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12V Adapter

  1. Oct 22, 2012 #1
    My router has an adapter with the output: +12V 0.7A

    And it does not work anymore. I have another adapter which has the output: +12V 2A.

    Is it okay to connect this adapter to the router? What does 2A and 0.7A indicate? Should the adapters have the same output impedance?
  2. jcsd
  3. Oct 22, 2012 #2


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    The router needs 0.7A. Your new PSU will provide 'up to 2A'. It will not 'insist on' putting more current than 0.7A into the router. It will be fine - as long as you check the polarity of the connector (+ and - on the correct pins).
  4. Oct 22, 2012 #3

    What happens if i connect the adapter to a resistor with resistance less than 6 ohms? The voltage drops from 12V?
  5. Oct 22, 2012 #4


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    You will be overloading the PSU. It may just overheat or it may have a cutout. Not worth trying unless you can afford to chuck it.
    Some of these PSUs are not very good (i.e. very cheap) and won't work flat out comfortably. But most devices don't actually (continuously) consume as much current as is marked on them .
  6. Oct 22, 2012 #5

    I want to take the opportunity and ask a relevant question.

    If I want to make an AC-DC converter of an output DC voltage of 6V, how can I figure out the maximum current that the circuit can supply? why is there a difference in current of the two adapters, one with 0.7A and another one with 2A, although they have the same output voltage?

    What happens if the router needs a rating of 12V/1A? will the PSU overheat and may burn? if so, then how can I determine the current that a load device, a router in this case, needs to operate, assuming that I don't have the current rating?

    Sorry for the too many questions, but I'm still confused about all of this!
  7. Oct 22, 2012 #6


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    You are jumping in with both feet, here I think. Think of a PSU as a battery. A 1.5V D cell will provide much more current than a 1.5V AA cell of the same type. Same volts - different current rating.
    You can find out the current taken by your router (or whatever) by putting an ammeter in series with the power lead. The plugs are moulded on, of course - so, if you want to do this non-invasively then you would need to make up a special short lead with a socket and plug and two wires. You can then put the meter in series in one of the wires,
    It will depend on the current carrying specification of various components - mainly the transformer. In 'wall wart' PSUs, everything is pared down to the minimum and their transformer windings are very thin wire so it can get hot just because of the resistance. An off the shelf transformer will cost at least as much as a suitable Power Supply. It's like trying to build a car at spare part prices. Sod's Law. Building your own PSU is bad value these days, btw. Building Anything yourself is only worth it for the experience and PSUs are the most boring things to start with.
  8. Oct 22, 2012 #7
    Is that how electronic manufacturers measure the current needed for an electronic device to operate?

    Let me assume the following scenario. I made a well-constructed electronic circuit and I want to power it using an AC-DC converter it so that it can operate. I have the following questions:

    1- How can I determine the voltage needed for the circuit to operate?

    2- How can I determine the current needed to supply the circuit so that it can operate?

    Assuming that I figured out the voltage and current ratings for the circuit to operate, how can I make sure that my AC-DC converter will be able to supply the current that's needed to supply the circuit?

    On the AC-DC converter, let's assume it starts with a step-down transformer, going through bridge rectifier, all the way through Zener Diode for shunt regulation.
  9. Oct 22, 2012 #8


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    If you made (i.e. Designed) such a circuit then you would not need to be asking those questions!

    You would buy a suitable transformer that was rated for an appropriate secondary voltage and current (you do not want to go down the road of designing one of them too - trust me) You would pick suitable diodes with an appropriate current capability and reverse breakdown voltage. You would also choose an appropriate reservoir capacitor which would limit the amount of ripple, for the current you required.
    Except if you wanted to power a very low power device, you would Never use a shunt regulator - certainly not a Zener, as it would need a series dropper resistor and it would dissipate so much power all the time.

    Like I said earlier - you are trying to go too far, too fast with this. Without some basics, you have no hope of starting from this point.
  10. Oct 22, 2012 #9
    Basics? If you mean the introductory circuit theory, or the advanced one which covers Laplace and Fourier, and in addition to the first Electronics course which covers diodes, BJT and MOSFET, then I have had all of those covered. Now, I'm going through digital design and an introductory course into communications. However, I still don't see that we've covered all the details for the previous courses, so I'm eager to learn them by myself.

    For the first course of electronics, we were tasked to design AC-DC converter with a given output DC voltage. It was too easy, but I'm still confused about the current rating of the converter, and the different current ratings of the various electronic devices out there.

    Still, I don't see that I've been given a direct and clear answer on this question. How can I figure out the maximum current that an AC-DC converter can supply?

    On step-down transformer, does the current rating indicate the maximum current that it can output into the circuit?

    Just these two questions and I will be thankful.
  11. Oct 22, 2012 #10
    If im guessing correctly, the AC-DC converter you were tasked to design was using rectifying diodes and then perhaps a smoothing capacitor. In that case, your current rating would be up to the AC source. However! There would of course be a limit depending on the diodes (which would have a maximum current they would be able to handle, before they break down).

    However, if you're dealing with SMPS (Switch-Mode Power Supplies), it will depend on your design. It's not really anything basic, so if you wanna know about it, you'll have to look up on SMPS. However, the answer to the question is that it will depend on your design and components.

    So you would need knowledge of MOSFETs, BJT's and diodes. PSU's are advanced stuff. Wait untill 3rd year :)
  12. Oct 22, 2012 #11


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    If you know what you say you know about all those topics then did you miss out on the earlier stuff about transformers (turns ratio, VA and the like)? Did you not do RC time constants and the Maximum Power theorem? Apart from the more arcane aspects of transformer design and specification, all the rest of PSU design is really basic stuff. It just needs a bit of application and sweat to work out everything but the transformer bit - and you just buy an appropriate one when you've decided what current you need. Why not look at some of the many designs of power supply on the web and see how they're done? Look at the parts list for the specs of the components and relate the values and specs to what you already know. You say your power supply design exercise "was too easy" but it clearly wasn't or you'd not have a problem here. In your design of power supply, did you calculate ripple? Did you really use a shunt Zener regulator and did no one point out the problems with it?
    You are asking for a "direct and clear answer" when I think what you are wanting is a specific design. There is no "maximum current" as such. Component specs depend on the particular application and the specific requirements. Remember, you can't make your transformer 'too beefy' as long as its secondary volts are right.
  13. Oct 23, 2012 #12

    Yes, my design was based on the 4 didoes configuration, smoothing capacitor and zener diode for shunt regulation. I used the famous 1N4007 diode for rectification, which breakdowns at a peak reverse voltage that exceeds 1000V according to its datasheet
    ("tinyurl.com/9264lrg" [Broken]).


    I covered all of that about transformers, RC constants and the rest.

    On the converter, I calculated the ripple voltage and all the needed calculations to guarantee an output DC voltage as required, 4.3V for me. For the zener, we were given its type and the test current, but in the design, I really had a problem which you can see (here). You even posted in the topic suggesting a three pin voltage regulator.

    Anyway, I will do AC-DC experiments to get things connected in my mind. That's the best way!
    Last edited by a moderator: May 6, 2017
  14. Oct 23, 2012 #13


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    The ratings of the transformer determine the voltage and current it can supply

    so if you want a DC power supply capable of 2A at a given voltage then you will be choosing a transformer that can supply a minimum of 2.5 - 3 Amps. You need some headroom so the transformer isnt running at max ratings else its gonna overheat and die.

  15. Oct 23, 2012 #14


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    Oh yes - I remember. And did your design work in practice? The point is that the ripple can be affected by the transformer if you go for the cheapest. Also, did you not learn from that exercise how much of a problem the Zener would be? I was not daft when I suggested using a proper regulator. But, as I now see, your design is hardly a 'Power Supply' if it has such a massive series resistance. Certainly no use for feeding a real world piece of electronics.
    Last edited by a moderator: May 6, 2017
  16. Oct 27, 2012 #15
    Yes, it did. I got the desired DC output, 4.3 V, of course not exactly, only with a difference in hundredth.

    On the zener, no, I don't know about how much of a problem it would be if it's used for regulation. Can you explain further on zener and about the problem of the series resistance?
  17. Oct 27, 2012 #16


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    The zener curcuit is fine if all you want is a few mA as a low power source but, if you want a lot of current, a series resistance will just get hot and waste power. Remember, it has to be passing the same total current (plus some) that the device you're driving will need (max), all the time and the zener needs to have the capacity to drain all that current when your device is taking very low current.
    A series regulator needs to dissipate virtually no power at all when on a low load and will be designed to be dropping as few volts as are practical (depending on how much you spend on the transformer, the secondary volts etc, etc,) to leave some in hand so that at the lowest point in the ripple, it is still a finite amount above the required output value. Altogether, it's a much more healthy approach - which is why it's always used!
    Using a shunt regulator, there is no point in using a particularly good transformer because you need loads of series resistance in any case (and that may as well be in the windings as any added series resistor).
    Power supplies are supposed to be voltage sources (of course) and to have as low a source resistance as possible and to supply as much current as may be needed without dropping volts. If you use a zener to do this, this high current needs to be passing all the time and is just shared between zener and load. Nasty nasty! You cannot just connect a zener across the source (rectifier or whatever) because it will just blow up if source volts are only just above zener voltage.
    Imagine the plumbing situation in a toilet, if you wanted to keep a cistern constantly topped up and had a constant supply flowing into it and just used an overflow to define the level of water in the cistern; that's like the zener solution. A ball cock that turns off when the cistern's full is the equivalent to the series regulator. Much cheaper to run and more elegant.
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