Voltage Divider wattage rating

Comments

• Franklin Posts: 4,747

  2007-02-09 19:34 edited 2007-02-09 19:34

  P = (E*E)/R (sorry, don't know how to make a 'squared' symbol) You can change the values of the resistors to limit the current across them.
  
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  - Stephen
• Dennis Ferron Posts: 480

  2007-02-09 19:41 edited 2007-02-09 19:41

  Ok so here's how you calculate that (assuming no load from the ADC input):
  
  Add the values of the two resistors together to get total resistance. The voltage drop across each resistor is going to be the fraction of the total resistance that each one is, times your total voltage. So one will be:
  
  V r1 = (680K / 900K) * 16
  
  And the other:
  
  V r2 = (220K / 900K) * 16
  
  Now that you have the voltage drop, you need to know the current. You could figure the current for each but since they are in series, the current should be the same, so you can just as easily find the current for the whole thing and it will be what you need. That will be total voltages divided by total resistance:
  
  Total Current = 16 / 900K
  
  Wattage will be current times voltage. So it will be:
  
  Watts r1 = (Volts r1) times Total Current
  Watts r2 = (Volts r2) times Total Current
  
  I can however tell you off the top of my head that the wattage you dissipate with your resistor values that high will be very, very small. In fact, if I were you I would even use resistors 10 times less ohms than what you have there - 68K and 22K, instead of 680K and 220K. Your resulting voltage division will act the same, but it will be less sensitive to loading and still not dissipate more than an eight of a watt I'm sure. The higher values you have will probably work fine, but remember that the higher you go, the less current the divider can source, and the more sensitive it is to loading from the ADC input. What you have specified is the highest I would recommend going. On the other hand if you went much below 2.2K and 6.8K then your wattage would start to go up, but I think at 22K and 68K it's pretty much a good compromise between the two.
• Bean Posts: 8,129

  2007-02-09 19:44 edited 2007-02-09 19:44

  James,
  · wattage is (voltage^2 / resistance) or (voltage * current) or (resistance * current^2)
  
  · 16V / 900,000·is 0.00001777 Amps.
  
  Using the·resistance * current^2 formula we get
  
  ·680,000 * (0.00001777^2) = 0.000214725572 Watts or 0.214 milliwatts for the 680K resistor
  ·220,000 * (0.00001777^2) = 0.0000689238 Watts or 0.0689 milliwatts for the 220K resistor
  
  
  Bean.
  
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  Post Edited (Bean (Hitt Consulting)) : 2/9/2007 7:56:35 PM GMT
• James Long Posts: 1,181

  2007-02-09 20:36 edited 2007-02-09 20:36

  Franklin, Dennis, and Bean,
  
  
  
  Thanks.....that insured what I was doing would work fine.
  
  
  
  Dennis,
  
  Thanks for the input....I'll drop those resistor values down about the power of ten....and put surface mount on there.
  
  
  
  Thanks all.......that did refresh my memory at all.....but did help the end result. Man...you really do lose it if you don't use it.
  
  James L