Greetings
My old 25W Antex soldering iron has just packed up and I was given an old Weller. The problem is, is that it's a 60W iron, and I don't want to burn components or lift tracks.
Our AC supply is 220V.
Can I use a 390R, 50W resistor in series with the iron's supply to drop the voltage? It doesn't have a transformer in it.
The math I used is as follows:
Voltage: 220V AC (required voltage is 110V AC)
Power: 60W (Which should drop this to 30W?)
Required voltage drop over resistor: 110V AC
Current drawn by iron at 110V AC, and power of 30W: I=P/V = 30/110 = .28A
Therefore R=V/I = 110/.28 = 393R or 390R.
And power=VI = 110x.28 = 31W
Is my math correct. and will it work?
Thanks
Max
My old 25W Antex soldering iron has just packed up and I was given an old Weller. The problem is, is that it's a 60W iron, and I don't want to burn components or lift tracks.
Our AC supply is 220V.
Can I use a 390R, 50W resistor in series with the iron's supply to drop the voltage? It doesn't have a transformer in it.
The math I used is as follows:
Voltage: 220V AC (required voltage is 110V AC)
Power: 60W (Which should drop this to 30W?)
Required voltage drop over resistor: 110V AC
Current drawn by iron at 110V AC, and power of 30W: I=P/V = 30/110 = .28A
Therefore R=V/I = 110/.28 = 393R or 390R.
And power=VI = 110x.28 = 31W
Is my math correct. and will it work?
Thanks
Max