MC34063 24Vin to 12Vout at 1.5A assistance needed - Page 1
Forum section: electronics / beginners

author: dummy
author_rank: Newbie
timestamp: April 26, 2012, 12:15:53 pm
content: Guys, I desperately need some advice in my design.I'm trying to build a 24V input to +12V output power supply with MC34063 that should be able to give me 1.5A.This current requires an external transistor as the peak current goes to 3A. First I decided to use a NPN like in the schematic:With this schematic the transistor overheats very quickly even at 0.4A. Obviously I have not selected the right power for it - it was a ZTX857.Then I decided to try with a MOSFET IRF540 like this:It turns out that this time I have made another error. The MOSFET gets very hot at 0.4A too.What to do now? Can you help me to build it?I have read the datasheet, used the online calculators and tried to calculate with the formulas, but obviously newbies always make mistakes.Oh, and one last question - what should be the power of the shunt resistor - is 0.5W enough?Thank you

author: Psi
author_rank: Super Contributor
timestamp: April 26, 2012, 12:36:17 pm
content: It's just a quick look at the circuit, but i can't see any obvious problems.Is your output cap low ESR?A picture of your design may help us see whats wrong.

author: amspire
author_rank: Super Contributor
timestamp: April 26, 2012, 01:11:45 pm
content: I think L2 is way to low. It needs to be about 220uH at 3A. It would be OK if you could increase the operating frequency, but you are on the limit for that IC. You need to run comfortably below 3 A and I suspect you are exceeding 4A peak - could well over 10A peak if the inductor is saturating.I think L2 is saturating, especially at startup, causing excess current in the switching transistors. If you have more of the 47uH inductors, put another 3 in series and see if it improves.Richard.

author: Kremmen
author_rank: Super Contributor
timestamp: April 26, 2012, 01:15:14 pm
content: Your fet heats up because it won't go into saturation in that circuit. This is the classical problem of high side switching. The Vgs needs to be several volts, especially with a traditional component like the 540. Now what happens is that when Vg goes high, Vs also jumps to near Vd. Or it would do so, only in this circuit it doesn't because Vg is limited to the supply voltage which also happens to be Vd. So you cannot really get a proper Vgs when Vs rises near Vd which in this case is Vg as well. Confused yet?The point is, to get the fet fully conducting you need a Vg supply which is several volts above the supply voltage seen by the drain of the fet. In this high side driver configuration it is typically done either by a transformer or a special high side driver such as IR2110 (well it is a half bridge driver really, containing both high and low side drivers). The driver circuits have a capacitive charge pump to raise the Vg supply circa 15 volts above Vd and then the fet will turn on properly.All your problems will go away if you change the topology to boost configuration where the fet source is grounded. That will turn it into a low side switch and you won't have this problem. But please check the data sheet for the correct circuit and component dimensions, i haven't checked that.

author: Jeff1946
author_rank: Regular Contributor
timestamp: April 26, 2012, 04:22:18 pm
content: 1. Use a suitable PNP power transistor. Use the output of the MC34063 to drive an NPN transistor (2N4401 or the like) to drive the base of the PNP. You will probably need 100-300 ma of drive current to the PNP to drive it to saturation.2. MOSFETs to be driven with a low impedence source to switch fast. I would suggest a special MOSFET driver like a Clare IXD-604, I have used these and driven them from pin 8 (which is connected to power with a 1k resistor) of the MC34063 With 24V you to need to be careful to not exceed the Vgs of the MOSFET.

author: amspire
author_rank: Super Contributor
timestamp: April 26, 2012, 09:36:25 pm
content: Kremmen and jeff1946 are right about the transistor drive, but you must make the inductor bigger as I mentioned.I hadn't realized that the Zetex transistor was a TO92 device - you need something that can dissipate 3 Watts along with a 220uH inductor. Going to a PNP transistor would lower the dissipation of the transistor, but the base drive resistors will need to dissipate several Watts. I would stick to a NPN with some heatsink attached.The IC is not really suitable for directly driving mosfets directly.If you want more efficiency, follow the advice in the two previous posts, but you will need to get the inductor right.Richard.

author: dummy
author_rank: Newbie
timestamp: May 02, 2012, 09:09:34 am
content: Many thanks to all of you guysIt seems I've fixed my problem now. I post the schematic that seems to work for me - I have used a P-MOSFET IRF4905 which have a resistance of 0.02 ohm. I tested it with up to 1A loads and without a heatsink. The temperature seems to be ok now. There is only a little bit of audible noise. If you can see any drawbacks, please, tell me before I make the PCB. I'm not so good in analog electronics and your answers were really helpful.

author: amspire
author_rank: Super Contributor
timestamp: May 02, 2012, 12:01:50 pm
content: Also add extra ceramic capacitors input and output to help reduce the ripple. At the moment, you have way more ripple current then your eletrolytics can handle so they will probably fail prematurely. Increasing the inductor will greatly reduce the ripple current for the 330uF.A 10uF ceramic cap would across the input will help a lot on the input electrolytic ripple. Increasing the inductor would halve the input ripple current as well.The audible noise means that the feedback loop of the regulator is unstable, but before worrying about that, I would get that inductor increased and then see what happens.Have a look at the specs for ripple current for the electrolytics. I suspect they are not just slightly out of spec - I think you will find they are disastrously underrated for the circuit you posted. With the 47uH inductor, your output ripple will be about 1A RMS and the input ripple about 0.5A RMS. Also the inductor will definitely saturate on startup which is very bad. If for some inexplicable reason you want to stick to the undersized 47uH inductor, increase the rating to at least 6A.Richard.

author: Jeff1946
author_rank: Regular Contributor
timestamp: May 02, 2012, 03:19:38 pm
content: Glad it's working. Note the max current for the 1n5818 is one amp. Use a 1n5821 or 1n5822 which are rated at 3A.

author: MarkBng
author_rank: Newbie
timestamp: October 26, 2012, 12:17:32 pm
content: Is it also possible to replace the schottky diode with a MOSFET (as an ideal diode) to enhance the efficiency a little bit more? If yes, how?

author: gxti
author_rank: Frequent Contributor
timestamp: October 26, 2012, 02:50:57 pm
content: Yes, it's called synchronous rectification. You're probably better off using a different IC. A quick search turns up NCP3030 as a good candidate. It has a builtin high-side driver, you just add two N-channel MOSFETs, an inductor, and some passives.

author: BravoV
author_rank: Super Contributor
timestamp: October 26, 2012, 05:11:24 pm
content: Quote from: AcHmed99 on May 02, 2012, 10:55:15 amAchmed, just interested with the mosfet driver section, mind explain what is the purpose of the pull-up R7 22K resistor ? I thought with Q2 when on is good enough ?

author: BravoV
author_rank: Super Contributor
timestamp: October 26, 2012, 06:13:19 pm
content: Quote from: AcHmed99 on October 26, 2012, 05:39:27 pmThanks for the confirmation, as an ee wannabe, my eyes spotted it and was wondering what I've missed, its just part of the learning.Yeah, if I was the op, I'd use the much simpler ic, and relatively cheap too like from simple switcher families such as LM2575 or even LM2576, they're much easier to use and less headache.