I have purchased a 300W Modified Sine Wave Inverter. I am also connecting a 12V 18A/hr SLA battery to it.  I am using it to power an RFID reader. The input power characteristics of the power block are: 100-250V, 50-60Hz, 0.5A.  The output characteristics are: +48V DC, 0.4A. Please note the power block connects to the RFID reader via an ethernet cable only. Hence I cannot just connect a DC supply directly to the RFID reader.  How would I calculate how long the battery will last before I need to charge it, do I use the input or output characteristics of the power block? (Is there an equation I could also use).  Also, I would also like to connect my laptop charger to the inverter while the inverter is powering the RFID reader. The input characteristics for the laptop charger is: 100-240V, 50-60Hz, 1.2A  Output characteristics are for the laptop charger are 19V DC, 2.73A.  Will it be okay to connect the laptop charger and the RFID reader at the same time, or will this exceed the limitations of the battery and the inverter?  I know how to calculate the duration of a battery given a DC to DC setup. But know I am going from DC to AC to DC again. Do I use the input characteristics of the power supply blocks i.e. the currents or the output characteristics of the power blocks?
27887
How do I calculate the approximated time for the Charging and Discharging of the battery? Is there any equation available for the purpose? If yes, then please provide me.
95935
I have ordered a Dual H-Bridge 12A continuous (30A peak) Polulu motor driver shield for my upcoming (first time ever, a 1.2m long tank made of wood) robot project.   I was thinking about how to supply the power to all the components in the circuit without wasting much power, but the only thing I can come up with is this simple voltage divider connected in parallel to the motor drive:    Knowing the Arduino can handle max 5V I did make sure to make it a bit lower.  But my wonders go to the fact that I am suspicious that connecting the motor driver and Arduino to the same power source isn't a great idea? Or am I paranoid?  And what about power waste by the voltage divider, how can I reduce/minimize it? Make the resistors smaller? (15/10 ohm - but more error and higher current)  Is there a formula to find the most effective R1 and R2?  Do I have to worry about current spikes?  Any other things I could miss in this simple-yet-suspicious circuit?  Also, if I wanted to simulate any possible scenario, which components should I choose in the circuit to keep it as simple as possible (for simulating in Multisim 13)?
63410
I am trying to measure voltage &amp; current off a solar charge controller from panel, battery &amp; load simultaneously with my arduino. I have managed to achieve this by tying all the negatives together and using it as GND, however, when I do that the charge controller doesn't function properly and when I switch to off-mode the loads still run. After checking the instruction manual further, I found out that the positives are shared common between Pan, Bat, Load. This was confirmed with a multimeter.   So now what I am thinking of doing is using the positive from the battery as the GND reference and then measuring all other voltages (through voltage dividers with respect to the GND). The voltages should then range between -17V to 0V.   However, I have read that the analog input pins are not able to read negative voltages, and with respect to the positive GND, the Analog inputs will all be negative.   EDIT: I was thinking of using an amplifier to invert positive ground to a negative ground, meaning I could measure all the positive sides of the voltages. But after getting answers from you guys it seems I may not have to use an amplifier and a voltage divider may suffice. I have currently included a 2 resistor voltage divider between each analog input, +5V source, Vmeasure which you can see in the diagram below.  According to my calculations, if this works, @-17V I should see 0.6V @ analog input. @0V I should see 4V at analog input.   I'd appreciate feedback on this circuit, and below in the second image I've also included a piece of another method suggested by @jonk (thanks), and would be interested in knowing what the benefits are of method B) over method A)  Thanks!  Ps. the ameters are there in place of ACS712 current sensors which also go into analog input pins.   Method A)        &ndash; Schematic created using   Method B (EDIT: This method has worked, I am still yet to test method A) thanks @Jonk))
74217
We've got a range of laptops in our office all macbook pros some of which are 13" and some of which are 15".  The 15" laptops came with 85w chargers, and the 13" came with 60w chargers.   I want to buy some additiaonl chargers to leave plugged in around the office for anyone to use, but im unsure as to which type i should get.    Would charging a 13" laptop on a 85w power supply damage it ?  Would charging a 13" laptop on a 60w power supply damage it, or just make it slow at charging ?
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.
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!
121668
I have used the above simple circuit for a constant source where \$V_{\text{in}} = -5\$V, \$R_{\text{in}} = 390\Omega\$, \$R_f\$ is the load which ranges from \$110\$ to \$170\Omega\$....would using this circuit for a constant current source for the load be recommended?  As current $$i = -\frac{V_{\text{in}}}{R_{\text{in}}}$$ $$V_{\text{out}} = -V_{\text{in}}\frac{Rf}{R_{\text{in}}}$$ so it should serve as a constant current source, but while browsing the Internet for a constant current source I typically found these sorts of circuits:        Is there a compulsion to include the transistor in my circuit? If yes, why wouldn't my circuit serve the purpose for producing a constant current?
112604
At an Not For Profit/Charity I volunteer for occasionally we were very fortunate to be provided (donated) with a brand new 3Kva APC UPS. This is total overkill for our modest server rack (four mid-range servers and a switch!) but hey, I'll take what I can get!  Pressing the TEST button on the front indicates that yes, the UPS does work. Brilliant. But it only tests it for about 15 seconds.  My question is - will I degrade the UPS by unplugging it from the wall to see how long it will last? My plan is to unplug it, and wait until the battery meter reaches its last LED before plugging it back in, so that I know about how long I will have in the event of a power outage.  Do people do this on a regular basis? I'm guessing no (Lead Acid is very different to Li-Ion batteries)... but what kind of harm would it do if this were to happen (on purpose) every 6 months?
44016
I am trying to design a circuit on a PCB whose supply voltage needs to be 3.3 V.  Currently I have two options:   Use three AA batteries of 1.5 V each to obtain 4.5 V, then use a converter/regulator to bring it down to 3.3 V Use a 9 V battery &amp; bring it down to 3.3 V   Which would be the better option, considering that I am making a PCB (not sure if the PCB matters)?  Cost is not a factor with respect to the above 2 options. My current requirement is around 100mA. What i have understood after reading about smps is that they can cause some interference, i am using a zigbee on the same PCB, so will the smps coil interfere with the zigbee?
10727
What would be the least expensive way to boost a DC voltage?   The aim is to convert 1.2&nbsp;V/1.5&nbsp;V (from an AA/AAA cell) to 3.3&nbsp;V to power a small 8-bit microprocessor, like Atmel ATtiny45 or ATtiny2313, and also (if possible) 6&nbsp;V to power a buzzer.  Also, what would be the maximum current one could draw safely from an alkaline battery, after boosting it to 3.3&nbsp;V/6&nbsp;V?  Finally, how I could compute the duration for which the alkaline battery would last, given a certain consumption?
94695
I have a powerbank with a 20,000 mAh capacity, from which I can power different devices.   The devices include an LED strip and a Raspberry Pi Zero W (RPI Zero W).   The LED strip consists of 60*5 LEDS = 300 LEDS, and will roughly draw 4.5 A with 50% PWM.   The Raspberry Pi requires about 170 mA.   The powerbank has 3 output ports, one capable of drawing 2 A, and the other two capable of drawing 1 A.   Would it be possible to somehow extend the current draw from these USB ports?  I could put them in series, such that I would have one port with 4 A, but how would that affect voltage? I need to power both my Raspberry Pi, and the LED strip.   Would the USB port be able to handle such a high current draw?
87416
I just started studying about batteries and battery chargers. I have this simple query so please help me out.  I am trying to charge my 3000mah li-ion battery. The battery voltage is around 3.6v and the charger regulates the 5v from the source to 4.2v. I connect the battery to the charger. Now what is the voltage at the output should I expect? Shouldn't the output voltage be at 4.2v so that the battery gradually charges to 4.2v ? I ask this because I am using Bq2057 IC and once I connect to the battery to the charger the output voltage is dropping to 3.63v.  My assumptions were, 1. I considered the charger as voltage source 2. Battery also as a voltage source  I experimented this buy connecting two DC voltage sources(5 and 3.3v) in parallel and measured the voltage at the node and it was 5v. So I was expecting the charger output to be 4.2v even when I connected my battery to it.    Thanks
107952
In the link below, this circuit I found is what I needed. Only the length of the pulse and the delay are not what I want.  When I want just 0.5sec before pin 3 is up and the pulse duration I want is also 0.5sec, can someone tell me how big my resistors and capacitors must be? Or can someone tell me how I can calculate this?  My source power is 16 Volts I have already found 555 timers that can operate with 16 Volts, so thats not the problem.  When someone has another idea with two 555-timers, or something else that does the same, it's also OK.  Thanks
111993
I have an eco worthy 4 solar panels (2 * 12 connected in series and 2 * 24 connected in parallel). I can confirm that on a good sunny day output voltage from panels (measured on the red and black cables is about 22 to 24V) produce the output expected. On the battery side, I have 4 * 12V 30AH batteries (li Ion) connected in series(2) and then parallel to produce output 24V. Issue When I connect solar panel in series mode across the 4 batteries(I tried even just 2 of them in series) they dont seem to charge since the solar output is just less than 24V.  On the other hand, if I connect the solar output in parallel on directly to one battery it charges fast (like an hour or so) to about &gt; 13.5V. And, then it is able to handle load for a few hours. My issue I am unable to get the batteries to charge in parallel from a solar array configured to produce 24V. On the battery side, I have 4 12V 30AH batteries (li Ion) connected in series(2) and then parallel to produce output 24V. Does anyone have any suggestions if solar panels expected to produce 24V are lesser than 24V expected(Live in a moderate weather not always bright sun)? My issue I am unable to get the batteries to charge in parallel from a solar array configured to produce 24V.
120583
I needed some basic advice about impedance matching. I have an RF amplifier with 50 ohm output impedance, capable of generating 400W output power at 500kHz. I am trying to deliver this to a 12 ohm load.  I looked online and figured out that I need to match the impedance of the amplifier to the load. What is the best way to achieve this? Would it be possible to simply use a capacitor and inductor and put an L section to the output of the amplifier? I saw that online stores have very cheap capacitor and inductors that are rated for very high voltages.  If that is not going to work, is it possible to buy any transformers that would do that? For instance I saw that there are many cheap 220V-110V voltage converters, which should have 2:1 coil ratio. So in theory that should convert 50 ohm to 12.5 ohm and matches to my load. Will these work at 500 kHz as required for my application? Or are there other type of transformers that I can use for this purpose (there are many many types of transformers and I got confused which ones are suitable for impedance matching)?
123059
In the power amplifier circuit shown if the input is an RF sine wave, with VDD=1.8V and Vgs = 1V, what will be the waveform at the drain of the MOSFET - sine or square wave? Is it possible to get a sine wave at the drain?
55034
I'm designing a project that controls 110VAC (plug in, socket out), and also has a microprocessor inside.  I'd really like this project to not need a wall wart to get it's 5V, so I want to include a small DC power supply inside.    Apple's USB power adapter looks like a great small design that I can fit inside my case.  I'm guessing this is a switching power supply but I can't find any teardowns or schematics anywhere.  Does anyone know how this thing works?  I know I could make my own 110V->5VDC supply from a bridge rectifier and some caps, but I'd like to also be super safe.  What's in the Apple brick?