Rectifying AC smoothly at 2 amps, and limiting current? - Page 1
Forum section: electronics / projects

author: rwgast_lowlevellogicdesin
author_rank: Frequent Contributor
timestamp: June 16, 2013, 11:41:57 pm
content: Ive been working on a digital solder station for a while, I already own a decent setup this is purely educational. Instead of driving the irons heater with AC and a Triac, I decided I wanted to go the DC/PWM route for a finer control.Im using a 24v 2.4 amp transformer, a long with a 60V 30AMP logic level N-Fet and Propeller Micro to control the heating element. Im having two issues though, the first is that the hakko iron wants to draw about 8 amps when its cold, which warms up the transformer lowers its voltage output. So what I need to do is limit the current draw down to about 2amps, Im not really sure how to do this though... My idea is to bias a Fet so that it will only allow two amps, but Im not sure if this is the best way, and Ive also noticed that when trying to build a constant current driver J-Fets are typically used over N or P. It seems to me an N fet should be just as suitable to limit current?Second problem is rectification! Im using 4 massive overkill shottky 400v 100amp diodes, this works well as they dont warm up a bit! But when I run the PWM at 100% and let the iron draw a full 2 amps I get massive ripple! Currently I just have a few good 2200uF NichiCon 50v caps set up after the bridge rectifier but it isnt enough to keep a smooth DC signal at 2amps! I searched online and found a few formulas for calculating a rectifiers filter cap, the problem is at 24v/2amps the cap I would need is close to the 1 farad range! I never see that much capacitance in bigger power supply's that output a few amps! So is there a trick to getting smooth DC voltages from a transformer when you need over an amp of current, besides uping the cap size?

author: mariush
author_rank: Super Contributor
timestamp: June 17, 2013, 12:36:43 am
content: 24v AC rms = 24*1.414 = 33.9V DC PEAK - 2-2.4v in bridge rectifier = ~ 31.5v2.6 A becomes about 0.62 x 2.6 = 1.6A DC .. so if you really have a 24V AC 60VA (2.6A), you can't really get 24v DC 2A from that.If you want 2A 24v DC , then you can use a linear regulator to keep that 24v regulated .. a LD1084 will do that with only 1.3v of voltage drop :http://uk.farnell.com/stmicroelectronics/ld1084v/v-reg-adj-1-25-28v-1084-to-220/dp/1087148If you use LD1084, add a few diodes in series or a resistor to drop about 3v and make sure the peak voltage will always be under 30v (the maximum the LD1084 supports).Now you only need to add enough capacitors so that input voltage will always be 24+1.3v or more ... let's say 25.5vC = current x 0.7 / (2 x mains frequency x allowed voltage drop) =2 x 0.7 / [2 x 50 x (29v max - 25.5 = 3.5v)] = 1.4 / 350 = 0.004 F or 4000 uF for 50Hz mains frequency2 x 0.7 / [2 x 60 x (29v max - 25.5 = 3.5v)] = 1.4 / 420 = 0.0033 F or 3333 uF for 60Hz mains frequencyBut I'm not sure you really need to use fixed 24v, you should be fine with just using enough capacitance to keep it between 24-30v.

author: Jebnor
author_rank: Regular Contributor
timestamp: June 17, 2013, 06:04:09 am
content: Quote from: mariush on June 17, 2013, 12:36:43 amCould you explain where these numbers come from?I understand Vdc = Vac * sqrt(2). And the C Calc based on Mains Frequency. Where did 0.62 come from? 1/sqrt(2) is ~0.707. I'm confused.

author: peter.mitchell
author_rank: Super Contributor
timestamp: June 17, 2013, 06:41:55 am
content: If you really want the 24v fixed, you'd be best just getting a 10a 24v SMPS.However, if you want to use the transformer, just throw more capacitance at it, thats the simplest way, if you want to go complicated you could put buck-boost on it, but that would probably cost more than the previously suggested 24v 10a smps.

author: mariush
author_rank: Super Contributor
timestamp: June 17, 2013, 10:47:26 am
content: Quote from: Jebnor on June 17, 2013, 06:04:09 amWell, the capacitance formula is a simplified one which works pretty good with bridge rectifiers, based on the basic equation for a capacitor I = C * dv/dtdv/dt just means the rate of change of the voltage, measured in volts/second. You know the maximum current I (he wants 2A), and you should choose a value for dv/dt that keeps the voltage high enough for your circuit to run in between peaks.For example, if you can afford to lose two volts, and your capacitor is being charged 100 times per second (full-wave rectification at 50 Hz), then the rate of change is 2V*100Hz = 200V/s. Then just solve for C:2 = C * 200V/s => C = 2/200 = 0.01 = 10.000 uF3.5v = C *200 => C = 2/350 = 0.0057 = 5700uFThe 0.7 factor I added just makes the value more realistic but it never hurts to add more capacitance than the determined value.As for the voltage and amps of transformer... it's pretty obvious ... if the transformer is rated for a particular VA number, you won't have the same DC current as the AC current, the voltage changes.24 v AC @ 60 VA ... 60/24 = 2.5A ... 24x1.414 = 33.93v ... 60va/33.9v = 1.77A ... but a much safer current value is by multiplying the current with 0.62-0.65, which gets you about 1.6ANow of course, that 0.62 is for bridge rectifier - you can get DC in different ways, using 2 diodes, or using inductors, it depends on your transformer and your needs, see attached picture:

author: richard.cs
author_rank: Super Contributor
timestamp: June 17, 2013, 10:59:41 am
content: I suspect you're trying to solve the problem the wrong way. I would suggest full wave rectifying without smoothing - it doesn't add anything in this application. Feed a small capacitor off this unsmoothed supply with a diode to get a low-current smoothed supply for the microcontroller that shares a -ve rail with the unsmoothed supply. You them connect the pwm output to the mosfet and use that to switch the unsmoothed supply to the soldering iron. You have 100 (or 120) Hz ripple with PWM superimposed on it. The soldering iron will be happy with that and you can have your fine control.For any calculations you do use the rms a.c. voltage minus the diode drops, all multiplied by your PWM fraction. No huge peak currents, no bulky expensive capacitors. I would be inclined to ignore the 8A switch-on, the transformer will handle it for the minute or so that's needed, and if you're worried simply monitor the transformer temperature and reduce the PWM or shut it off completely if it starts to get hot.

author: Rufus
author_rank: Super Contributor
timestamp: June 17, 2013, 01:21:03 pm
content: Quote from: rwgast_lowlevellogicdesin on June 16, 2013, 11:41:57 pmThe finer control isn't going to buy you anything useful (for soldering anyway) and you are seeing the drawbacks of a chopped DC drive.How long does the iron take to get to some temperature? Divide the temperature change by the number of half mains cycles in that time and that is roughly how 'fine' your control will be using a triac or something to switch complete half mains cycles.

author: gxti
author_rank: Frequent Contributor
timestamp: June 17, 2013, 11:51:31 pm
content: If one did want to go the route of using a switching "rectifier" instead of a rectifier+capacitor, how does one go about implementing it? I know that's how "active PFC" stages work, but those are meant to run at mains voltage. Do you still use the same ICs, or would a conventional boost converter work well enough if you don't care about power factor?

author: rwgast_lowlevellogicdesin
author_rank: Frequent Contributor
timestamp: June 18, 2013, 03:38:22 am
content: Active PFC stage?, as in the positive temp coefficient thermistor? The hakko heater is designed for 24v AC, but DC will heat it just as well. The rest of the chips and sensors are designed for low DC voltages. The PTC can be monitored in many different ways with a 5 or even 3.3v supply.As far as using DC PWM instead of Phase Angle AC, you may be right but this is the jist of the experiment.The heating element will handle 30v just fine, Im not concerned with regulating the rectified AC. I just want it to smooth out to at least 100mv of ripple...In the equation given by Mariush, the voltage drop is the same as the ripple voltage number. At 3v drop/ripple a 3333uf cap is fine, my problem is if you want somewhere around 100ma of ripple the cap size grows enormously to 110000 farads!!!Im thinking a switching regulator (or LDO) is going to be the best bet if one wants a fairly decent DC signal?

author: mariush
author_rank: Super Contributor
timestamp: June 18, 2013, 03:54:09 am
content: In the equation given by Mariush, the voltage drop is the same as the ripple voltage number. At 3v drop/ripple a 3333uf cap is fine, my problem is if you want somewhere around 100ma of ripple the cap size grows enormously to 110000 farads!!!You probably mean mV ... and yes, you'd be right. You'd need close to infinite amount of uF to have the output of the transformer perfectly smooth DC line. That's not how it's typically done.Too much capacitance causes problems, the capacitors behave as a short circuit so for a few us-ms the transformer and the bridge rectifier will see huge current peaks which can blow fuses if you're not careful.You put as much capacitance so that the voltage drop will be smaller than the minimum a linear regulator needs... in your case 24v + about 2v, so you must not let the voltage drop under 26v.If you want as low ripple as possible, there's linear regulators for that, which use the linear mode of transistors and a feedback loop to constantly adjust the output voltage and keep the voltage ripple very low.But again... for this application you really don't need fixed 24 volts...

author: richard.cs
author_rank: Super Contributor
timestamp: June 18, 2013, 09:09:15 am
content: rwgastWhy do you want less than 100 mV ripple? On a resistive load with a thermal time constant of many seconds you can feed it pretty much any waveform you like and only the rms value will matter.gxtiAssuming you're meaning what I suggested to drive a resistive load a conventional switching IC would work so long as you give it a separate power supply for its internal logic and the feedback network is slow enough that it doesn't try to follow the ripple, or not too much anyway. If you give the heater 23% pwm of 24V dc or 23% pwm of any other waveform that is 24V rms it will perform exactly the same.Getting offtopic a little, if instead you are trying to generate a regulated voltage rail with minimal input capacitance you simply build a conventional switcher but calculate the input capacitor to be just big enough that you never fall below the your minimum input requirement. So you might be making 12V with a buck regulator that can accept 15-50V in. You would then size your transformer for nearly 50V peak, say use a 30V transformer, then size your capacitor for about 25V ripple. The buck regulator then sees its input swinging between 20 and 45V say and can always produce 12V output. You only need a small amount of capacitance to ride through the shor time the sine wave it under 20V, but it needs to survive large ripple currents.