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I decided to make a battery charger for the phone. I realized that the voltage needed is less than or equal to 5 V, like a conventional charger. Here we used 4 batteries of 1.5 V 4 * 1.5 = 6 And a resistor of 2 Ohms and then it turns out: 6-2 = 4 And we get a voltage of 4 V? The point is to do this if then you can take 3 batteries of 4.5 V? Or another resistor, for example 1 ohm?
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Author 0
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admin,
Well, this is a small error, but then it turns out better to take a resistor that reduces the voltage by 0.5 V?
And I checked, there is voltage at the ends of the connector!
0
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Could it happen that the voltage has become lower due to the diode?

But what about, there is a voltage drop on the diode, somewhere in the region of 0.2-0.5V.
Author 0
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I have not measured it yet, there is no equipment at home. Could it happen that the voltage has become lower due to the diode? By the way, the diode is soldered correctly 100%
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And what happened? What is the voltage?
Author 0
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So, okay, now what's wrong? The phone writes "Removable device mode", the type thinks it is connected to the computer. No charge
0
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I have to charge different batteries and 3.7 V, and 6.5 V and 12.6 V
A stabilized voltage (adjustable) is taken higher by 3 ... 5 V. Then, through a diode and a bulb (sequentially on), it is connected to the battery (counter). The bulbs are different depending on the charging current (3.5V x 0.15A, 6VX0.3A, etc.). And the rechargeable battery is immediately visible on the light bulb - they fade, those current decreases (this is necessary for the correct charge)
Author 0
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Then can do for 3 and 4 batteries at the same time? Only benefit from this will be? I drew a diagram for this, but it makes no sense ... Is it possible to replace the resistor with a bulb? Or is there a diode with a luminous diode?
+1
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Cyril 517,
Try with a diode, then write off what happened.
Author 0
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Even if you use a diode? Am I planning a diode and 4 batteries or is it still possible 3? Yes, and which diode to choose? . More economical ...
0
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It makes no sense 3 batteries to use. They will not be able to give all their charge, maybe 10 minutes and charge, and then sit down and that's it.
Author 0
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Clearly, it means that you need exactly 2 ohms and 4 batteries! And 3 batteries just do not fit?
And why did I decide to take physics? I don’t understand anything ... At least there is still a week ...
+1
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Quote: Cyril 517
And the resistor is 2 Ohms and then it turns out: 6-2 = 4 And we get a voltage of 4 V?

Why on earth? The resistor is calculated differently, according to Ohm's law and the second law of Kirchhoff. In general terms, the formula for this task will be as follows:

R = (Uin-Uout) / Iout,

where Uin - source voltage, at the input;
Uout - voltage after the resistor, at the output;
Iout - current output.

Then we get: R = (6V-5V) /0.5A=2Ohm

The current is taken standard in USB 500mA.
Author 0
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And can you do the same with two batteries?

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