MC34063 - Page 1
Forum section: electronics / beginners

author: volyaire
author_rank: Contributor
timestamp: December 10, 2017, 08:16:38 pm
content: I have built a step-up converter with the MC34063 to convert 7.5v to 12v using the circuit on page 5 of the datasheet with no extra components added. At no load it gives 11.4v which is correct for the resistors I'm using (R2=18k, R1=2k2), but when a 100 ohm load is added the voltage drops to 7.5v. Why?

author: Zero999
author_rank: Super Contributor
timestamp: December 10, 2017, 10:58:18 pm
content: Is the inductor rated for the peak current? It could have too higher series resistance or be saturating.What construction method have you used to make the circuit? Please post a picture.

author: ovnr
author_rank: Frequent Contributor
timestamp: December 10, 2017, 11:08:39 pm
content: inb4 "it's a pile of wires on a breadboard".

author: volyaire
author_rank: Contributor
timestamp: December 12, 2017, 12:39:36 am
content: It's built on a 1 inch x 0.75 inch piece of veroboard. The 170uH inductor is like a resistor, 10mm long x 3mm diameter.

author: westfw
author_rank: Super Contributor
timestamp: December 12, 2017, 02:21:07 am
content: Quotetypically inductors that look like resistors are for signal-processing, and have too high a DC resistance to make good SMPS.The inductor you use with an MC34063 should look like a coil!

author: volyaire
author_rank: Contributor
timestamp: December 12, 2017, 03:13:54 am
content: Thanks for solving this mystery. Can you recommend a suitable inductor to buy, or the dimensions and wire guage so I can wind it myself?

author: volyaire
author_rank: Contributor
timestamp: December 12, 2017, 07:35:57 am
content: I don't fully follow that theory. I am making 3 versions of this circuit: one 7.5v to 12v at about 20mA, one 7.5v to 40v at 6mA, and also a buck from 7.5v to 5v at about 150mA. I don't understand the relationship between current and inductance - does higher current require higher inductance or lower? I can't find that video, and I haven't any equipment to plunder. I could easily wind one if I knew the dimensions and wire guage and number of turns. Or if I buy one do I need to specify inductance, saturation current, or what?

author: Buriedcode
author_rank: Super Contributor
timestamp: December 12, 2017, 03:35:39 pm
content: Quote from: volyaire on December 12, 2017, 07:35:57 amSome suggestions:The maximum voltage on the switch node (collector) of the MC34063 is 40, so I don't think you will achieve your 7.5 to 40V directly - but as it only requires 6mA you could use a charge pump /voltage doubler on the output for this.The 7.5 to 12V converter only needs 20mA, IMO whilst the MC34063 is cheap and available, it isn't well suited to very low output currents, and is somewhat over kill. There are plenty of cheap chinese boost converter modules available for a couple of US dollars that will be more efficient, and have lower noise output than any MC34063 circuit.Again, the 7.5V to 5V would be about 66% efficient using a linear regulator. Using an MC34063 for a relatively small step down wouldn't fair much better.A handy design tool that makes a for a goodstarting pointis here:http://www.nomad.ee/micros/mc34063a/Note the 'Ipeak' is the peak current in the inductor - any inductor you use must have a saturation current greater than this. Its DC resistance will impact on efficiency, but for the most part isn't particularly critical, just make sure if the Ipeak is say 400mA, you get one with a Isat of >600mA, preferably with a larger margin for error. Designing switching converters isn't always straight forward, not only do parts have to meet certain specs (like using inductors designed for power electronics, rather than filtering) but also, the layout can be critical, keeping high current paths short and low impedance - don't run long thin wires for the switching path. Copying a reference schematic is a good starting point, but also look at the PCB layout and component specs.If your 7.5V supply isn't coming from a battery - where you want to maximise battery life - then efficiency isn't as critical, especially at such low currents, I would suggest a linear regulator for the 5V, and a single boost converter to convert 7.5V to 15V, then use a 12V linear regulator to get a *clean* 12V from this 15V line. Also, adding a voltage tripler (5 cheap diodes, and 5 cheap caps) to the boost converter would get you 3*15 (minus diode drops) = ~42V @ <10mA. Similar to this setup:http://www.clivemaxfield.com/area51/do-not-delete/max-4-john-01.jpgThat would mean you only need one boost converter and one linear regulator (cheap) to produce all three rails. Less parts, so it'll be cheaper, smaller, and less to go wrong, but it would require a bit of experimenting with the charge-pump bit. If you decide to give it a try I'll try and run some tests in LTspice to give you an idea of component values.

author: volyaire
author_rank: Contributor
timestamp: December 12, 2017, 05:42:40 pm
content: The guy in that video is quite funny. I am using a battery so I really do want high efficiency. I'm looking at LTC1779 for the 7.5v to 5v converter - any advice?

author: westfw
author_rank: Super Contributor
timestamp: December 13, 2017, 07:26:28 am
content: QuoteHuh. I played with it some. How come the desired inductor value goes UP with reduced current requirements? I'd sort-of expect bigger inductors to store more energy, and therefore be more useful at higher output currents...

author: volyaire
author_rank: Contributor
timestamp: December 13, 2017, 12:27:45 pm
content: Buriedcode: that development aid is useful. The voltage tripler looks good. I'd like to use a more efficient buck for the 7.5v to 5v because I'm using a battery - I'm thinking of LTC1779?

author: Buriedcode
author_rank: Super Contributor
timestamp: December 13, 2017, 07:55:18 pm
content: Quote from: westfw on December 13, 2017, 07:26:28 amBecause I believe it tries to keep the converter in continuous mode to reduce output ripple, which is wise. But this means the peak current should be closer to the average output current, so for a given operating frequency - on period - you want a slower current rise for a given input voltage. That means, higher inductance. A smaller inductor would reach its peak current very quickly, and given a smaller output load, current would take longer to decrease requiring a very short duty cycle.Notice you said 'bigger inductors', not 'higher inductance'. This is partially true, because for higher output currents, the peak inductor current is higher, requiring a higher saturation current = larger core. But it also means the DCR should be lower which means thicker gauge windings, which also increases size. SO "bigger" inductor for higher output currents? yes. Higher inductance? no.

author: chris_leyson
author_rank: Super Contributor
timestamp: December 13, 2017, 08:33:17 pm
content: Try one of those radial leaded bobbin chokes, panasonic ELC08D181 for example or anything rated at a few hundred mA or even 1/2A should be good for the circuit shown in fig. 9. Besides, they're not too expensive and you can buy a few different values to try out. The little "resistor like" inductor you used is an RF choke and is designed for a completely diferent job i.e. filtering and not energy storage, it might only be rated for a few 10s of mA and probably has a very high series resistance as well. It's an easy mistake to make for beginers. Don't worry we've all done it when starting out and I've even seen a "professional engineer" put an 0805 RF choke into a buck converter and he wondered why it didn't workJust get some cheap radial bobbin chokes and play around to get an intuative feel for what is going on, when you've some sort of idea formulated then go look at the theory and it will make a lot more sense. You might want to get a few more MC34063s just in case

author: volyaire
author_rank: Contributor
timestamp: December 14, 2017, 07:19:41 pm
content: I fitted an ELC10D181E 180uH 1.3A 0.18 ohm choke to the MC34063 boost circuit and now it works really well! Thanks to all for your help.

author: Buriedcode
author_rank: Super Contributor
timestamp: December 16, 2017, 04:52:16 am
content: Quote from: volyaire on December 13, 2017, 12:27:45 pmI realise you've probably settled on the MC34063, but I forgot to hit 'post' the other day when I replied to this bit.The LTC1779 seems rather expensive - like most Linear Tech offerings. Although I cannot fault their products, they do tend to do an excellent job. That particular device, although mentions 250mA output, does not show any graphs for that output current at 5V. This isn't to say its incapable, just that often converters with internal switches state the maximum current through the switch, rather than the actual output current - which depends on the input and output voltage.For 7.5V to 5V @150mA there are many options. If you wish to make it 'from scratch' (as in PCB layout, SMD parts etcc) then Digikey, Farnell, mouser etc.. have hundreds of cheaper suitable switching converters that will require little more than an inductor and a couple of caps (at 150mA output current they can easily include the switch and diode internally, the term 'synchronous buck' will help the search). As with the LTC1779 though, you will have to use a PCB, veroboard or prototype board *can* work for switching power supplies, but not recommended.What narrows down selection is usually the current requirements. You said 150mA max, but what about min? Some converters will just use up 30mA even when you don't draw any output current, and for battery powered applications this isn't great (as in, with smaller loads the efficiency drops drastically). For a beginner, searching for parts is often frustrating because you can enter all the restrictions and still end up with 200+ matches. But there is also IC package. No-one likes hand soldering QFN, DFN, or SC70, but DIP is becoming less common, so go for something like SOIC.tl;dr: nice device, Expensive, and there are many cheaper ones that require less parts and have better efficiency.

author: volyaire
author_rank: Contributor
timestamp: December 17, 2017, 12:30:09 am
content: Thanks for the reply. The MC34063 is working well and I'll probably use it for the boost to 15v then use a linear regulator to drop it to 12v as you recommended. The efficiency isn't too important here because the current is only about 10 or 20 mA. I've already bought an LTC1779 for the 7.5v to 5v and built most of the board using veroboard, but keeping all connections very short and chopping off the tracks where they are not needed so hopefully there will be no problems with stray inductance and capacitance. The board is only 1.5" x 0.5"The 5v supply is for powering a Tascam mini sound recorder which takes a bit over 100mA which doesn't vary too much. I haven't soldered the chip in yet - it is going to be tricky because it has microscopic little pins coming out of it and I don't want to overheat it.