XL6009 DC Step-up converter - Page 1
Forum section: electronics / beginners

author: bitman
author_rank: Unknown
timestamp: June 20, 2017, 03:30:15 am
content: I have a "simple" circuit using a XL6009 (http://www.electrodragon.com/product/xl6009-step-dc-boost-adj-module-4a/). I'm providing it 9V and extracting about 14V on the output. It works just fine, however I have a question about the current load on the converter.I know these by no means are efficient and there's a lot of wasted energy. However, when I measure amps on power support with the converter ON I get about 1.8Amps. When I measure amps directly on the load connected to the 14Volts I get about 0.6 Amps. It seems a rather large difference? I'm trying to understand why and if I screwed up somewhere.

author: not1xor1
author_rank: Frequent Contributor
timestamp: June 20, 2017, 04:00:17 am
content: Quote from: bitman on June 20, 2017, 03:30:15 amin a boost converter input current is always more than output one (principle of conservation of energy)in your case the rate is much higher than one would expect from an efficient circuit, but that is probably due to errors in the measure methodI do not understand what you mean for "measure amps on power support"If you are using a multimeter in series with the PSU then you should subtract the DMM burden from the input voltagebesides that you should consider that the input current is not DC current but more likely DC with superimposed high frequency, high value current so your DMM might read a wrong valueI never tried this kind of measure myself so this is just an educated guess of mine

author: bitman
author_rank: Unknown
timestamp: June 20, 2017, 04:12:03 am
content: Quote from: not1xor1 on June 20, 2017, 04:00:17 amI use a 9v battery for input. I have one meter in series with the positive side of the battery on the circuit's "plus" power line. The second volt meter also measuring amps, sits in series on the output of the boost converter, to the load (led strip) again on the positive side, not that I think that matters.I'm not sure what you mean by DMM.I know the amps are high as the wires on the 9v battery become warn/hot when the full load is on. So the 1.8Amps (at one point I had more LEDs on the strip and the load was 2.2Amps). When I run the strip by itself on 9v, it has a 0.33Amps - so this is what I'm trying to understand.That's for the ideas ... once I get DMM I may be able to try to adjust for that. However, since the wires are getting hot I'm pretty sure the measurement isn't that far off.

author: not1xor1
author_rank: Frequent Contributor
timestamp: June 20, 2017, 05:10:22 am
content: Quote from: bitman on June 20, 2017, 04:12:03 amDMM is digital multi meteryou can't measure current with a volt meter, so I guess you are using the current functions of 2 different DMMsthe current, unless you are using a clamp meter, is measured by checking the voltage across a resistor of stable and known valueso, in your case, you have a (small value) resistor (inside the multimeter) in series with the input of the voltage booster circuit that so might even get as low voltage as 8.5V or lessthe lower the input voltage the higher will be the current needed to boost the output voltageyou can directly connect the load to the voltage boost circuit and use the other DMM (don't forget to move the banana jack to the Voltage position) to read the actual voltage at the input of the circuitin any case I think the greatest part of the error comes from the spikes in the input current

author: MagicSmoker
author_rank: Super Contributor
timestamp: June 20, 2017, 11:49:21 am
content: Quote from: bitman on June 20, 2017, 04:12:03 amThe actual terminal voltage of a typical good quality alkaline 9V battery will drop quite a bit if asked to deliver 1.8A, perhaps down to 6V or less, and it won't tolerate this kind of abuse for very long, since the typical capacity is around 500-600mAh so a 1.8A load will drain in flat in less than 20 minutes (if it doesn't fail from overheating first).

author: bitman
author_rank: Unknown
timestamp: June 20, 2017, 05:47:05 pm
content: Quote from: MagicSmoker on June 20, 2017, 11:49:21 amAhh, but I'm not using that6 AA batteries = 9vSorry for the confusion.

author: not1xor1
author_rank: Frequent Contributor
timestamp: June 21, 2017, 04:33:07 am
content: Quote from: bitman on June 20, 2017, 05:47:05 pmin that case you have both the internal impedances of the batteries and the contact impedances plus the instrument inner shunt resistance plus wire impedance so your circuit input average voltage is probably more like 5-6V or lessas I already wrote just check the real voltage with a multimeter

author: bitman
author_rank: Unknown
timestamp: June 21, 2017, 09:08:11 pm
content: Quote from: not1xor1 on June 21, 2017, 04:33:07 amThanks not1xor1 - I think there's a bit of terminology confusion here. I unfortunately wrote volt-meter - I use volt/multi meter interchangable all the time - most of the times I just say "meter". That's probably a bad habit. t measures both amps, volts and a bunch more. I have a few of them, in varying quality. I'm measuring directly on the bread-board - not on the terminals of the battery. I've also used a small display for volts while I was measuring amps. I'm very sure that very close to 9v is being applied to the circuit because of that. Not that the wires doesn't have impedence and all that other stuff, but since I'm measuring AFTER the battery wires those wires shouldn't be of a concern?And note, without a load on the booster I only get 20-30 milliamps which is definitely within expectations - so it's not as if the wires totally are messing up things?

author: not1xor1
author_rank: Frequent Contributor
timestamp: June 23, 2017, 04:19:04 am
content: Quote from: bitman on June 21, 2017, 09:08:11 pmbradboards usually are rated for a maximum of 100mAwhich part of the circuit is on breadboard?can't you just post some clear photos of your circuit or some sketches so one can be sure of what you mean?

author: bitman
author_rank: Unknown
timestamp: June 23, 2017, 11:38:24 pm
content: Quote from: not1xor1 on June 23, 2017, 04:19:04 amDo you mind explaining how you get to that number? There's a lot of sites out there contradicting that, even eevblog:https://www.eevblog.com/forum/projects/maximum-vac-and-amp-on-breadboard/Also, if I exceed it's capacity I should see magic smoke or hot/melting plastic. I'm not. In any way, the circuit works. The only thing that gets hot (not melting hot, or so hot you cannot hold it) is the wires from the battery to the bread-board, and it's a very thing wire to begin with. It's also the wires I see 1.6->2.0amps on. Everything else is much less (as I described initially).QuoteEverything? Except for the batteries which are connected to the bread-board's power +/- lines. Even the boost converter is on the bread board. And I connected the LED strip directly to the bread-board where the output from the boost converter is. I have no idea how to diagram the boost converter - if I did I would have posted one.QuoteI'll see if I can get a picture taken this weekend.

author: bitman
author_rank: Unknown
timestamp: June 25, 2017, 12:55:59 am
content: Quote from: not1xor1 on June 23, 2017, 02:19:04 PMI've taken a picture of the breadboard.Nothing really exciting about it. A simple relay, the boost converter and what you cannot see is the PIR connected that triggers the relay (those are the 3 colored wires).Now, I changed the battery pack today. I unfortunately shorted the 9v volt with the 6 AA batteries and it took me way too long to realize what I had done. So instead of replacing all 6 I thought I would try with just 6 volts and see how high the amp would go since the stepup was now more than double the incoming voltage (6-14). To my amazement it totals out at 1Amp down and slowly goes down from there the longer the light is on. I'm baffled - I expected the opposite. NOW I'm thinking there may be a short in the 9v battery pack - I'm not sure. I'm just as confused as before, albeit 1 Amp was sorta around my expectation.