Hi..

Going to change my 749 to handle 8 p/h injectors now, I have the drivers and some 0.05 1% 2W resistors, but after reading about Funsticks conversion and he used 0.05 1% 5W. I guess that both options work but what would be best to use?

/N.

 

Where might one find this article?

 

as long as the impedence is the same using a higher wattage resistor shouldnt cuase any trouble. you can use pretty much any wattage the issue is going to be in fitting in on the board.

 

http://mywebpages.comcast.net/p-928/749mods.html

 

So I wont risk to burn the resistors if I use 2watts?

/N.

 

yeah the 2watts should be fine. i dont think id go any lower however. the inductive kickback might burn a lower wattage resistor out.

 

Quote:

So I wont risk to burn the resistors if I use 2watts?

As funstick mentioned, 2W should be fine. The formula for calculating wattage dissipation is P=IE where P is power is Watts, I is current in Amps and E is Voltage in Volts. Add Ohms law and you get P=E^2/R where R is resistance in Ohms. I looked at the LM1949 data sheet and found the peak voltage across the sense resistor is .385 volts and the hold voltage is .094 volts, so worst case (if the driver were to short in peak mode) is 2.96W and in normal operation at 90% duty cycle is less than 1W. BTW the formula for sense resistor wattage in the LM1949 data sheet is wrong. HTH

 

I simply doubled up the original resistors with two additional OEM take outs added in parallel.