Hello everyone,

I am stuck in my experience. I would like to be able to charge my phone on a 12V battery (7 Ah). To do such a thing, I use a DC/ DC Car Power Converter to adapt voltage from 12V to 5V. Then, I placed a diode to isolate the circuit. Because I will add an other 5V input (from a cell phone charger), and i want to isolate the 12V battery from this second input. (The board I want to build will be able to charge my phone when it is connected to the cell phone charger or the 12V battery.) And last but not least, I put a red LED and its resistor (165 ohms) in parallel in order to know when the phone receives power from the battery.

The problem is that the voltage decreases to 4.47V which is insufficient to charge the phone (nominal voltage is 5V). I know that the LED normally induces a drop of voltage when placed in series, but that's why i placed it in parallel.



So, here are my questions:
Does a LED in parallel induces adrop of voltage? Or the voltage drop is due to another thing?
If the LED really induces a drop of voltage, how could I obtain a 4.92V to charge my phone, and keep my LED in the board?

 

The problem is not the LED, the problem is the series diode.

A diode such as you use will not provide any isolation. All I can ever see it do is prevent accidents should the DC-DC ever get connected backwards, which can only happen once when you assemble the unit.

Just remove that diode and you will get the 5V to charge your unit and light that LED.

And welcome to the forums!

 

Welcome to AAC!

The voltage drop is coming from the steering diode, not the LED.

In the top circuit, you're measuring 4.92V with no load. With a load, the diode will drop 0.6-0.7V (depending on load).

In the bottom circuit, I assume you have the LED wired correctly. In the schematic, it's wired backwards.

 

Hello everyone,

I am stuck in my experience. I would like to be able to charge my phone on a 12V battery (7 Ah). To do such a thing, I use a DC/ DC Car Power Converter to adapt voltage from 12V to 5V. Then, I placed a diode to isolate the circuit. Because I will add an other 5V input (from a cell phone charger), and i want to isolate the 12V battery from this second input. (The board I want to build will be able to charge my phone when it is connected to the cell phone charger or the 12V battery.) And last but not least, I put a red LED and its resistor (165 ohms) in parallel in order to know when the phone receives power from the battery.


The problem is that the voltage decreases to 4.47V which is insufficient to charge the phone (nominal voltage is 5V). I know that the LED normally induces a drop of voltage when placed in series, but that's why i placed it in parallel.

View attachment 114257

So, here are my questions:
Does a LED in parallel induces adrop of voltage? Or the voltage drop is due to another thing?
If the LED really induces a drop of voltage, how could I obtain a 4.92V to charge my phone, and keep my LED in the board?

A 1N5819 diode has about half the voltage drop as a 1N4004.
You do not need to drive the LED at that high a current. Figure 2 mA instead. Depending on the efficiency of the LED. Adjust the resistor for minimum acceptable brightness. And yes, it is backwards in the original drawing.
Yes, the diode is a good idea. If the charger is turned off but still connected to the phone the diode isolates the charger from the phone. Not knowing what the circuit of the phone is like it may or may not be necessary.

 

Thank you for all your answers! I'll test your solution in the afternoon, and I'll go back to this wonderful forum if needed.