# How do JBC calculate the power rating of their soldering stations?

I am having trouble understanding how a JBC power rating is calculated. For example, this soldering iron uses the T245-A handle which uses the C245 cartridges. The peak power is listed as 130W. We also know that the station uses a 23.5V transformer, and the C245 cartridges are about 2.5 ohms resistance. These stations work by passing full AC waves to the cartridge when it needs to heat up.

So, to find the power of the cartridge it would be V=IR I=V/R 23.5V/2.5ohm = 9.4A

P=(I^2)R 9.4A^2)*2.5ohm = 220.9W

So the calculated power of 220.9W is well above the 130W JBC says. Does JBC just say 130W and never let the cartridge continuously draw above that by limiting the average power to the cartridge? Or perhaps I am missing something.

Edit I should add that JBC stations do not use phase angle control for the heater. They pass half or full waves of AC from the 23.5V transformer to the heater. When heating is required some AC will be passed and when no heating is requred nothing will be passed. There is no current adjustable current limit besides the resistance of the cartridge. Also, JBC irons have a separate pin for the thermocouple, so the heating current is not passed through the thermocouple.

Source for image

• Is AC power the same as DC power? Feb 1, 2021 at 21:44
• The resistance increases with temperature, the heater doesn't get 100% because it is used also as thermocouple for measuring the actual temperature, so it is PWM-ed or used AC with phase angle control with SCR. Feb 1, 2021 at 21:49

P = $$\\frac{(23.5V\cdot 0.85-1V)^2}{2.5\Omega\cdot 1.15)}\approx\$$ 125W