Boost converters without load - Page 1
Forum section: electronics / beginners

author: chrisabird
author_rank: Newbie
timestamp: July 15, 2015, 11:40:26 am
content: After over a decade hiatus from electronics in the world of software I'm getting my hands dirty again. And my very very rusty skills need help already...I'm currently looking at the MC34063 to boost a 3.7v/2200mAh/2C battery to 12v and 500mA, if my maths is correct this looks okay. But the part of my design that requires the 12v (solenoid) is only turned on for <600ms at a time and many days apart. I'm struggling to find details on what current draw may occur with no load on the boost converter, as I'd like to leave this device sitting for weeks at a time without having to replace the battery.The other thought that crosses my mind is that, I have a hammer shaped boost converter and this problem looks very much like a nail. Are there alternative solutions to this type of problem?I don't see this type of information in the Datasheet, but I could be mis-reading it. I suspect my next step is to build and measure (as soon I get my test equipment back out of storage)

author: Simon
author_rank: Unknown
timestamp: July 15, 2015, 11:44:57 am
content: Don't use that particular crusty chip unless you use an external switch. I tried and just made a really hot thing that failed, the problem is the internal switch has such a voltage drop that in boost mode because more current goes in that comes out it's really bad on efficiency.How about turning the power regulator on when the valve needs switching ?

author: rs20
author_rank: Super Contributor
timestamp: July 15, 2015, 12:04:21 pm
content: Have you considered putting the switch driving the solenoidbeforethe step-up converter? That way you don't really care how much current draw occurs while the thing is off, you'll have the entire step-up converter shut down. You can get converters with an "enable" input pin, which makes it even easier. (Ed: repeating what Simon says).Regardless, it's reasonably straightforward to figure out what your losses will be; just plug the current draw of the output voltage divider into the formulae and find out?Quote from: Simon on July 15, 2015, 11:44:57 amNot to be blunt, but did you do calculations behind the scenes for that particular chip in the OP's particular use case, or did you just criticize a perfectly serviceable chip becauseyouonce attempted to use it in an inappropriate use-case?

author: bktemp
author_rank: Super Contributor
timestamp: July 15, 2015, 12:30:47 pm
content: Quote from: Simon on July 15, 2015, 11:44:57 amThe 34063 may not be the best ic, but it works fine if you use it within its limits. You probably did something wrong or simply overloaded the circuit. The transistor is rated for 1.5A switching current. The avarage output current is much lower. If you need more, the 34063 is simply the wrong choice.Quote from: chrisabird on July 15, 2015, 11:40:26 amThere are two types of losses:The 34063 itself has a quiescent current of a few mA and the feedback divider discharges the output cap. The boost converter has to compensate this loss.If you need a lower quiescent current, there are better ics with lower quiscent currents that allow the usage of higher impedance feedback dividers.Does the boost converter need to be on all time? Would it be possible to switch the converter only on when needed? This could reduce the current to some uA on most modern stepup ics.

author: Simon
author_rank: Unknown
timestamp: July 15, 2015, 12:48:40 pm
content: ok the switch is a BJT with 0.2V drop do the math and look at the datasheet, the efficiency for step up is poor and I'm sure that if you work out what that translates into in watts it soon exceeds the pakages power handling. For low power it will probably work, I did my best to follow the guidlines on the datasheet. But this chip was not meant for any amount of power for stepping up from 3.7V to 12V. An expernal switch would probably solve it.

author: bktemp
author_rank: Super Contributor
timestamp: July 15, 2015, 12:59:46 pm
content: Quote from: Simon on July 15, 2015, 12:48:40 pmEven at 1.6A the transistor has only 0.6V voltage drop. That is ~1W at peak current. The avarage dissipation will be much lower, probably half of that or less. That is well within specs. Since 3.7V is a quite low input voltage, my guess is you made the base resistor too large. The output transistor has a current gain of ~50 at higher currents. So for the full 1.5A you need around 60ohms at pin 8.Calculating all the parts around the 34063 can be quite difficult. The 34063 will work in most cases even when badly designed, but for a good efficiency everything needs to be right: Current limit, inductor, switching frequency and base drive resistor.

author: chrisabird
author_rank: Newbie
timestamp: July 15, 2015, 01:05:42 pm
content: Thanks everyone.The converter only needs to be on for that 600ms period, so I think the running theme of switching the converter is the right thing to try. Now I'm just kicking myself for not thinking of that earlier.Thanks also for pointers on working out the losses, I'm going to work all that out anyway, so many many things I've forgotten

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 16, 2015, 03:08:56 am
content: Quote from: rs20 on July 15, 2015, 12:04:21 pmBoth.I personally disregard 34063 out of hand. It's old, it SUCKS, it's inefficient, it's slow, shall I continue?Anything TPS54xxx will do a fine job; follow the application information and you'll get a good circuit. There are parts in there with low quiescent current, that will do a fantastic job on battery powered applications. You can go with internal or external switch, up to reasonable power levels (20W?), beyond which you need external.Tim