Reducing voltage 5 to 4.2 V

[Tricarico]

Hi guys, I'm an electronic newbie, reading some books and I thought I had finally understood the basis of circuitry, but no... I didn't. I'm a pic programmer and began on arduino a few years ago. I saw a hole in all this "arduino stuffs" that only electronics could fill so I decided to study it.

In arduino forums, what I call "5v forum" :razz: they say that you can reduce voltage using resistors and ohms law, this made my mind to fly. I made some basic circuits using only resistors and everything worked like a charm.

So I got a tablet that cannot turn on anymore - a very cheap and simple model - to fix it. I tested the battery (3.7 v) and the measures said 0 v, completely killed. I used an dc power supply at 4.2 volts to get it to life again and it worked for a while. But not enough to start the tablet, it doesn't even blinks.

The external connector seens not get any load, it's broken. So I had to made all the tests on the battery wire. I applied 4.2 v on the battery output and the tablet was on, like magic, my life turned blue again. So I decided to leave the battery and use a simple wire to an usb 5 v wall source. All I had to do was using a resistor to reduce the 5v to 4.2 and voilá! But no... again... Actually, that's where my problem begins.

Using ohms law, r = (5-4.2) / 0.8
So I have to use 1 ohm resistor to do that? Is that correct guys?

I don't have 1 ohm resistor, I don't know even if it's possible. I took a potentiometer in the lowest resistance level witch gave me 2.2 ohms. Very near I right? But no results, no response from tablet. All the tests I made here was direct on the battery wire.
I want to know your opinion about this guys, about what should I do?

Well, I don't like to give up so I read some posts in other forums, someone came with a suggestion to use 2 diodes in series. I had two diodes from an old tv (eic ba 159). Measuring the voltage between the two diodes, gave me perfect 4.2 volts! I understood the drop voltage on diodes but nothing happens. I thought my wall dc source was giving low current, it says 2A but I'm not sure if it's true (I'm in Brazil, we have many crap in here...). I tried with my dc power supply (I'm sure it can give up to 2A) and this time tablet blinks and shutdown 1 second later.

So, my big question:

Why doesn't work? An I using a wrong diode? Should I use a resistor? (when should I use resistor or not?). Did I said something wrong on my electronics learning?

I know it's a long question guys, thank you for reading.

 

[KlausST]

Hi,

Yes long question. ;-)

r = (5-4.2) / 0.8

Ohm's law is: R = U / I.
U is correct. It is the voltage ycross the resistor.
But the 0.8? Is it I?

The problem is, that I is unknown, it may be above 1A when active, and itvmay be only some mA when idle.

So a resistive solution is not the best way.

Your idea with a diodes is better, because the voltage drop across a diode is about constant, independent of I.
A typical diode like the 1N4003 wil drop about 550mV in the mA range, and about 0.9V with 1A.
Therefore i'd try with a single diode instead of two.

Maybe two schottky diodes in series are better.
But first you need to know the range of current. Without ... it's just guessing.

Klaus

 

[Tricarico]

Thank you Klaus.

Yes, I = 0.8 on a max peek.

Using the DC power supply (4.2v with 2A max) I got readings from 0.46 to 0.8 A. This 0.8 A current seems to occur twice in a fraction of a second, most time is about 0.46A.

I will try this 1N4003. Thanks!

 

[Audioguru]

You are playing with fire that might burn down your home. Look up Lithium Battery Dangers on the internet.

1) A Lithium battery cell must NEVER be charged higher than 4.20V and you are charging it to 5V. Your series resistor limits the current but does not limit the voltage unless it has a high current load.
2) A Lithium battery charger must limit the current to the recommended charging current but your current is MUCH too high.
3) The charger must be disconnected from a charging Lithium battery when the current drops to a low amount.
4) A Lithium battery that has discharged to a voltage too low must NEVER be recharged.

 

[KlausST]

Hi,

I agree with Audioguru.

Is the battery still connected, or do you replace the battery with your 4.2V circuit?

If the battery is connected, you need a true charging circuit.

Klaus

 

[Tricarico]

Thanks Audioguru, I didn't know it was so dangerous. But I removed completely the battery, it's dead. I'm only using it's wires because external jack doesn't work either.

 

[c_mitra]

The battery, even if it is half-dead, acts as a decent load and buffer. It must have some charge left to run the tablet for 5-10 seconds after the charger is removed.

Without the battery, some laptops (I guess tablets are similar) figure out that the battery is missing and refuse to start up. In fact, it does start up and sees that the battery is missing and shuts down.

My HP laptop refuse to run only on the charger if the battery is not installed. But one student brought her Sony laptop that happily ran only on the charger (no battery)- but it has other serious problems.

While running only on the charger, laptops are very sensitive to noise and spikes and resets every now and then.

 

[Tricarico]

The battery, even if it is half-dead, acts as a decent load and buffer. It must have some charge left to run the tablet for 5-10 seconds after the charger is removed.

Without the battery, some laptops (I guess tablets are similar) figure out that the battery is missing and refuse to start up. In fact, it does start up and sees that the battery is missing and shuts down.

My HP laptop refuse to run only on the charger if the battery is not installed. But one student brought her Sony laptop that happily ran only on the charger (no battery)- but it has other serious problems.

While running only on the charger, laptops are very sensitive to noise and spikes and resets every now and then.

The battery is dead but I don't care so much because my intention is to use the tablet as an Arduino compiler and upload only. For me it's ok to use a wall power source.

I know the tablet works fine with a 4.2 v on the batteries wire because I tested it using a DC power supply. I noticed peeks of +800mA with an average 500mA. Maybe the battery is necessary when external power jack is used to bust this 300mA extra load. But my external jack seems to be broken because there's no response at all on the battery wires which supposed to be charging. The same test using my DC Power Supply didn't work switching to 5 v, 1 or 2 diodes. I didn't tried yet 1N4003 diode recommended by Klaus.

 

[c_mitra]

Test that the wall jack can give 5V at 1A DC and then connect a regular silicon diode in series and also an inductor and a capacitor (to act as filter). It should work without fuss...

 

[Tricarico]

Test that the wall jack can give 5V at 1A DC and then connect a regular silicon diode in series and also an inductor and a capacitor (to act as filter). It should work without fuss...

Wow, thanks Mitra but capacitors and inductors are far away from my acknowledgments. I believe my wall jack cannot handle 1A even saying 2A on the specifications. The diode tests I did with a true "table" DC power supply and it worked perfectly at 4.2 V. But when I switched to 5v with 1 and 2 diodes it didn't work.

Could you please point me some documentation about capacitors and inductors please? Thank you

 

[c_mitra]

I do not know the make and model of the tablet, but I am sure that it will drink power like a fish!

The wall adapters that give 5V at 2A are possibly SMPS based (feels light weight) or perhaps transformer based (bit heavy).

Perhaps the power available with the desk power supply is not enough. With high current, the power supply may even shut down as safety measure.

I suggested a capacitor (2200 uF or more) so that the tablet can get some surge current from the capacitor. I also suggested an inductor so that the voltage at the power supply should not dip too much to shut down the tablet...

 

[ads-ee]

I'm wondering why nobody noticed that the OP actually hooked up the 5V USB source to the tablet through a simple resistor. If the tablet was on a low current (which it was at startup) it saw the entire 5V on the input, I'd say it's possible the OP damaged the device with the initial over voltage test.

Have you attempted to power up the tablet with the original 4.2V bench supply to get a baseline working condition?

Actually you should probably run the tablet at 4.0V or 3.7V as that is the nominal voltage.

 

[Tricarico]

I'm wondering why nobody noticed that the OP actually hooked up the 5V USB source to the tablet through a simple resistor. If the tablet was on a low current (which it was at startup) it saw the entire 5V on the input, I'd say it's possible the OP damaged the device with the initial over voltage test.

Have you attempted to power up the tablet with the original 4.2V bench supply to get a baseline working condition?

Actually you should probably run the tablet at 4.0V or 3.7V as that is the nominal voltage.

Guys, excuse me the delay to answer. Ads, the original wall supply is 5V but plugging it to the external power jack seems to have no effect. I believe the jack is broken.

These are the different behaviors by power supply and voltage (at the battery wires):

Simple 5V Wall Supply (1A?): Tablet blinks and shutdown after 1 second

Simple 5V Wall Supply and 1 resistor: Nothing happens

Desk Power Supply at 5V: Tablet turns on, load Android system completely (welcome screen), after loading complete it shutdown

Desk Power Supply at 4.2V: Works perfectly, battery icon shows 90%

Desk Power Supply at 5V and 1 diode (eic ba 159): Tablet blinks and shutdown after 1 second

Desk Power Supply at 5V and 2 diodes (eic ba 159): Nothing happens.

I didn't tried 1N4003 diode recommended by Klaus yet, my job requires 12+ hours per day. Thank you!

- - - Updated - - -

Hi Mitra, yes, it feels light weight. The desk power supply is very different, I'ts very different:

https://produto.mercadolivre.com.br/MLB-697625153-fonte-de-alimentaco-digital-yaxun-ps-1502dd-110v-_JM

I'm learning about "this magic component" called capacitors, they seems to serve a lot of purposes. If I take one of this as you recommended, would it work like an extra charge? I'm confused because capacitors don't allow dc current, I can imagine that current will flow until the capacitor is full loaded, after that current will stop, what would be the next step? I mean, the circuit have to be open so that capacitor may discharge, is this correct? What is the capacitor rule in all this?

Thank you

 

[c_mitra]

The following conclusions can be safely made:

1. Your tablet is fine - because it is getting up and running.

2. The original power supply was a Li-ion battery that has a full charged potential of 3.7V; with 10% margin, it will be 3.7+/-0.4V.

3. There is overvoltage protection within the tablet and it is not accepting 5V as such. This is apparently under software control and comes into play only after complete booting.

4. The diode (BA159) drops more than 1.2V at 1A. Not satisfactory.

5. Two diodes in series is out of the question. The drop will be 2.4V at 1A and the tablet sure takes more than 1A.

6. If the wall 5V supply is light weight, it is SMPS based and can give more than 1A (hopefully) if rated for 1A (I hope the designed was conservative)

The capacitor need to connected across the input, the positive side connected to the positive rail and the negative end connected to the negative end.

The inductor need to be connected in series. You connect one end of the inductor to the 5V and the the other end of the inductor goes to the load.

You are correct that capacitors stores charge and acts like a battery. Only thing is that it has only a very tiny capacity.

 

[hobbyckts]

how much is the current withdrawing when you use the desk power supply? If you want to use the Simple wall supply better use a capacitor which is suggested on Post #11

 

[Tricarico]

Mitra, thank you so much for your information. I got it perfectly, the voltage drop makes sense now. In fact, If i use 4V it doesn't turn on. Hobby, the current keeps on 450mA and twice 850mA for a quarter of a second (peeks). Maybe more current is being hide from the desk power supply because it's to fast?