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Can this be done?

Discussion in 'General Electronics Discussion' started by mkp, Mar 30, 2013.

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  1. mkp

    mkp

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    Mar 30, 2013
    I'm looking for a transistor device (I think), but instead of turning on when the gate gets above a certain voltage, it will turn on when the gate voltage drops below a certain voltage (ideally 3V or so).

    Basically, I'm looking for a low cost, low power, no voltage overhead way to drop resistance when the voltage drops to 3V. The easiest way I can imagine this to be done is a switch that would turn on when the voltage drops below a certain point to bypass a resistor.

    Any help would be greatly appreciated.
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    One way to do this is to have a transistor with a voltage divider going to the base so that it turns on when the input voltage hits about 3V (I assume it's not critical that it changes state at exactly 3V or does so with a very sharp transition).

    This transistor can then be used to turn off another transistor.

    Of course, we need to know something about that resistor you want to short, and whether one end of it is ground with respect to the 3V signal.
     
  3. mkp

    mkp

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    Mar 30, 2013
    Below is an image of a simplified circuit with this special transistor that turns on when the voltage drops below 3V effectively shorting R2 and reducing the resistance of the circuit.


    [​IMG]
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    What is the aim of this circuit?
     
  5. mkp

    mkp

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    Mar 30, 2013
    Unfortunately, I can't disclose any details as to why I need this but I think something like this could have many applications.

    For instance, say my source is a 12V battery that drops voltage with use (typical alkaline). At some point, the current running through the circuit has become too low for practical use. So, in order to bump up the current, we need to drop resistance.

    Thank you for thinking about this. Any input is greatly appreciated.
     
  6. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Have you considered a constant current source? Or a voltage regulator?

    If your application is secret, it's really hard to help.

    Placing resistors in series with batteries to power a generic "load" that requires a lower voltage is beginner thinking for the most part.
     
  7. mkp

    mkp

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    Mar 30, 2013
    I understand. I guess the quick answer is no it can not be done easily. I didn't know if I was missing something obvious (like a special magic transistor I haven't heard of).

    I could put a voltage regulator into the circuit but this bumps cost and complexity up an order of magnitude. If you disagree, please point me to one that can output a constant 4V from a source that will decrease from 6V to 2V. Current is about 120mA. Cost should be less than a $1 for parts.

    Thanks, and yes, I'm a novice at best.

    Kind regards,
    Mark
     
  8. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Well, a resistor won't work when the input voltage falls below the desired output voltage.

    It sounds like a voltage regulator is what you need, but you have a very tight constraint on cost.

    If your budget stretched to $2 then you could use something like this.

    It would be much more efficient than a resistor, and would allow you to use a higher voltage input source and retain efficiency.

    You could replace the entire set of resistors with a transistor, a resistor and a zener diode -- this would require the input voltage be at least 0.7 volts higher than your desired output voltage, but would be very inefficient at both low currents and higher input voltages.

    If you're using a 12V battery to power a 6V load, the regulator I pointed to you would be ideal. In addition to it maintaining the voltage at a constant value (until the input falls to less than 6V), your battery would last much longer -- possibly up to twice as long.

    If we had some idea of the range of voltages your load can operate at and the range of currents it requires, we could make more recommendations.
     
  9. mkp

    mkp

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    Mar 30, 2013
    Steve, I have actually tested that exact unit (from amazon). On amazon it says as low as 3V input and Ebay says 4V. It works quite nicely and I haven't seen it for that cheap so that now becomes an interesting option. I'm curious now how much power this consumes. I will need to do some more testing with it.

    In regards to our application, we are making an LED light. We are considering 3 to 6 AA batteries. The load is several LED's in parallel (probably 12, 5mm LEDs or maybe 2 higher power smd). So, ideally we have them at 3V and about 10mA.

    Here is a typical battery performance curve. So if we are using 3 AA batteries, the voltage will start at 4.5V but then quickly go to 3.3V with a significant amount of energy still in the battery. If we use a resistor like most LED lights use, then the light will be super bright for about 2 hours and then start the slow decline. So, my thought was I could start with say 100ohms and then reduce it to 50ohms at some later voltage bumping up the voltage to the LEDs.
     
  10. BobK

    BobK

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  11. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    One issue is that you really shouldn't place LEDs in parallel.

    See https://www.electronicspoint.com/got-question-driving-leds-t256849.html

    Since you seem not to need a particular number of LEDs, I would suggest a single high powered LED connected via a current source to your battery.

    For multiple LEDs you may want to consider placing them in series and using a boost SMPS to raise the battery voltage..

    It all really depends on whether you need 1 LED or many, and how bright it needs to be (i.e will it be 10mA, or 250mA)

    Also, knowing the colour of the LEDs you plan on using (presumably a single colour) would be helpful.
     
  12. mkp

    mkp

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    Mar 30, 2013
    Steve,

    That is a good write up. Thank you for putting that together.

    To simplify our discussion, let’s assume we are using this white light 5mm LED for the multiple, low power, LED light..

    I'm not sure I agree that putting 20+ of these LEDs in parallel with a single resistor is a bad idea. Go to Harbor Freight and buy the 24 LED light. It costs $2.60. If one of the LEDs is different and has thermal runaway and burns out, how is that a problem? There are still 23 left and the current increase is less than 5%. I believe for the most part the LEDs are operating at low current (about 2-5mA) so there is less likelihood of a problem. I bet that 99% of the lights last more than 2,000 hrs which is plenty of life time for such cheapos. The main problem with this type of light is that it is nice and bright for about 30% the life of the battery. That's what got me thinking that if I could reduce the resistors resistance and allow more current then I could get more usefulness out of the battery.

    The other option is to use a single LED such as the one in this unit. I was running it at 120mA and the light output was great (130° angle). It can also handle 300mA (blinding) which is the higher output mode (approximate, my multimeter caps at 200mA). Strange thing is that this unit has a complex circuit (I will post photos later) but I think it is only to switch between power modes. I tried varying the voltage to it and the light dimmed so it wasn't trying to give constant current to the LED. An ideal case here would be this LED with 4AA batteries operating at 120-160mA, maximum circuit efficiency and maximum depletion of the batteries energy.

    Bob, thank you for your reply. Can you elaborate on a JFET circuit that would work here? Do you a part number in mind?
     
  13. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    You might not be, but the manufacturers of LEDs are.

    The only exception to this is when all the LEDs are in very good thermal contact so that their junction temperatures track very closely. COB LEDs are an example of this.
     
  14. BobK

    BobK

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    The circuit is in the link I gave you. I don't know how practical it is, or what part to use, but it has only a JFET and a resistor. Try googling to find more info.

    Bob
     
  15. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    The JFET solution only works for currents up to Idss for the device. If you're lucky you can get 20mA. Many jfets have a much lower Idss. It is also very variable between devices, so while a collection of jfets you buy might have all very similar Idss, the absolute value could vary over a large range.

    There are ICs designed to produce a constant current, as well as "diodes" that give a constant current (these are jfets with 2 leads!). In both cases you'll get more consistency and (in the case of the IC) higher currents.

    conclusion: the jfet circuit is really only suited to low currents.
     
  16. mkp

    mkp

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    Mar 30, 2013
    So the LED in the Rayovac head lamp was a Cree XLamp XR-C or E (looks similar). The price of these in bulk are around a $1 each. Combine this with the LM2596 DC-DC converter and we have a good solution. It would be cool to have a custom dial for the trimming pot to adjust intensity and somehow restrict it to a safe range. A thumb dial would be perfect like those on old portable tape players which turned the unit off/on and controlled volume (0-10). Any idea on where to get those? I looked through mouser.com briefly but no luck.
     
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