Hi, I am using a buck boost converter MC34063 to bring 14.8 V from four Li Ion batteries in series down to three different voltages 12V, 5V and 3.3V. I am planning to use the step - down converter given in the datasheet (image below) to do this but specifically how do modify the circuits components (the resistors) to get the voltage I want? Please help me with the calculations know-how...
MC34603 datasheet

 

R-1 & R-2 set the Feedback-Voltage.
When the Voltage between the 2 Resistors reaches ~1.25-Volts, the Output shuts-down.
.
.
.

 

What you have there is a buck converter. The MC34063 needs an additional transistor in order to be used as a buck-boost but from your description you only need the buck function.

For the kind of power you intend, you will need an external transistor as the switch.


The circuit above can be tweaked to provide the voltages you want.

I, for one would rather just make a separate regulator for each voltage - a lot simpler to get up and running.

I have attached and old Motorola application note which details the circuit above and presents lots of ideas for ways to use the MC34063.

 

Also worth a look is the MAX5035.

 

Using a Buck - Boost converter

The circuit in post #1 is a buck.

 

How much current is required from the 3.3V supply? It may be more efficient to derive the 3.3V supply from the 5V supply using a low-dropout linear regulator such as LD1117 or LP2951.
This would produce a cleaner 3.3V supply.
Also, using the LP2951 would produce a 3.3V supply that is sufficiently accurate to be used as a voltage reference.

 

R-1 & R-2 set the Feedback-Voltage.
When the Voltage between the 2 Resistors reaches ~1.25-Volts, the Output shuts-down.
.
.
.
View attachment 279970

Thank you for your quick response, I have found a similar formula in TI's datasheet

 

What you have there is a buck converter. The MC34063 needs an additional transistor in order to be used as a buck-boost but from your description you only need the buck function.

For the kind of power you intend, you will need an external transistor as the switch.
View attachment 279972

The circuit above can be tweaked to provide the voltages you want.

I, for one would rather just make a separate regulator for each voltage - a lot simpler to get up and running.

I have attached and old Motorola application note which details the circuit above and presents lots of ideas for ways to use the MC34063.

Yes, apparently I have made a grave mistake, thanks for pointing it out for me. I've always wanted to make a buck boost circuit for my line following color sorting robot project where the 4 wheeled robot carries the colored package to the desired destination, but it seems to me that buck - boost IC like the ones TI make are not available in Vietnam.
Since I'm gonna have to go with a buck converter anyway, I might as well go with the LM2596 to minimize external components. Thanks for your datasheet.

 

How much current is required from the 3.3V supply? It may be more efficient to derive the 3.3V supply from the 5V supply using a low-dropout linear regulator such as LD1117 or LP2951.
This would produce a cleaner 3.3V supply.
Also, using the LP2951 would produce a 3.3V supply that is sufficiently accurate to be used as a voltage reference.

Well here is a list of components I plan to use for my line following color sorting robot.
The current in purple are assumed values, I measured the current with a multimeter in real life but the current is unacceptably low, as in the Servo motor draws only 150mA when measured but the datasheet says it draws about 500 - 900mA.

Also, I'm switching to a buck only LM2596 with an additional capacitor for noise filtering...(says so in the datasheet)? So I guess it would also provide a clean, stable 3V3 and 5V.

While we're on the subject, did I make myself clear that I'm drawing a 14.8 V power supply from 4 Li ion batteries in series and dividing it using 3 LM2596 down to 3 separate voltages? Does the plan sound good or doable?

 

Complete LM2596 modules can be purchased online very inexpensively so it's hard to argue against them. See https://www.ebay.co.uk/itm/255123504691

But to keep things simple I'd be tempted to use an L7812 linear regulator for your 12V, the LM2596 for your 5V where you get the most benefit from the efficiency of a buck converter (as the drop from battery voltage is significant) and linear regulator from 5V down to 3.3V where you will be assured of a clean supply as suggested by Ian0

In case your Atmega328 is part of a processor board - for example an Arduino UNO, you may have a 3.3V output pin on that anyway?

Maybe the servo motor draws more current under load?

 

Hi, I am using a buck boost converter MC34063 to bring 14.8 V from four Li Ion batteries in series down to three different voltages 12V, 5V and 3.3V. I am planning to use the step - down converter given in the datasheet (image below) to do this but specifically how do modify the circuits components (the resistors) to get the voltage I want? Please help me with the calculations know-how...
MC34603 datasheet
View attachment 279968

Go here, and look for his vid(s) on using the MC34063

https://www.youtube.com/c/EevblogDave

 

Complete LM2596 modules can be purchased online very inexpensively so it's hard to argue against them. See https://www.ebay.co.uk/itm/255123504691

But to keep things simple I'd be tempted to use an L7812 linear regulator for your 12V, the LM2596 for your 5V where you get the most benefit from the efficiency of a buck converter (as the drop from battery voltage is significant) and linear regulator from 5V down to 3.3V where you will be assured of a clean supply as suggested by Ian0

In case your Atmega328 is part of a processor board - for example an Arduino UNO, you may have a 3.3V output pin on that anyway?

Maybe the servo motor draws more current under load?

My professors wouldn't approve premade module so the usual LM2596 module can't be used.
Can you explain more about your choice of ICs? Does buck converter have an advantage over linear regulators when there is a significant change in voltage? Why? And does linear regulators produce a cleaner output than the buck converters? The datasheet advises me to add an extra capacitor to clean the output... I think? Here's the schematic and the 3D view of the PCB I'm planning to make, please tell me if I have anything wrong with the schematic or component placement.

 

Change "R-3" to ~20-Ohms,
( for a Blue LED with a Power-Supply of 3.3-Volts ).
============================================================

One of the 5-Volt-Regulators will probably need a larger Inductor ......
~14-Volts Input @ ~1-Amp Output-Current = 68uh Inductor
See the Spec-Sheet for details.
.

============================================================

The 3.3-Volt Regulator could benefit from having an additional Filtering-Stage on its Output.
.

.
.
.

 

Change "R-3" to ~20-Ohms,
( for a Blue LED with a Power-Supply of 3.3-Volts ).
============================================================

One of the 5-Volt-Regulators will probably need a larger Inductor ......
~14-Volts Input @ ~1-Amp Output-Current = 68uh Inductor
See the Spec-Sheet for details.
.
View attachment 280372
============================================================

The 3.3-Volt Regulator could benefit from having an additional Filtering-Stage on its Output.
.
View attachment 280376
.
.
.

Yes, using the TI's datasheet, I have reselected the main components to be
L1 = 33uH, Cout = 330uF/50V, Cin = 680uF/25V and three 1N5822 Schottky for all three branches,. Originally each branch has its own inductor (12V branch => 47mH; 5V and 3.3V branch => 68mH) but from the quick look up table 1 I have decided to go with the above values.
I will try to implement the additional filtering stage.
Thanks a lot.