MC34063 buck converter, the shortly, and inductor wattage

I'm trying to wrap my head around the buck converter.

It seems that voltage will actually be controlled by the divider between Vout and ground that feeds to the comparator input (pin 5), the frequency by the timing capacitor connected to pin 3, and that all other values simply need to be "adequate".

A small resistor between Vin and the oscillator, drive collector, and switch collector shorted together (all I see is .6ohm)--but why? And what kind of wattage--if an amp is passing through it, at .6v, won't that mean .6 watts, regardless of input voltage?

I think I understand everything I need except for the wattage of the inductor.

I assume that the wattage is for the heat it dissipates, but how much will that be? I'm not finding anything about the resistance of the inductors.

I'm assuming 3.3v out, and planning for 1A (but I shouldn't be near that), so with a 3.3W draw, would a 1/4W inductor be enough? Or do I need something heavier?

And shat about the shortly on the switch emitter before the inductor? I assume it's there over spikes from Vin? So in case of a spike, it should have a current rating of at least as much as the output will draw? Does it really matter if it's a 20v, 30v, or 40v diode?

Without a schematic this post makes absolutely no sense. Either post a link or post a screenshot of the page you are referring to.

This one is off of TI’s data sheet:

Screen Shot 2019-04-09 at 3.32.57 PM.png

It seems electrically identical to every other implementation as a buck converter on the web (save for the high current variant) on the datasheet itself)

This one, though, uses .33ohm for Rsc, rather than the calculator I found offering .6 (which seems even harder to find . . .)

1.- It seems that voltage will actually be controlled by the divider between Vout and ground that feeds to the comparator input (pin 5),

The voltage at the comparator input is 1.2V when VOUT =5.00V

Voltage Divider

(1.25*(1+(3800/1200))=5.208V, which is the expected UNLOADED voltage of a 500mA 5V supply.

2.- the frequency by the timing capacitor connected to pin 3,

It is labeled “CT”, which one would logically assume stands for “timing cap”

3.- and that all other values simply need to be “adequate”.

I have no idea what this refers to. all WHAT other values ?

4.- A small resistor between Vin and the oscillator, drive collector, and switch collector shorted together (all I see is .6ohm)–but why?

The resistor is clearly 0.33 ohm and it is the current sense resistor, (obviously from the label “RSC”)

At 500mA, the voltage drop across the resistor is (V=IR=0.50.33=0.165V.
The input to the oscillator is labeled “IPK” which obviously means Ipeak
which means PEAK CURRENT. Thus , if the current drawn by the load exceeds 500mA, the voltage
drop across the resistor will increase and the resultance voltage (downstream) from the 25V input
will DECREASE because MORE voltage is being dropped across the resistor, thus lowering the reference voltage at the oscillator input to LESS than the nominal 24.835V (25V-0.165V), causing a
the oscillator to switch phase . If the voltage exceeds 5.20V the comparator input will exceed the
1.25 voltage reference resulting in a LOW on the AND gate input thus shutting down the transistor.

5.- And what kind of wattage–if an amp is passing through it, at .6v, won’t that mean .6 watts, regardless of input voltage?

The regulator is rated for 500mA so it would never output an amp and there is no 0.6 ohm resistor.

  1. -I think I understand everything I need except for the wattage of the inductor.

Since the regulator is rated for .5A @ 5V , and P= IV, then Nominal power is 0.5A5V =2.5W
so the inductor would need to be at least 20% higher rating so we could expect a rating of 3W.

  1. -I assume that the wattage is for the heat it dissipates, but how much will that be? I’m not finding
    anything about the resistance of the inductors.

8.- The resistance of the inductor is what it is for that rating. An inductor is a SHORT at DC. The inductor
is part of the low pass filter and would not heat up.

The 220uH/470uF LC output Low Pass Filter has a cutoff frequency of 494 Hz.

    • I’m assuming 3.3v out, and planning for 1A (but I shouldn’t be near that), so with a 3.3W draw, would a 1/4W inductor be enough? Or do I need something heavier?

Obviously this makes no sense for a 5V regulator rated for 500mA.

And what about the schottky diode on the transistor switch emitter before the inductor? I assume it’s there over spikes from Vin? So in case of a spike, it should have a current rating of at least as much as the output will draw?

1N5819 SCHOTTKY BARRIER DIODE

Ideally suited for use as rectifiers in
low−voltage, high−frequency inverters, free wheeling diodes, and
polarity protection diodes.

10.- Does it really matter if it’s a 20v, 30v, or 40v diode?
That depends on what voltage your circuit operates at.

Have a look at this

and this on line calculator

The MC34063 is an adjustable regulator. You have to choose the resistors R1 and R2 in your voltage divider such that you get 1.25V on pin 5 at the desired output voltage. So if you want 3.3V out, then 3.3 * R1 / (R1 + R2) = 1.25, so R2 = R1 * (3.3 / 1.25 - 1)

When Q1 is on, current flows through Rsc, Q1, L to Cout. Because of the inductance L, it ramps up. Then when Q1 is off, current through L ramps down, but doesn't stop immediately. So then it flows through that Schottky and L to Cout. So the current to your output always flows through L, but part of the time it comes from Vin through Rsc and Q1, and part of the time it comes from GND through the Schottky.

The inductor and the Schottky must be rated for your output current. The Schottky must be rated for your input voltage.

raschemmel: 8.- The resistance of the inductor is what it is for that rating. An inductor is a SHORT at DC. The inductor is part of the low pass filter and would not heat up.

Inductors have resistance, and at high frequencies the resistance can be significantly larger than the DC value due to the skin effect.

Inductors can and do heat up significantly in switch-mode power supplies - partly due to resistive losses but mainly due to magnetic hysteresis losses.

The 220uH/470uF LC output Low Pass Filter has a cutoff frequency of 494 Hz.

Its not viewed as a filter, its a switch mode circuit, there is no notion of cutoff frequency, only of current ripple on the inductor and voltage ripple on the capacitor. The switching means the circuit is not analyzable as a (linear) filter as the circuit changes topology on every half-cycle, meaning the design equations are different from a filter, and in general non-linear. The inductance and capacitor are treated as energy storage devices.

Thanks for the correction Mark