What would be the most efficient way to step down and deliver 1A from a battery? - Page 1
Forum section: electronics / projects

author: WarFreak131
author_rank: Contributor
timestamp: September 16, 2020, 02:17:56 pm
content: I am building a project to regulate the temperature of my Hakko FX-901 portable soldering iron. In it's stock configuration, it puts 6 V across a heater of 4.8 Ohms. Assuming you're using a fresh set of 4x AA batteries (so the voltage is close to 6.4 V), the temperature immediately jumps up to about 950 F and then decreases linearly to about 575 F over a period of 70 minutes, at which point the batteries drain to a point of being unusable.I don't like this setup. First, 950 F will wear out the tips (which are expensive) quickly, and it spends about half of it's life at higher temperatures (further reducing tip life). Second, 70 minutes isn't a lot. Third, I don't want to have to buy AA batteries that much.To resolve #3, I bought two 1100 mAh lithium ion AA-size batteries. My approach to this is to step down the 7.4-8.4 V from the Li-ions to whatever voltage will cause the iron to heat up to 350 C only. I used my power supply and found that the tip has a negative resistance coefficient, and will settle at 350 C @ 1.01 A with a resistance of about 4.3 Ohms. Keep in mind, these tips haveno feedback mechanism. I cannot simply read a thermistor/thermocouple value and adjust accordingly. All this testing was done empirically with a knock off Hakko tip thermometer.My initial idea was to use an NPN transistor as a current source, but that would dissipate a lot of heat in the transistor. My second idea was to use a MOSFET, but due to the negative resistance, the voltage that I set the gate to won't be sufficient once the heater resistance changes. Icouldhave a digital potentiometer that dynamically changes the voltage at the gate, but that would require a microprocessor and a current sense resistor.My third idea is to use a buck LED driver. As opposed to a typical voltage-mode buck converter, these are current-mode buck converters, which will step down the voltage as much as is necessary to deliver 1.01 A, usually at around 90% efficiency. This should automatically account for the negative resistance changes in the tip. My only hesitation is that since the tips have a rather low thermal mass, they respond to current changes quickly, so would I need a large output capacitor to smooth the current out?Does anyone else have any suggestions on simple and effective ways to source current accurately and efficiently?

author: ledtester
author_rank: Super Contributor
timestamp: September 16, 2020, 03:43:36 pm
content: I think you're asking a lot out of such a simple soldering iron and AA batteries.Any kind of of constant voltage or constant current regulation is going to result in a lot of inefficiency. What would really help is constant temperature regulation, but that needs a temperature sensor in the iron - preferably the tip.If you want a better battery-powered iron I would look at the TS100 which lots of people power from lithium-ion battery packs or the TS80 which can be powered from a QC 3.0 power bank. Just search for "TS100 battery" or "TS80 battery".

author: WarFreak131
author_rank: Contributor
timestamp: September 16, 2020, 04:23:19 pm
content: I considered that, but this is purely for my own edification and enjoyment, and as a challenge. I already have a decent soldering station.

author: MosherIV
author_rank: Super Contributor
timestamp: September 16, 2020, 04:31:37 pm
content: HiIf you just want pure fixed voltage or current then option3 (some kind of buck converter) is your best bet.The other option is to build your own temperature control system but use pulse width modulation to vary the power. In effect you would be designing your own TS1000

author: ledtester
author_rank: Super Contributor
timestamp: September 16, 2020, 05:14:56 pm
content: Maybe replacing the slide switch with a push button would be a good mod to give you more control of how hot the iron gets.

author: WarFreak131
author_rank: Contributor
timestamp: September 16, 2020, 05:19:41 pm
content: Quote from: MosherIV on September 16, 2020, 04:31:37 pmUsing PWM was one of my original ideas, but someone brought up the point that having doing so may cause inductive or capacitive coupling at the tip which would put the tip at a (possibly) high enough potential to damage sensitive components.Quote from: ledtester on September 16, 2020, 05:14:56 pmThat's an interesting idea, I may just roll with that.

author: David Hess
author_rank: Super Contributor
timestamp: September 18, 2020, 05:15:55 am
content: Low frequency pulse width modulation will be the most efficient method, and it is simpler than a buck regulator since it requires no inductor or second switch.

author: magic
author_rank: Super Contributor
timestamp: September 18, 2020, 05:41:37 am
content: Adding an inductor is series with the heater will smoothen the switching edges without much energy loss. Be sure to add a flyback diode. And by the way, you now have almost a buck converter.There is a possibility of using the heater's thermal coefficient for sensing. I have seen a station which used PWM and sensed heater resistance during off time by pulling it up with a resistor paralleled with the power switch. If you choose variable DC drive, I suppose you could add a small resistor in series with the heater and compare a fraction of heater+resistor voltage against resistor voltage.I might be the one who scared you about voltage transients at the tipRecently somebody provided compelling evidence that they may not be much of a problem if there is no long cable and the controller is in the iron itself. The only way to know for sure is to measure it on your unit. Easily done if you have a scope.

author: WarFreak131
author_rank: Contributor
timestamp: September 18, 2020, 04:57:16 pm
content: Quote from: magic on September 18, 2020, 05:41:37 amYes! It was you who told me about inductive coupling. XDYou'll have to explain a little more, perhaps with a schematic, of your explanation about testing the thermal coefficient. But for now, I think I found something to solve my design issue.https://www.ebay.com/itm/1-8A-SXD-Super-X-Drive-Laser-Driver-M140-PLTB450B-NDB7875-NDG7475/183726019793?ssPageName=STRK%3AMEBIDX%3AIT&_trksid=p2060353.m1438.l2649It's a programmable current source that is small enough to fit into the battery compartment of the soldering iron. I think I may just buy this. But sometime in the future, I may make something that measures the voltage/current to work out a resistance, which can be mapped to a temperature.

author: magic
author_rank: Super Contributor
timestamp: September 18, 2020, 06:54:41 pm
content: Wheatstone bridge, basically.The two voltages are equal when heater resistance is exactly 4.3Ω.The lower resistor will dissipate a bit of heat, tough luck.

author: David Hess
author_rank: Super Contributor
timestamp: September 18, 2020, 07:34:40 pm
content: Quote from: WarFreak131 on September 18, 2020, 04:57:16 pmThe heating element has a rather high temperature coefficient of resistance so if powered with pulse width modulation, the heating element resistance can be measured and sampled during the off period.

author: WarFreak131
author_rank: Contributor
timestamp: September 18, 2020, 07:46:36 pm
content: Quote from: magic on September 18, 2020, 06:54:41 pmWouldn't this mean that once the soldering iron reaches temperature, both the heating element and the resistor are both dissipating ~4.3W?Quote from: David Hess on September 18, 2020, 07:34:40 pmWhat method do you suggest to measure the resistance using a uC like an Atmega chip? Off the top of my head I can imagine passing a known current through it via a voltage regulator in current mode and measuring the voltage at the output of the regulator.

author: David Hess
author_rank: Super Contributor
timestamp: September 20, 2020, 11:11:21 pm
content: Quote from: WarFreak131 on September 18, 2020, 07:46:36 pmThat is what I would do, with the provision that the low sensing current can be in parallel with the PWM drive so there is no need to turn it on and off.

author: ledtester
author_rank: Super Contributor
timestamp: September 21, 2020, 12:45:06 am
content: If you're going to employ a uC, maybe use a thermocouple and a MAX6675:https://youtu.be/RxZJujS1znU

author: magic
author_rank: Super Contributor
timestamp: September 21, 2020, 05:03:03 am
content: Quote from: WarFreak131 on September 18, 2020, 07:46:36 pmWhat makes you think so? Power dissipated in a resistor is current times voltage. Current through the two series resistors is equal but voltages very much are not.Quote from: WarFreak131 on September 18, 2020, 07:46:36 pmNo need for a regulated current source, just a resistor. Measure what fraction of VCC is the voltage at the heater then, do the math.

author: WarFreak131
author_rank: Contributor
timestamp: September 22, 2020, 02:06:43 am
content: Quote from: ledtester on September 21, 2020, 12:45:06 amThanks, I'll give this a watch.Quote from: magic on September 21, 2020, 05:03:03 amIn the diagram you sent (ignoring the small resistors, and I'm assuming they're supposed to both be 100m?), there's 4R3 resistor and the heater. Once the voltage across the bridge equals zero, that means that the heater has reached the proper temperature, due to the current passed through it at some voltage V. But if the heating element reaches 4R3 when its passing 1 amp, that must mean that the other leg of that bridge is also passing 1 amp and I2R = 12*4R3 = 4.3W.Scratch that, I understand. I'll give it a shot.

author: magic
author_rank: Super Contributor
timestamp: September 22, 2020, 10:00:41 pm
content: Note that this was for driving the heater with variable DC when power is always applied to it.With PWM, just connect 1kΩ or something like that from VCC to the heater and measure output of the resulting divider when power to the heater is off.

author: WarFreak131
author_rank: Contributor
timestamp: September 23, 2020, 08:24:40 pm
content: Quote from: magic on September 22, 2020, 10:00:41 pmCan you provide a diagram? I'm not quite following what you're saying. VCC>1k>heater? Wouldn't that cause the total current to drop, so then I'd never make it to 1A?

author: ledtester
author_rank: Super Contributor
timestamp: September 23, 2020, 11:02:26 pm
content: Here's an idea:Code:ADC|Vbatt --- heater -+- MOSFET --- GND|\/ R\|GPIONormally the GPIO is an input. When the MOSFET is off, turn the GPIO into an output pin with a value of LOW. Then take an ADC reading. Finally turn the GPIO back into an input.

author: WarFreak131
author_rank: Contributor
timestamp: September 24, 2020, 02:47:36 am
content: Quote from: ledtester on September 23, 2020, 11:02:26 pmGot it, thank you. What is a safe way to determine whether the PWM value is low, and will remain low for long enough to read the voltage correctly? The Atmega328P PWM runs at 490 Hz, and I believe it's independent of the rest of the code. The PWM period is 2ms, and the low period will be half of that, 1 ms. The analog read function takes 100 us, so I can take 10 reads in the period of the low phase of the PWM signal. There's a reasonable chance that the voltage will eventually read during the high phase if I were to just chance it.

author: ledtester
author_rank: Super Contributor
timestamp: September 24, 2020, 03:31:11 am
content: IIRC, on the atmega328 automatic PWM is done by setting an output compare register to a certain value. When the timer hits that value a GPIO pin is toggled.You can also generate an interrupt when the timer hits that output compare value. So you should be able to tell exactly when the GPIO pin is set to zero by the timer unit. All of the details are in the 300 page datasheet - a very worthwhile read.In any case it probably would be a good idea to disable the MOSFET gate pin while you are doing the ADC reading. The reading itself might only take 20 microseconds so even if you attempt a read when the MOSFET is supposed to be on it won't affect the heater that much.

author: magic
author_rank: Super Contributor
timestamp: September 24, 2020, 07:19:04 am
content: Don't bother with the GPIO, wire the resistor permanently to ground.That means there will always be a few mA flowing through the heater so it will neverreally, fullyturn off, but who cares

author: Haenk
author_rank: Super Contributor
timestamp: September 24, 2020, 06:02:45 pm
content: If you have a defined targetvoltage and targetcurrent, maybe this dirt cheap LED supply board will do:qs-2405ccbd-3a(version with 3 potentiometers)It's not large and will well go with some batteries.Might be worth a try and see how it behaves. Advantage: It works on a large range of input voltage.

author: WarFreak131
author_rank: Contributor
timestamp: September 25, 2020, 05:55:11 pm
content: Thank you everyone for the input. If I'm driving this by PWM through a MOSFET, what is the heat dissipated through the MOSFET? Is it just I2RDsON?

author: ledtester
author_rank: Super Contributor
timestamp: September 25, 2020, 07:01:42 pm
content: My understanding is that RdsON is the resistance when the MOSFET is ON and thereby represents the minimal value. The actual power dissipation will depend on how fast the MOSFET is driven ON and OFF by the gate signal.

author: WarFreak131
author_rank: Contributor
timestamp: October 02, 2020, 03:50:50 am
content: Shouldn't it theoretically be possible to complete this all with purely active/passive components and no uC?Let me know if this would work. I set up an op amp as a non inverting amplifier. Since the tip settles at 4.395V @ 1.01A, that means I can set V+ to ~4.4V. Then I can set the heating element, and a 100m resistor as R1 and R2 in this picture respectivelyhttps://en.wikipedia.org/wiki/Operational_amplifier_applications#/media/File:Op-Amp_Non-Inverting_Amplifier.svg. The op amp will try to maintain 4.4V across the heater.If the tip temperature drops (because I put it against a large thermal load), resistance will increase slightly due to negative coefficient, which will decrease Vout. But if Vout decreases, V- decreases, so it pumps up the output voltage to bring it back up to 4.4V.The same would happen in reverse. If the tip temperature overshot, resistance will decrease slightly, which will increase Vout and cause V- to be larger than V-. So the op amp will decrease the output voltage to bring it back down to 4.4V.

author: IanB
author_rank: Super Contributor
timestamp: October 02, 2020, 04:56:32 am
content: OK, maybe simplest is best?My solution is to use rechargeable cells like Eneloops:1. They support high current drains, so they work better in high power devices like soldering irons.2. They have a lower voltage range between fully charged and empty, so they are inherently self-regulating.3. They spend most of their time between 1.2 V and 1.3 V, so they avoid the "super-hot tip" that happens with fully charged alkaline cells.4. You can recharge them hundreds of times, so you don't have to keep buying new batteries.Simple, reliable, effective.