Here is the circuit of cellphone charger. I have seen the input voltage range for a cellphone charger to be 100-230 V but the output voltage to be fixed to the rated value(like ~4.5 V). So, my question is how the output voltage is fixed to the optimized value if I provide different mains supply voltage. Like, in India I give 220 V RMS value while in Japan I provide 100 V RMS.The grid frequency is also different in different countries as well as the RMS voltge. I am attaching a picture of the rating of an ac adapter(cellphone charger).-
123976
I would like to know about the workings of mobile phone battery chargers. I know the basic principle, you take a transformer to bring it down to 9/6/3.7 volts then a bridge rectifier circuit and a capacitor input filter to rectify it into DC. Then you apply it to the battery. But I know this should not be enough, I need at least two things,   Protection against back current flow Switch off power when the battery is fully charged.   Can someone point me to something simple (apart from this )
64648
I don't understand how cells behave when connected in parallel. So what if I have two Li-Ion cells: one ist "rated" at 15 amps (which means that the cell is able to discharge at 15 amps without problems , right?) The other cell ist rated to 5 amps. Now assume both have appropriate same Voltage to connect them both in parallel. What happens if you connect a load which would draw 10 amps now? Would both cells try to discharge with the same 10A (and thus the 5A-cell would overheat and fail...), or would the current be spread up eavenly? Is it actually forseeable at all what would happen, since it depends on the internal resistance of both cells, doesn't it? Does someone know what would happen?  Then another question: what if you put two cells of different capacity in parallel? Isn't the cell with the higher capacity supposed to give more current? Because the discharge curve of the high-capacity-cell is above the discharge curve of the low-capacity cell. So the high-capacity-cell would have to discharge more of its current until it decreases its voltage to the same level as the low-capacity-cell? So would that mean that the internal resistance gets lower with a higher capacity, is it proportional?  Hopefully my explanations aren't to confusing and hopefully there are people who are bothered by the same question :) Thank you!
22406
I am supposed to come up with some control schemes for the converter before understanding the standard PWM control scheme.   Isn't it just enough if the output of the converter is compared with a fixed reference voltage to drive the output to required voltage level? Although I'm not sure there is a lot more to it, this seems to work pretty well.   I've tried simulating this circuit with L,C and R values from an example.  Schematic:    The required output voltage is 6V. S2 is a voltage controlled switch, controlled by the output of comparator.    V2(green) is the output of comparator and V1(red) is the output of buck converter.
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The rating of a real transformer is 15 kVA , 2300/230-V then can we say that the primary rms is 2300 V or can we say that the secondary rms is 230 V or both.  To clarify further, suppose to solve for certain characteristics of this transformer like primary current and secondary current i use the equivalent circuit equation of the transformer referred to the primary side as:  Vp = aVs + Req*Ip + jXeq*Ip  Then what should i use for Vp or Vs in the above equation looking at the rating of the transformer.  In the above equation looking at the rating data given above is Vp= 2300 or Vs = 230 ?As mentioned earlier the transformer is mentioned to be real.
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let me say if I have a pack of lithium batteries connected in series and parallels, and i want to connect it to an Inverter, BUT...BEFORE I connect it to inverter, lets say it gives out 12 volts "hypothetically" some informations like 3.7 volt batteries, and 2500 mAh,  lets say then the pack gives 12 volts in DC and hypothetically again I calculate it ((still in DC)) is .....  W = V* Amp ... then it means wattage = 11.1 V * 5 Ah which is 55.5 W. then we got 55.5 watts, THEN....after we want to change the dc into ac it gives you 120 volts "please hypothetically" just answer me this, ARE the Wattage And Ampere are remains the same as DC? what would it be the result after AC?
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I need to amplify a voltage, which comes from a DAC and is 0-2.5V and scale it 0-25V in an operational amplifier. After amplified, I need to offset it -1.25V, so the final voltage is -1.25 to 23.75V. In other words, being an operational amplifier, I want to do the y=ax+b operation. with a=10, and b=-1.25V  I thought of 2 different ways of doing this.   The first idea I had is to use a second amplifier as a summing amplifier, with a fixed -1.25V voltage.  The second idea is to use a 1.2V zener in series with the output, along with a biasing resistor to -V.   Why do I want this? I want to set the ADJ pin in an LM338 voltage regulator from an MCU. LM338 out is ADJ+1.25V. If ADJ is -1.25V then the output is 0V.
71951
If a power supply says the input power is 100-240V, ~1.5A, I'm assuming that the "1.5A" draw would be at 100V, not at 240V, right?  I.e. would the total max watts be 150W regardless of the input voltage?    The context of this is trying to determine the maximum draw for something based on the input power specs on its power supply to ensure a circuit doesn't get overloaded if too much is on it.  I want to make sure to get the max draw correct whether this is in the US at 110V or Europe at 220V.    As a side question, the output rating on this supply is 19V, 3.42A.  Does that mean that it's pretty inefficient?
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I've heard about laptops such as the new Chromebooks that are charged via a wall charger that connects to a USB-C port. I'm quite happy that this will supposedly standardize laptop chargers but I'm a little unclear about how this works.   Existing USB ports provide a 5 volt source, but the laptop chargers provide up to 20 volts. is there some kind of higher voltage line or are the USB-C powered laptops running on a lower voltage?  All this information I've seen gives a fairly vague idea of providing more power, providing up to 100 watts rather than 10 watts. Even so, my laptop is not the most powerful machine and the charger still outputs fairly near 100 watts, I can imagine more chargers for power laptops providing much more than 100 watts. Could a general purpose USB-C charger really power all these machines?
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Many of us have lots of devices that uses AC mains (110V/220V) to DC (typically 12VDC, 9VDC and sometimes 15VDC) adapters. There's my Wi-Fi router, my powered speakers, printer, scanner, etc.  So I would like to combine all these into a single device; do the AC/DC conversion once and have everyone tap the DC line, instead of each doing the conversion themselves.  Regarding to this, I have two questions:   is there any commercial solution out there to do this? (bench power supply for example)? suppose I have one of these beauties powering my stuff and this power supply can deliver 100W of power. Will the device always be consuming 100W of power or will the consume increase as I hang more devices?    thanks.
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This DAC has an output range of 1-3 volts. Here's the datasheet:    I'm getting pulses with different amplitudes, frequencies, duty cycles and widths varying from 0.2 ms to 2 s from the DAC. I want an output range of 0-4 volts instead (Vcc=5v). I tried a high-pass RC filter, and then amplified it with an op-amp, but the output doesn't look like a pulse anymore. Also, I tried the circuit suggested in the datasheet (page 8) with no luck. I'm not sure what values I should assign to this circuit, since I don't understand how it functions. How should I go about this?
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Related:        From what I've heard:   110 V (or 220 V; household voltage pretty much) is dangerous (i.e. can kill you) I think there's consensus on this, no need to try :) 60 V (old telephone lines) is supposedly dangerous (never tried, only heard it once... probably won't try)   From what I know first-hand:   9 V is not dangerous (I've put a 9-V battery on my tongue, nbd... actually it kinda hurt!) 1.5 V can indeed be quite shocking with enough current (fell for one of those "Do you want some gum?" tricks back in high school...), but they sometimes do not use 1.5 V with the low amperage levels, some use a DC motor to vibrate and complete the trick.   So I guess there's two parameters here, voltage and current... but are there rough numbers on how much of each (or in combination, which I guess would be power) would be considered hazardous?  No old telephone lines have always been 48vDC well at least since from 1950s, if your skin is wet you can feel it slightly, like on your forearm.  Now the ring voltage is 90-110vAC with a 2 on 4 sec off cycle (USA).  It will ring your bell but good, should you be touching the wires when someone calls. The ring voltage rides on top of the 48vDC, so its present on the same two conductors that the voice voltage(DC) is on. Luckily it's 4 seconds off will give you a chance to get off the conductors with a scream (of pain).
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AFAIK each of the world leading cordless power tools manufacturers produces several "product lines" of cordless power tools with different voltages. For example, Bosch currently produces tools with Li-Ion batteries with 10,7V, 14,4V, 18V and 36V output and the higher the voltage the more powerful a tool is.  Now those tools are powered by batteries that are assembled from cells with lower voltage (something like 3,7 volts for Li-Ion cells I guess) and cells are connected in sequence until the target voltage is reached.  They could instead connect cells in parallel. They would have the same voltage, but higher current and that would again yield higher power.  Why do they choose higher voltage over same voltage and higher current to get higher power in electric tools?
111547
I have made a solar panel charger, it consists of a 1.2 watt panel connected to a buck converter, then to the phone.  Now, when I connect it, it charges the phone and in the battery settings it shows it is charging as "plugged ac" and slow charging.  I wanted to test if it can slow down the discharge rate, so I came up with a test procedure where I:   Downloaded "battery drainer" app (which essentially turns on cpu stress, gps, vibrator, full brightness etc);   Measure the time it take for the phone to get to 90% from 100% with solar panel attached and without solar panel attached.   But the time it takes to discharge is almost the same. I am wondering what am I missing. Wouldn't the solar charging increase the discharge time?  When I measured the solar panel output it was 5 V and around 110 mA.   Thank you, any help will be appreciated.
110230
The AC adapter supplied with my laptop is designed to accept voltage in the range of 100-240VAC at 50-60Hz, but the AC cable that plugs into the adapter unit itself is only rated to 125VAC 7.5A 2.5A.  I know the adapter can accept 220V power, but is it safe to use this cable with the AC adapter to plug into a 220V socket?  Edit: The part in question is a 90W HP Smart AC Adapter.  The laptop does not cause the AC adapter to draw more than about 80 watts from the socket (as measured with a Kill A Watt meter).  Edit 2: The amperage rating of the cable as marked on the attached label is 2.5A, not 7.5A.
121308
I have an interesting mathematical/electrical dilemma and I'm not sure where to start.  I am considering buying  (specs on page).  I will be hopefully buying a battery pack with it to support a long-road trip (a few months) wherein I'll keep medication in the fridge (powered by the battery pack when parked and by the car when driving). The medication needs to be keep between 2 and 8 degrees C at all times.  Here's the  I'm thinking of getting:  Can someone please help me understand what I need to do in order to calculate whether or not this fridge + battery combo will work, and if so, how much time will I be able to power the fridge for.  Ideally, I need a combination of fridge and battery which will allow me to have the fridge constantly powered, assuming I sleep in the car for approx. 8 hours a night and drive for the rest of the day (although, there will be times which I'm away from the car during the day for several hours + times where I hope to have access to a wall outlet).   Specs:   Fridge:   Input voltage (AC 220-240 V Input frequency 50/60 Hz Rated input power (AC) 64 W Rated input power (DC) 46W DC12V/50W DC24V W Input voltage (DC) 12/24 V   Battery:   Capacity: 453Wh Battery Type: Lithium-ion Input Voltage: AC 110-220 V, 50/60 Hz; Solar Power: 18 V; Car Input: 12 V Output Voltage: AC 110V 60 Hz; DC Output: 4 * 12V 4 A (Max.10 A); USB Output: 4 ports: 2* 5V 2.1 A and 2* 5V 1 A Power Output: Rate 500 W Peak/Surge 1000 W
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I have the following circuit, where I put R52 = R53 = 2700 Ohms :    Triggering levels are of 1V and 2V approximately according to my computations, that is okay according to the following figure showing input in green and output in blue :    But what I find really strange is that the maximum voltage output is around 3.3V and not at 5V. How would you explain that ? And last detail : how would you explain that the output is not exactly at 0V when the input is high?  Thank you very much for your help.
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I am going for my higher studies to USA. Since I am from India, my laptop, electric shaver and mobile charger all run on 220V.  Power rating of laptop charger: 65W (100V - 240V ~ 1.5A)  What are some good converters, that can run all the above mentioned appliances?