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AC Laptop Adapter -> How to make 19v output into 5v?

Discussion in 'Power Electronics' started by tonydizzle, Oct 13, 2019.

  1. tonydizzle

    tonydizzle

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    Oct 13, 2019
    Hello everyone!

    I have what I feel has to be a simple question, but I just can't seem to work it out !!

    I have the following 90w 19v adapter, that I would like to convert to output 5v (at an increased amperage of course)

    I am assuming that, it being chinese and all, that I will be lucky to get 75W actual and the rest will be lost as heat, which is ok. That leaves the amp rating at 75/5=15a. That'll do.
    I think I might even be able to get away with more at it's only going to be pushing that current for 10-20 seconds at a time, on an on-off basis. Between this and the load, the circuit will be controlled by a mosfetted hold-down switch, so it will default to off.

    However.

    But I can't find the IC in this circuit. I also can't identify which resistor is responsible for this, and I have reached the point where don't know what I am doing.

    Is there any chance someone can spare a couple minutes to help me out?

    Here are some photos I took: imgur/a/UXr0KMC
    you'll see some damage to the tracing on the left - I will just have to solder that across I think, so that should be ok. I'll do it when I work out what else I'll need to repeal and replace.

    Thanks :)
     
    Last edited: Oct 13, 2019
  2. tonydizzle

    tonydizzle

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    Oct 13, 2019
  3. WHONOES

    WHONOES

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    May 20, 2017
    I would suggest using a secondary buck converter as being the easiest option rather than try to mod your regulator.
     
    davenn likes this.
  4. tonydizzle

    tonydizzle

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    Oct 13, 2019
    I actually agree with you to be fair - I have a buck on the way (ebay) that I'll be soldering a lenovo female slim tip adapter to the back of to accommodate the much better regulated/official lenovo adapter.

    This would essentially cut out the buck though, and make it so that I could carry effectively a laptop charger around, rather than a full setup.
    Especially if it just means replacing a couple resistors/rewiring the transformer, which, based on a video I found on youtube, seems onerous but not impossible.

     
    Last edited: Oct 13, 2019
  5. Alec_t

    Alec_t

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    The present output rating is implicitly 90W/19V = 4.7A. It is most unlikely a simple resistor change would safely allow any greater current, because the designed current capacity of any components such as the high-frequency transformer secondary, the output smoothing choke and power diode would then be exceeded.
     
  6. tonydizzle

    tonydizzle

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    Oct 13, 2019
    That is indeed the correct current output rating. So what you're saying is that I in fact have two problems:
    -how to reduce the voltage down to 5v, and
    -how to adjust the circuit to take ~15a; about 10 more than it was designed to handle.

    So then presumably rewiring the transformer (as in the video) is a no-go? Same for the choke - that too would get rewired. My problem is just that I'm not too sure how.

    Aside from the complexity, it's starting to look like it might be dangerous to use!

    What's the final verdict here? Too much for a beginner?
     
  7. davenn

    davenn Moderator

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    Sep 5, 2009
    reducing to 5V is easy with an external buck convertor as other have stated

    You are NOT going to get 15A out of a circuit designed for ~ 4.7A, it's not going to happen. It would call for a totally new PSU with components rated accordingly


    It's even beyond that issue, see above comment.

    Just buy a proper PSU for the required job


    Dave
     
    tonydizzle and thedoc298 like this.
  8. tonydizzle

    tonydizzle

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    Oct 13, 2019
    Dave, Alec, Whonoes,
    Gentlemen, your input is much appreciated. I will go find something properly suited to the task at hand.
    Shouldn't necessarily believe everything you YouTube these days I suppose.
    Thanks again - you might have just saved me blowing my whiskers off.
     
    davenn likes this.
  9. Bluejets

    Bluejets

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    Forget the laptop p/s and use say a 350W atx supply.
    You should be able to get your 15A @ 5V fairly easily directly out of a standard unit.
    Just need a bridge on green to black to switch the unit on.(PG ....Power Good)
     
  10. BobK

    BobK

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    Simply adding a buck converter might not work either. In a buck converter, when the switch is on, the power supply must supply all the current used by the load + some more due to losses. Thus the power supply will need to supply roughly 4 times its max rated current 1/4 of the time. This will most likely over stress it and may cause it to fail.

    Bob
     
  11. tonydizzle

    tonydizzle

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    Oct 13, 2019
    @Bluejets @BobK

    Just logged in to say that I'm not ignoring you gentlemen - just waiting for a step down buck to get here from eBay. I'm going to try and give that a shot, on the basis that I can put both circuits (i.e. the AC and the buck) in the same box.

    The ATX point is a very interesting one, although it makes the box larger and therefore somewhat more inconvenient.
    Ideally, with the ATX being what it is, I'd prefer to aggregate all the rails (or 5v + 12v at least) into one, ideally adjustable, channel. The end result being 5v at however many amps it works out to - like ~40 or something.
    I found an old Dell PSU, so I might give that a shot if the buck doesn't work, or if, as Bob says, it blows.

    Also Bob, your point is interesting but surely is just a math question?
    Let's assume my load is a simple heat generator like a radiator or toaster.
    All I would need to do is ensure that the total amps drawn by my load (taking into account ohms, etc.) is not more than the amount provided by the initial power supply, minus the inherent inefficiency of both the buck and the supply itself.
    So:
    90w PSU = ~81w actual. (10%)
    ->81w Buck = 70w actual (15%ish)
    Allowing 5w headroom, I should ensure that the heater does not draw more than 65w. Nor, indeed, more than the 20a rating on the buck.

    Regardless, I must wait for the buck to arrive before I can proceed, so will update you all in due course.
     
    Last edited: Oct 17, 2019
  12. Bluejets

    Bluejets

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    Dell atx may not work with simple PG jumper.
     
  13. tonydizzle

    tonydizzle

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    Oct 13, 2019
    Ok, so, quick update:

    The buck arrived. I plugged in the 90w AC laptop for my lenovo laptop into the buck's input and wired out a couple of 16 gauge wires. Due to terrible soldering on a male DC plug (I stripped the metal off the positive pin xD) I had to revise my initial design.

    Currently, its a male/female DC connector to the end of my heating element, which double as a switch (i.e. i pull the plug and it goes off). I have the switch already from an old PC case but doubt it will take the amperage of the circuit. Got a pack of 5 mosfets coming in from China. Will wrap those in a resistor and bind the mechanical switch to the gate - effectively creating an electronic (ish) switch.

    I had some challenges around the heating element, but that's because I confused the law of Ohm with the laws of men and thought it was breakable; it is not, lol.
    All fine now.

    I don't have a volt/ammeter, so just going by the colour of the element and the trimmers on the buck. An lcd meter is coming (no prizes for guessing where from), so I'll shove that just after the buck and rewire accordingly.


    On a tangentially related note: does anyone know of a ready-made circuit for a remote on/off switch? Basically, I want to be able to turn on the buck from the end of the electrical cable currently attached to the output. So essentially, when I pick up the heating element (the housing, not directly), I want to have a sensor/button on there that wirelessly communicated to the buck that it's time to wake up.
    I'm guessing this is unlikely, but still. A sensor that detects a change in air pressure automatically and does the same thing could also work, but I'd prefer the former, as that way I retain control.

    sorry for the length, but thanks all for your time and for helping me achieve my goals.
     
  14. BobK

    BobK

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    You are missing my point entirely.

    To make the math simpler, lets say your power supply can provide 5A at 20V, which is 100W.

    And let us say that your heater takes 15A at 5V for 75W.

    No problem, right, because The heater needs only 75% of the power your power supply can provide?

    The problem is that, when you use a buck converter to go from 20V to 5V, the supply is only providing current 1/4 if the time, and the current it has to provide during that period is 15A, which is way above the supply’s current rating. So it is within the power spec at 20 x 15 / 4 = 75W, but it is not within the max current spec.

    Now, it is possible it will work if the PWM frequency is high enough and there is enough capacitance on the supply output to provide the needed current, but this is bu no means guaranteed.

    Bob
     
    tonydizzle likes this.
  15. tonydizzle

    tonydizzle

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    Oct 13, 2019
    I was in fact missing your point entirely. I was under the impression that the buck would facilitate the increased current draw by relying on the fact that I am not pulling as many volts. Interesting, so the buck serves to reduce the voltage from X to Y basically, having little to no impact on the ability of the original supply to actually provide the required current?

    I see. It appears to be working, but no doubt is probably also stressing the capacitors more than they were designed to handle. I will keep and eye on it and see how I get on.

    Thanks for clarifying though - didn't realise that the buck wouldn't be able use the voltage decrease to scale up the current. Looks like I'll shortly be looking at a replacement AC adapter ! Might try the PSU idea.
     
  16. davenn

    davenn Moderator

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    Sep 5, 2009
    correct, it the buck ( or boost) regulator will just pass the current that the source can supply
    So if the source can only supply 5A max, then that is all you have to work with regardless of the voltage your buck converter is putting out



    As above .... there isn't a current to scale up, you have 5A max :)
     
  17. Muleywagon

    Muleywagon

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    Oct 17, 2019
    Old school? A voltage divider lowers the volts using resistors, and an amplifier increases the amps. An amp uses collector, emmitter and base to double the amps each pass, until the power is high enough to pass a cap or aaspecific resistance. An OP amp does not fire power circuits, that is done with solenoids. When you lower the volts, you have to ground each resistor separately, and if you amp one of the lower voltage circuits, your divider voltages will change across the circuit. You can hack anything once!
     
  18. davenn

    davenn Moderator

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    Sep 5, 2009
    Again, you cannot generate current out of nothing .... if you have a 5A supply, that is all you have to play with
     
  19. hevans1944

    hevans1944 Hop - AC8NS

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    Jun 21, 2012
    If you have some power (a few watts) to spare, a circuit could be designed that accepted 19 V DC and produced 5 V DC at a current limited by the available input power, not the maximum output current. So, assuming the maximum power is 90 watts delivered at 19 volts, then you should be able to design and build a power oscillator circuit that accepts 19 volts input and produces 5 volts output (under load) while consuming, say, 75 watts of power. That means 15 amperes at 5 volts (75 watts) is available for your load with the new circuit, while 15 watts is "wasted" as heat to power the oscillator, transformer, and voltage regulating circuitry. At 19 volts output from the original power supply, 15 watts is about 0.789 amperes and 75 watts is 3.947 amperes, the sum of which (4.736 amperes) is within the original 4.74 ampere maximum rating for the original power supply, providing 19 volts at 90 watts or 90/19=4.74 amperes.

    Throwing away fifteen watts to obtain seventy-five watts of well-regulated 5 V DC power may seem extravagant. It is, and no power engineer would do a new design that way. Instead, they are given an input voltage range (typically 90 to 260 VAC, a frequency range (typically 50, 60 or up to 400 Hz), an output voltage (typically 5 VDC) and an output current (typically 50 to 100 A), and perhaps two or more "auxiliary" power supply outputs at lesser current ratings. The actual "design" is pretty much "cook book" now, with standardized components used by everyone, all designed to fit inside a "standard" ATX-sized case and mount inside a "standard" desktop cabinet. Hobbyists have been quick to re-purpose these inexpensive power supplies for all sorts of other uses, and because designs are close to the limits of reliability, mean-time-before-failure (MTBF) has been decreasing. But so has cost to build and purchase. So most folks just brick failed PSUs instead of trying to repair them.
     
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