How to decrease the output current and voltage from this DC to DC converter?

This my device and I am using it as a cell phone charger. I tested its current and voltage during charging my NOKIA cell phone and got 6V and 700mA output. I think 5V and around 450mA output is more better for charging Li-ion cell for its long life. So I want to decrease its output to 5V, 450mA. Which components should I have to modify here. I guessed R1 (5 ohms, 2W) is responsible here. I didn't replaced and didn't measured because everytime if I replace components I damage the board (copper tracks) due to lack of desoldering devices!

Hello there Willen,

Are you sure that is the correct schematic, ie where did you get that from?
I ask because according to the data sheet and related app notes, that would only put out about 50ma or something small like that, and that is due to that 5 Ohm resistor which going by the data sheet would have to be more like 0.5 Ohms. Also the inductor values seems way too small, with the typical lowest value maybe 200uH not 20uH, and 2mH would be typical. We could look at this more though.

With that in mind and the fact that you dont like to desolder parts on PC boards, maybe it would be better if you just tried to add an input series resistor to lower the output current to the cell to be charged. This would not be hard to do but would require a power resistor, and you may have to try a couple different values.
If the input voltage is really 12v then a 10 ohm resistor would drop 5 volts at 500ma so that would only allow 7v to reach the chip now, which may just be enough. Unfortunately if the input power supply voltage changes then so does the current, so hopefully it doesnt change too much, and if it does not change too much it would still be ok.

If 10 ohms is too much then of course lower it a little to like 9 ohms then try again. If 10 ohms is not enough then try 12 ohms and try again, etc., etc.

Hopefully this does the trick and you wont have to modify the actual PC board which i see you dont like to do and i dont blame you. If you do have to do that though, the resistor to change would be that "5 Ohm" resistor but after looking at the data sheet i cant see yet how it can really be 5 ohms. If it was 0.5 ohms that would make more sense, and then you could try raising it to 1 ohm and that would halve the output current from what it is now, and of course a little less than that would mean a little more current than half.

The schematic isn't correct. The non-earth side of R3 should be connected to pin 5 of the IC.

R2 and R3 form a potential divider, so that when the output is at the desired voltage, pin 5 of the IC is at 1.25 V.

To reduce the voltage, reduce R2 to 3 kOhms. You could do that by adding a 13 kOhm resistor in parallel with the 3.9 kOhm one that is already there.

Increasing the 5 Ohm resistor would decrease the current limit, but it is probably not a good way to do so, as that resistor is there as a sense resistor, to keep the inductor from saturating.

Do you know where the charging current is controlled? I guess that it would be controlled in the phone, and the current limit in the charger may not matter. A 2.2 Ohm resistor in series with the output of the charger would ensure that you don't charge at more than around 450 mA, as long as the voltage is only 5 V and you are charging a lithium-ion battery at around 4 V.

You still need something to cut the charging off at 4.2 V, but the phone almost certainly does that.

Hi,

Yes that schematic is bogus. But it should be ok to limit the current by changing the resistor labeled "5 ohms" to a higher value but first noting that it can not be 5 ohms it must be 0.5 ohms or even something different.
That does however mean changing the PC board itself.
It should not hurt to increase that resistor because then the inductor sees less current. It would not be good to decrease that resistor however as that would mean the inductor would see more current. The basic function of that resistor is to adjust the peak current of the inductor, and that peak current reflects to the output as an average current so the peak current setting also sets the maximum output current.

Using a resistor on the output may limit current but it will also increase charging time. That's because the output impedance goes up substantially, and that could mean a substantial increase in the charge time. Using a resistor on the input is different because it does not change the output impedance over the whole operation period, only during the current limit phase which does not change the charge time. 2.2 ohms is a very significant change to the output impedance, and the effect in a normal Li-ion charger would be quite significant, while in this application it may be a little less because it appears that this circuit is going to be used to feed another circuit which actually regulates the charging. This in turn means the charge time then partly depends on the dropout voltage of the following charge circuit when the extra resistor is present.

Ops! I made fault while drawing!

I made the schematics by looking real device which I have. Yes, R1 is not 5 ohms, it is 0.5 ohm. And Yes, 5th pin of IC has 1k resistor to ground! Inductor has almost 30 or 40 turns on a ferrite core so I simply guessed hee hee hee. Now I used 200uH (or?) in schematic. Thank you both MrAl and Driver300! I edited the circuit now.

Suggested by MrAl, increase R1 little to get low output. wow simple, but it is not easy to find 0.6 or 0.7ohm 1 watt (or more) resistor, so may be it is little hard to get little lower output as my wish (5V, 450mA). So I think I have to do as your idea- few parallel resistors to get right value and power.

And you were saying datasheet has Max I is 50mA. My device has 8pin DIP IC marked MC34063A SUM11F02XX. I tested output current while charging my 99 percent charged cell phone and I measured 400mA current which was decreasing quickly. Just with few second cell phone was consuming 500mA, after a minute 400mA and after another few minutes current was 300mA. Then I used half charged another cell phone, I measured 860mA few second then after a minute current was 750mA and so on. After few minutes current was little more stable in 650 to 700mA. (less charged battery consumed more current.)

I don't know why datasheet is saying just '50mA'. IC number has 'A' letter after 34063, may be it is responsible, isn't it? During charging it is HOT and it has no heatsink. But my uncle is using it as a regular charger from couple of year and it is not destroyed till now.

Suggested by Driver300, Decreasing R3 from 3.9k to 3k will decrease the output. Wow then it's cool! Does it really work well? If I used 3k instead of 3.9k will it decrease output V and Amp? Nice!

Actually I don't know where current is limited. My friend said he have a charger which charges his cell very faster than other. So guessed current is limited on charger. But I think phone also has a limiter and protection circuit.

I bought this DC charger circuit but it was not working. Cell phone said "Charger not supported!" then I opened and looked the circuit then I found the fault made by manufacturer- 12V DC was bypassed to output! I felt like shock! My cell phone didn't burn with even 12V input! It means- phone has much more circuits like high input detector, limiter and protector too. But again I don't know exactly.

Hi again,

Hey, that circuit is looking much more believable now <chuckle>.

The way most cell phones work is they require a steady 5v input, and they usually accept a wall wart that can put out a full 1 amp even though they dont use that much normally. 700ma is typical.

So you should probably just lower the voltage to start with and go from there. A 3k in place of the 3.9k should take care of that as Diver suggested. If the current is still too high with a very discharged cell phone battery, then go ahead and try to limit the current with a resistor somewhere or change the 0.5 ohm to 1 ohm to start.
You can try 1 ohm first because the current may not go down to half of what it was, it may be a bit higher. If it is not high enough then the following parallel resistors makes the following total resistance:
10 ohms in parallel makes 0.9 ohms approximately,
5 ohms in parallel makes 0.8 about,
2.5 ohms in parallel makes 0.7 about.
So it's not hard to do, just a single 1 ohm resistor (1 watt rating) in parallel with one of the above, but only if the 1 ohm does not work out well enough.

Let us know how it works

Willen,
You said here that you want to charge a Li-Ion battery cell, not a cell phone.
You have a 6V POWER SUPPLY, not a charger. The charger circuit is inside the phone.
If you apply 6V or 5V to a Li-Ion cell then it will probably explode then catch on fire without having a proper charging circuit.

The circuit shouldn't have a 1N4007 as the flyback diode. It has a reverse recovery time of 2 us. With a 5 V supply and the 200 uH inductor that means that the current will build up to around 50 mA before being stopped by the diode. That is far too big. A Schottky diode should be used, as Schottky diodes have far shorter reverse recovery times.

If you have an existing converter, the diode will probably be Schottky already. The "1N4007" could just be another mistake.

This my device and I am using it as a cell phone charger. I tested its current and voltage during charging my NOKIA cell phone and got 6V and 700mA output. I think 5V and around 450mA output is more better for charging Li-ion cell for its long life.

Click to expand...

As mentioned by others the actual charging and control of the battery is handled by the phone itself. Most cell phones have 5 volt nominal input voltage rating meaning that honestly anywhere close to 5 volts or a bit above is just fine for them. 6 volts input is by my judgement well within the safe limits of what any decent band of cell phone will handle.

That said I do agree that since changing it to 5 volt output is simple single restor change go for it. Power limiting wise I would look at what the factory charger puts out and consider anything close to that good enough. The phone is only going to take the power it need anyway so why force it to starve if its not necessary especially considering that if you use a common USB charger cord your 5 volt USB port allows for at least 1 full amp of available current.

Why would you want to change the charging rate?
Your phone should have the battery supervisor chip built in and set for the battery in the phone.
The power supply just gives it more power than it needs.
If you give it less the battery supervisor chip may not work at all.

Thanks MrAl for tricky alternatives for 0.6 to 0.9 ohm resistor.

Driver300, hee hee hee I am really fool, actually device has 1N5819 schottky diode. I thought schottky diode has 'glass package' like 1N4148. I have few schottky 1N5711 in glass package so. But this one has black package same as 1N4007 so simply I thought it is also 1N4007.

And also I actually don't know how much inductance is there on this inductor. So It is (200uH) is just my guess.

Audioguru, if 5V will burn the Li-ion cell then what might be the ideal charger voltage and current for Li-ion. In near future I am going to cut laptop battery pack and will charge its cell so.

As others say, if cell phone has its own limiter then I won't modify the charger. (just now have no more time for electronics preparation).

Willen said:

Audioguru, if 5V will burn the Li-ion cell then what might be the ideal charger voltage and current for Li-ion. In near future I am going to cut laptop battery pack and will charge its cell so.

Click to expand...

Go to www.batteryuniversity.com and read about rechargeable batteries. Most Li-Ion and Li-Po cells have an absolute maximum charge voltage of 4.2V plus or minus 0.05V.
When the battery voltage reaches 4.2V then it is about 70% fully charged. Keep charging it until its charging current drops to a few mA then the charging circuit must be disconnected.

Somebody in another thread or on another website forum posted a Chinese ad for FAKE protection circuits for Li-Ion battery cells. They look identical to the protection circuits used in some battery cells but they do not do anything.

And another question, almost all this types of charger (power supply) circuit (220V AC to 5V DC) has a 1N5819 schottky diode at its output, why this actually? I was thinking that output AC has 50Hz too, But seeing this fast diode I guessed this type of power supply has higher frequency output, then they use fast diode, isn't it?

What whould be the difference if I used 50Hz (using linear transformer) AC rectifed, 5V output DC charger and What whould be the difference if I used high frequency AC (using SMPS like cell phone charger) rectified, 5V DC output? For cell phone charge or any devices?

The Schottky diode has a smaller forward voltage drop than an ordinary diode. Its speed does not matter when it is in series with the output of a power supply.
The diode prevents damage caused by very high current if the power supply is connected to a product that uses the opposite polarity of the power supply output.

Here in my computer room I have 13 AC-DC wall warts. 9 of them are plugged into the AC power and are being used. They all have different polarities and most of them have the same output plug.
I put a label on only one of them. It is easy to plug the wrong one into a product. A Schottky diode in series with the output of each one would prevent damage if the wrong one was plugged into a product.

The charging circuit expects a 5VDC input. It does not know and does not care if it came from a 50Hz rectified and filtered transformer or from a high frequency SMPS rectified and filtered converter.

If it is not related to frequency then why all manufracturer use 1N5819 schokttky at the output series? But one NOKIA charger has FR202 silicone diode but it is also rated as Fast Diode. Why actually 'Fast Diode' there? (It is just a single rectier there on every power supply for output.) What whould it be with slow diode like 1N4007 there instead?

Willen said:

If it is not related to frequency then why all manufracturer use 1N5819 schokttky at the output series? But one NOKIA charger has FR202 silicone diode but it is also rated as Fast Diode. Why actually 'Fast Diode' there? (It is just a single rectier there on every power supply for output.) What whould it be with slow diode like 1N4007 there instead?

Click to expand...

Look at the datasheet of a Schottky diode and see that its forward voltage drop is only about 0.3V while it is 0.7V for a 1N400x rectifier diode. That is why nit is used in series with the output of the power supply. Its high speed is not used.

EDIT: A 1N5817 has a lower voltage drop than a 1N5819.

Hello Willen,

Are you charging a cell phone battery or a cell phone? These are very different:
1. Cell Phone Battery
2. Cell Phone

If you are charging a CELL PHONE then 5v output is the acceptable standard because the cell phone has a charger build in it in most cases. This of course assumes that the original wall wart had a 5v output too.
But you have to be careful when you go over 5v because the internal circuit could very well be a linear charger that depends on a fixed input voltage of 5v, and the calculated max power dissipation for the charging chip inside the cell phone is based on a 5v input, not a 6v input, so you have to be more careful about that. If the charge is 500ma with 5v then it is 500ma with 6v input also, so that's an extra 1/2 watt to get rid of which may be too much extra power.

But if you are charging a CELL PHONE BATTERY ie a Li-on battery, then your circuit has to be designed differently. It has to put out a maximum of 4.2 volts (better is 4.15v) or else the battery could be damaged and explode. The max current also has to be limited as you noted.

So you should explicitly specify what it is you want to charge, and if you want to charge both then we need two circuits or at least one circuit with a modification that can be used as the other circuit.
Note you can use a switch to make the circuit work for both, but if you do that you have to make sure that the "OPEN" position of the switch is for 4.2v charging and the "CLOSED" position is for 5v charging for safety.

MrAl said:

Hello Willen,

Are you charging a cell phone battery or a cell phone? These are very different:
1. Cell Phone Battery
2. Cell Phone

Click to expand...

Hi,
I was trying to charge cell phone so wanted to set power supply to 5 volts. I will do as you and Driver300 suggested (decreasing 3.9k to almost 3k) to get lower output.

And now I want to charge Li-ion cell to. I will make another circuit. Some time ago after your and alec_t discussion, he designed a Ni-Cad 3.6V (or 2.4V what) charger (limiter) circuit which has input of 8 or 9V. I made the circuit and charged 2.4V Ni-cad battery. In the charger, 10 ohms emitter resistor gave me 180mA, 15 ohms gave 115mA and 33 ohms gave only 57mA.

My queation is- is the circuit able to charge a Li-ion cell to 4.2V? What is the safe current level? Or what modification needed here? https://www.electro-tech-online.com/attachments/shuntregbasedcharger-gif.75069/

A Lithium rechargeable battery has its life shortened if it is over-charged. The charger should limit the charging current to maybe 1/4 of its mAh rating and limit the voltage to 4.20V for one cell.
A current-monitor circuit should sense that the charging current has dropped to about 3% of the maximum charging current then it must disconnect the battery from the charger.

The charger circuit you found DOES NOT LIMIT THE CHARGING VOLTAGE! Your Lithium battery will EXPLODE!
Use a charger IC that is made to charge a Lithium cell.

Li-ion charger IC................ OK then I will be busy in gardening. Hehe

Will battery explode with charging voltage up to 4.2V, then can't use this (above) circuit and monitor its charging voltage mannually and disconnect after it reaches 4.2V?

-But almost every Li-ion cell has its own built-in tiny chipset look like this (uploaded below, sorry for bad pic. It was a round shape Li-ion cell by LG). Is it a Li-ion charger chip you were talking about? Then can I apply any supply of 5V in the chip then will it limit everything needed for its Li-ion cell?


-But I think each cells inside laptop battery pack has no its own differents chip. I guessed all cells have single chipset, called battery system circuit.