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Resistor Wattage Help


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#1 Shift838 ONLINE  

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Posted Sun Apr 8, 2018 2:29 PM

I'm not real clear on  how to calculate the required wattage of resistor required when wiring up a voltage divider.

 

I know how to wire one up and I have tested it with 1/4 watt resistors.  But I want to make sure I have the correct wattage.

 

I wire up

 

12v --> 470ohm-->150 ohm --> ground

                           |

                           |

                          2.9v  

                        

I'll get 2.9 volts from the circuit, exactly what I want.

 

so do I need 1/4 or 1/2 watt resistors?  well i'm not real clear.

 

Any one?



#2 FALCOR4 OFFLINE  

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Posted Sun Apr 8, 2018 3:10 PM

Shift838, use the voltage drop across the 470ohm resistor for that calculation and the the node voltage at your takeoff node (2.9v) for the 15ohm resistor.  So; (12-2.9)^2 / 470 = .18 watts.  (2.9)^2 / 150 = .06 watts.  If I did my math correctly. Hope this helps.



#3 Osgeld OFFLINE  

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Posted Sun Apr 8, 2018 3:31 PM

how much current are you sucking though it 

 

http://www.bowdensho...its.info/r2.htm


Edited by Osgeld, Sun Apr 8, 2018 3:35 PM.


#4 Stuart OFFLINE  

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Posted Sun Apr 8, 2018 5:11 PM

What are you using the 2.9V for? It will measure as 2.9V if you measure it with a multimeter, but if you draw any appreciable current from it then it will upset the voltage.



#5 FALCOR4 OFFLINE  

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Posted Sun Apr 8, 2018 7:11 PM

Ah, good point!  I assumed that is was in place and drawing current.  :?



#6 senior_falcon ONLINE  

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Posted Sun Apr 8, 2018 7:52 PM

You might consider using a voltage regulator instead of a voltage divider. With a voltage regulator you get a constant voltage regardless of the load. As noted above, a voltage divider will give a different output depending on how much current is drawn by the load.  There are many possibilities; here are two 3 volt regulators:

LP2950-30LP                 up to 100mA output

MCP1827S-3002E/AB          up to 1.5A output

The data sheets show the use of capacitors. In some applications the regulators are a little unstable and the caps control that. 



#7 Osgeld OFFLINE  

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Posted Sun Apr 8, 2018 8:29 PM

yea a divider is ok for making a reference voltage but anything but a tiny load and you start dealing with watts of power and the voltage will start going all over the place 



#8 Shift838 ONLINE  

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Posted Sun Apr 8, 2018 9:07 PM

Shift838, use the voltage drop across the 470ohm resistor for that calculation and the the node voltage at your takeoff node (2.9v) for the 15ohm resistor.  So; (12-2.9)^2 / 470 = .18 watts.  (2.9)^2 / 150 = .06 watts.  If I did my math correctly. Hope this helps.

 

your math is right.  That's what I came up with, but since I was not sure I wanted a few extra brains to help.

 

So with those figures a 1/4 watt for each of the resistors should be sufficient, correct?



#9 Osgeld OFFLINE  

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Posted Sun Apr 8, 2018 9:09 PM

 

your math is right.  That's what I came up with, but since I was not sure I wanted a few extra brains to help.

 

So with those figures a 1/4 watt for each of the resistors should be sufficient, correct?

 

yes with no current draw on the 2.9v output



#10 Ed in SoDak OFFLINE  

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Posted Sun Apr 8, 2018 11:50 PM

.18 is nudging close to .25, so a half-watter there might be better. Oversize current capacity doesn't matter in resistors, unless space is a concern. If the smaller one runs warm, upgrade it. For the 150 Ohm drawing .06, even 1/8-watt should be fine, or use what ya got in the bin.

-Ed


Edited by Ed in SoDak, Sun Apr 8, 2018 11:50 PM.


#11 apersson850 OFFLINE  

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Posted Mon Apr 9, 2018 5:27 AM

If you don't want to go all in with a voltage regulator, then adding a simple Zener diode to the circuit improves it too.

 

Example



#12 senior_falcon ONLINE  

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Posted Mon Apr 9, 2018 8:38 AM

If you don't want to go all in with a voltage regulator, then adding a simple Zener diode to the circuit improves it too.

 

Yep, that is a very good way to do it. Even easier than the voltage reg and should work just as well.



#13 FALCOR4 OFFLINE  

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Posted Mon Apr 9, 2018 3:58 PM

Curious, to what circuit/device are you providing the 2.9v?  I just want to make sure not to mislead you because what Osgeld and Stuart are saying as a caution are very true.  For example, if you ground that node then your 470ohm resistor will have to dissipate .3 watts, that's a worst case scenario.  It will be a toasty warm 1/4 watt resistor in that instance.



#14 Shift838 ONLINE  

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Posted Mon Apr 9, 2018 8:49 PM

I ordered 1/2 watts for the 470ohm just to be safe.  the 2.9 volts is going to the SCART interface.  it requires 1 to 3v of power to put the monitor in RGB mode.

 

I did a recalculation and dropped the 150 ohm to a 100 ohm to pull the voltage down to about 2.1 volts.  



#15 Shift838 ONLINE  

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Posted Mon Apr 9, 2018 8:59 PM

Curious, to what circuit/device are you providing the 2.9v?  I just want to make sure not to mislead you because what Osgeld and Stuart are saying as a caution are very true.  For example, if you ground that node then your 470ohm resistor will have to dissipate .3 watts, that's a worst case scenario.  It will be a toasty warm 1/4 watt resistor in that instance.

 

1/4 watt should be ok for the 100 ohm though I believe?



#16 phoenixdownita OFFLINE  

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Posted Mon Apr 9, 2018 9:41 PM

Not sure what you are trying to do but just in case:

http://members.optus...t/gamescart.htm



#17 Shift838 ONLINE  

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Posted Mon Apr 9, 2018 9:44 PM

Not sure what you are trying to do but just in case:

http://members.optus...t/gamescart.htm

 

something similar but making a board to  for the myarc Geneve 9640 to be able to hook up to SCART monitors and SCART Upscalers with standard cabling.  No more soldering!

 

you can read and see the final retentions of the prototype design that has been sent off this week to get made.

 

here






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