MC34063 regulator (mp3 player charger circuit)-output current

[humvivz]

Condition:
    It is a mp3 player charger circuit. The mp3 player charger is replaced 10 ohm load for hardware testing. However only output voltage reaches 4.46 V ( 5 V for actual value), while the output current only becomes nearly zero. The maximum output current should reach 660m A in order to charge the mp3. The circuit is only able to charge a wireless modem but not mp3.
    MC34063 only can operate when its 5th pin and 4th pin exchange their connections.

    Thus, what should i do to enable the circuit to charge mp3?

[c4757p]

Use the datasheet, Luke!

That schematic symbol has pins swapped. Check the pinout.

[fcb]

    MC34063 only can operate when its 5th pin and 4th pin exchange their connections.

You need to sort this out first!
Pin4=Vss
Pin5=Inv. Input

[Lurch]

I like the posted screenshot of an image in an image.

[mariush]

That Rsc seems a bit high at 0.33 ohm (3x 1 ohm resistors in parallel) ... my math says 0.15-0.2 ohm should be more suitable.
The inductor choice also seems a bit weird .. 200uH is a lot... for 12v in, 5v out ...  online calculators say 33-47uH so 100 uH should be a lot.

See for example: http://dics.voicecontrol.ro/tutorials/mc34063/

The MC34063 is also capable of only about 750mA output, with 10 ohm resistor you're basically drawing 500mA so it should work.. but again, it depends on inductor and components.

Also note that some devices only charge if they get some exact voltage on the data lines. If you just put 5v and ground and the data lines are floating, some devices won't charge. Esp. Apple devices ... they detect how much the charger can provide by the voltage value on the d+ and d- lines

2.0 v  and 2.0 v  -> 500mA
2.7 v  and 2.0 v  ->1000mA

So you can use some resistors to get 2.7v and 2.0v on d- and d+  to signal devices that your "charger" supports up to 1A even though it can only do about 750mA, or you could just use 2.0v on both d- and d+ to limit it to 500mA.

[humvivz]

I did followed the pin connection by first. However, MC34063 became super hot and output voltage cannot reach around 5 V.
But, after i swapped the the connection of pin 5th and 4th, it worked and output voltage reached 4.42 V even though output current reached nearly zero.

[humvivz]

The MC34063 is also capable of only about 750mA output, with 10 ohm resistor you're basically drawing 500mA so it should work.. but again, it depends on inductor and components.

550m A is gained due to 10 ohm resistance (I=V/R) but mp3 or modem wont become 10 ohm. Yes, i think the inductor plays the important role but my inductor becomes useless. Because output current remains nearly zero with changing the inductance value.

So you can use some resistors to get 2.7v and 2.0v on d- and d+  to signal devices that your "charger" supports up to 1A even though it can only do about 750mA, or you could just use 2.0v on both d- and d+ to limit it to 500mA.

I am sorry. I don't understand. Would you mind to explain it in detail?
Do you mean i should change the output voltage to get 2.7 V?

[fcb]

I did followed the pin connection by first. However, MC34063 became super hot and output voltage cannot reach around 5 V.
But, after i swapped the the connection of pin 5th and 4th, it worked and output voltage reached 4.42 V even though output current reached nearly zero.

Ignore the positions shown on the diagram - go by the PIN NUMBERS & functions.

[humvivz]

Ignore the positions shown on the diagram - go by the PIN NUMBERS & functions.

Does it mean it is wrong if I swap the pin connection ? Pin 4th should be connected to ground and pin 5th is connected to other part. But if I follow the arrangement of PIN NUMBER, the ic chip becomes super hot. I am afraid that the ic chip may break down.
However i will rearrange the PIN CONNECTION again for hardware testing.

[humvivz]

I like the posted screenshot of an image in an image.

Okay. Thanks for advices. 

[humvivz]

That Rsc seems a bit high at 0.33 ohm (3x 1 ohm resistors in parallel) ... my math says 0.15-0.2 ohm should be more suitable.
The inductor choice also seems a bit weird .. 200uH is a lot... for 12v in, 5v out ...  online calculators say 33-47uH so 100 uH should be a lot.

I check the datasheet of MC34063. It shows the inductance value and circuit sample.
http://www.ti.com/lit/an/slva252b/slva252b.pdf

[mariush]

Sigh...

CHECK the polarity of the wires coming from your power supply. You may have connected the wires the other way around so you have -12v hence why it heats up.

Read the DATASHEET of the MC34063 and DOUBLE CHECK the pinout of the IC ... the pins of the IC may not be in the same order as the ones in the Spice simulation.

Here's one, for example, but your IC may be made by another company : http://www.onsemi.com/pub_link/Collateral/MC34063A-D.PDF

I check the datasheet of MC34063. It shows the inductance value and circuit sample.
http://www.ti.com/lit/an/slva252b/slva252b.pdf

Inductance value and circuit sample FOR WHAT OUTPUT?  What technical specs for the inductor (peak current, frequency etc... there's lots of inductors out there)?

You linked to the application report ... did you actually calculate all the values using the formulas for the BUCK regulator version - because MC34063 works in 3 modes, buck (from high voltage to low)  , boost (from low to high) , inverter (from positive to negative)? Are you sure your inductor is suitable (see selecting the right induction section)

-

USB has four wires ... GND ,  D- , D+ , V 

GND is ground, V is 5v .. D- and D+ are for transmitting data.
Normally, the charger has to have some brains and the device tries to talk to the charger and learn how much power the charger can give.  BUT, since such chargers are more expensive due to having to use an IC to talk to devices, device manufacturers chose to implement a method of figuring out how much power the charger can provide without using an IC to talk to.
The way this is done is by outputting some voltage on the D- and D+ wires. If D- has 2.7v on it, and D+ has 2v, then an Apple device "learns" that the charger can give it up to 1A ... if D- and D+ both have 2v, then device "learns" the charger can only give 0.5A.
You can generate those voltages from the 5v using simple voltage dividers, two resistors.  For example, if you use two resistors of same value, you get half the voltage between the resistors:   

V ----  R1 -------x-------R2---------GND

If  V = 5v, R1 = R2, then at x you have 2.5v.   

So you can use a calculator like the one here to get 2.7v and 2.0v  and connect those to D- and D+:  http://www.raltron.com/cust/tools/voltage_divider.asp

ex 820ohm and 1kohm  will get you 2.74v which should be close enough for 2.7v    820ohm and 560ohm should get you 2.029v

Apple for example, sometimes refuses to charge

[humvivz]

So,I need to correct circuit arrangement
                     review the components' rating
                     check voltage required of D+ D- for mp3 player

Actually, the link below is my original circuit. But i modify it for circuit simulation. The picture shown is my modified circuit. Does the modification make the circuit fails to charge mp3?
http://electronics-diy.com/schematics/ipod_charger.htm

[humvivz]

I have rearranged the pin connection.
The output voltage becomes zero now.
For sure, output current equals to zero too.
What should I do?

[Jeff1946]

Here is what you need to calculate values and correct pin numbers:  http://dics.voicecontrol.ro/tutorials/mc34063/

Good to use a low resistance inductor and schotky diode for max efficiency.