Please bare with me, i am new to electronics and am trying to understand it bit by bit

I downloaded a datasheet for a white led step up converter, in the description it says:

"The AP5724 switches at 1.2MHz that allows the use of tiny external components."

What exactly is the use/benefit of this high switching?
Does this switching affect the output signal and does it affect other devices on the same power line?
If yes, What happens if I connect multiple devices with different (operating) frequencies?

 

The chip is a onboard pwm oscillator working at 1.2Mhz, and dims the leds using high frequency switching for smooth flicker free operation with current limit built in.

 

It has to do with the frequency relationship with inductance and capacitance. Plus this frequency is for the DC conversion, not directly feeding the LEDs.

 

Besides what the other members have stated, I find the interesting part is that it steps up the voltage to run up to six white LED's. The nominal voltage for white LED's is 3 volts DC. If the input voltage to the device is within specifications, about 3 volts it will supply about 18 volts to the light the 6 white LED's in series. The inductor is charged by the device and then the load (LED's) discharge the inductor.

 

Related to what BR-549 noted, is's mostly related to the size the required inductors and capacitors.
For a given power out, the required size of both inductors and capacitors in a switching power supply goes down with an increase in frequency, so as high a frequency as practical is used for switching power supplies to minimize the size of these components.
The limit is that the losses in the switching devices go up with frequency so that usually sets the upper limit.

 

Thank you all for the explanations.

This is a typical application circuit from the datasheet.

I am trying to wrap my head around this schematic with the knowledge i got.

Is it correct to assume that:

• The 2 capacitors close to the input and output are smoothing caps to get rid of noise (especially the output cap because of the fast switching).
• The inductor is charged and discharged via the leds @1.2MHz (1 200 000 times per second).
  
• The SW pin is opening and closing @1.2MHz creating the switching/pump effect.
• Because the device and external inductor are so tiny a high rate of 1.2MHz is necessary to achieve the pump effect to get the low input voltage from +/- 3.3V to a high voltage +/-18V (6 leds * 3V).

 

Cin is actually called a Bypass Capacitor. It has to be located very close to the power pin of the IC. It compensates the for the parasitic inductance of the PC traces.

Cout filters.

Rset is for current feedback.

The fundamental equaton for the voltage across an inductor is v9t) = L* di/dt. di/dt is the derivative or instantaneous slope. So, ways to make v bigger is to increase di/dt or increase L. Big L's are really huge. It's easy to increase frequency. Staying out of the audio range is a good thing.
Not interfering with communications is another good thing.

 

Thank you all for the explanations.

This is a typical application circuit from the datasheet.

I am trying to wrap my head around this schematic with the knowledge i got.

Is it correct to assume that:

• The 2 capacitors close to the input and output are smoothing caps to get rid of noise (especially the output cap because of the fast switching).
• The inductor is charged and discharged via the leds @1.2MHz (1 200 000 times per second).
  
• The SW pin is opening and closing @1.2MHz creating the switching/pump effect.
• Because the device and external inductor are so tiny a high rate of 1.2MHz is necessary to achieve the pump effect to get the low input voltage from +/- 3.3V to a high voltage +/-18V (6 leds * 3V).

Hi,

Your questions can be answered mostly by understanding how the inductor works with a switching action.

The SW terminal is connected to GND (roughly) when the internal switch is 'on', and open circuit when 'off'.
The inductor behavior follows this little equation:
V=L*di/dt

and i an hoping you understand algebra at least a little as that will help here.
In that little equation, V is the voltage across the inductor, di is the change in current, and dt is the change in time, and L is the inductance.

Now when the switch is 'on', that connects the inductor between Vin and GND. Since V is now the drive, we solve that equation for di and get:
di=V*dt/L

This is the change in current in the inductor during the 'on' switch time. Since V is constant dt will be roughly constant too, so we have something that we can estimate as:
di=K/L

where K is some constant.

The second fact we have to consider is that the inductor stores energy based on the current:
W=(1/2)*L*i^2

and from this we see that the energy W is mostly dependent on 'i' when L is constant, so the more current we have the more energy we have.

Now to keep the LED's lit we need a certain amount of energy W and that means we need a certain amount of current in the inductor. We know that the current changes as:
di=K/L

and so to get the most increase in current (which gives us the most increase in energy) we find that a lower value L produces a higher value of 'di' because L is in the denominator. Thus, a smaller L gives us a faster rise in energy. Since it also gives us a faster rise in current, we have to be careful not to go too low in inductance or the switch will burn out. Thus we are limited on the value of the inductor L and the time dt in di=V*dt/L.

Since the time dt and value of L are related, we have to select L to match the switch time and the peak current allowed for the switch. If we have a long switch time then we need a larger inductor, but if we have a short switch time (higher frequency) then we can use a smaller value inductor (same peak current).
If we take this to the limit, we find we could use a very very high frequency and thus need a very very tiny inductor, except we hit another limit like Crutschow was talking about. That's the frequency where the losses due to various secondary phenomena kick in such as skin effect and proximity effect and possible high core loss, so we reach a frequency where going higher causes more undesirable effects than staying lower. We end up settling on about 1MHz. Around that frequency we find that various physical attributes of the inductor and swtich show decent performance without degrading too much. Whether or not this is optimum though is questionable because if we go lower (300kHz) we may find that we have less losses. Since the physical size of the inductor has a certain number of turns a smaller inductor would have less turns than a bigger inductor and/or a core that has less core loss, so there are a number of factors that would have to go into an optimization, but sometimes the physical size has a dominant importance in an application because it just has to be small, period, even if not truly optimum.

So the value of the inductor is related to the switch time and the faster the switch the smaller the inductor, and the smaller the inductor the smaller the circuit, and the smaller the circuit the more consumer devices it can fit inside of so the circuit becomes useful for even very small devices like a Tablet or Cell Phone.

 

Excellent explanation. Thank you for that.

 

Excellent explanation. Thank you for that.

Hello again,

You're welcome. I had to go back and correct a number of typos though