.  reverse current I am not sure this has much to do with photogenerated current, but the bigger the reverse current, the bigger the area, which is a good thing.  breakdown voltage I do not think this has a relationship with photogeneration.  My question finally is: Do you agree with my analysis, and do you think I am missing something? or better What are the parameters to look at when searching for a photovoltaic panel in the LED section of a supplier?
35418
I'm trying to control a solenoid using the RasPi 2's GPIO pins. Since the solenoid needs more power than the RasPi can provide, I'm using a 9V battery to provide the power.  I started with just wiring the circuit and checking if it works by connecting and disconnecting the 3v3 output to the board.  First, here's the wiring I'm currently using:    The diode accross the solenoid inputs is a standard flyback diode, and the resistor is a 1kOhm resistor.  Apparently the RasPi GPIO output pins produce 3.3V at a maximum draw of 16mA.  The inductor represents my solendid, which I got from . As far as I can tell, it requires 12V / 1A to operate, but connecting the 9V battery straight to it (skipping the transistor) works nicely enough. Even the RasPi 5V output provides a weak activation.  However, when going through the transistor, the solenoid does not activate properly. I have noticed that when the base pin is on though, it will "hold" the end position if pushed manually, just not push towards there by itself. It does not do this if the base pin is not connected.  I tried using a BC547 NPN transistor, then I figured the output current was not enough, so I switched to a BC639 (which should have up to 1-1.5A current output), but that didn't help either.  My question is, what transistor should I use? Am I doing anything else wrong? I've seen other people use a TP120, which seems to have an even higher output current.  Since I'm setting up a circuit with 24 of these, I'd like to know for sure before ordering another batch of transistors.
93688
I am building a mini pinball machine, which is about 10x22cm.  To drive the flippers I plan to use this 5v mini push/pull solenoid:  , along with momentary SPST buttons and a 5V 2A wall power adapter.  Those are the only electronic components in the machine. I am not using an Arduino or any other controller.  In this scenario, is it advisable to use a diode (or any other components) to protect the power supply?  If diodes or other components are advisable, I will be grateful if you could explain at a very basic level why they are needed and where they should be inserted into the circuit.  The schematic below is pretty much the current extent of my electronics knowledge.  I am interested in expanding my knowledge but find many of the technical terms used on this site overwhelming.  Here is the circuit that I have so far:
113023
I've been working on a data logging device that will eventually go into my personal vehicle.  I've scoured the interwebs and built what, in theory, should be a fairly stout power supply capable to handle fast transients and undervoltage and the fabled load dump.  My problem is, though... I want to actually test the power supply.  In previous questions that I've asked related to the design of the power supply portion, I've been told that the "right way" to simulate a load dump is with a proper, preofessional testing device.  I'm curious, though... what's stopping me from rigging up something that is "close enough"?  My idea is simply this: get a low-voltage DC power source, such as an old PC power supply.  Get a DC-DC converter to step that 12V up to something in the 80V range.  Assemble a bank of capacitors to get me in the neighborhood of a theoretical 12V load dump..  which from what I can tell is in the hundreds of joules range.  Once charged, simply connect it to the battery input of the device - which wouldn't be connected to an actual power source lest I want to potentially screw that power source up - and see if explodes, and if not, see if it works normally afterwards.  Does this sound like the dumbest idea you've ever heard?  Is it a somewhat decent idea that simply needs a lot of planning and number checking to in order to work roughly as intended?
81403
I'm new to EE so my theory isn't that strong. I want to trigger a  5v relay with my Arduino using a PN2222 transistor and a flyback diode. I plan on using the schematic below:         &ndash; Schematic created using   Now my question is: If I'm powering the Arduino via USB, will the current draw be too much for the Arduino's Vcc pin, thus requiring external power? I believe I'm reading that the relay will draw ~90mA and the Vcc pin from the Arduino can handle up to 200mA total. However, the transistor drives how much current is actually drawn, correct? Which is a function of the digtal pin voltage, R1 and the Hfe rating of the transistor? So if you were to mess that up, the current flowing to the relay would be incorrect and would not trigger it?   Thanks!
95271
I'm trying to do a simple, low power, low cost power "switch" for a custom board (I know, it will be complicated).  The device will be powered by default by a coin cell battery (3V3 and lower) and can also be powered by USB (5V). I will use a boost converter () for the battery, it will allow a constant 3V3, with a bypass for sleep mode controlled by a MCU.  This is a part of the power rails I designed:         &ndash; Schematic created using   The battery is protected by the schottky diode (some nA can flow into it, but it is accepted by the manufacturer)  If the VUSB is not present, the diode D2 will avoid reverse current in USB connector. D4 is protecting the LDO itself from the reverse current by bypassing it.  The boost bypass will be activated by the MCU when it is in active or when USB is plugged.  Do you think it will work correctly?