Undefined resistor wattage in an application note

Discussion in 'General Electronics Chat' started by DMahalko, Jun 23, 2012.

  1. DMahalko

    Thread Starter Active Member

    Oct 5, 2008
    Is there some generic resistor wattage to use, for when a circuit doesn't specify it?

    I've pretty much settled on the TL783 as doing what I need to do.

    Figure 19, page 10 looks like a 6 amp regulator circuit:


    (For my case the input would be bridge-rectified AC, and the 3.3 kOhm changed to 8.2 kOhm for 125v DC output. An AC line isolation transformer will be used on the input side.)

    I can read schematics, but this app note leaves out some important details, such as the wattage rating of all but one resistor.

    Can I just assume 1/2 watt is fine if the schematic doesn't specify it and there's no general value listed elsewhere in the application note?
  2. jimmy101

    New Member

    Jun 23, 2012
    I would assume 1/2 watt is OK for all unspecified resistors. The only reason why that one resistor has a wattage spec is because it is unusual.

    Of course, to be a bit more precise, you could always calculate the currents and voltages for all the resistors to make sure they are less than 1/2 watt.
  3. Dodgydave

    Distinguished Member

    Jun 22, 2012
    As the spec of the regulator says its max o/p current is 700mA , the 10 Ohm resistor should be 5W min>>

    W = I2 R
    W= 0.7 X 0.7 x10
    = 4.9W
  4. DMahalko

    Thread Starter Active Member

    Oct 5, 2008
    From looking at the TL783 specification, it technically doesn't need anything more than the two bottom resistors to function properly. The 3.3 kohm on the ADJ side sets the reference voltage, and the 82 ohm provides a minimum power draw to keep the regulator stable.

    The upper two transistors and three resistors appear to provide supplemental current that follows the output of the regulator. There's probably a name for that transistor/resistor component arrangement but I am not an EE so I don't know just by looking at it what the arrangement is or does.
  5. Jony130

    AAC Fanatic!

    Feb 17, 2009
    The general rule says that all resistors are 1/4W. And if resistor need to have a higher wattage. Then we put this on the diagram.
    So all resistor are 1/4W = 0.25W except 3.3K resistor that need to be 1W resistor. P = V^2 / R ≈ 50^2/3.3k ≈ 0.76W so you need 1W resistor.
  6. ifixit

    Well-Known Member

    Nov 20, 2008
    Hi DMahalko,

    The TIPL762 handles most of the load current. You cannot get 6 Amps from the output without severely overheating the 762. Depending on how good your heat sink is you can expect to get 0.5 Amp max. for the 50V output and 125V input.

    Conservative design...

    For 125VDC input and 50V out the delta is 75V. Theta JC is 1.25C per Watt. If we limit the junction temp to 105C and the case temperature can be held at or below 60C then the max watts is 105 - 60 = 45C / 1.25 = 36 Watts. With Vout = 50V then I max is 36 / 75 = .48 Amps

    For 170VDC input and 125V out the delta is 45V, which is a better situation. Theta JC is 1.25C per Watt. If we limit the junction temp to 105C and the case temperature can be held at or below 60C then the max watts is 105 - 60 = 45C / 1.25 = 36 Watts. With Vout = 125V then I max is 36 / 45 = .8 Amps.

    Should the output ever get shorted to ground you will have 170V across a 125V device:eek: Perhapes a circuit like Figure 18 would be better.

    How much current do you need to get?

    Your 8.2K resistor needs to be 2W power resistor. Expect it to get hot.

  7. cork_ie


    Oct 8, 2011

    Sink resistor (3.3K in schematic) is now 8.2 KΩ.

    From the datasheet for the TL783 Vo=Vref ( 1 + R2/R1)

    From circuit proposed by OP:
    Vo = 125V R2=8.2K , R1=82Ω

    R2/R1 = 100. +1 = 101 , => Vref = 125(Vo) / 101 = 1.24 Volts. =>
    If R1 drops 1.24 Volts then R2 drops 123.76 Volts

    Current in R2= 123.76/8200 = 15mA => R2 will dissipate 1.89 Watt,

    R1 drops 1.24V so it will also have a current of 15 mA. and dissipate 18.9 mW.

    Is there any internal current flowing between the IN and ADJ terminals of the regulator which would cause even more current to flow through R2?
    Last edited: Jun 24, 2012
  8. takao21203

    Distinguished Member

    Apr 28, 2012
    Commercially 1/8W resistors have been used since the mid 1980s as standard resistor.

    Now largely SMD only.

    CFLs use 1/4W where possible and yes this is the standard to use if not something else is specified.

    Note at high voltage carbon film tend to become high-Z after a while.

    For this reason sometimes metal film is used or two resistors in series.

    Other standard sizes are 1/2W and 1W. You can get assortments for these on eBay!

    1W is pretty decent already.

    For the circuit you maybe don't consider the total power that is dropped accross the regulator.

    TO220 can archieve some 1W to 2W without cooling, and about 50W max. with good (forced) cooling. With a small heatsink you can get out about 5W.

    for 50V delta without cooling = 1/50 A = 20mA!

    What is the intended application??

    I would rather suggest switching regulator instead of expensive cooling system.

    Most high-power/high voltage circuits are at first built as prototypes, for instance to estimate heat developement. If resistors begin to smoke/discolor, they will require higher wattage! And in general it is desireable to keep heat devleopement at 50C , 70 to 80C absolute maximum.