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I want to make a converter that takes in 24v and charges a phone by giving it 5v. According to my AC phone charger the output given to my phone is 5v and 0.7A. Therefore I have to lower the voltage by 19V. So, V/I = 5/19 = 27.1 = R. I then make a circuit connecting the plus from the 24v power source to one end of the 27.1 Ohm resistor, the other end of the resistor to the plus of the phone charger (a simple transformer that allows a phone to be plugged into the circuit), the minus of the phone charger into the minus of the power source.

Assuming that that is correct the power that the resistor would have to withstand would be I^2*R = 0.7^2 * 27.1 = 13.28 W. 13.28W is huge. I think that I made a mistake. If so can you tell me where?

Shabab
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Daniel
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  • No mistake. 19V * 0.7 A = 13.3W – user28910 Sep 27 '13 at 14:20
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    What does V/I = 5/19 = 27.1 = R mean? Looks like a typo – Roman Susi Sep 27 '13 at 14:22
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    DO NOT USE A VOLTAGE DIVIDER TO CHARGE YOUR CELL PHONE. – scld Sep 27 '13 at 14:37
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    Anyway, yes, your resistor would dissipate 13.3 watts - which is why linear regulators (the workable version of this approach, where the resistance is a semiconductor adjusted by a feedback circuit so that the voltage drop yields the desired output voltage) are a poor choice for conversions of this type. Such a solution is about 20% charger - and 80% space heater. – Chris Stratton Sep 27 '13 at 14:39
  • Yes, it is a repeat of my previous question. I was answered that my approach was imposable yet I thought I found a way to make it possible. I don't see what caused you to write in bold (anger) I was asking a sincere question. – Daniel Sep 27 '13 at 15:05
  • It's not anger, it's emphasis. We don't want to see you break your own property or, worse, hurt yourself.

    Also, the other question is the appropriate place to expand on that question, not a new topic. This allows all relevant information to be contained in one area.

     – scld Sep 27 '13 at 15:17
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    @Daniel - you were explicitly told why it would not work in the answers to the other question, but you chose to ignore this. Your calculation here is accurate, but that is irrelevant to your problem, and you're being rather impolite in posting this while ignoring the answer you were already given which explains why load variation makes a resistor solution unworkable. – Chris Stratton Sep 27 '13 at 15:37
  • This is what your car/phone will look like if you use a resistor to divide the 24V to 5: http://static.seekingalpha.com/uploads/2009/8/5/saupload_exploding_iphone.jpg – jippie Sep 27 '13 at 16:12

1 Answers1

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That's correct ; since it is a series circuit you would indeed have the same current in the resistor and the load ; and the power is then split according to voltage drops (19V vs 5V).

An other problem with your circuit is that you would indeed have 5V output only @0.7A. Any lower current draw would increase the voltage across your phone well over 5V. You could solve that problem using a linear regulator (like lm7805) but then you still would have to lose (and evacuate) 19/24 of the power.

A better solution would be a DC-DC switching step-down regulator : this basically uses 100% of your 24V to charge up some energy store (typically an inductor) for a small amount of time, and then disconnects from your source to use this energy over a longer time (at a lower 5V voltage).

Nicolas D
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