trying to figure out what wattage rating I need for a certain resistor

Discussion in 'Power Electronics' started by loosecannon, Jul 16, 2016.

  1. loosecannon

    Thread Starter New Member

    Aug 13, 2013
    12
    0
    Hi all.

    I need to learn a little electronics math.

    Here's what i'm trying to do:

    I am supplying a circuit 14 volts DC supply voltage. This circuit will draw up to about 10 amps peak, and will draw around 4 amps continually while in use.

    What i would like to do is reduce the voltage to this circuit to around 12 volts DC or so (not critical) using a resistor in series with the power input.
    I cannot reduce the voltage at it's source as it is powering other circuits that do need the 14 volts.

    so i am looking for the value of the resistor i might want to use, and how many watts it will need to handle.

    something tells me i may be going about this the wrong way by trying to add a resistor in series.
    If that is true, will someone please show me the error of my ways, and a better way to accomplish my goal.

    thanks for any and all input.
     
  2. SLK001

    Well-Known Member

    Nov 29, 2011
    810
    224
    Don't do it this way. A resistor will have to be 14Vx10A watts to handle your power. You don't want to use resistors as voltage regulators.
     
  3. DickCappels

    Moderator

    Aug 21, 2008
    2,647
    632
    2 volts/10 amps = 0.2 ohms, power loss would be 2 volts x 10 amps = 20 watts. An example is at the URL below.
    http://www.digikey.com/product-detail/en/vishay-dale/TMC050R2000FE02/TMC50-.20-ND/269957

    Perhaps a better solution is to put this (URL below) 15A rectifier in series with your load.
    http://www.digikey.com/product-deta...onductor/FFPF15S60STU/FFPF15S60STU-ND/1755362

    Regardless of the method chosen you will probably have to heatsink the part in series with the load.

    There may be other considerations, such as the needs of the load in terms of regulation and impedance of the input voltage. If you can tell us more about the power source and load we might be able to give more extensive advice.
     
  4. crutschow

    Expert

    Mar 14, 2008
    12,986
    3,224
    Dick has a good idea but you need two diodes in series, which will drop close to 2 volts @ 4A.
    For that you could use two legs of a diode bridge such as one of these.
    Also as Dick noted, you will need to mount it on a heat sink capable of dissipating about 8W.
     
  5. loosecannon

    Thread Starter New Member

    Aug 13, 2013
    12
    0
    thanks for the responses guys.

    the load for this power is a MOSFET transistor in amplifier mode.

    This power feeds the drain lead through an inductor to keep any RF off of this line.

    I am finding the mosfet running hot, and am attempting to lower it's gain a bit by reducing the voltage supply to it.

    I am honestly a little embarrassed for the way i posed this question, as i am aware of ohms law, and had calculated previously that i would need a 140 watt resistor. Obviously this seemed like a ridiculous value for my circuit, which should have told me right away that i was attempting something dumb, but i just thought i was misusing the formula or forgetting some factor.

    I would love to hear of any other ideas on how to best accomplish my goal.

    I had considered the diode idea, but thought that i would need about three of them in series to get close to a 2 volt drop.
    also they would all have to handle 10 amps. again, i feel like im barking up the wrong tree.

    It sounds to me like i am going to need to build a voltage regulator circuit, and the fact that it will have to handle 10 amps means it will have to be quite robust.

    i am open to any and all ideas.
    thanks again!
     
  6. DickCappels

    Moderator

    Aug 21, 2008
    2,647
    632
    The diode I referenced is specified to drop no more 2.3 volts at 15 amps so one might be enough.

    The power dissipation in watts is the number of volts lost across the device times the current in amps.

    Is the amplifier one you made yourself?

    Maybe a better question revolves around your MOSFET overheating and whether the circuit itself is what needs to be addressed. Most MOSFETs that I have used have maximum operating junction temperatures of 150°C, though for reliability this should be derated. That would correspond to a case temperature that you be very painful.
     
  7. DickCappels

    Moderator

    Aug 21, 2008
    2,647
    632
    Whoops! I missed the "4 Amps continuously part". That explains why you would probably need two of the diodes, as crutschow pointed out.

    Still, the question might have become those of whether there really is a problem that needs to be solved and whether the power supply voltage is the best place to deal with the problem. Let us know if you want assistance looking further into this.
     
  8. loosecannon

    Thread Starter New Member

    Aug 13, 2013
    12
    0
    wanted to thank you guys for the responses.

    I do realize that i should be trying to tame the MOSFET at it's source, and should really be looking at the bias circuit and not the power supply.

    This is a logistical thing, as i cannot access the parts in the MOSFET amp without completely disconnecting it from the circuit.
    this means that i have to guess at parts values every time i take it off and put it back.

    This little amp is not going to stand for this kind of abuse for too much longer, so i thought i might try sourcing it with less voltage since i can change that while the amp is hooked up.

    I think i will try the rectifier that was suggested before that will handle 15 amps. sinking it will be no problem, as i have the space and aluminum fins to make that happen.

    thanks to all for the responses so far.
    all suggestions are welcome.
     
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