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supplying power to 12v 1.4a, 3v 3a with a diy power supply


chylld

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heyas,

i have 3 motors that draw a maximum (under load/stall) of 1A@3V. is it possible to run all of these simultaneously off a 12V 1A ac adapter? i figured that the motors will be drawing 9W total max which is under the 12W the ac adapter can deliver, but i'm not sure if it works this way.

i plan to drop the voltage to the motors using some kind of 1A-rated regulator arrangement.

if it's not possible to draw 3A@3V off a 12V/1A ac adapter, how do i best go about this?

thanks in advance  :)

edit: topic changed from "does an ac adapter's max current rating increase with decreased voltage?" to reflect the latest content of this thread

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ahh ok, so when a regulator drops the voltage, it loses it in heat, and the current stays the same?

is the current drawn at this stage equal to the current drawn from the mains? i'm pretty sure drawing 3A@3V means less than 3A at 230V AC! :) i take it that once the inefficiency of the power supply is taken into account, the power(W) drawn at the 3V DC end will be roughly equal to the power drawn at the 230V AC end?

unfortunately the closest thing i can find to a 5V/3A adapter is a 5V/2.5A adapter, which won't be able to sustain all 3 motors at the same time, so it looks like i'll have to build one myself. as seems to be common amongst first-time power supply builders, i'm having trouble picking the right transformer. with this one:

# Secondary Output: 0 - 6.3 - 7.5 - 8.5 - 9.5 - 12.6 15V
# Secondary Current: 2A


does this mean that i can pull 6.3V at 2*15/6.3=4.7A? i keep reading that transformers are rated on VA, so is the VA for the above transformer 15*2 = 30VA? also, would i be able to take 6.3V off it in one circuit and then 9.5V off it in another circuit at the same time? (i have 300mA worth of other things that i wish to run at 9V)

i know i'm asking a lot but i've honestly tried all day to better understand these things, i'm slowly coming to grips with it so please bare with me for now! :)
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Your multiple output voltages transformer is overbuilt and probably not suitable for your requirements.
Its secondary current is rated for only 2A max probably due to the small size of its secondary wire. At 6.3V it can pass only 12.6VA.
If you rectify and filter its 6.3V then the peak voltage would be 8.9V. The 12.6VA is divided by the 8.9V peak to result in only 1.414A continuous DC output. The rectifier diodes have a 2V drop and therefore the filtered DC output voltage will be 6.9V.
The voltage is too high and the current is too low.

Depending on how the transformer is made then you might be able to use the remainder of its secondary to power other things. Maybe you could get 7.5V between its 7.5V output and its 15V output. If the secondary has layers of turns on top of each other then it would overheat.

What is cooling your 3 stalled motors? 9W is a lot of heat if they are small. Maybe you need a regulator with foldback current limiting then a transformer with a lower current rating will be fine. ;D

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ahhh ok... that makes a lot more sense. i managed to find another type of transformer, it's more expensive but it looks like it'll do the job:


TYPE 60VA TRANSFORMER
VOLTAGE@ CURRENT
[email protected], [email protected]
12V@ 5A, 21V@ 2.9A
15V@ 4A, 24V@ 2.5A


so using your maths, if i take power from the 9v "bit" the peak voltage will be 12.73v and max continuous current 60/that = 4.7A, which is more than the 3A i need.

i think regulating this voltage down to 3v is a bit of an ask for the regulator, but the thing is the motors will only be used for a few seconds at a time, and usually one at a time, under normal conditions which is somewhere under 1A (and over 300mA). so the regulator won't be working continuously which should help it from burning up?

also, can i power my 9v 300mA load from that same transformer? should i take the power off in parallel just after the transformer (and fuse)? or after the smoothing capacitor?
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Hi, your maths are correct. ;D

Why use an expensive transformer that has a continuous power rating of 60VA and many voltages up to 24V when you will be using only 9VA for a few seconds at a time at only a few volts?
Why use a voltage regulator for your electric motors when their speed and torque depends on their load?

You could take an inexpensive 6.3V center-tapped transformer (3.15V - 0V - 3.15V) that is rated at only a few VA. It will use only 2 rectifier diodes that reduce its 4.45V peak voltage to 3.45VDC. The smoothing capacitor will be big enough to power your 3 stalled motors for a few seconds at a time. The transformer will be big enough to have pretty good voltage regulation, with a small load the DC voltage might increase only to 4V and your motors won't care.

Your 9V/300mA requirement must be analysed separately. Does it need voltage regulation?   

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sorry for the delayed reply audioguru, my internet conked out and only came back this evening.

before it went out though i did as much reading up on transformers as i was able to, and in the downtime i think i managed to figure out how they work, or rather, how i can use them. like... if i get it properly, the secondary coil in many transformers is divided up, each giving a portion of the total voltage. an amp-rated unit (as opposed to a VA-rated unit, like the 60VA unit i mentioned earlier) delivers its amp rating across secondary coil voltage range, e.g. 6v @ 1A for the 0-6v portion, or 12v @ 1A for the 0-12v portion. and if the 0-6v portion is wound adjacent (as opposed to on top of) the 6-12v portion, then 1A can be taken from each of the parts simultaneously, since the current limit is mainly based on heat?

also, if i take off from the 0v and 6v 'take-offs' from the secondary coil on the transformer, what AC voltage will i actually be reading there? from your calculations (and what my mechatronic engineering friend tells me) it'll be root(2) times that =~ 1.414*6v = 8.484v; and then minus 1.4v for each of the 2 diodes that the current passes through. however the explanation on this page (towards the end) suggests that taking off from the 0-12v portion will give 12v-1.4v=10.6v AC, i.e. no root(2) peak. what gives?

regarding my 9V/300mA requirement - it has a couple of main duties:

1) power a timer circuit (2x 555 IC's, 22mA draw) which trips a relay (+35mA)
2) power 3 fans which draw ~90mA @ 9V

i didn't mention the relay before so the peak draw of this circuit will be around 327mA. i also decided to bump it up from 9v to 12v, point 1) above accounts for this and point 2) will be achieved through the use of 4 in-series 1n4004 diodes.

the thing is, there's another device that will be incorporated into this project that i bought commercially, it runs off an unregulated 12V 1A ac adapter which i figure i can replace with my own unit. so, i found this transformer for less than AUD10:


General Purpose AC/AC Transformer 12 - 30V @ 2 Amps
Secondary output: 0-12-15-17.5-20-24-27.5 and 0-30 volts.


what i plan to do is:

- take off 0-12V, rectify and smooth that (470uF) and give the smoothed output to the device originally running off its own ac adapter. the peak sustainable current for the 0-12V bit is (12*2)/(12*sqrt(2)) = 1414mA, 1000mA of which can go to this thing, the rest can go to the next, which is...
- feed the smoothed input also to a 7812 (12v 1A) regulator (with 0.1uF capacitors either side) and have that power the 12v circuit above (which in turn powers the fans). this bit draws around 327mA peak as calculated above.
- take off 15-17.5, 17.5-20, 27.5-30 (i.e. 3x 2.5v each) and rectify and smooth those and dump that straight onto the motors. i figure it'll give sqrt(2)*2.5-1.4 =~ 2.1v which is ideal because i think 3v will be a bit too much for what i want it to do (also the specified voltage range of the motor is 1.5-3v). i'm taking your advice here and not bothering to regulate the power going to the motors, it's overkill and quite silly actually!

so... how am i doing? :)
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That multi-output voltage transformer is designed to provide a single output. If you take multiple outputs from it then maybe its primary might be overloaded.
1) 12VAC when rectified and filtered is barely enough input voltage for a 7812 regulator. If the transformer or the mains voltage is a little low, and the 7812 is a little high then it won't work!
2) The voltage drop of a rectifier bridge is about 1V for each of its 2 diodes in series. There will also be quite a few volts lost in the main filter capacitor due to ripple, since 470uF is much too small for your high current requirement.
3) Your "page" shows a rectified output without filtering. Its voltage fluctuates wildly from 0V to the AC peak voltage minus the rectifier drop. Their stated voltage is its average.

Good luck. ;D

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hmmm i thought that taking off from the 12v ac part of the transformer would be ok? i read on this page:


Note that smoothing significantly increases the average DC voltage to almost the peak value (1.4 × RMS value). For example 6V RMS AC is rectified to full wave DC of about 4.6V RMS (1.4V is lost in the bridge rectifier), with smoothing this increases to almost the peak value giving 1.4 × 4.6 = 6.4V smooth DC.


so 12v ac in = 12-1.4 (rectifier diodes) = 10.6, 10.6*1.4 (sin peak) = 14.8, and smoothing to within 10% of this peak voltage would result in a minimum of 13.3v which should be enough for the 7812? i do understand that feeding less than 12v to the 7812 is pointless :)

one other thing that's been bothering me is the size of the smoothing capacitor, as you pointed out it's way too small. so i used the formula (from the same page) Smoothing capacitor for 10% ripple, C = (5 x Io)/(Vs x f), where Io = output current, Vs = supply voltage and f = frequency. so C = (5 x 1.4)/(14.8 x 50) = 9500uF! isn't that quite big? or am i doing something wrong there?

overall i'm getting much closer to a final solution though  :P
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The 7812 has a maximum dropout voltage of 2.5V at 1.5A, 13.3V is far too low to ensure the output will be 12v.


ahh ok didn't know that. if i take off 0-15V AC, that'll give me 17V minimum (10% ripple) which is ok for the 7812, right? for that i'll need a 7300uF cap...

also, assuming it's ok to use the 0-30V 2A transformer i pointed out earlier - did i mention how HUGE that thing is?? it's like 8cm x 7cm x 7cm! why is it so big? is there a 'smaller' way to make a power supply? (cheaply)?
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Hi Chylid,
You used 50Hz in the formula for the filter capacitor. The full wave bridge rectifier doubles the frequency, so a capacitor half the value of the one you calculated will be fine.

The rectifier diodes conduct only when the voltage of the rectifier exceeds the capacitor's voltage, for each half-cycle of the mains. The filter capacitor is powering the load the whole time and is charged with a massive current (maybe 10 times the load's current) for only a short time. The massive current makes-up for the short duty-cycle. With 14A short-duration pulses in the rectifier diodes, their voltage drop will be way higher than the 0.7V at much lower currents, maybe 1.25V each.

Also, with 14A pulses coming from the transformer, its voltage will be "flat-topped" and won't reach its peak. Maybe another volt of loss.
It looks like 0-15VAC will be a good source for the 7812. ;D

If you try to use a 0-30V transformer for a 7812 then its rectified and filtered voltage of 41V (with low output current) far exceeds the absolute max input voltage rating of 35V of a 7812.
Also, with the loaded input of the 7812 at about 38V, then at 1.4A the power dissipated by the 7812 would be 36.4W which far exceeds its absolute max limit of 20W if it had an infinite heatsink. :(

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so it looks like i'll need a 3650uF cap - it's all starting to sound a lot more sensible, thanks!

by 0-30V transformer i meant the transformer thing itself (in general), not the 0-30V portion of the secondary coil! sorry i didn't make that clear. i'm stupid but believe me i'm not that stupid :)

so that pretty much makes things happy on the 12v side of my project, now back to the motors - assuming i can rectify some ac source (be it the same transformer or another one), is it ok give them unsmoothed output? or would it be better to give them smoothed output? the cap i'll need for each will be roughly (5 x 1)/(2 x 100) = 25mF - !?

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