Ignition loads current verses input voltage.

It was stated in the Evolution 58cc dying thread by a member that higher battery voltage results in lower current draw. Quote “Check out ohms law!!! The higher the voltage the lower the current draw, FACT!!! “

This is backwards. A higher source voltage with a constant load always means a higher current draw.

First you need to know what the load resistance is. To do that ohms law says R=E/I . Lets use 4.8 volts for the voltage. For the current I’ll use what bcchi stated (since he builds what most consider the finest ignition systems on the market).
“The RCEXL for example a clone of mine. On average will draw 400MA at 8000 RPM,4.8 volts ,500 MA at 8000RPM 6 volts,600MA 8000RPM.7.4 volts.Go figure.I just went and checked 10 new CHRCEXL ignitions..This was average on all of them,not all exactly the same but close.”

So lets use 500 mA or .5 amps. The load resistance is therefore 4.8/.5 or 9.6 ohms.

If we use a 6 volt battery pack the current draw is I=E/R or 6/9.6= .625 or 625 mA current draw.

Higher voltage results in higher current. Sorry guys but this was really bugging me.

Later,
John