14-06-2010, 07:28 PM
Hi I have had a go at the 'drop shunt' calculation so see what you think, please go easy on me as it is only really my first go at something like this, but feed back is greatly received.
Drop away current=PU current x 68% of PU current
121x0.68=82.28mA
Drop away voltage=DA current x relay resistance
0.08228x20=1.645 volts
Voltage at feed end=feed voltage-drop away voltage/feed resistance
10-1.64/7=1.193 volts
Current through feed resistance=voltage at feed end/feed+relay resistance
1.193/27=44mA
Ballast current at this time=voltage at relay end/ballast resistance
1.645/2.9=567mA
Current through 'drop shunt'=567-44-82.28=440.72mA
Value of drop shunt=Drop away voltage/current through shunt
1.645/440.72=3.7 Ohms
Like I have already mentioned this is my first real attempt so go 'easy.' but it does seem like a pretty reasonable answer for a 'drop shunt' in my experience.'
Drop away current=PU current x 68% of PU current
121x0.68=82.28mA
Drop away voltage=DA current x relay resistance
0.08228x20=1.645 volts
Voltage at feed end=feed voltage-drop away voltage/feed resistance
10-1.64/7=1.193 volts
Current through feed resistance=voltage at feed end/feed+relay resistance
1.193/27=44mA
Ballast current at this time=voltage at relay end/ballast resistance
1.645/2.9=567mA
Current through 'drop shunt'=567-44-82.28=440.72mA
Value of drop shunt=Drop away voltage/current through shunt
1.645/440.72=3.7 Ohms
Like I have already mentioned this is my first real attempt so go 'easy.' but it does seem like a pretty reasonable answer for a 'drop shunt' in my experience.'