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I am looking to use a Texas Instruments MC33063 Buck/Boost regulator. I have never used a switching regulator before so this may be a conceptual issue.

My input voltage will probably be anywhere from 6 to 12V and I would like to use the regulator to buck down to 5V. However, I get confused when I read through the data sheet. Here's the layout for a step down from the sheet:

enter image description here

So the data sheet shows that if I were to input 25V, given this setup, I can get 5V. However what about 6 to 12V for a Vin?

Also, the optional filter is shown that \$V_{out} = 1.25(1 + R_2/R_1)\$. Is this independent of the input voltage? If so, what is the advantages of varying input voltages or is it simply flexibility?

Here's the link to the datasheet: http://www.ti.com/lit/ds/symlink/mc33063a.pdf

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Most buck converters (including this one) have an internal voltage reference and a feedback loop, which regulates the output voltage. The output voltage is set by the resistors R1 and R2. The output voltage is independent of the input voltage fluctuations. The output voltage formula is not related the optional filter.

Switch mode converters can be layout sensitive, so it's usually a good idea to follow a reference design. On the other hand, this one has a top frequency of only 250kHz, so it may be more forgiving than switchers with higher frequencies.

There's more details about the principles of operation of buck converter here and here.

edit: A somewhat odd thing, however, is that the resistor values in the drawing don't quite check with the output voltage

\$V_{out}=1.25\left( 1 + \cfrac{R_2}{R_1} \right)=1.25\left( 1 + \cfrac{3.8}{1.2} \right) = 5.2 V\$

should be 5.0V. I wonder if there's a reason for the extra 0.2V?

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There are several calculator tools you can use to make the calculations (or you can use the equations of the datasheet). One of them is this spreadsheet available from ONsemi.

If you set the output to 5v and the input to 6v then you will get an error because the calculated duty cycle becomes more than 84%. For 5v in step down mode you'll need about 7.5v minimum input.

the optional filter is shown that \$ Vout = 1.25(1 + \frac{R2}{R1})\$

That equation is not related to the filter but to the output voltage of the step down circuit, the filter is just used to reduce the ripple of the output.

If so, what is the advantages of varying input voltages or is it simply flexibility?

If the output voltage changed when the input voltage changed (within limits of course) then this wouldn't really be a voltage regulator.

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  • \$\begingroup\$ How did you calculate the 84%? Is that ton/toff? This is really interesting. Also, what are the ramifications when the duty cycle goes beyond 84%? \$\endgroup\$ – Nick Williams Jan 2 '14 at 19:25
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    \$\begingroup\$ The 84% is written in the spreadsheet I linked. I found it mentioned in note 6 in this [application note](www.onsemi.com/pub_link/Collateral/AN920-D.PDF) . 'Note that the ratio of ton/(ton + toff) does not exceed the maximum of 6/7 or 0.857' \$\endgroup\$ – alexan_e Jan 2 '14 at 20:35
  • \$\begingroup\$ Wow, this is a really nice spreadsheet. \$\endgroup\$ – Nick Williams Jan 2 '14 at 21:48
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If you look at the output transistors Q1 and Q2, the base of Q2 needs to be about 1.2 to 1.4 volts higher than the emitter of Q1 for it to switch on. On this basis alone, the output voltage that can be achieved is going to be no-greater than input voltage level minus 1.2 to 1.4 volts.

If you are looking to regulate 6V to 5V then this circuit won't work. As the input voltage gets a little below 6.5 volts the output will start to reduce.

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  • \$\begingroup\$ Have a look at at Linear Technologies offerings - they have some devices that run from internal transistors that can probably get down to 6V (maybe 1A output at 5V). Also take note that they utilize a technique called bootstrapping which, in effect provides a higher supply voltage to the output driver - equivalent to raising pin 8 higher than pin 1 on your circuit. \$\endgroup\$ – Andy aka Jan 2 '14 at 21:39
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    \$\begingroup\$ In addition to what Andy have said. There are buck converters, which use P-channel MOSFET as a switch. They can also work with small headroom. [Nick W, welcome to the wonderful world of switch mode regulators, by the way.] \$\endgroup\$ – Nick Alexeev Jan 2 '14 at 22:01
  • \$\begingroup\$ So would using an external pFET or pnp -- as in figure 9b in the datasheet of the TI MC33063 mentioned by Nick Williams -- be adequate to buck 6V down to 5V ? \$\endgroup\$ – davidcary Jan 4 '14 at 3:58

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