I have used an mc34063a as a step-down from 12V to 5V. The circuit is straight out of the datasheet with no modifications at all. It's been working very well for me in various projects. I have a project now where the supply voltage is only 7.5V. Is this enough? Also, at the other end of the scale, what is the max voltage it will take? Data sheet says 40V max but refers to the chip I presume and not specifically this reference design. Load is max 100mA. Thanks!

 

You need to look at the data sheet specification under "Minimum supply voltage", or possibly Minimum voltage drop. If you can locate the application notes they will be quite explicit about that limitation.

 

Thanks for the reply. The data sheet says minimum voltage is 3V but that won't apply to that particular circuit as it's a buck circuit and output voltage is 5V. I haven't been able to find anything in the datasheet. I could probably spend a bit more time to understand it fully but still wouldn't be 100% sure I would be correct.

 

I have used an mc34063a as a step-down from 12V to 5V. The circuit is straight out of the datasheet with no modifications at all. It's been working very well for me in various projects. I have a project now where the supply voltage is only 7.5V. Is this enough? Also, at the other end of the scale, what is the max voltage it will take? Data sheet says 40V max but refers to the chip I presume and not specifically this reference design. Load is max 100mA. Thanks!

If you understand how a switching regulator works, you know that the output voltage of a buck regulator is controlled by the duty cycle of the switch. For a buck regulator the duty cycle is given approximately by Vout/Vin. In the case of the reference design that would be 5/12 or 41.7%. Generally speaking, unless the datasheet says otherwise, the duty cycle of a buck regulator should be kept at less than 50%. As you can see from the design equations on page 11, the actual duty cycle will involve Vout and Vin as well as other factors. So as a matter of practical reality I would put the minimum input voltage at 10 Volts for a 5 volt output. You can try 7.5 Volts but you will need to make other component value adjustments. Note the computation of the inductor and capacitor values on page 11. A working design will require more information than is contained in the datasheets. Consult the referenced application notes and prepare yourself for disappointment if it does not work.

One more thing. This part is over 3 decades old and there may be better alternatives.

 

I have a project now where the supply voltage is only 7.5V. Is this enough?

I'd say you should be able to make it work.

The datasheet specifies the maximum voltage drop for Q1 and Q2 when configured as a Darlington is 1.3V at 1A:



If Rsc is 0.33 ohms, you shouldn't have trouble with drop out voltage. If you connect Q1 and Q2 as a split Darlington, you'll get back around 0.7V.

 

Don't forget to read the Application Notes and the footnotes as you head into uncharted territory.

 

The MC34063 is a dreadful old device. It has a dropout voltage of 1.3V, so to step down from 7.5V to 5V it's not worth the bother - use a LDO linear regulator, and it will probably be as efficient once the running current has been taken into consideration. Alternatively, get a modern, new device. They are much more efficient and there are hundreds to choose from.

 

Add an external PNP transistor to convert the MC34063 buck converter to a low dropout buck converter. The resistor connected to pin 3 of the MC34063 sets the base current of the PNP. It may help to use a choke (say 470 uH to 1 mH) in place of the resistor that goes from the base of Q1 to its emitter to improve its turn-off speed at high frequencies.