PWM Buck DC/DC Converter

[JerryLee]

Was finding a suitable DC/DC converter to be used in my project recently to supply a 12V input  from an external 24V.

As attached is an example converter's datasheet that I came across but not quite understand.

Correct me if I'm wrong, I know the feedback voltage is to compare the output voltage with a reference voltage but I just couldn't understand how they calculate the feedback voltage. Like how they came up with the rating 1.23V as in the attachment.

I'm new to DC-DC converter and hopes someone could enlighten me, thanks.

[mariush]

Read the datasheet, page 3 (the internal blocks of the chip) :
http://www.diodes.com/datasheets/AP1509.pdf

There's an internal voltage reference built inside the chip.

Keep in mind that this particular IC ( AP1509) seems to only handle up to 22v according to datasheet, so it wouldn't be a good choice if you have 24v input (give or take a couple of volts for safety margins).

[JerryLee]

Ahh I see, din notice that thanks man!

Do you have any efficient method of determining the input and output capacitance for the converter?

Been searching for an easy way on the net but what I've found so far is a couple of generic equations from TI.

[mariush]

Look at various datasheets for various regulator ICs ... some manufacturers have better datasheets, with plenty of information and formulas to figure out what technical specs the capacitors need to fulfill. 

See Linear Technology, Texas Instruments, Intersil, ON Semi,  from memory these usually have good datasheets.  Knock yourself out: http://www.digikey.com/product-search/en/integrated-circuits-ics/pmic-voltage-regulators-dc-dc-switching-regulators/2556570

[ez24]

I also am learning about bucks.

I got this to play with:

http://www.aliexpress.com/item/Wholesale-1pcs-DC-Step-Down-Converter-DC-4-0-38V-to-1-25V-36V-5A-75W/32285636716.html

It uses a XL4015 chip    http://www.xlsemi.com/datasheet/XL4015%20datasheet.pdf

My source is a laptop 19v power supply.  Yesterday I ran it at about 2 amps at 3.3 v with no problems.

Today I am running it (and taking pictures) at about 2 amps 5 volts, and the surprise is it is running cooler.  I can press and keep my finger on the heat sink.  At 3.3 v  I could hold only 3 secs, go figure.

Tomorrow I plan to see if it can run 12v at 2 amps (my l limits are 2 amps because I am going to use breadboards).   This module can be calibrated and V and I limited.  My limits are my resistors.   I scoped about 70 mVpp ripple both at 3.3 and 5 v at about 2 amps each.  I ordered some 317s to see if I can reduce the ripple with them. 

Playing with this module was easier than adjusting my B&H bench ps and puts out more amps (I think there is something wrong with the B&H).

I just ordered a couple of these   http://www.aliexpress.com/snapshot/6668866069.html?orderId=67255429203851
because they look like they can put out more power.  They do not say what chips they use.  I want to use this for a variable source.  I plan to replace the pots.

But one interesting fact they state is "power tube temperature exceeds 65°C please add cooling fan".  I guess "power tube" is the chip that gets the hottest.

 I have been measuring the temp of my chip and I do not know what the limit is.  The datasheet says the max junction temp is 125C, so 65C seems safe.  Topic of new post.

I will use 65C as my T limit to set my I limit, so far my max has been 40C. 

[JerryLee]

Thanks guys for the recommendations, decided to go with TI LM2575 chip and its recommended application due to project deadline.

[JerryLee]

Currently I'm trying to find out whether or not this regulator is producing the current it's meant to make, which is 2A.

I supply 5V through a lab DC supply, then connect a 2 ohm 5 watt resistor and a multimeter in series to its inductor to measure its output current.

3.3V/1.56A = 2.11Ohm

The efficiency of the regulator is 78% which I believe will give me a 1.56A.

I wanted some confirmation whether my method is correct or it's wrong cause I'm getting much lesser than that.

[HackedFridgeMagnet]

I wanted some confirmation whether my method is correct or it's wrong cause I'm getting much lesser than that.

It's not really clear how much current you are measuring, or what you are expecting.
Also not really sure what is your method.

Better restate these.

[JerryLee]

Here's some description to clear things up.

Please do enlighten me on any mistake or problem. Thanks ya!

[HackedFridgeMagnet]

I think it is working fine.

Your power supply is designed to give 3.3 volts out at up to 2 amps.
Every power supply has voltage and current limitations but are generally not as ideal as a rectangle.
I have drawn your psu ideally in the image as the red line.

I have drawn 3 loads. Two are blue and one green.
Notice they are a linear relationship between voltage and current.

Where the load lines intersect with the red lines is the operating point. This is the voltage and current you will be able to read off.
The blue lines are loads that draw less than 2 amps.
The green line would draw more than 2 amps at 3.3 volts but is current limited. So the output voltage is more like 2.8volts.

The only way you will get exactly 3v3 and 2amps out of your psu simultaneously is by using a 1.65 ohm resistor.
Hope this helps.

[LukeW]

Was finding a suitable DC/DC converter to be used in my project recently to supply a 12V input  from an external 24V.

As attached is an example converter's datasheet that I came across but not quite understand.

Correct me if I'm wrong, I know the feedback voltage is to compare the output voltage with a reference voltage but I just couldn't understand how they calculate the feedback voltage. Like how they came up with the rating 1.23V as in the attachment.

I'm new to DC-DC converter and hopes someone could enlighten me, thanks.

There's an internal voltage reference inside the chip, in this case it's 1.23V. It's generated by a bandgap voltage reference in the chip.

It will be some small fixed voltage, but it's different for different chips, so you should check this value in the datasheet.

You take the voltage from the output and use a voltage divider to divide it, and this voltage enters the controller IC where it is compared against the reference voltage - so it's  a feedback loop, like a servo mechanism, it makes the feedback voltage equal to the reference voltage, and the output voltage is whatever it has to be, higher or lower until that the feedback voltage matches.

So if the reference is 1.23 volts, and the voltage divider is giving a 1:10 division (in other words, say R1 = 9k and R2 = 1k) then the output voltage will be 12.3 volts.

VFB = Vout * R2 / (R1 + R2), because it's just a ordinary voltage divider.

VFB  = VRef

.: Vout = VRef * (1 + (R1/R2))

You might see the same equation like that given in the datasheet for many different voltage regulators - it's the same principle.

[Zero999]

The efficiency makes not difference to the output current.

The efficiency is a measure of how much of the power in is converted to useful power out.

If the output power is 5W and the efficiency is 78% then the input power will be 5/0.78 = 6.41W, so with 5V in the input current will be 1.28A.

In real life it's not that simple, since the efficiency will depend on the inductor losses, input voltage, output current and even the temperature of the IC.

[JerryLee]

Thanks HackedFridgeMagnet, Hero999 and LukeW for your detail explanation about the regulator, I get it now.

To measure its output current, I used a 3.3 ohm, 7 W resistor in series with the output inductor then to ground.

But to my surprise I measure only 0.5V over the resistor where I'd expect 3.3V.

This result in a current of 0.15A only which is way too low of what I want. Correct me if I'm wrong but I think the component itself could not drive the high load with only 5V input.

[HackedFridgeMagnet]

Have you got the fixed voltage one or the adjustable? Your 2 circuits are different.

Check your circuit matches your converter chip.

[JerryLee]

Check and confirm the chip I bought was fixed type regulator and the circuit layout was also the same with the data sheet 

[HackedFridgeMagnet]

Correct me if I'm wrong but I think the component itself could not drive the high load with only 5V input.

The datasheet for AP1509-3.3V  says that for:
4.75V <VIN<22V 
0.2A <ILOAD<2A

the output voltage should be 3.3v.
So the dropout voltage is not the issue.
So the load current is not the issue.

I would say there is a mistake in the breadboarding, if that is how you have connected it, or the chip itself is damaged.

When I get into a situation like this, go back to what you know works, ie the previous circuit and use it to check the chip.
Also for any quick check increase the input voltage to say 12v.

Check the voltage on the feedback pin. Check voltage on the ground pins, all of them. Still no joy, Check voltages on all nodes, and see if you can make sense of it.

Is the diode in the correct orientation?

[JerryLee]

The funny thing is before I connect the resistor to test the current it produced, I measured the output voltage and it showed 3.3V.

This isn't normal right? Or I'm mistaken.

[HackedFridgeMagnet]

Internal to the chip is a small load via the feedback voltage divider.
I would say it is normal that 3.3v appears on the output with no external load.
But it is not normal that the output voltage drops to 0.5v with a 3.3 ohm resistor as load. 

So try the 2 ohm resistor again and you should get your 1.56 amps.
Then if this is the case find out what loads work and what load don't. Maybe the 3.3 ohm resistor isn't good.

Again when I get into a situation like this, go back to what you know works. Then start changing things one at a time.